6915 lines
463 KiB
Plaintext
6915 lines
463 KiB
Plaintext
处理第 1/821 条数据...
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[
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{
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"question": "How does a coiled metal strip temperature indicator work when the temperature increases?",
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"answer": "Bimetallic materials are produced by bonding two materials having different coefficients of thermal expansion to one another, forming a laminar composite. When the temperature changes, one of the materials will expand or contract more than the other material. This difference in expansion or contraction causes the bimetallic material to change shape; if the original shape is that of a coil, then the device will coil or uncoil, depending on the direction of the temperature change."
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},
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{
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"question": "From what kind of material would the temperature indicator be made?",
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"answer": "The temperature indicator is made from bimetallic materials, which consist of two bonded materials with different coefficients of thermal expansion."
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},
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{
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"question": "What are the important properties that the material in the temperature indicator must possess?",
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"answer": "In order for the material to perform well, the two materials must have very different coefficients of thermal expansion and should have high enough modulus of elasticity so that no permanent deformation of the material occurs."
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}
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]
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处理第 2/821 条数据...
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[
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{
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"question": "What properties should the head of a carpenter's hammer possess?",
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"answer": "The striking face and claws of the hammer should be hard-the metal should not dent or deform when driving or removing nails. Yet these portions must also possess some impact resistance, particularly so that chips do not flake off the striking face and cause injuries."
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},
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{
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"question": "How would you manufacture a hammer head for a carpenter's hammer?",
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"answer": "The head for a carpenter's hammer is produced by forging, a metalworking process; a simple steel shape is heated and formed in several steps while hot into the required shape. The head is then heat treated to produce the required mechanical and physical properties."
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}
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]
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处理第 3/821 条数据...
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It's a single issue.
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处理第 4/821 条数据...
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[
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{
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"question": "Using the density (11.36 g/cm3) and atomic weight (207.19 g/mol), calculate the number of atoms per cubic centimeter in lead.",
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"answer": "(11.36 g/cm3)(1 cm3)(6.02 x 10^23 atoms/mol) / 207.19 g/mol = 3.3 x 10^22 atoms/cm3"
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},
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{
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"question": "Using the density (0.534 g/cm3) and atomic weight (6.94 g/mol), calculate the number of atoms per cubic centimeter in lithium.",
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"answer": "(0.534 g/cm3)(1 cm3)(6.02 x 10^23 atoms/mol) / 6.94 g/mol = 4.63 x 10^22 atoms/cm3"
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}
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]
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处理第 5/821 条数据...
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It's a single issue.
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处理第 6/821 条数据...
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It's a single issue.
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处理第 7/821 条数据...
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```json
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[
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{
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"question": "In order to plate a steel part having a surface area of 200 in.² with a 0.002 in. thick layer of nickel, how many atoms of nickel are required?",
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"answer": "Volume = (200 in.²)(0.002 in.)(2.54 cm/in.)³ = 6.555 cm³\n(6.555 cm³)(8.902 g/cm³)(6.02 × 10²³ atoms/mol) / (58.71 g/mol) = 5.98 × 10²³ atoms"
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},
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{
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"question": "In order to plate a steel part having a surface area of 200 in.² with a 0.002 in. thick layer of nickel, how many moles of nickel are required?",
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"answer": "Volume = (200 in.²)(0.002 in.)(2.54 cm/in.)³ = 6.555 cm³\n(6.555 cm³)(8.902 g/cm³) / (58.71 g/mol) = 0.994 mol Ni required"
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}
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]
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```
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处理第 8/821 条数据...
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It's a single issue.
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处理第 9/821 条数据...
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It's a single issue.
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处理第 10/821 条数据...
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It's a single issue.
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处理第 11/821 条数据...
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[
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{
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"question": "Calculate the atomic radius in cm for a BCC metal with a0=0.3294 nm and one atom per lattice point.",
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"answer": "For BCC metals, r=((sqrt(3))a0)/4=((sqrt(3))(0.3294 nm))/4=0.1426 nm=1.426×10^(-8) cm."
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},
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{
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"question": "Calculate the atomic radius in cm for an FCC metal with a0=4.0862 Å and one atom per lattice point.",
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"answer": "For FCC metals, r=((sqrt(2))a0)/4=((sqrt(2))(4.0862 Å))/4=1.4447 Å=1.4447×10^(-8) cm."
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}
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]
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处理第 12/821 条数据...
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[
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{
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"question": "Determine the crystal structure for a metal with a0=4.9489 AA, r=1.75 AA and one atom per lattice point.",
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"answer": "We want to determine if x in the calculations equals sqrt(2) (for FCC) or sqrt(3) (for BCC): (x)(4.9489 AA)=(4)(1.75 AA). x=sqrt(2), therefore FCC."
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},
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{
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"question": "Determine the crystal structure for a metal with a0=0.42906 nm, r=0.1858 nm and one atom per lattice point.",
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"answer": "We want to determine if x in the calculations equals sqrt(2) (for FCC) or sqrt(3) (for BCC): (x)(0.42906 nm)=(4)(0.1858 nm). x=sqrt(3), therefore BCC."
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}
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]
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处理第 13/821 条数据...
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```json
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[
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{
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"question": "The density of potassium, which has the BCC structure and one atom per lattice point, is 0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate the lattice parameter.",
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"answer": "Using Equation 3-5:\n0.855 g/cm3 = (2 atoms/cell)(39.09 g/mol) / (a0)^3(6.02 x 10^23 atoms/mol)\na0^3 = 1.5189 x 10^-22 cm3 or a0 = 5.3355 x 10^-8 cm"
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},
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{
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"question": "The lattice parameter of potassium with BCC structure is 5.3355 x 10^-8 cm. Calculate the atomic radius of potassium.",
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"answer": "From the relationship between atomic radius and lattice parameter:\nr = (sqrt(3))(5.3355 x 10^-8 cm) / 4 = 2.3103 x 10^-8 cm"
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}
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]
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```
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处理第 14/821 条数据...
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```json
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[
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{
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"question": "The density of thorium, which has the FCC structure and one atom per lattice point, is 11.72 g/cm³. The atomic weight of thorium is 232 g/mol. Calculate the lattice parameter.",
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"answer": "From Equation 3-5: 11.72 g/cm³ = (4 atoms/cell)(232 g/mol) / (a₀³)(6.02 × 10²³ atoms/mol). a₀³ = 1.315297 × 10⁻²² cm³ or a₀ = 5.0856 × 10⁻⁸ cm."
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},
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{
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"question": "The density of thorium, which has the FCC structure and one atom per lattice point, is 11.72 g/cm³. The atomic weight of thorium is 232 g/mol. The lattice parameter is 5.0856 × 10⁻⁸ cm. Calculate the atomic radius of thorium.",
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"answer": "From the relationship between atomic radius and lattice parameter: r = (√2)(5.0856 × 10⁻⁸ cm) / 4 = 1.7980 × 10⁻⁸ cm."
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}
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]
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```
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处理第 15/821 条数据...
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It's a single issue.
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处理第 16/821 条数据...
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It's a single issue.
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处理第 17/821 条数据...
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It's a single issue.
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处理第 18/821 条数据...
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[
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{
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"question": "Bismuth has a hexagonal structure, with a0=0.4546 nm and c0=1.186 nm. Determine the volume of the unit cell.",
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"answer": "The volume of the unit cell is V=a1^2 c cos30. V=(0.4546 nm)^2(1.186 nm)(cos 30)=0.21226 nm^3=2.1226×10^-22 cm^3."
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},
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{
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"question": "Bismuth has a density of 9.808 g/cm^3 and an atomic weight of 208.98 g/mol. The volume of the unit cell is 2.1226×10^-22 cm^3. Determine the number of atoms in each unit cell.",
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"answer": "If x is the number of atoms per unit cell, then: 9.808 g/cm^3=(x atoms/cell)(208.98 g/mol)/((2.1226×10^-22 cm^3)(6.02×10^23 atoms/mol)). x=6 atoms/cell."
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}
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]
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处理第 19/821 条数据...
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[
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{
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"question": "Gallium has an orthorhombic structure, with a0=0.45258 nm, b0=0.45186 nm, and c0=0.76570 nm. The atomic radius is 0.1218 nm. The density is 5.904 g/cm3 and the atomic weight is 69.72 g/mol. Determine the number of atoms in each unit cell.",
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"answer": "The volume of the unit cell is V=a0 b0 c0 or V=(0.45258 nm)(0.45186 nm)(0.76570 nm)=0.1566 nm3 =1.566×10^-22 cm3. From the density equation: 5.904 g/cm3 = (x atoms/cell)(69.72 g/mol)/(1.566×10^-22 cm3)(6.02×10^23 atoms/mol). Solving for x gives x=8 atoms/cell."
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},
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{
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"question": "Gallium has an orthorhombic structure, with a0=0.45258 nm, b0=0.45186 nm, and c0=0.76570 nm. The atomic radius is 0.1218 nm. The density is 5.904 g/cm3 and the atomic weight is 69.72 g/mol. Determine the packing factor in the unit cell.",
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"answer": "The volume of the unit cell is V=a0 b0 c0 or V=(0.45258 nm)(0.45186 nm)(0.76570 nm)=0.1566 nm3. The packing factor (PF) is calculated as: PF=(8 atoms/cell)(4π/3)(0.1218 nm)^3/0.1566 nm3=0.387."
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}
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]
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处理第 20/821 条数据...
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```json
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[
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{
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"question": "Beryllium has a hexagonal crystal structure, with a0=0.22858 nm and c0=0.35842 nm. The atomic radius is 0.1143 nm, the density is 1.848 g/cm3, and the atomic weight is 9.01 g/mol. Determine the number of atoms in each unit cell.",
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"answer": "V=(0.22858 nm)2(0.35842 nm) cos 30=0.01622 nm3=16.22×10−24 cm3\nFrom the density equation:\n1.848 g/cm3 = (x atoms/cell)(9.01 g/mol) / (16.22×10−24 cm3)(6.02×1023 atoms/mol)\nx = 2 atoms/cell"
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},
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{
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"question": "Beryllium has a hexagonal crystal structure, with a0=0.22858 nm and c0=0.35842 nm. The atomic radius is 0.1143 nm, the density is 1.848 g/cm3, and the atomic weight is 9.01 g/mol. Determine the packing factor in the unit cell.",
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"answer": "The packing factor (PF) is:\nPF = (2 atoms/cell)(4π/3)(0.1143 nm)3 / 0.01622 nm3 = 0.77"
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}
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]
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```
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处理第 21/821 条数据...
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[
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{
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"question": "A typical paper clip weighs 0.59 g and consists of BCC iron. The lattice parameter for BCC iron is 2.866 × 10^-8 cm. Calculate the number of unit cells in the paper clip, given the density of iron is 7.87 g/cm^3.",
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"answer": "The volume of a unit cell is (2.866 × 10^-8 cm)^3 = 2.354 × 10^-23 cm^3. The number of unit cells is: number = (0.59 g) / [(7.87 g/cm^3)(2.354 × 10^-23 cm^3/cell)] = 3.185 × 10^21 cells."
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},
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{
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"question": "A typical paper clip weighs 0.59 g and consists of BCC iron. Given there are 2 atoms per unit cell in BCC iron and the number of unit cells is 3.185 × 10^21, calculate the number of iron atoms in the paper clip.",
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"answer": "The number of atoms is: number = (3.185 × 10^21 cells)(2 atoms/cell) = 6.37 × 10^21 atoms."
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}
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]
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处理第 22/821 条数据...
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```json
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[
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{
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"question": "Determine the planar density for BCC lithium in the (100) plane (a_o = 3.5089 Å).",
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"answer": "planar density = 1 / (3.5089 × 10^-8 cm)^2 = 0.0812 × 10^16 points/cm^2"
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},
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{
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"question": "Determine the packing fraction for BCC lithium in the (100) plane (a_o = 3.5089 Å).",
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"answer": "packing fraction = π[√3 a_o / 4]^2 / a_o^2 = 0.589"
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},
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{
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"question": "Determine the planar density for BCC lithium in the (110) plane (a_o = 3.5089 Å).",
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"answer": "planar density = 2 / √2 (3.5089 × 10^-8 cm)^2 = 0.1149 × 10^16 points/cm^2"
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},
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{
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"question": "Determine the packing fraction for BCC lithium in the (110) plane (a_o = 3.5089 Å).",
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"answer": "packing fraction = 2 √3 a_o / 4 / √2 a_o^2 = 0.833"
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},
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{
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"question": "Determine the planar density for BCC lithium in the (111) plane (a_o = 3.5089 Å).",
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"answer": "planar density = 1/2 / 0.866 (3.5089 × 10^-8 cm)^2 = 0.0469 × 10^16 points/cm^2"
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},
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{
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"question": "Determine the packing fraction for BCC lithium in the (111) plane (a_o = 3.5089 Å).",
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"answer": "packing fraction = 1/2 / 0.866 a_o^2 √3 a_o / 4"
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},
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{
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"question": "Which, if any, of the (100), (110), and (111) planes in BCC lithium is close packed?",
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"answer": "There is no close-packed plane in BCC structures."
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}
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]
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```
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处理第 23/821 条数据...
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It's a single issue.
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处理第 24/821 条数据...
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It's a single issue.
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处理第 25/821 条数据...
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```json
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[
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{
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"question": "Determine the minimum radius of an atom that will just fit into the tetrahedral interstitial site in FCC nickel (σ_c = 3.5167 Å).",
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"answer": "For the tetrahedral site in FCC nickel (σ_c = 3.5167 Å): r_Ni = (√2 × 3.5167 Å) / 4 = 1.243 Å. r / r_Ni = 0.225 for a tetrahedral site. Therefore: r = (1.243 Å)(0.225) = 0.2797 Å."
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},
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{
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"question": "Determine the minimum radius of an atom that will just fit into the octahedral interstitial site in BCC lithium (a_o = 3.5089 Å).",
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"answer": "For the octahedral site in BCC lithium (a_o = 3.5089 Å): r_Li = (√3 × 3.5089 Å) / 4 = 1.519 Å. r / r_Li = 0.414 for an octahedral site. Therefore: r = (1.519 Å)(0.414) = 0.629 Å."
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}
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]
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```
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处理第 26/821 条数据...
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It's a single issue.
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处理第 27/821 条数据...
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```json
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[
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{
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"question": "Using the ionic radii, determine the coordination number expected for Y2O3.",
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"answer": "rY3+/rO2- = 0.89/1.32 = 0.67, CN = 6"
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},
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{
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"question": "Using the ionic radii, determine the coordination number expected for UO2.",
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"answer": "rU4+/rO2- = 0.97/1.32 = 0.73, CN = 6"
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},
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{
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"question": "Using the ionic radii, determine the coordination number expected for BaO.",
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"answer": "rO2-/rBa2+ = 1.32/1.34 = 0.99, CN = 8"
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},
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{
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"question": "Using the ionic radii, determine the coordination number expected for Si3N4.",
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"answer": "rN3-/rSi4+ = 0.15/0.42 = 0.36, CN = 4"
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},
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{
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"question": "Using the ionic radii, determine the coordination number expected for GeO2.",
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"answer": "rGe4+/rO2- = 0.53/1.32 = 0.40, CN = 4"
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},
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{
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"question": "Using the ionic radii, determine the coordination number expected for MnO.",
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"answer": "rMn2+/rO2- = 0.80/1.32 = 0.61, CN = 6"
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},
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{
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"question": "Using the ionic radii, determine the coordination number expected for MgS.",
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"answer": "rMg2+/rS2- = 0.66/1.84 = 0.36, CN = 6"
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},
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{
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"question": "Using the ionic radii, determine the coordination number expected for KBr.",
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"answer": "rK+/rBr- = 1.33/1.96 = 0.68, CN = 6"
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}
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]
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```
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处理第 28/821 条数据...
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```json
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[
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{
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"question": "Would you expect NiO to have the cesium chloride, sodium chloride, or zinc blende structure? (Given: r_Ni+2 = 0.69 Å, r_O-2 = 1.32 Å, r_Ni+2 = 0.52, CN = 6)",
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"answer": "A coordination number of 8 is expected for the CsCl structure, and a coordination number of 4 is expected for ZnS. But a coordination number of 6 is consistent with the NaCl structure."
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},
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{
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"question": "Based on the NaCl structure for NiO, determine the lattice parameter. (Given: r_Ni+2 = 0.69 Å, r_O-2 = 1.32 Å)",
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"answer": "a0 = 2(0.69) + 2(1.32) = 4.02 Å"
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},
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{
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"question": "Based on the NaCl structure for NiO, determine the density. (Given: a0 = 4.02 Å, atomic weights: Ni = 58.71 g/mol, O = 16 g/mol)",
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"answer": "ρ = (4 of each ion(cell)(58.71 + 16 g/mol)) / ((4.02 × 10^-8 cm)^3 (6.02 × 10^23 atoms/mol)) = 7.64 g/cm^3"
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},
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{
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"question": "Based on the NaCl structure for NiO, determine the packing factor. (Given: a0 = 4.02 Å, r_Ni+2 = 0.69 Å, r_O-2 = 1.32 Å)",
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"answer": "PF = (4π/3)(4 ions/cell)[(0.69)^3 + (1.32)^3] / (4.02)^3 = 0.678"
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}
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]
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```
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处理第 29/821 条数据...
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```json
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[
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{
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||
"question": "Would you expect UO2 to have the sodium chloride, zinc blende, or fluoride structure? Given the ionic radii r_U+4 = 0.97 Å and r_O-2 = 1.32 Å, and the valence of U = +4 and O = -2.",
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"answer": "The radius ratio r_U+4/r_O-2 = 0.97/1.32 = 0.735 predicts a coordination number of 8. To balance the charge, there must be twice as many oxygen ions as uranium ions. The fluorite structure satisfies these requirements, with U in FCC positions (4) and O in tetrahedral positions (8)."
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},
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{
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"question": "Determine the lattice parameter (a0) for UO2 with the fluorite structure, given r_U+4 = 0.97 Å and r_O-2 = 1.32 Å.",
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"answer": "For the fluorite structure, sqrt(3) * a0 = 4(r_U+4 + r_O-2) = 4(0.97 + 1.32) = 9.16 Å. Therefore, a0 = 5.2885 Å."
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},
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||
{
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"question": "Calculate the density (ρ) of UO2 with the fluorite structure, given a0 = 5.2885 Å, atomic weight of U = 238.03 g/mol, and atomic weight of O = 16 g/mol.",
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"answer": "ρ = [4(238.03 g/mol) + 8(16 g/mol)] / [(5.2885 × 10^-8 cm)^3 * (6.02 × 10^23 atoms/mol)] = 12.13 g/cm^3."
|
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},
|
||
{
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||
"question": "Determine the packing factor (PF) for UO2 with the fluorite structure, given a0 = 5.2885 Å, r_U+4 = 0.97 Å, and r_O-2 = 1.32 Å.",
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"answer": "PF = [(4π/3)(4(0.97)^3 + 8(1.32)^3)] / (5.2885)^3 = 0.624."
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}
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||
]
|
||
```
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处理第 30/821 条数据...
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[
|
||
{
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||
"question": "Would you expect BeO to have the sodium chloride, zinc blende, or fluorite structure? (Given: r_Be+2 = 0.35 Å, r_O-2 = 1.32 Å)",
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"answer": "r_Be/r_O = 0.265, CN = 4, therefore Zinc Blende structure."
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},
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{
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"question": "Determine the lattice parameter of BeO with zinc blende structure. (Given: r_Be+2 = 0.35 Å, r_O-2 = 1.32 Å)",
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"answer": "sqrt(3) a_0 = 4 r_Be+2 + 4 r_O-2 = 4(0.35 + 1.32) = 6.68 or a_0 = 3.8567 Å."
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},
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{
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"question": "Calculate the density of BeO with zinc blende structure. (Given: a_0 = 3.8567 Å, atomic weights: Be = 9.01 g/mol, O = 16 g/mol)",
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"answer": "ρ = 4(9.01 + 16 g/mol) / (3.8567 × 10^-8 cm)^3 (6.02 × 10^23 atoms/mol) = 2.897 g/cm^3."
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},
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{
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"question": "Calculate the packing factor of BeO with zinc blende structure. (Given: a_0 = 3.8567 Å, r_Be+2 = 0.35 Å, r_O-2 = 1.32 Å)",
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"answer": "PF = (4π/3)(4)(0.35)^3 + 8(1.32)^3 / (3.8567)^3 = 0.684."
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||
}
|
||
]
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处理第 31/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Would you expect CsBr to have the sodium chloride, zinc blende, fluorite, or cesium chloride structure? (Given r_Cs+1=1.67 Å, r_Br-1=1.96 Å)",
|
||
"answer": "r_Cs+1=1.67 Å, r_Br-1=1.96 Å. The radius ratio r_Cs+1/r_Br-1=0.852, coordination number=8, therefore CsCl structure."
|
||
},
|
||
{
|
||
"question": "Determine the lattice parameter for CsBr with the cesium chloride structure. (Given r_Cs+1=1.67 Å, r_Br-1=1.96 Å)",
|
||
"answer": "For CsCl structure, √3 a0=2 r_Cs+1+2 r_Br-1=2(1.96+1.67)=7.26 Å or a0=4.1916 Å"
|
||
},
|
||
{
|
||
"question": "Calculate the density of CsBr with the cesium chloride structure. (Given r_Cs+1=1.67 Å, r_Br-1=1.96 Å, a0=4.1916 Å)",
|
||
"answer": "ρ=((4.1916×10^-8 cm)^3 (6.02×10^23 atoms/mol))/((4π/3)[(1.96)^3+(1.67)^3])=4.8 g/cm^3"
|
||
},
|
||
{
|
||
"question": "Calculate the packing factor for CsBr with the cesium chloride structure. (Given a0=4.1916 Å)",
|
||
"answer": "PF=(4.1916)^3/(4.1916)^3=0.693"
|
||
}
|
||
]
|
||
```
|
||
处理第 32/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Sketch the ion arrangement on the (110) plane of ZnS (with the zinc blende structure).",
|
||
"answer": "The ion arrangement on the (110) plane of ZnS (zinc blende structure) consists of alternating Zn²⁺ and S²⁻ ions arranged in a rectangular pattern. The Zn²⁺ ions have a radius of 0.074 nm and the S²⁻ ions have a radius of 0.184 nm. The lattice parameter a₀ for ZnS is calculated as 0.596 nm."
|
||
},
|
||
{
|
||
"question": "Sketch the ion arrangement on the (110) plane of CaF₂ (with the fluorite structure).",
|
||
"answer": "The ion arrangement on the (110) plane of CaF₂ (fluorite structure) consists of Ca²⁺ ions surrounded by F⁻ ions in a more complex rectangular pattern. The Ca²⁺ ions have a radius of 0.099 nm and the F⁻ ions have a radius of 0.133 nm. The lattice parameter a₀ for CaF₂ is calculated as 0.536 nm."
|
||
},
|
||
{
|
||
"question": "Compare the ion arrangement on the (110) plane of ZnS (zinc blende structure) to that of CaF₂ (fluorite structure).",
|
||
"answer": "The (110) plane of ZnS shows alternating Zn²⁺ and S²⁻ ions in a simpler rectangular arrangement, while the (110) plane of CaF₂ shows Ca²⁺ ions surrounded by F⁻ ions in a more complex pattern. The ZnS arrangement has two types of ions (Zn²⁺ and S²⁻), whereas CaF₂ has one type of cation (Ca²⁺) and two types of anion positions (F⁻)."
|
||
},
|
||
{
|
||
"question": "Calculate the planar packing fraction (PPF) for the (110) plane of ZnS (zinc blende structure).",
|
||
"answer": "The planar packing fraction (PPF) for the (110) plane of ZnS is calculated as follows: PPF = [2π(0.074 nm)² + 2π(0.184 nm)²] / [√2 × (0.596 nm)²] = 0.492."
|
||
},
|
||
{
|
||
"question": "Calculate the planar packing fraction (PPF) for the (110) plane of CaF₂ (fluorite structure).",
|
||
"answer": "The planar packing fraction (PPF) for the (110) plane of CaF₂ is calculated as follows: PPF = [2π(0.099 nm)² + 4π(0.133 nm)²] / [√2 × (0.536 nm)²] = 0.699."
|
||
},
|
||
{
|
||
"question": "Compare the planar packing fraction (PPF) on the (110) planes for ZnS (zinc blende structure) and CaF₂ (fluorite structure).",
|
||
"answer": "The planar packing fraction (PPF) for the (110) plane of ZnS is 0.492, while for CaF₂ it is 0.699. This indicates that the (110) plane of CaF₂ has a higher packing density compared to ZnS."
|
||
}
|
||
]
|
||
```
|
||
处理第 33/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Determine the planar density for the (111) plane of MgO, which has the sodium chloride structure and a lattice parameter of 0.396 nm.",
|
||
"answer": "The area of the (111) plane is 0.866 a_o^2. Given a_o = 3.96 Å (0.396 nm), the planar density (P.D.) is calculated as: P.D. = 2 Mg / (0.866)(3.96 × 10^-8 cm)^2 = 0.1473 × 10^16 points/cm^2."
|
||
},
|
||
{
|
||
"question": "Determine the planar packing fraction for the (111) plane of MgO, which has the sodium chloride structure and a lattice parameter of 0.396 nm.",
|
||
"answer": "The planar packing fraction (PPF) for the (111) plane is calculated as: PPF = 2 π(0.66)^2 / (0.866)(3.96)^2 = 0.202."
|
||
},
|
||
{
|
||
"question": "Determine the planar density for the (222) plane of MgO, which has the sodium chloride structure and a lattice parameter of 0.396 nm.",
|
||
"answer": "The planar density (P.D.) for the (222) plane is the same as for the (111) plane: P.D. = 0.1473 × 10^16 points/cm^2."
|
||
},
|
||
{
|
||
"question": "Determine the planar packing fraction for the (222) plane of MgO, which has the sodium chloride structure and a lattice parameter of 0.396 nm.",
|
||
"answer": "The planar packing fraction (PPF) for the (222) plane is calculated as: PPF = 2 π(1.32)^2 / (0.866)(3.96)^2 = 0.806."
|
||
},
|
||
{
|
||
"question": "What ions are present on the (111) plane of MgO, which has the sodium chloride structure?",
|
||
"answer": "The (111) plane of MgO contains Mg^2+ ions."
|
||
},
|
||
{
|
||
"question": "What ions are present on the (222) plane of MgO, which has the sodium chloride structure?",
|
||
"answer": "The (222) plane of MgO contains O^2- ions."
|
||
}
|
||
]
|
||
```
|
||
处理第 34/821 条数据...
|
||
It's a single issue.
|
||
处理第 35/821 条数据...
|
||
It's a single issue.
|
||
处理第 36/821 条数据...
|
||
It's a single issue.
|
||
处理第 37/821 条数据...
|
||
[
|
||
{
|
||
"question": "The density of a sample of HCP beryllium is 1.844 g/cm³ and the lattice parameters are a₀=0.22858 nm and c₀=0.35842 nm. Calculate (a) the fraction of the lattice points that contain vacancies.",
|
||
"answer": "V_u,c=(0.22858 nm)²(0.35842 nm) cos 30=0.01622 nm³ =1.622×10⁻²³ cm³ 1.844 g/cm³=(x)(9.01 g/mol)/(1.622×10⁻²³ cm³)(6.02×10²³ atoms/mol) x=1.9984 fraction=(2-1.9984)/2=0.0008"
|
||
},
|
||
{
|
||
"question": "The density of a sample of HCP beryllium is 1.844 g/cm³ and the lattice parameters are a₀=0.22858 nm and c₀=0.35842 nm. Calculate (b) the total number of vacancies in a cubic centimeter.",
|
||
"answer": "number=0.0016 vacancies/uc / 1.622×10⁻²³ cm³=0.986×10²⁰ vacancies/cm³"
|
||
}
|
||
]
|
||
处理第 38/821 条数据...
|
||
[
|
||
{
|
||
"question": "BCC lithium has a lattice parameter of 3.5089 x 10^-8 cm and contains one vacancy per 200 unit cells. Calculate the number of vacancies per cubic centimeter.",
|
||
"answer": "(1 vacancy) / (200 x (3.5089 x 10^-8 cm)^3) = 1.157 x 10^20 vacancies/cm^3"
|
||
},
|
||
{
|
||
"question": "BCC lithium has a lattice parameter of 3.5089 x 10^-8 cm and contains one vacancy per 200 unit cells. Calculate the density of Li given that in 200 unit cells there are 399 Li atoms.",
|
||
"answer": "Atoms per cell = 399 / 200. Density = ((399 / 200) x 6.94 g/mol) / ((3.5089 x 10^-8 cm)^3 x 6.02 x 10^23 atoms/mol) = 0.532 g/cm^3"
|
||
}
|
||
]
|
||
处理第 39/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "FCC lead has a lattice parameter of 0.4949 nm and contains one vacancy per 500 Pb atoms. Calculate the density.",
|
||
"answer": "The number of atoms/cell = (499 / 500)(4 sites/cell)\n\nρ = (499 / 500)(4)(207.19 g/mol) / (4.949 × 10^-8 cm)^3(6.02 × 10^23 atoms/mol) = 11.335 g/cm^3"
|
||
},
|
||
{
|
||
"question": "FCC lead has a lattice parameter of 0.4949 nm and contains one vacancy per 500 Pb atoms. Calculate the number of vacancies per gram of Pb.",
|
||
"answer": "The 500 Pb atoms occupy 500 / 4 = 125 unit cells:\n\n(1 vacancy / 125 cells) × (1 / 11.335 g/cm^3) = 5.82 × 10^18 vacancies/g"
|
||
}
|
||
]
|
||
```
|
||
处理第 40/821 条数据...
|
||
It's a single issue.
|
||
处理第 41/821 条数据...
|
||
It's a single issue.
|
||
处理第 42/821 条数据...
|
||
It's a single issue.
|
||
处理第 43/821 条数据...
|
||
[
|
||
{
|
||
"question": "For the Fe-C alloy with one carbon atom introduced for every 100 iron atoms in an interstitial position in BCC iron and a lattice parameter of 0.2867 nm, find the density.",
|
||
"answer": "There is one carbon atom per 100 iron atoms, or 1 C / 50 unit cells, or 1/50 C per unit cell: ρ = ((2)(55.847 g/mol) + (1/50)(12 g/mol)) / ((2.867 × 10^-8 cm)^3 (6.02 × 10^23 atoms/mol)) = 7.89 g/cm^3"
|
||
},
|
||
{
|
||
"question": "For the Fe-C alloy with one carbon atom introduced for every 100 iron atoms in an interstitial position in BCC iron and a lattice parameter of 0.2867 nm, find the packing factor.",
|
||
"answer": "Packing Factor = (2(4π/3)(1.241)^3 + (1/50)(4π/3)(0.77)^3) / (2.867)^3 = 0.681"
|
||
}
|
||
]
|
||
处理第 44/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "The density of BCC iron is 7.882 g/cm³ and the lattice parameter is 0.2886 nm when hydrogen atoms are introduced at interstitial positions. Calculate the atomic fraction of hydrogen atoms.",
|
||
"answer": "7.882 g/cm³ = (2(55.847 g/mol) + x(1.00797 g/mol)) / ((2.866 × 10⁻⁸ cm)³(6.022 × 10²³ atoms/mol)). Solving gives x = 0.0081 H atoms/cell. The total atoms per cell include 2 Fe atoms and 0.0081 H atoms. Thus: f_H = 0.0081 / 2.0081 = 0.004."
|
||
},
|
||
{
|
||
"question": "The density of BCC iron is 7.882 g/cm³ and the lattice parameter is 0.2886 nm when hydrogen atoms are introduced at interstitial positions. Calculate the number of unit cells required on average that contain hydrogen atoms.",
|
||
"answer": "Since there is 0.0081 H/cell, then the number of cells containing H atoms is: cells = 1 / 0.0081 = 123.5 or 1 H in 123.5 cells."
|
||
}
|
||
]
|
||
```
|
||
处理第 45/821 条数据...
|
||
[
|
||
{
|
||
"question": "Suppose one Schottky defect is present in every tenth unit cell of MgO. MgO has the sodium chloride crystal structure and a lattice parameter of 0.396 nm. Calculate the number of anion vacancies per cm3.",
|
||
"answer": "In 10 unit cells, we expect 40 Mg + 40 O ions, but due to the defect: 40 O - 1 = 39. 1 vacancy / (10 cells)(3.96 x 10^-8 cm)^3 = 1.61 x 10^21 vacancies/cm3."
|
||
},
|
||
{
|
||
"question": "Suppose one Schottky defect is present in every tenth unit cell of MgO. MgO has the sodium chloride crystal structure and a lattice parameter of 0.396 nm. Calculate the density of the ceramic.",
|
||
"answer": "(39/40)(4)(24.312 g/mol) + (39/40)(4)(16 g/mol) / (3.96 x 10^-8 cm)^3 (6.02 x 10^23 atoms/mol) = 4.205 g/cm3."
|
||
}
|
||
]
|
||
处理第 46/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "ZnS has the zinc blende structure. If the density is 3.02 g/cm³ and the lattice parameter is 0.59583 nm, determine the number of Schottky defects per unit cell.",
|
||
"answer": "Let x be the number of each type of ion in the unit cell. There normally are 4 of each type. 3.02 g/cm³ = (x(65.38 g/mol) + x(32.064 g/mol)) / ((5.9583 × 10⁻⁸ cm)³ × (6.02 × 10²³ ions/mol)) → x = 3.9465. 4 - 3.9465 = 0.0535 defects/u.c."
|
||
},
|
||
{
|
||
"question": "ZnS has the zinc blende structure. If the density is 3.02 g/cm³ and the lattice parameter is 0.59583 nm, determine the number of Schottky defects per cubic centimeter.",
|
||
"answer": "Number of unit cells/cm³ = 1 / (5.9583 × 10⁻⁸ cm)³ = 4.704 × 10²¹. Schottky defects per cm³ = (4.704 × 10²¹)(0.0535) = 2.517 × 10²⁰."
|
||
}
|
||
]
|
||
```
|
||
处理第 47/821 条数据...
|
||
[
|
||
{
|
||
"question": "Calculate the length of the Burgers vector in BCC niobium with a lattice parameter of 3.294 Å",
|
||
"answer": "The repeat distance, or Burgers vector, is half the body diagonal, or: b = (sqrt(2)) * (sqrt(3)) * (3.294 Å) = 2.853 Å"
|
||
},
|
||
{
|
||
"question": "Calculate the length of the Burgers vector in FCC silver with a lattice parameter of 4.0862 Å",
|
||
"answer": "The repeat distance, or Burgers vector, is half of the face diagonal, or: b = (sqrt(2)) * (sqrt(2)) * (4.0862 Å) = 2.889 Å"
|
||
},
|
||
{
|
||
"question": "Calculate the length of the Burgers vector in diamond cubic silicon with a lattice parameter of 5.4307 Å",
|
||
"answer": "The slip direction is , where the repeat distance is half of the face diagonal: b = (sqrt(2)) * (sqrt(Σ)) * (5.4307 Å) = 3.840 Å"
|
||
}
|
||
]
|
||
处理第 48/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "A single crystal of an FCC metal is oriented so that the direction is parallel to an applied stress of 5000 psi. Calculate the resolved shear stress acting on the (111) slip plane in the [110] slip direction.",
|
||
"answer": "phi=54.76 degrees, lambda=90 degrees, tau=5000 cos 54.76 cos 90, tau=0 psi"
|
||
},
|
||
{
|
||
"question": "A single crystal of an FCC metal is oriented so that the direction is parallel to an applied stress of 5000 psi. Calculate the resolved shear stress acting on the (111) slip plane in the [011] slip direction.",
|
||
"answer": "phi=54.76 degrees, lambda=45 degrees, tau=5000 cos 54.76 cos 45, tau=2040 psi"
|
||
},
|
||
{
|
||
"question": "A single crystal of an FCC metal is oriented so that the direction is parallel to an applied stress of 5000 psi. Calculate the resolved shear stress acting on the (111) slip plane in the [101] slip direction.",
|
||
"answer": "phi=54.76 degrees, lambda=45 degrees, tau=5000 cos 54.76 cos 45, tau=2040 psi"
|
||
},
|
||
{
|
||
"question": "Which slip system(s) will become active first for the given single crystal of an FCC metal under 5000 psi applied stress?",
|
||
"answer": "The slip systems with [011] and [101] slip directions will become active first, both having a resolved shear stress of 2040 psi."
|
||
}
|
||
]
|
||
```
|
||
处理第 49/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "A single crystal of a BCC metal is oriented so that the direction is parallel to the applied stress. If the critical resolved shear stress required for slip is 12,000 psi, calculate the magnitude of the applied stress required to cause slip to begin in the [1,-1,1] direction on the (110) slip plane.",
|
||
"answer": "CRSS = 12,000 psi = σ cos φ cos λ; λ = 54.76°; φ = 90°; σ = 12,000 psi / (cos 90° cos 54.76°) = ∞"
|
||
},
|
||
{
|
||
"question": "A single crystal of a BCC metal is oriented so that the direction is parallel to the applied stress. If the critical resolved shear stress required for slip is 12,000 psi, calculate the magnitude of the applied stress required to cause slip to begin in the [1,-1,1] direction on the (011) slip plane.",
|
||
"answer": "CRSS = 12,000 psi = σ cos φ cos λ; λ = 54.76°; φ = 45°; σ = 12,000 psi / (cos 45° cos 54.76°) = 29,412 psi"
|
||
},
|
||
{
|
||
"question": "A single crystal of a BCC metal is oriented so that the direction is parallel to the applied stress. If the critical resolved shear stress required for slip is 12,000 psi, calculate the magnitude of the applied stress required to cause slip to begin in the [1,-1,1] direction on the (101) slip plane.",
|
||
"answer": "CRSS = 12,000 psi = σ cos φ cos λ; λ = 54.76°; φ = 45°; σ = 12,000 psi / (cos 45° cos 54.76°) = 29,412 psi"
|
||
}
|
||
]
|
||
```
|
||
处理第 50/821 条数据...
|
||
[
|
||
{
|
||
"question": "The strength of titanium is found to be 65,000 psi when the grain size is 17 x 10^-6 m and 82,000 psi when the grain size is 0.8 x 10^-6 m. Determine the constants in the Hall-Petch equation.",
|
||
"answer": "65000 = σ_o + K / (sqrt(17 x 10^-6)) = σ_o + 242.5 K\n82000 = σ_o + K / (sqrt(0.8 x 10^-6)) = σ_o + 1118.0 K\nBy solving the two simultaneous equations:\nK = 19.4 psi / sqrt(d)\nσ_o = 60,290 psi"
|
||
},
|
||
{
|
||
"question": "Using the Hall-Petch equation constants determined previously (σ_o = 60,290 psi and K = 19.4 psi / sqrt(d)), find the strength of the titanium when the grain size is reduced to 0.2 x 10^-6 m.",
|
||
"answer": "σ = 60,290 + 19.4 / sqrt(0.2 x 10^-6) = 103,670 psi"
|
||
}
|
||
]
|
||
处理第 51/821 条数据...
|
||
[
|
||
{
|
||
"question": "For an ASTM grain size number of 8, calculate the number of grains per square inch at a magnification of 100.",
|
||
"answer": "N=2^(n-1) N=2^(n-1)=2^7=128 grains/in.^2"
|
||
},
|
||
{
|
||
"question": "For an ASTM grain size number of 8, calculate the number of grains per square inch with no magnification.",
|
||
"answer": "No magnification means that the magnification is 1: (27)(100/1)^2=1.28x10^6 grains/in.^2"
|
||
}
|
||
]
|
||
处理第 52/821 条数据...
|
||
It's a single issue.
|
||
处理第 53/821 条数据...
|
||
It's a single issue.
|
||
处理第 54/821 条数据...
|
||
It's a single issue.
|
||
处理第 55/821 条数据...
|
||
It's a single issue.
|
||
处理第 56/821 条数据...
|
||
It's a single issue.
|
||
处理第 57/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "The diffusion coefficient for Cr+3 in Cr2O3 is 6 × 10^-15 cm²/s at 727°C and is 1 × 10^-9 cm²/s at 1400°C. Calculate the activation energy (Q).",
|
||
"answer": "6 × 10^-15 / 1 × 10^-9 = D0 exp[-Q/(1.987)(1000)] / D0 exp[-Q/(1.987)(1673)]\n6 × 10^-6 = exp[-Q(0.000503 - 0.00030)] = exp[-0.000203 Q]\n-12.024 = -0.000203 Q or Q = 59,230 cal/mol"
|
||
},
|
||
{
|
||
"question": "The diffusion coefficient for Cr+3 in Cr2O3 is 6 × 10^-15 cm²/s at 727°C and is 1 × 10^-9 cm²/s at 1400°C. Calculate the constant D0.",
|
||
"answer": "1 × 10^-9 = D0 exp[-59,230/(1.987)(1673)] = D0 exp(-17.818)\n1 × 10^-9 = 1.828 × 10^-8 D0 or D0 = 0.055 cm²/s"
|
||
}
|
||
]
|
||
```
|
||
处理第 58/821 条数据...
|
||
[
|
||
{
|
||
"question": "A 0.2-mm thick wafer of silicon is treated so that a uniform concentration gradient of antimony is produced. One surface contains 1 Sb atom per 10^8 Si atoms and the other surface contains 500 Sb atoms per 10^8 Si atoms. The lattice parameter for Si is 5.407 Å. Calculate the concentration gradient in (a) atomic percent Sb per cm.",
|
||
"answer": "Δc/Δx = ((1/10^8 - 500/10^8)/0.02 cm) × 100% = -0.02495 at%/cm"
|
||
},
|
||
{
|
||
"question": "A 0.2-mm thick wafer of silicon is treated so that a uniform concentration gradient of antimony is produced. One surface contains 1 Sb atom per 10^8 Si atoms and the other surface contains 500 Sb atoms per 10^8 Si atoms. The lattice parameter for Si is 5.407 Å. Calculate the concentration gradient in (b) Sb atoms/cm^3·cm.",
|
||
"answer": "a0 = 5.4307 Å, Vunit cell = 160.16 × 10^-24 cm^3; c1 = (8 Si atoms/unit cell)(1 Sb/10^8 Si)/160.16 × 10^-24 cm^3/unit cell = 0.04995 × 10^16 Sb atoms/cm^3; c2 = (8 Si atoms/unit cell)(500 Sb/10^8 Si)/160.16 × 10^-24 cm^3/unit cell = 24.975 × 10^16 Sb atoms/cm^3; Δc/Δx = (0.04995 × 10^16 - 24.975 × 10^16)/0.02 cm = -1.246 × 10^19 Sb atoms/cm^3·cm"
|
||
}
|
||
]
|
||
处理第 59/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "When a Cu-Zn alloy solidifies, one portion of the structure contains 25 atomic percent zinc and another portion 0.025 mm away contains 20 atomic percent zinc. The lattice parameter for the FCC alloy is 3.63 x 10^-8 cm. Determine the concentration gradient in (a) atomic percent Zn per cm.",
|
||
"answer": "Delta c / Delta x = (20% - 25%) / (0.025 mm x 0.1 cm/mm) = -2000 at% Zn/cm"
|
||
},
|
||
{
|
||
"question": "When a Cu-Zn alloy solidifies, one portion of the structure contains 25 atomic percent zinc and another portion 0.025 mm away contains 20 atomic percent zinc. The lattice parameter for the FCC alloy is 3.63 x 10^-8 cm. Determine the concentration gradient in (b) weight percent Zn per cm.",
|
||
"answer": "wt% Zn = (20 x 65.38 g/mol) / (20 x 65.38 + 80 x 63.54) x 100 = 20.46; wt% Zn = (25 x 65.38 g/mol) / (25 x 65.38 + 75 x 63.54) x 100 = 25.54; Delta c / Delta x = (20.46% - 25.54%) / 0.0025 cm = -2032 wt% Zn/cm"
|
||
}
|
||
]
|
||
```
|
||
处理第 60/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "A 0.001-in. BCC iron foil is used to separate a high hydrogen gas from a low hydrogen gas at 650 degrees C. 5 x 10^8 H atoms/cm^3 are in equilibrium with the hot side of the foil, while 2 x 10^3 H atoms/cm^3 are in equilibrium with the cold side. Determine the concentration gradient of hydrogen.",
|
||
"answer": "Delta c / Delta x = (2 x 10^3 - 5 x 10^8) / ((0.001 in)(2.54 cm/in)) = -1969 x 10^8 H atoms/cm^3.cm"
|
||
},
|
||
{
|
||
"question": "A 0.001-in. BCC iron foil is used to separate a high hydrogen gas from a low hydrogen gas at 650 degrees C. 5 x 10^8 H atoms/cm^3 are in equilibrium with the hot side of the foil, while 2 x 10^3 H atoms/cm^3 are in equilibrium with the cold side. Determine the flux of hydrogen through the foil.",
|
||
"answer": "J = -D(Delta c / Delta x) = -0.0012 exp[-3600 / (1.987)(923)] (-1969 x 10^8) = 0.33 x 10^8 H atoms/cm^2.s"
|
||
}
|
||
]
|
||
```
|
||
处理第 61/821 条数据...
|
||
[
|
||
{
|
||
"question": "Given a 1-mm sheet of FCC iron used to contain nitrogen in a heat exchanger at 1200°C, with nitrogen concentrations of 0.04 atomic percent at one surface and 0.005 atomic percent at the second surface, calculate the concentration gradient (Δc/Δx) in N atoms/cm³·cm.",
|
||
"answer": "Δc/Δx = (0.00005 - 0.0004)(4 atoms per cell)/(3.589 × 10^-8 cm)^3 / (1 mm)(0.1 cm/mm) = -3.03 × 10^20 N atoms/cm³·cm"
|
||
},
|
||
{
|
||
"question": "Using the concentration gradient calculated previously (-3.03 × 10^20 N atoms/cm³·cm) for nitrogen diffusion through a 1-mm FCC iron sheet at 1200°C, determine the flux of nitrogen (J) in N atoms/cm²·s. The diffusion coefficient (D) is given by D = 0.0034 exp[-34,600/(1.987)(1473)] cm²/s.",
|
||
"answer": "J = -D(Δc/Δx) = -0.0034 exp[-34,600/(1.987)(1473)] × (3.03 × 10^20) = 7.57 × 10^12 N atoms/cm²·s"
|
||
}
|
||
]
|
||
处理第 62/821 条数据...
|
||
It's a single issue.
|
||
处理第 63/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Calculate the concentration of hydrogen atoms (c1) at the first surface of the BCC iron structure, given that there are 0.05 H atoms per unit cell and the lattice parameter is 2.866×10⁻⁸ cm.",
|
||
"answer": "c1 = 0.05 H / (2.866×10⁻⁸ cm)³ = 212.4×10¹⁹ H atoms/cm³"
|
||
},
|
||
{
|
||
"question": "Calculate the concentration of hydrogen atoms (c2) at the second surface of the BCC iron structure, given that there are 0.001 H atoms per unit cell and the lattice parameter is 2.866×10⁻⁸ cm.",
|
||
"answer": "c2 = 0.001 H / (2.866×10⁻⁸ cm)³ = 4.25×10¹⁹ H atoms/cm³"
|
||
},
|
||
{
|
||
"question": "Determine the concentration gradient (Δc/Δx) of hydrogen atoms across the iron structure, given the concentrations c1 and c2, and the unknown thickness Δx.",
|
||
"answer": "Δc/Δx = (4.25×10¹⁹ - 212.4×10¹⁹) / Δx = -2.08×10²¹ / Δx"
|
||
},
|
||
{
|
||
"question": "Convert the allowable hydrogen loss rate (50 g/cm² per year) into a flux (J) in H atoms/cm²·s, given the atomic weight of hydrogen (1.00797 g/mol) and Avogadro's number (6.02×10²³ atoms/mol).",
|
||
"answer": "J = (50 g/cm²·y)(6.02×10²³ atoms/mol) / (1.00797 g/mol)(31.536×10⁶ s/y) = 9.47×10¹⁷ H atoms/cm²·s"
|
||
},
|
||
{
|
||
"question": "Calculate the minimum thickness (Δx) of the iron structure required to limit the hydrogen flux to 9.47×10¹⁷ H atoms/cm²·s, given the diffusion coefficient for hydrogen in BCC iron at 400°C (D = 0.0012 exp[-3600/(1.987)(673)]).",
|
||
"answer": "J = -D(Δc/Δx) → 9.47×10¹⁷ = (2.08×10²¹/Δx)(0.0012 exp[-3600/(1.987)(673)]) → Δx = 0.179 cm"
|
||
}
|
||
]
|
||
```
|
||
处理第 64/821 条数据...
|
||
It's a single issue.
|
||
处理第 65/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Calculate the diffusion rate of oxygen ions (D_O^-2) in Al2O3 at 1500°C.",
|
||
"answer": "D_O^-2 = 1900 exp[-152,000/(1.987)(1773)] = 3.47 × 10^-16 cm^2/s"
|
||
},
|
||
{
|
||
"question": "Calculate the diffusion rate of aluminum ions (D_Al^+3) in Al2O3 at 1500°C.",
|
||
"answer": "D_Al^+3 = 28 exp[-114,000/(1.987)(1773)] = 2.48 × 10^-13 cm^2/s"
|
||
},
|
||
{
|
||
"question": "Compare the diffusion rates of oxygen ions and aluminum ions in Al2O3 at 1500°C and explain the difference based on their ionic radii.",
|
||
"answer": "The diffusion rate of aluminum ions (2.48 × 10^-13 cm^2/s) is higher than that of oxygen ions (3.47 × 10^-16 cm^2/s). The ionic radius of the oxygen ion is 1.32 Å, compared with the aluminum ionic radius of 0.51 Å; consequently it is much easier for the smaller aluminum ion to diffuse in the ceramic."
|
||
}
|
||
]
|
||
```
|
||
处理第 66/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Calculate the diffusion coefficient of carbon in BCC iron at the allotropic transformation temperature of 912°C (1185 K).",
|
||
"answer": "D_BCC = 0.011 exp [-20,900 / (1.987)(1185)] = 1.51 × 10^-6 cm^2/s"
|
||
},
|
||
{
|
||
"question": "Calculate the diffusion coefficient of carbon in FCC iron at the allotropic transformation temperature of 912°C (1185 K).",
|
||
"answer": "D_FCC = 0.23 exp [-32,900 / (1.987)(1185)] = 1.92 × 10^-7 cm^2/s"
|
||
},
|
||
{
|
||
"question": "Explain why the diffusion coefficient of carbon in BCC iron is higher than in FCC iron at the allotropic transformation temperature.",
|
||
"answer": "Packing factor of the BCC lattice (0.68) is less than that of the FCC lattice; consequently atoms are expected to be able to diffuse more rapidly in the BCC iron."
|
||
}
|
||
]
|
||
```
|
||
处理第 67/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Iron containing 0.05 % C is heated to 912°C in an atmosphere that produces 1.20 % C at the surface and is held for 24 h. Calculate the carbon content at 0.05 cm beneath the surface if the iron is BCC.",
|
||
"answer": "t=(24 h)(3600 s/h)=86,400 s\nD_BCC=0.011 exp [-20,900 /(1.987)(1185)]=1.54 × 10^-6 cm^2/s\nBCC: (1.2-c_x)/(1.2-0.05)=erf[0.05 /(2 sqrt((1.54 × 10^-6)(86,400)))]=erf[0.0685]=0.077\nc_x=0.95 % C"
|
||
},
|
||
{
|
||
"question": "Iron containing 0.05 % C is heated to 912°C in an atmosphere that produces 1.20 % C at the surface and is held for 24 h. Calculate the carbon content at 0.05 cm beneath the surface if the iron is FCC.",
|
||
"answer": "t=(24 h)(3600 s/h)=86,400 s\nD_FCC=0.23 exp [-32,900 /(1.987)(1185)]=1.97 × 10^-7 cm^2/s\nFCC: (1.2-c_x)/(1.2-0.05)=erf[0.05 /(2 sqrt((1.97 × 10^-7)(86,400)))]=erf[0.192]=0.2139\nc_x=0.95 % C"
|
||
},
|
||
{
|
||
"question": "Explain the difference in carbon content between BCC and FCC iron structures.",
|
||
"answer": "Faster diffusion occurs in the looser packed BCC structure, leading to the higher carbon content at point 'X'."
|
||
}
|
||
]
|
||
```
|
||
处理第 68/821 条数据...
|
||
It's a single issue.
|
||
处理第 69/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is the value of the error function argument (z) when a 0.15% C steel is carburized at 1100°C to achieve 0.35% C at 1 mm beneath the surface with a surface composition of 0.90% C?",
|
||
"answer": "The error function argument is calculated as (0.9-0.35)/(0.9-0.15) = 0.733 = erf[0.1 / 2√(Dt)]."
|
||
},
|
||
{
|
||
"question": "What is the value of 0.1 / 2√(Dt) corresponding to the error function value of 0.733?",
|
||
"answer": "0.1 / 2√(Dt) = 0.786."
|
||
},
|
||
{
|
||
"question": "What is the value of √(Dt) derived from the equation 0.1 / 2√(Dt) = 0.786?",
|
||
"answer": "√(Dt) = 0.0636."
|
||
},
|
||
{
|
||
"question": "What is the value of Dt calculated from √(Dt) = 0.0636?",
|
||
"answer": "Dt = 0.00405 cm²."
|
||
},
|
||
{
|
||
"question": "What is the diffusion coefficient (D) for carbon in steel at 1100°C?",
|
||
"answer": "D = 0.23 exp[-32,900 / (1.987)(1373)] = 1.332 × 10⁻⁶ cm²/s."
|
||
},
|
||
{
|
||
"question": "What time (t) is required to achieve the desired carbon concentration profile, given Dt = 0.00405 cm² and D = 1.332 × 10⁻⁶ cm²/s?",
|
||
"answer": "t = 0.00405 / 1.332 × 10⁻⁶ = 3040 s = 51 min."
|
||
}
|
||
]
|
||
```
|
||
处理第 70/821 条数据...
|
||
It's a single issue.
|
||
处理第 71/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is the diffusion coefficient (D) for a 1.2% C tool steel held at 1150°C?",
|
||
"answer": "D = 0.23 exp(-32,900 / (1.987)(1423)) ≈ 0.234 × 10^-6 cm²/s"
|
||
},
|
||
{
|
||
"question": "What is the total time (t) in seconds for the steel exposed to oxygen for 48 hours?",
|
||
"answer": "t = (48 h)(3600 s/h) = 17.28 × 10^4 s"
|
||
},
|
||
{
|
||
"question": "What is the value of √(Dt) for the given diffusion coefficient and time?",
|
||
"answer": "√(Dt) = 0.5929"
|
||
},
|
||
{
|
||
"question": "To what depth (x) will the steel be decarburized to less than 0.20% C, given the surface carbon content is zero and the initial carbon content is 1.2% C?",
|
||
"answer": "(0 - 0.2)/(0 - 1.2) = 0.1667 ∴ x / 2√(Dt) = 0.149\nx = (0.149)(2)(0.5929) = 0.177 cm"
|
||
}
|
||
]
|
||
```
|
||
处理第 72/821 条数据...
|
||
It's a single issue.
|
||
处理第 73/821 条数据...
|
||
It's a single issue.
|
||
处理第 74/821 条数据...
|
||
It's a single issue.
|
||
处理第 75/821 条数据...
|
||
It's a single issue.
|
||
处理第 76/821 条数据...
|
||
It's a single issue.
|
||
处理第 77/821 条数据...
|
||
[
|
||
{
|
||
"question": "A ceramic part made of MgO is sintered successfully at 1700 degrees Celsius in 90 minutes. To minimize thermal stresses during the process, we plan to reduce the temperature to 1500 degrees Celsius. Which will limit the rate at which sintering can be done: diffusion of magnesium ions or diffusion of oxygen ions?",
|
||
"answer": "Diffusion of oxygen is the slower of the two, due to the larger ionic radius of the oxygen."
|
||
},
|
||
{
|
||
"question": "A ceramic part made of MgO is sintered successfully at 1700 degrees Celsius in 90 minutes. To minimize thermal stresses during the process, we plan to reduce the temperature to 1500 degrees Celsius. What time will be required at the lower temperature?",
|
||
"answer": "D_1760 = 0.000043 exp[-82,100 /(1.987)(1973)] = 3.455 x 10^-14 cm^2/s. D_1560 = 0.000043 exp[-82,100 /(1.987)(1773)] = 3.255 x 10^-15 cm^2/s. t_1560 = D_1760 t_1760 / D_1560 = (3.455 x 10^-14)(90) / 3.255 x 10^-15 = 955 min = 15.9 h."
|
||
}
|
||
]
|
||
处理第 78/821 条数据...
|
||
[
|
||
{
|
||
"question": "A 850-lb force is applied to a 0.15-in. diameter nickel wire having a yield strength of 45,000 psi. Determine whether the wire will plastically deform.",
|
||
"answer": "First determine the stress acting on the wire: sigma = F / A = 850 lb / (pi / 4)(0.15 in)^2 = 48,100 psi. Because sigma is greater than the yield strength of 45,000 psi, the wire will plastically deform."
|
||
},
|
||
{
|
||
"question": "A 850-lb force is applied to a 0.15-in. diameter nickel wire having a tensile strength of 55,000 psi. Determine whether the wire will experience necking.",
|
||
"answer": "First determine the stress acting on the wire: sigma = F / A = 850 lb / (pi / 4)(0.15 in)^2 = 48,100 psi. Because sigma is less than the tensile strength of 55,000 psi, no necking will occur."
|
||
}
|
||
]
|
||
处理第 79/821 条数据...
|
||
[
|
||
{
|
||
"question": "A force of 100,000 N is applied to a 10 mm × 20 mm iron bar having a yield strength of 400 MPa. Determine whether the bar will plastically deform.",
|
||
"answer": "First determine the stress acting on the bar: σ = F / A = 100,000 N / (10 mm)(20 mm) = 500 N/mm² = 500 MPa. Because σ is greater than the yield strength of 400 MPa, the bar will plastically deform."
|
||
},
|
||
{
|
||
"question": "A force of 100,000 N is applied to a 10 mm × 20 mm iron bar having a tensile strength of 480 MPa. Determine whether the bar will experience necking.",
|
||
"answer": "First determine the stress acting on the bar: σ = F / A = 100,000 N / (10 mm)(20 mm) = 500 N/mm² = 500 MPa. Because σ is greater than the tensile strength of 480 MPa, the bar will experience necking."
|
||
}
|
||
]
|
||
处理第 80/821 条数据...
|
||
It's a single issue.
|
||
处理第 81/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "A force of 20,000 N will cause a 1 cm × 1 cm bar of magnesium to stretch from 10 cm to 10.045 cm. Calculate the strain.",
|
||
"answer": "The strain e is ε=(10.045 cm−10 cm)/10 cm=0.0045 cm/cm"
|
||
},
|
||
{
|
||
"question": "A force of 20,000 N is applied to a 1 cm × 1 cm bar of magnesium. Calculate the stress in MPa.",
|
||
"answer": "The stress σ is σ=20,000 N/(10 mm×10 mm)=200 N/mm²=200 MPa"
|
||
},
|
||
{
|
||
"question": "Using the calculated stress of 200 MPa and strain of 0.0045 cm/cm, determine the modulus of elasticity in GPa.",
|
||
"answer": "E=σ/ε=200 MPa/0.0045 cm/cm=44,444 MPa=44.4 GPa"
|
||
},
|
||
{
|
||
"question": "Convert the modulus of elasticity from 44,444 MPa to psi.",
|
||
"answer": "E=(44,444 MPa)(145 psi/MPa)=6.44×10⁶ psi"
|
||
}
|
||
]
|
||
```
|
||
处理第 82/821 条数据...
|
||
It's a single issue.
|
||
处理第 83/821 条数据...
|
||
It's a single issue.
|
||
处理第 84/821 条数据...
|
||
It's a single issue.
|
||
处理第 85/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "A 0.4-in. diameter, 12-in-long titanium bar has a yield strength of 50,000 psi, a modulus of elasticity of 16 x 10^6 psi, and Poisson's ratio of 0.30. Determine the length of the bar when a 500-lb load is applied.",
|
||
"answer": "The stress is σ=F/A=500 lb/(π/4)(0.4 in.)^2=3,979 psi. The applied stress is much less than the yield strength; therefore Hooke's law can be used. The strain is ε=σ/E=3,979 psi/(16 x 10^6 psi)=0.00024868 in./in. (ℓ_f-ℓ_0)/ℓ_0=(ℓ_f-12 in.)/12 in.=0.00024868 in./in. ℓ_f=12.00298 in."
|
||
},
|
||
{
|
||
"question": "A 0.4-in. diameter, 12-in-long titanium bar has a yield strength of 50,000 psi, a modulus of elasticity of 16 x 10^6 psi, and Poisson's ratio of 0.30. Determine the diameter of the bar when a 500-lb load is applied.",
|
||
"answer": "From Poisson's ratio, μ=-ε_lat/ε_long=0.3. ε_lat=-(0.3)(0.00024868)=-0.0000746 in./in. (d_f-d_0)/d_0=(d_f-0.4 in.)/0.4 in.=-0.0000746 in./in. d_f=0.39997 in."
|
||
}
|
||
]
|
||
```
|
||
处理第 86/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "A three-point bend test is performed on a block of ZrO2 that is 8 in. long, 0.50 in. wide, and 0.25 in. thick and is resting on two supports 4 in. apart. When a force of 400 lb is applied, the specimen deflects 0.037 in. and breaks. Calculate the flexural strength.",
|
||
"answer": "flexural strength = 3 F L / 2 w h^2 = (3)(400 lb)(4 in) / (2)(0.5 in)(0.25 in)^2 = 76,800 psi"
|
||
},
|
||
{
|
||
"question": "A three-point bend test is performed on a block of ZrO2 that is 8 in. long, 0.50 in. wide, and 0.25 in. thick and is resting on two supports 4 in. apart. When a force of 400 lb is applied, the specimen deflects 0.037 in. and breaks. Calculate the flexural modulus, assuming that no plastic deformation occurs.",
|
||
"answer": "flexural modulus = F L^3 / 4 w h^3 delta = (400 lb)(4 in)^3 / (4)(0.5 in)(0.25 in)^3(0.037 in) = 22.14 x 10^6 psi"
|
||
}
|
||
]
|
||
```
|
||
处理第 87/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "A three-point bend test is performed on a block of silicon carbide that is 10 cm long, 1.5 cm wide, and 0.6 cm thick and is resting on two supports 7.5 cm apart. The sample breaks when a deflection of 0.09 mm is recorded. Calculate the force that caused the fracture. The flexural modulus for silicon carbide is 480 GPa. Assume that no plastic deformation occurs.",
|
||
"answer": "The force F required to produce a deflection of 0.09 mm is F = (flexural modulus)(4 w h^3 δ) / L^3 F = (480,000 MPa)(4)(15 mm)(6 mm)^3(0.09 mm) / (75 mm)^3 F = 1327 N"
|
||
},
|
||
{
|
||
"question": "A three-point bend test is performed on a block of silicon carbide that is 10 cm long, 1.5 cm wide, and 0.6 cm thick and is resting on two supports 7.5 cm apart. The sample breaks when a deflection of 0.09 mm is recorded. Calculate the flexural strength. The flexural modulus for silicon carbide is 480 GPa. Assume that no plastic deformation occurs.",
|
||
"answer": "Flexural strength = 3 F L / 2 w h^2 = (3)(1327 N)(75 mm) / (2)(15 mm)(6 mm)^2 = 276 MPa"
|
||
}
|
||
]
|
||
```
|
||
处理第 88/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "A thermosetting polymer containing glass beads is required to deflect 0.5 mm when a force of 500 N is applied. The polymer part is 2 cm wide, 0.5 cm thick, and 10 cm long. If the flexural modulus is 6.9 GPa, determine the minimum distance between the supports.",
|
||
"answer": "The minimum distance L between the supports can be calculated from the flexural modulus. L^3 = 4 w h^2 δ (flexural modulus) / F. L^3 = (4)(20 mm)(5 mm)^3(0.5 mm)(6.9 GPa)(1000 MPa/GPa) / 500 N. L^3 = 69,000 mm^3 or L = 41 mm."
|
||
},
|
||
{
|
||
"question": "Will the polymer fracture if its flexural strength is 85 MPa? Assume that no plastic deformation occurs.",
|
||
"answer": "The stress acting on the bar when a deflection of 0.5 mm is obtained is σ = 3 F L / 2 w h^2 = (3)(500 N)(41 mm) / (2)(20 mm)(5 mm)^2 = 61.5 MPa. The applied stress is less than the flexural strength of 85 MPa; the polymer is not expected to fracture."
|
||
}
|
||
]
|
||
```
|
||
处理第 89/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "The flexural strength of alumina is 46,000 psi. A bar of alumina 0.3 in. thick, 1.0 in. wide, and 10 in. long is placed on supports 7 in. apart. Determine the force required to break the bar.",
|
||
"answer": "F = (2)(1 in.)(0.3 in.)^2(46,000 psi)/(3)(7 in.) = 394 lb"
|
||
},
|
||
{
|
||
"question": "The flexural modulus of alumina is 45 x 10^6 psi. A bar of alumina 0.3 in. thick, 1.0 in. wide, and 10 in. long is placed on supports 7 in. apart. Determine the amount of deflection at the moment the bar breaks, assuming that no plastic deformation occurs. The force required to break the bar is 394 lb.",
|
||
"answer": "δ = (394 lb)(7 in.)^3/(4)(1 in.)(0.3 in.)^3(45 x 10^6 psi) = 0.0278 in."
|
||
}
|
||
]
|
||
```
|
||
处理第 90/821 条数据...
|
||
It's a single issue.
|
||
处理第 91/821 条数据...
|
||
It's a single issue.
|
||
处理第 92/821 条数据...
|
||
It's a single issue.
|
||
处理第 93/821 条数据...
|
||
It's a single issue.
|
||
处理第 94/821 条数据...
|
||
It's a single issue.
|
||
处理第 95/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is the maximum applied stress for the ceramic part given a yield strength of 75,000 psi and a safety factor of one third?",
|
||
"answer": "The applied stress is sigma = (1/3) * 75,000 psi = 25,000 psi."
|
||
},
|
||
{
|
||
"question": "What is the critical flaw size (a) for the ceramic part given a plane strain fracture toughness of 5,000 psi sqrt(in), a geometric factor f of 1.4, and the applied stress of 25,000 psi?",
|
||
"answer": "a = (1/pi) * [K_IC / (f * sigma)]^2 = (1/pi) * [5,000 psi sqrt(in) / (1.4 * 25,000 psi)]^2 = 0.0065 in. The length of internal flaws is 2a = 0.013 in."
|
||
},
|
||
{
|
||
"question": "Does the nondestructive test that detects flaws greater than 0.05 in. long have the required sensitivity given the critical flaw size of 0.013 in.?",
|
||
"answer": "The nondestructive test can detect flaws as small as 0.05 in. long, which is not smaller than the critical flaw size of 0.013 in. required for failure. Thus, the NDT test is not satisfactory."
|
||
}
|
||
]
|
||
```
|
||
处理第 96/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is the fatigue strength, or maximum stress amplitude, required to survive for one million cycles under conditions that provide for equal compressive and tensile stresses?",
|
||
"answer": "From the figure, the fatigue strength at one million cycles is 22 MPa."
|
||
},
|
||
{
|
||
"question": "What are the maximum stress, the minimum stress, and the mean stress on the part during its use under conditions that provide for equal compressive and tensile stresses?",
|
||
"answer": "The maximum stress is +22 MPa, the minimum stress is -22 MPa, and the mean stress is 0 MPa."
|
||
},
|
||
{
|
||
"question": "What effect would the frequency of the stress application have on your answers regarding the fatigue strength, maximum stress, minimum stress, and mean stress?",
|
||
"answer": "A high frequency will cause heating of the polymer. As the temperature of the polymer increases, the fatigue strength will decrease. If the applied stress is not reduced, then the polymer will fail in a shorter time."
|
||
},
|
||
{
|
||
"question": "The high-strength steel in Figure 7-21 is subjected to a stress alternating at 200 revolutions per minute between 600 MPa and 200 MPa (both tension). Calculate the growth rate of a surface crack when it reaches a length of 0.2 mm in both m/cycle and m/s. Assume that f=1.0.",
|
||
"answer": "For the steel, C=1.62×10^-12 and n=3.2. The change in the stress intensity factor ΔK is ΔK−√A S √α u=(1.2)(600 MPa−200 MPa)√α(0.0002 m)=12.03 MPa√m. The crack growth rate is da/dN=1.62×10^-12(ΔK)^3.2=1.62×10^-12(12.03)^3.2=4.638×10^-9 m/cycle. da/dt=(4.638×10^-9 m/cycle)(200 cycles/min)/60 s/min=1.55×10^-8 m/s."
|
||
},
|
||
{
|
||
"question": "The high-strength steel in Figure 7-21, which has a critical fracture toughness of 80 MPa√m, is subjected to an alternating stress varying from -900 MPa (compression) to +900 MPa (tension). It is to survive for 10^5 cycles before failure occurs. Calculate (a) the size of a surface crack required for failure to occur and (b) the largest initial surface crack size that will permit this to happen. Assume that f=1.",
|
||
"answer": "(a) Only the tensile portion of the applied stress is considered in Δσ. Based on the applied stress of 900 MPa and the fracture toughness of 80 MPa√m, the size of a surface crack required for failure to occur is K=fσ√πa_c or a_c=(1/π)[K/fσ]^2. a_c=(1/π)[80 MPa√m/(1)(900 MPa)]^2=0.0025 m=2.5 mm. (b) The largest initial surface crack tolerable to prevent failure within 10^5 cycles is N=10^5 cycles=2[(0.0025 m)^(2-3.2)/2−a_i^(2-3.2)/2]/(2-3.2)(1.62×10^-12)(1)^3.2(900)^3.2(π)^3.2/2. 10^5=2[36.41−(a_i)^-0.60]/(−1.2)(1.62×10^-12)(1)(2.84×10^9)(6.244). (a_i)^-0.6=1760. a_i=3.9×10^-6 m=0.0039 mm."
|
||
}
|
||
]
|
||
```
|
||
处理第 97/821 条数据...
|
||
It's a single issue.
|
||
处理第 98/821 条数据...
|
||
It's a single issue.
|
||
处理第 99/821 条数据...
|
||
[
|
||
{
|
||
"question": "Using the data in Figure 7-27 for an iron-chromium-nickel alloy, determine the activation energy Qr for rupture in the temperature range 980 to 1090 degrees Celsius.",
|
||
"answer": "The appropriate equation is tr=K sigma'' exp(Qr/RT). From Figure 7-27(a), we can determine the rupture time versus temperature for a fixed stress, sigma=1000 psi: tr=2,400 h at 1090 degrees Celsius, tr=14,000 h at 1040 degrees Celsius, tr=100,000 h at 980 degrees Celsius. From this data, the equation becomes tr=K' exp(Qr/RT) and we can find Qr by simultaneous equations or graphically. Qr=117,000 cal/mol."
|
||
},
|
||
{
|
||
"question": "Using the data in Figure 7-27 for an iron-chromium-nickel alloy, determine the constant m for rupture in the temperature range 980 to 1090 degrees Celsius.",
|
||
"answer": "The appropriate equation is tr=K sigma'' exp(Qr/RT). We can determine the rupture time versus applied stress for a constant temperature, 1090 degrees Celsius: tr=10^5 h for sigma=450 psi, tr=10^4 h for sigma=800 psi, tr=10^3 h for sigma=1200 psi, tr=10^2 h for sigma=2100 psi. With this approach, the equation becomes tr=K'' sigma^m, where m is obtained graphically or by simultaneous equations: m=3.9."
|
||
}
|
||
]
|
||
处理第 100/821 条数据...
|
||
It's a single issue.
|
||
处理第 101/821 条数据...
|
||
It's a single issue.
|
||
处理第 102/821 条数据...
|
||
[
|
||
{
|
||
"question": "A 2-in.-diameter copper rod is reduced to 1.5 in. diameter, then reduced again to a final diameter of 1 in. Calculate the % CW for this case.",
|
||
"answer": "% CW = ((2)^2 - (1)^2) / (2)^2 * 100 = 75%"
|
||
},
|
||
{
|
||
"question": "A 2-in.-diameter copper rod is reduced in one step from 2 in. to a 1 in. diameter. Calculate the % CW for this case.",
|
||
"answer": "% CW = ((2)^2 - (1)^2) / (2)^2 * 100 = 75%"
|
||
}
|
||
]
|
||
处理第 103/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Suppose that liquid nickel is undercooled until homogeneous nucleation occurs. Calculate the critical radius of the nucleus required. Assume that the lattice parameter of the solid FCC nickel is 0.356 nm.",
|
||
"answer": "From Table 9-1, ΔT_max = 480°C\nr^s = (2)(255×10^-7 J/cm^2)(1453+273)/(2756 J/cm^3)(480) = 6.65×10^-8 cm"
|
||
},
|
||
{
|
||
"question": "Suppose that liquid nickel is undercooled until homogeneous nucleation occurs. Calculate the number of nickel atoms in the nucleus. Assume that the lattice parameter of the solid FCC nickel is 0.356 nm.",
|
||
"answer": "a_0 = 3.56 Å, V = 45.118×10^-24 cm^3\nV_nucleus = (4π/3)(6.65×10^-8 cm)^3 = 1232×10^-24 cm^3\nnumber of unit cells = 1232/45.118 = 27.3\natoms per nucleus = (4 atoms/cell)(27.3 cells) = 109 atoms"
|
||
}
|
||
]
|
||
```
|
||
处理第 104/821 条数据...
|
||
[
|
||
{
|
||
"question": "Suppose that liquid iron is undercooled until homogeneous nucleation occurs. Calculate the critical radius of the nucleus required. Assume that the lattice parameter of the solid BCC iron is 2.92 Å.",
|
||
"answer": "r^s = (2)(204×10^-7 J/cm^2)(1538+273)/(1737 J/cm^3)(420) = 10.128×10^-8 cm"
|
||
},
|
||
{
|
||
"question": "Suppose that liquid iron is undercooled until homogeneous nucleation occurs. Calculate the number of iron atoms in the nucleus. Assume that the lattice parameter of the solid BCC iron is 2.92 Å.",
|
||
"answer": "V = (4π/3)(10.128)^3 = 4352 Å^3 = 4352×10^-24 cm^3. V_lc = (2.92 Å)^3 = 24.897 Å^3 = 24.897×10^-24 cm^3. Number of unit cells = 4352/24.897 = 175. Atoms per nucleus = (175 cells)(2 atoms/cell) = 350 atoms"
|
||
}
|
||
]
|
||
处理第 105/821 条数据...
|
||
It's a single issue.
|
||
处理第 106/821 条数据...
|
||
It's a single issue.
|
||
处理第 107/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Calculate the fraction of solidification that occurs dendritically when iron nucleates at 10°C undercooling. The specific heat of iron is 5.78 J/cm³·°C and the heat of fusion is 1737 J/cm³.",
|
||
"answer": "f = (c ΔT) / ΔH_f = (5.78 J/cm³·°C * 10°C) / 1737 J/cm³ = 0.0333"
|
||
},
|
||
{
|
||
"question": "Calculate the fraction of solidification that occurs dendritically when iron nucleates at 100°C undercooling. The specific heat of iron is 5.78 J/cm³·°C and the heat of fusion is 1737 J/cm³.",
|
||
"answer": "f = (c ΔT) / ΔH_f = (5.78 J/cm³·°C * 100°C) / 1737 J/cm³ = 0.333"
|
||
},
|
||
{
|
||
"question": "Calculate the fraction of solidification that occurs dendritically when iron nucleates homogeneously. The specific heat of iron is 5.78 J/cm³·°C and the heat of fusion is 1737 J/cm³.",
|
||
"answer": "f = (c ΔT) / ΔH_f = (5.78 J/cm³·°C * 420°C) / 1737 J/cm³ = 1.4, therefore, all solidification occurs dendritically."
|
||
}
|
||
]
|
||
```
|
||
处理第 108/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Calculate the fraction of solidification that occurs dendritically when silver nucleates at 10°C undercooling. The specific heat of silver is 3.25 J/cm³·°C and the heat of fusion is 965 J/cm³.",
|
||
"answer": "f = (c ΔT) / ΔH_f = (3.25 J/cm³·°C)(10°C) / 965 J/cm³ = 0.0337"
|
||
},
|
||
{
|
||
"question": "Calculate the fraction of solidification that occurs dendritically when silver nucleates at 100°C undercooling. The specific heat of silver is 3.25 J/cm³·°C and the heat of fusion is 965 J/cm³.",
|
||
"answer": "f = (c ΔT) / ΔH_f = (3.25 J/cm³·°C)(100°C) / 965 J/cm³ = 0.337"
|
||
},
|
||
{
|
||
"question": "Calculate the fraction of solidification that occurs dendritically when silver nucleates homogeneously. The specific heat of silver is 3.25 J/cm³·°C and the heat of fusion is 965 J/cm³.",
|
||
"answer": "f = (c ΔT) / ΔH_f = (3.25 J/cm³·°C)(250°C) / 965 J/cm³ = 0.842"
|
||
}
|
||
]
|
||
```
|
||
处理第 109/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Calculate the mold constant in Chvorinov's rule for a 2-in. cube that solidifies in 4.6 min. Assume that n=2.",
|
||
"answer": "We can find the volume and surface area of the cube: V=(2)^3=8 in^3, A=6(2)^2=24 in^2, t=4.6=B(8/24)^2. B=4.6/(0.333)^2=41.48 min/in^2."
|
||
},
|
||
{
|
||
"question": "Calculate the solidification time for a 0.5 in. x 0.5 in. x 6 in. bar cast under the same conditions, using the mold constant B=41.48 min/in^2. Assume that n=2.",
|
||
"answer": "For the bar: V=(0.5)(0.5)(6)=1.5 in^3, A=2(0.5)(0.5)+4(0.5)(6)=12.5 in^2. t=(41.48)(1.5/12.5)^2=0.60 min."
|
||
}
|
||
]
|
||
```
|
||
处理第 110/821 条数据...
|
||
It's a single issue.
|
||
处理第 111/821 条数据...
|
||
[
|
||
{
|
||
"question": "Calculate the volume of the cylindrical riser required to prevent shrinkage in a 4 in. × 10 in. × 20 in. casting if the H/D of the riser is 1.5.",
|
||
"answer": "(V A)_C = (4)(10)(20) / [2(4)(10) + 2(4)(20) + 2(10)(20)] = 800 / 640 = 1.25\n(V A)_F = [(π/4)D²H] / [2(π/4)D² + πDH] = [(π/4)(3/2)D³] / [(π/2)D² + (3π/2)D²] = (3D/8) / 2 = 3D/16 ≈ 1.25\nD ≥ 6.67 in., H ≥ 10 in., V ≥ 349 in.³"
|
||
},
|
||
{
|
||
"question": "Calculate the diameter of the cylindrical riser required to prevent shrinkage in a 4 in. × 10 in. × 20 in. casting if the H/D of the riser is 1.5.",
|
||
"answer": "(V A)_C = (4)(10)(20) / [2(4)(10) + 2(4)(20) + 2(10)(20)] = 800 / 640 = 1.25\n(V A)_F = [(π/4)D²H] / [2(π/4)D² + πDH] = [(π/4)(3/2)D³] / [(π/2)D² + (3π/2)D²] = (3D/8) / 2 = 3D/16 ≈ 1.25\nD ≥ 6.67 in., H ≥ 10 in., V ≥ 349 in.³"
|
||
},
|
||
{
|
||
"question": "Calculate the height of the cylindrical riser required to prevent shrinkage in a 4 in. × 10 in. × 20 in. casting if the H/D of the riser is 1.5.",
|
||
"answer": "(V A)_C = (4)(10)(20) / [2(4)(10) + 2(4)(20) + 2(10)(20)] = 800 / 640 = 1.25\n(V A)_F = [(π/4)D²H] / [2(π/4)D² + πDH] = [(π/4)(3/2)D³] / [(π/2)D² + (3π/2)D²] = (3D/8) / 2 = 3D/16 ≈ 1.25\nD ≥ 6.67 in., H ≥ 10 in., V ≥ 349 in.³"
|
||
}
|
||
]
|
||
处理第 112/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Calculate the volume of the casting with dimensions 1 in. × 6 in. × 6 in.",
|
||
"answer": "V = (1)(6)(6) = 36 in^3"
|
||
},
|
||
{
|
||
"question": "Calculate the surface area of the casting with dimensions 1 in. × 6 in. × 6 in.",
|
||
"answer": "A = 2(1)(6) + 2(1)(6) + 2(6)(6) = 96 in^2"
|
||
},
|
||
{
|
||
"question": "Determine the volume-to-surface area ratio (V/A) of the casting.",
|
||
"answer": "(V/A)_C = 36 / 96 = 0.375"
|
||
},
|
||
{
|
||
"question": "Calculate the minimum diameter (D) of the cylindrical riser with H/D = 1.0 to prevent shrinkage, given (V/A)_F ≥ (V/A)_C.",
|
||
"answer": "(V/A)_F = D/6 ≥ 0.375 → D ≥ 2.25 in."
|
||
},
|
||
{
|
||
"question": "Calculate the minimum height (H) of the cylindrical riser with H/D = 1.0.",
|
||
"answer": "H = D ≥ 2.25 in."
|
||
},
|
||
{
|
||
"question": "Calculate the minimum volume (V) of the cylindrical riser with D = 2.25 in. and H = 2.25 in.",
|
||
"answer": "V = (π/4)D^2H = (π/4)(2.25)^2(2.25) ≥ 8.95 in^3"
|
||
}
|
||
]
|
||
```
|
||
处理第 113/821 条数据...
|
||
[
|
||
{
|
||
"question": "For a 4-in-diameter sphere of liquid copper (Cr) allowed to solidify, producing a spherical shrinkage cavity in the center of the casting, calculate the volume and diameter of the shrinkage cavity. Given: Shrinkage percentage for Cr = 5.1%, r_sphere = 2 in.",
|
||
"answer": "V_shrinkage = (4π/3)(2)^3(0.051) = 1.709 in^3. Diameter of shrinkage cavity = 1.30 in."
|
||
},
|
||
{
|
||
"question": "For a 4-in-diameter sphere of liquid iron (Fe) allowed to solidify, producing a spherical shrinkage cavity in the center of the casting, calculate the volume and diameter of the shrinkage cavity. Given: Shrinkage percentage for Fe = 3.4%, r_sphere = 2 in.",
|
||
"answer": "V_shrinkage = (4π/3)(2)^3(0.034) = 1.139 in^3. Diameter of shrinkage cavity = 1.06 in."
|
||
}
|
||
]
|
||
处理第 114/821 条数据...
|
||
It's a single issue.
|
||
处理第 115/821 条数据...
|
||
[
|
||
{
|
||
"question": "A 2 cm × 4 cm × 6 cm magnesium casting is produced. After cooling to room temperature, calculate the volume cavity at the center of the casting. The density of the magnesium is 1.738 g/cm3.",
|
||
"answer": "V_initial = (2)(4)(6) = 48 cm3. V_initial = 80 g / 1.738 g/cm3 = 46.03 cm3."
|
||
},
|
||
{
|
||
"question": "A 2 cm × 4 cm × 6 cm magnesium casting is produced. After cooling to room temperature, calculate the percent shrinkage that must have occurred during solidification. The density of the magnesium is 1.738 g/cm3.",
|
||
"answer": "% shrinkage = (48 - 46.03) / 48 × 100% = 4.1%."
|
||
}
|
||
]
|
||
处理第 116/821 条数据...
|
||
[
|
||
{
|
||
"question": "A 2 in. x 8 in. x 10 in. iron casting is produced and, after cooling to room temperature, is found to weigh 43.9 lb. The density of the iron is 7.87 g/cm3. Determine the percent shrinkage that must have occurred during solidification.",
|
||
"answer": "v_actual = (43.9 lb)(454 g) / 7.87 g/cm3 = 2532.5 cm3\nV_intended = (2)(8)(10) = 160 in3 x (2.54 cm/in)3 = 2621.9 cm3\nshrinkage = (2621.9 - 2532.5) / 2621.9 x 100% = 3.4%"
|
||
},
|
||
{
|
||
"question": "A 2 in. x 8 in. x 10 in. iron casting is produced and, after cooling to room temperature, is found to weigh 43.9 lb. The density of the iron is 7.87 g/cm3. Determine the number of shrinkage pores in the casting if all of the shrinkage occurs as pores with a diameter of 0.05 in.",
|
||
"answer": "V_shrinkage = 2621.9 cm3 - 2532.5 cm3 = 89.4 cm3\nV_pore = (0.05 in / 2)(2.54 cm/in) = 0.0635 cm\n# pores = 89.4 cm3 / [(4π/3)(0.0635 cm)3] = 83,354 pores"
|
||
}
|
||
]
|
||
处理第 117/821 条数据...
|
||
It's a single issue.
|
||
处理第 118/821 条数据...
|
||
[
|
||
{
|
||
"question": "A liquid cast iron has a density of 7.65 g/cm3. Immediately after solidification, the density of the solid cast iron is found to be 7.71 g/cm3. Determine the percent volume change that occurs during solidification.",
|
||
"answer": "(1 / 7.65 - 1 / 7.71) / (1 / 7.65) * 100% = (0.1307 cm3 - 0.1297 cm3) / 0.1307 cm3 * 100% = 0.77%"
|
||
},
|
||
{
|
||
"question": "Does the cast iron expand or contract during solidification?",
|
||
"answer": "The casting contracts."
|
||
}
|
||
]
|
||
处理第 119/821 条数据...
|
||
It's a single issue.
|
||
处理第 120/821 条数据...
|
||
[
|
||
{
|
||
"question": "In the unary phase diagram for SiO2 shown in Figure 10-19, locate the triple point where solid, liquid, and vapor coexist and give the temperature and the type of solid present.",
|
||
"answer": "The solid-liquid-vapor triple point occurs at 1713 degrees Celsius; the solid phase present at this point is beta-cristobalite."
|
||
},
|
||
{
|
||
"question": "In the unary phase diagram for SiO2 shown in Figure 10-19, what do the other triple points indicate?",
|
||
"answer": "The other triple points describe the equilibrium between two solids and a vapor phase."
|
||
}
|
||
]
|
||
处理第 121/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Based on Hume-Rothery's conditions, would the Au-Ag system be expected to display unlimited solid solubility? Explain.",
|
||
"answer": "r_Au=1.442 v=+1 FCC, r_Ag=1.445 v=+1 FCC, radius difference=0.2%. The Au-Ag system has the required radius ratio, the same crystal structures, and the same valences. It might be expected to display complete solid solubility. The Au-Ag system does have isomorphous phase diagrams."
|
||
},
|
||
{
|
||
"question": "Based on Hume-Rothery's conditions, would the Al-Cu system be expected to display unlimited solid solubility? Explain.",
|
||
"answer": "r_Al=1.432 v=+3 FCC, r_Cu=1.278 v=+1 FCC, radius difference=10.7%. The Al-Cu system does not meet the Hume-Rothery conditions for unlimited solid solubility."
|
||
},
|
||
{
|
||
"question": "Based on Hume-Rothery's conditions, would the Al-Au system be expected to display unlimited solid solubility? Explain.",
|
||
"answer": "r_Al=1.432 v=+3 FCC, r_Au=1.442 v=+1 FCC, radius difference=0%. The Al-Au system does not meet the Hume-Rothery conditions for unlimited solid solubility due to different valences."
|
||
},
|
||
{
|
||
"question": "Based on Hume-Rothery's conditions, would the U-W system be expected to display unlimited solid solubility? Explain.",
|
||
"answer": "r_U=1.38 v=+4 Ortho, r_W=1.371 v=+4 FCC, radius difference=0.7%. The U-W system does not meet the Hume-Rothery conditions for unlimited solid solubility due to different crystal structures."
|
||
},
|
||
{
|
||
"question": "Based on Hume-Rothery's conditions, would the Mo-Ta system be expected to display unlimited solid solubility? Explain.",
|
||
"answer": "r_Mo=1.363 v=+4, r_Ta=1.43 v=+5, radius difference=4.7%. The Mo-Ta system has the required radius ratio, the same crystal structures, and similar valences. It might be expected to display complete solid solubility. The Mo-Ta system does have isomorphous phase diagrams."
|
||
},
|
||
{
|
||
"question": "Based on Hume-Rothery's conditions, would the Nb-W system be expected to display unlimited solid solubility? Explain.",
|
||
"answer": "r_Nb=1.426 v=+4, r_W=1.371 v=+4, radius difference=3.9%. The Nb-W system does not meet the Hume-Rothery conditions for unlimited solid solubility due to different crystal structures."
|
||
},
|
||
{
|
||
"question": "Based on Hume-Rothery's conditions, would the Mg-Zn system be expected to display unlimited solid solubility? Explain.",
|
||
"answer": "r_Mg=1.604 v=+2, r_Zn=1.332 v=+2, radius difference=17%. The Mg-Zn system does not meet the Hume-Rothery conditions for unlimited solid solubility due to large radius difference."
|
||
},
|
||
{
|
||
"question": "Based on Hume-Rothery's conditions, would the Mg-Cd system be expected to display unlimited solid solubility? Explain.",
|
||
"answer": "r_Mg=1.604 v=+2, r_Cd=1.490 v=+2, radius difference=7.1%. The Mg-Cd system has the required radius ratio, the same crystal structures, and the same valences. It might be expected to display complete solid solubility. The Mg-Cd alloys all solidify like isomorphous alloys, though solid-state phase transformations complicate the diagram."
|
||
}
|
||
]
|
||
```
|
||
处理第 122/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Suppose 1 at% of the following elements is added to copper (forming a separate alloy with each element) without exceeding the solubility limit. Which one would be expected to give the higher strength alloy? (a) Au (b) Mn (c) Sr (d) Si (e) Co",
|
||
"answer": "The Cu-Sr alloy would be expected to be strongest (largest size difference)."
|
||
},
|
||
{
|
||
"question": "Is any of the alloying elements expected to have unlimited solid solubility in copper? (a) Au (b) Mn (c) Sr (d) Si (e) Co",
|
||
"answer": "The Cu-Au alloy satisfies Hume-Rothery's conditions and might be expected to display complete solid solubility-in fact it freezes like an isomorphous series of alloys, but a number of solid-state transformations occur at lower temperatures."
|
||
}
|
||
]
|
||
```
|
||
处理第 123/821 条数据...
|
||
[
|
||
{
|
||
"question": "Suppose 1 at% of the following elements is added to aluminum (forming a separate alloy with each element) without exceeding the solubility limit. Which one would be expected to give the least reduction in electrical conductivity? For aluminum: r=1.432 AA (FCC structure with valence of 3). Consider the following elements: (a) Li: r=1.519 AA, phi r=6.1%, BCC valence=1; (b) Ba: r=2.176 AA, phi r=-52.0%, BCC valence=2; (c) Be: r=1.143 AA, phi r=-20.2%, HCP valence=2; (d) Cd: r=1.49 AA, phi r=4.1%, HCP valence=2; (e) Ga: r=1.218 AA, phi r=14.9%, Orthorhombic valence=3",
|
||
"answer": "The cadmium would be expected to give the smallest reduction in electrical conductivity, since the Cd atoms are most similar in size to the aluminum atoms."
|
||
},
|
||
{
|
||
"question": "Is any of the alloy elements (Li, Ba, Be, Cd, Ga) expected to have unlimited solid solubility in aluminum? For aluminum: r=1.432 AA (FCC structure with valence of 3). Consider the following elements: (a) Li: r=1.519 AA, phi r=6.1%, BCC valence=1; (b) Ba: r=2.176 AA, phi r=-52.0%, BCC valence=2; (c) Be: r=1.143 AA, phi r=-20.2%, HCP valence=2; (d) Cd: r=1.49 AA, phi r=4.1%, HCP valence=2; (e) Ga: r=1.218 AA, phi r=14.9%, Orthorhombic valence=3",
|
||
"answer": "None are expected to have unlimited solid solubility, due either to difference in valence, atomic radius, or crystal structure."
|
||
}
|
||
]
|
||
处理第 124/821 条数据...
|
||
It's a single issue.
|
||
处理第 125/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is the required weight percentage of nickel (Ni) in the alloy to achieve 50 wt% α phase at 1300°C, given the phase compositions L: 46 wt% Ni and α: 58 wt% Ni?",
|
||
"answer": "The required weight percentage of nickel (Ni) in the alloy is calculated using the lever rule: (x - 46)/(58 - 46) * 100 = 50% α, solving for x gives x = 52 wt% Ni."
|
||
},
|
||
{
|
||
"question": "How many grams of nickel (Ni) must be added to 500 grams of copper (Cu) to produce an alloy with 52 wt% Ni?",
|
||
"answer": "The number of grams of nickel (Ni) required is calculated by: (x)/(x + 500) * 100% = 52, solving for x gives x = 541.7 g Ni."
|
||
}
|
||
]
|
||
```
|
||
处理第 126/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is the molecular weight of MgO and NiO?",
|
||
"answer": "MW_MgO = 40.312 g/mol, MW_NiO = 74.71 g/mol"
|
||
},
|
||
{
|
||
"question": "What mole percentage of MgO is needed to achieve a solidus temperature of 2200°C in the MgO-NiO ceramic system?",
|
||
"answer": "38 mol% MgO is needed to obtain the correct solidus temperature."
|
||
},
|
||
{
|
||
"question": "How to calculate the weight percentage of MgO required in the mixture, given the molecular weights and mole percentage?",
|
||
"answer": "wt% MgO = (38 * 40.312) / (38 * 40.312 + 62 * 74.71) * 100% = 24.9%"
|
||
},
|
||
{
|
||
"question": "How many grams of MgO must be added to 1 kg of NiO to achieve 24.9 wt% MgO in the mixture?",
|
||
"answer": "x / (x + 1000) * 100% = 24.9% or x = 332 g of MgO"
|
||
}
|
||
]
|
||
```
|
||
处理第 127/821 条数据...
|
||
It's a single issue.
|
||
处理第 128/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "A Nb-W alloy held at 2800 degrees Celsius is partly liquid and partly solid. If possible, determine the composition of each phase in the alloy.",
|
||
"answer": "L: 49 wt% W; alpha: 70 wt% W"
|
||
},
|
||
{
|
||
"question": "A Nb-W alloy held at 2800 degrees Celsius is partly liquid and partly solid. If possible, determine the amount of each phase in the alloy.",
|
||
"answer": "Not possible unless we know the original composition of the alloy."
|
||
}
|
||
]
|
||
```
|
||
处理第 129/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "A Nb-W alloy contains 55% alpha at 2600 degrees C. Determine the composition of each phase.",
|
||
"answer": "L: 22 wt% W, alpha: 42 wt% W"
|
||
},
|
||
{
|
||
"question": "A Nb-W alloy contains 55% alpha at 2600 degrees C. Determine the original composition of the alloy.",
|
||
"answer": "0.55 = (x - 22)/(42 - 22) or x = 33 wt% W"
|
||
}
|
||
]
|
||
```
|
||
处理第 130/821 条数据...
|
||
[
|
||
{
|
||
"question": "Suppose a 1200-lb bath of a Nb-40 wt% W alloy is held at 2800°C. How many pounds of tungsten can be added to the bath before any solid forms?",
|
||
"answer": "Solid starts to form at 2800°C when 49 wt% W is in the alloy. In 1200 lb of the original Nb-40% W alloy, there are (0.4)(1200)=480 lb W and 720 lb Nb. The total amount of tungsten that must be in the final alloy is: 0.49 = x/(x+720) or x=692 lb W total or 692-480=212 additional pounds of W must be added."
|
||
},
|
||
{
|
||
"question": "Suppose a 1200-lb bath of a Nb-40 wt% W alloy is held at 2800°C. How many pounds of tungsten must be added to cause the entire bath to be solid?",
|
||
"answer": "To be completely solid at 2800°C, the alloy must contain 70 wt% W. The total amount of tungsten required in the final alloy is: 0.70 = x/(x+720) or x=1680 lb W total or 1680-480=1200 additional pounds of W must be added."
|
||
}
|
||
]
|
||
处理第 131/821 条数据...
|
||
It's a single issue.
|
||
处理第 132/821 条数据...
|
||
[
|
||
{
|
||
"question": "A NiO-60 mol% MgO ceramic is allowed to solidify under equilibrium conditions. Determine the composition of the first solid to form.",
|
||
"answer": "1st α: 80% MgO"
|
||
},
|
||
{
|
||
"question": "A NiO-60 mol% MgO ceramic is allowed to solidify under equilibrium conditions. Determine the composition of the last liquid to solidify.",
|
||
"answer": "Last L: 35% MgO"
|
||
}
|
||
]
|
||
处理第 133/821 条数据...
|
||
[
|
||
{
|
||
"question": "A Nb-35% W alloy is allowed to solidify under equilibrium conditions. Determine the composition of the first solid to form.",
|
||
"answer": "1st alpha: 55% W"
|
||
},
|
||
{
|
||
"question": "A Nb-35% W alloy is allowed to solidify under equilibrium conditions. Determine the composition of the last liquid to solidify.",
|
||
"answer": "Last L: 18% W"
|
||
}
|
||
]
|
||
处理第 134/821 条数据...
|
||
It's a single issue.
|
||
处理第 135/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Consider a Pb-15% Sn alloy. During solidification, determine the composition of the first solid to form.",
|
||
"answer": "8% Sn"
|
||
},
|
||
{
|
||
"question": "Consider a Pb-15% Sn alloy. Determine the liquidus temperature, solidus temperature, solvus temperature, and freezing range of the alloy.",
|
||
"answer": "liquidus = 290 degrees C, solidus = 240 degrees C, solvus = 170 degrees C, freezing range = 50 degrees C"
|
||
},
|
||
{
|
||
"question": "Consider a Pb-15% Sn alloy. Determine the amounts and compositions of each phase at 260 degrees C.",
|
||
"answer": "L: 30% Sn; alpha: 12% Sn; % L = ((15 - 12)/(30 - 12)) x 100% = 17%; % alpha = 83%"
|
||
},
|
||
{
|
||
"question": "Consider a Pb-15% Sn alloy. Determine the amounts and compositions of each phase at 183 degrees C.",
|
||
"answer": "alpha: 15% Sn; 100% alpha"
|
||
},
|
||
{
|
||
"question": "Consider a Pb-15% Sn alloy. Determine the amounts and compositions of each phase at 25 degrees C.",
|
||
"answer": "alpha: 2% Pb; beta: 100% Sn; % alpha = ((100 - 15)/(100 - 2)) x 100 = 87%; % beta = 13%"
|
||
}
|
||
]
|
||
```
|
||
处理第 136/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Consider a Pb-35% Sn alloy. Determine if the alloy is hypoeutectic or hypereutectic.",
|
||
"answer": "hypoeutectic"
|
||
},
|
||
{
|
||
"question": "Consider a Pb-35% Sn alloy. Determine the composition of the first solid to form during solidification.",
|
||
"answer": "14% Sn"
|
||
},
|
||
{
|
||
"question": "Consider a Pb-35% Sn alloy. Determine the amounts and compositions of each phase at 184°C.",
|
||
"answer": "α: 19% Sn, L: 61.9% Sn; % α = (61.9 - 35)/(61.9 - 19) × 100% = 63%, % L = 37%"
|
||
},
|
||
{
|
||
"question": "Consider a Pb-35% Sn alloy. Determine the amounts and compositions of each phase at 182°C.",
|
||
"answer": "α: 19% Sn, β: 97.5% Sn; % α = (97.5 - 35)/(97.5 - 19) × 100% = 80%, % β = 20%"
|
||
},
|
||
{
|
||
"question": "Consider a Pb-35% Sn alloy. Determine the amounts and compositions of each microconstituent at 182°C.",
|
||
"answer": "primary α: 19% Sn, % primary α = 63%; eutectic: 61.9% Sn, % eutectic = 37%"
|
||
},
|
||
{
|
||
"question": "Consider a Pb-35% Sn alloy. Determine the amounts and compositions of each phase at 25°C.",
|
||
"answer": "α: 2% Sn, β: 100% Sn; % α = (100 - 35)/(100 - 2) × 100% = 66%, % β = 34%"
|
||
}
|
||
]
|
||
```
|
||
处理第 137/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Consider a Pb-70% Sn alloy. Determine if the alloy is hypoeutectic or hypereutectic.",
|
||
"answer": "hypereutectic"
|
||
},
|
||
{
|
||
"question": "Consider a Pb-70% Sn alloy. Determine the composition of the first solid to form during solidification.",
|
||
"answer": "98% Sn"
|
||
},
|
||
{
|
||
"question": "Consider a Pb-70% Sn alloy. Determine the amounts and compositions of each phase at 184°C.",
|
||
"answer": "β: 97.5% Sn L: 61.9% Sn % β=(70-61.9)/(97.5-61.9)×100%=22.8% % L=77.2%"
|
||
},
|
||
{
|
||
"question": "Consider a Pb-70% Sn alloy. Determine the amounts and compositions of each phase at 182°C.",
|
||
"answer": "α: 19% Sn β: 97.5% Sn % α=(97.5-70)/(97.5-19)×100%=35% % β=65%"
|
||
},
|
||
{
|
||
"question": "Consider a Pb-70% Sn alloy. Determine the amounts and compositions of each microconstituent at 182°C.",
|
||
"answer": "primary β: 97.5% Sn % primary β=22.8% eutectic: 61.9% Sn % eutectic=77.2%"
|
||
},
|
||
{
|
||
"question": "Consider a Pb-70% Sn alloy. Determine the amounts and compositions of each phase at 25°C.",
|
||
"answer": "α: 2% Sn β: 100% Sn % α=(100-70)/(100-2)×100%=30% % β=70%"
|
||
}
|
||
]
|
||
```
|
||
处理第 138/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Consider an Al-4% Si alloy. Determine if the alloy is hypo eutectic or hyper eutectic.",
|
||
"answer": "hypo eutectic"
|
||
},
|
||
{
|
||
"question": "Consider an Al-4% Si alloy. Determine the composition of the first solid to form during solidification.",
|
||
"answer": "1% Si"
|
||
},
|
||
{
|
||
"question": "Consider an Al-4% Si alloy. Determine the amounts and compositions of each phase at 578 degrees Celsius.",
|
||
"answer": "α: 1.65% Si, L: 12.6% Si. % α = (12.6 - 4)/(12.6 - 1.65) = 78.5%, % L = 21.5%"
|
||
},
|
||
{
|
||
"question": "Consider an Al-4% Si alloy. Determine the amounts and compositions of each phase at 576 degrees Celsius.",
|
||
"answer": "α: 1.65% Si, β: 99.83% Si. % α = (99.83 - 4)/(99.83 - 1.65) = 97.6%, % β = 2.4%"
|
||
},
|
||
{
|
||
"question": "Consider an Al-4% Si alloy. Determine the amounts and compositions of each microconstituent at 576 degrees Celsius.",
|
||
"answer": "primary α: 1.65% Si, % primary α = 78.5%. eutectic: 12.6% Si, % eutectic = 21.5%"
|
||
},
|
||
{
|
||
"question": "Consider an Al-4% Si alloy. Determine the amounts and compositions of each phase at 25 degrees Celsius.",
|
||
"answer": "α: 0% Si, β: 100% Si. % α = 100 - 4 = 96%, % β = 4%"
|
||
}
|
||
]
|
||
```
|
||
处理第 139/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Consider a Al-25% Si alloy. Determine if the alloy is hypo eutectic or hyper eutectic.",
|
||
"answer": "hyper eutectic"
|
||
},
|
||
{
|
||
"question": "Consider a Al-25% Si alloy. Determine the composition of the first solid to form during solidification.",
|
||
"answer": "100% Si"
|
||
},
|
||
{
|
||
"question": "Consider a Al-25% Si alloy. Determine the amounts and compositions of each phase at 578°C.",
|
||
"answer": "β: 99.83% Si, L: 12.6% Si\n% L = (99.83 - 25)/(99.83 - 12.6) = 85.8%, % β = 14.2%"
|
||
},
|
||
{
|
||
"question": "Consider a Al-25% Si alloy. Determine the amounts and compositions of each phase at 576°C.",
|
||
"answer": "α: 1.65% Si, β: 99.83% Si\n% α = (99.83 - 25)/(99.83 - 1.65) = 76.2%, % β = 23.8%"
|
||
},
|
||
{
|
||
"question": "Consider a Al-25% Si alloy. Determine the amounts and compositions of each microconstituent at 576°C.",
|
||
"answer": "primary β: 99.83% Si, % primary β = 14.2%\neutectic: 12.6% Si, % eutectic = 85.8%"
|
||
},
|
||
{
|
||
"question": "Consider a Al-25% Si alloy. Determine the amounts and compositions of each phase at 25°C.",
|
||
"answer": "α: 0% Si, β: 100% Si\n% α = (100 - 25)/(100 - 0) = 75%, % β = 25%"
|
||
}
|
||
]
|
||
```
|
||
处理第 140/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "A Pb-Sn alloy contains 45% α and 55% β at 100°C. Determine the composition of the alloy.",
|
||
"answer": "45% α = (98.0 - x)/(98.0 - 5) × 100 or x = 56.15% Sn"
|
||
},
|
||
{
|
||
"question": "Is the Pb-Sn alloy with 56.15% Sn hypoeutectic or hypereutectic?",
|
||
"answer": "Hypoeutectic"
|
||
}
|
||
]
|
||
```
|
||
处理第 141/821 条数据...
|
||
[
|
||
{
|
||
"question": "An Al-Si alloy contains 85% α and 15% β at 500°C. Determine the composition of the alloy.",
|
||
"answer": "85 = (100 - x) / (100 - 1) × 100 or x = 15.85% Si"
|
||
},
|
||
{
|
||
"question": "Is the alloy hypoeutectic or hypereutectic?",
|
||
"answer": "Hypereutectic"
|
||
}
|
||
]
|
||
处理第 142/821 条数据...
|
||
It's a single issue.
|
||
处理第 143/821 条数据...
|
||
It's a single issue.
|
||
处理第 144/821 条数据...
|
||
[
|
||
{
|
||
"question": "Observation of a microstructure shows that there is 28% eutectic and 72% primary β in an Al-Li alloy. Determine the composition of the alloy and whether it is hypoeutectic or hypereutectic.",
|
||
"answer": "28=(20.4-x)/(20.4-9.9)×100 or x=17.46% Li Hypereutectic"
|
||
},
|
||
{
|
||
"question": "How much α and β are in the eutectic microconstituent?",
|
||
"answer": "% αEst=(20.4-9.9)/(20.4-4)×100%=64% and % βEst=36%"
|
||
}
|
||
]
|
||
处理第 145/821 条数据...
|
||
[
|
||
{
|
||
"question": "Calculate the total amount of alpha and beta in a Pb-50% Sn alloy at 182°C.",
|
||
"answer": "alpha_total = (97.5 - 50)/(97.5 - 19) × 100% = 60.5%, beta_Total = 39.5%"
|
||
},
|
||
{
|
||
"question": "Calculate the amount of each microconstituent (primary alpha and eutectic) in a Pb-50% Sn alloy at 182°C.",
|
||
"answer": "alpha_Primary = (61.9 - 50)/(61.9 - 19) × 100% = 27.7%, Eutectic = 72.3%"
|
||
},
|
||
{
|
||
"question": "What fraction of the total alpha in the alloy is contained in the eutectic microconstituent for a Pb-50% Sn alloy at 182°C?",
|
||
"answer": "alpha_in eutectic = 60.5 - 27.7 = 32.8%, f = 32.8/60.5 = 0.54"
|
||
}
|
||
]
|
||
处理第 146/821 条数据...
|
||
[
|
||
{
|
||
"question": "Recommend an artificial age-hardening heat treatment for a Cu-1.2% Be alloy (see Figure 12-34). Include appropriate temperatures.",
|
||
"answer": "For the Cu-1.2% Be alloy, the peritectic temperature is 870 degrees C; above this temperature, liquid may form. The solvus temperature is about 530 degrees C. Therefore: 1) Solution treat between 530 degrees C and 870 degrees C (780 degrees C is typical for beryllium copper alloys) 2) Quench 3) Age below 530 degrees C (330 degrees C is typical for these alloys)"
|
||
},
|
||
{
|
||
"question": "Compare the amount of the gamma2 precipitate that forms by artificial aging at 400 degrees C with the amount of the precipitate that forms by natural aging for a Cu-1.2% Be alloy.",
|
||
"answer": "We can perform lever law calculations at 400 degrees C and at room temperature. The solubility of Be in Cu at 400 degrees C is about 0.6% Be and that at room temperature is about 0.2% Be: gamma2 at 400 degrees C = (1.2-0.6)/(11.7-0.6)*100 = 5.4% gamma2 at room T = (1.2-0.2)/(12-0.2)*100 = 8.5%"
|
||
}
|
||
]
|
||
处理第 147/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For an Fe-0.35% C alloy, determine the temperature at which austenite first begins to transform on cooling.",
|
||
"answer": "795°C"
|
||
},
|
||
{
|
||
"question": "For an Fe-0.35% C alloy, determine the primary microconstituent that forms.",
|
||
"answer": "Primary α-ferrite"
|
||
},
|
||
{
|
||
"question": "For an Fe-0.35% C alloy, determine the composition and amount of each phase present at 728°C.",
|
||
"answer": "α: 0.0218% C, %α = (0.77 - 0.35)/(0.77 - 0.0218) × 100 = 56.1%; γ: 0.77% C, %γ = 43.9%"
|
||
},
|
||
{
|
||
"question": "For an Fe-0.35% C alloy, determine the composition and amount of each phase present at 726°C.",
|
||
"answer": "α: 0.0218% C, %α = (6.67 - 0.35)/(6.67 - 0.0218) × 100 = 95.1%; Fe3C: 6.67% C, %Fe3C = 4.9%"
|
||
},
|
||
{
|
||
"question": "For an Fe-0.35% C alloy, determine the composition and amount of each microconstituent present at 726°C.",
|
||
"answer": "Primary α: 0.0218% C, % primary α = 56.1%; Pearlite: 0.77% C, % Pearlite = 43.9%"
|
||
}
|
||
]
|
||
```
|
||
处理第 148/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For an Fe-1.15% C alloy, determine the temperature at which austenite first begins to transform on cooling.",
|
||
"answer": "880 degrees C"
|
||
},
|
||
{
|
||
"question": "For an Fe-1.15% C alloy, determine the primary microconstituent that forms.",
|
||
"answer": "primary Fe3C"
|
||
},
|
||
{
|
||
"question": "For an Fe-1.15% C alloy, determine the composition and amount of each phase present at 728 degrees C.",
|
||
"answer": "Fe3C: 6.67% C, % Fe3C = (1.15-0.77)/(6.67-0.77) x 100 = 6.4%; gamma: 0.77% C, % gamma = 93.6%"
|
||
},
|
||
{
|
||
"question": "For an Fe-1.15% C alloy, determine the composition and amount of each phase present at 720 degrees C.",
|
||
"answer": "alpha: 0.0218% C, % alpha = (6.67-1.15)/(6.67-0.0218) x 100 = 83%; Fe3C: 6.67% C, % Fe3C = 17%"
|
||
},
|
||
{
|
||
"question": "For an Fe-1.15% C alloy, determine the composition and amount of each microconstituent present at 726 degrees C.",
|
||
"answer": "primary Fe3C: 6.67% C, % primary Fe3C = 6.4%; pearlite: 0.77% C, % Pearlite = 93.6%"
|
||
}
|
||
]
|
||
```
|
||
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|
||
[
|
||
{
|
||
"question": "A steel contains 8% cementite and 92% ferrite at room temperature. Estimate the carbon content of the steel.",
|
||
"answer": "alpha=0.92=(6.67-x)/(6.67-0); x=0.53% C"
|
||
},
|
||
{
|
||
"question": "Is the steel hypoeutectoid or hypereutectoid?",
|
||
"answer": "Hypoeutectoid"
|
||
}
|
||
]
|
||
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|
||
[
|
||
{
|
||
"question": "A steel contains 18% cementite and 82% ferrite at room temperature. Estimate the carbon content of the steel.",
|
||
"answer": "alpha=0.82=(6.67-x)/(6.67-0), x=1.20%C"
|
||
},
|
||
{
|
||
"question": "Is the steel hypoeutectoid or hypereutectoid?",
|
||
"answer": "Hypereutectoid"
|
||
}
|
||
]
|
||
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|
||
[
|
||
{
|
||
"question": "A steel contains 18% pearlite and 82% primary ferrite at room temperature. Estimate the carbon content of the steel.",
|
||
"answer": "primary α=0.82=(0.77-x)/(0.77-0.0218)\nx=0.156% C"
|
||
},
|
||
{
|
||
"question": "Is the steel hypoeutectoid or hypereutectoid?",
|
||
"answer": "Hypoeutectoid"
|
||
}
|
||
]
|
||
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|
||
[
|
||
{
|
||
"question": "A steel contains 94% pearlite and 6% primary cementite at room temperature. Estimate the carbon content of the steel.",
|
||
"answer": "Pearlite = 0.94 = (6.67 - x) / (6.67 - 0.77). Solving for x gives x = 1.124% C."
|
||
},
|
||
{
|
||
"question": "Is the steel hypoeutectoid or hypereutectoid?",
|
||
"answer": "Hypereutectoid"
|
||
}
|
||
]
|
||
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|
||
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|
||
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||
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|
||
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|
||
[
|
||
{
|
||
"question": "A steel is heated until 40% austenite, with a carbon content of 0.5%, forms. Estimate the temperature.",
|
||
"answer": "In order for γ to contain 0.5% C, the austenitizing temperature must be about 760°C (from the tie line)."
|
||
},
|
||
{
|
||
"question": "A steel is heated until 40% austenite, with a carbon content of 0.5%, forms. Estimate the overall carbon content of the steel.",
|
||
"answer": "At 760°C: 0.4 = (x - 0.02) / (0.5 - 0.02) x = 0.212% C"
|
||
}
|
||
]
|
||
处理第 156/821 条数据...
|
||
[
|
||
{
|
||
"question": "An isothermally transformed eutectoid steel is found to have a yield strength of 410 MPa. Estimate the transformation temperature.",
|
||
"answer": "transformation temperature = 615 degrees C"
|
||
},
|
||
{
|
||
"question": "An isothermally transformed eutectoid steel is found to have a yield strength of 410 MPa. Estimate the interlamellar spacing in the pearlite.",
|
||
"answer": "1 / lambda = 60,000 or lambda = 1.67 x 10^-5 cm"
|
||
}
|
||
]
|
||
处理第 157/821 条数据...
|
||
[
|
||
{
|
||
"question": "Determine the required transformation temperature and microconstituent if an eutectoid steel is to have a hardness value of HRC 38",
|
||
"answer": "600 degrees Celsius, pearlite"
|
||
},
|
||
{
|
||
"question": "Determine the required transformation temperature and microconstituent if an eutectoid steel is to have a hardness value of HRC 42",
|
||
"answer": "400 degrees Celsius, bainite"
|
||
},
|
||
{
|
||
"question": "Determine the required transformation temperature and microconstituent if an eutectoid steel is to have a hardness value of HRC 48",
|
||
"answer": "340 degrees Celsius, bainite"
|
||
},
|
||
{
|
||
"question": "Determine the required transformation temperature and microconstituent if an eutectoid steel is to have a hardness value of HRC 52",
|
||
"answer": "300 degrees Celsius, bainite"
|
||
}
|
||
]
|
||
处理第 158/821 条数据...
|
||
[
|
||
{
|
||
"question": "What is the hardness in Rockwell C scale (HRC) of an eutectoid steel that has been heated to 800 degrees Celsius for 1 hour, quenched to 350 degrees Celsius and held for 750 seconds, and finally quenched to room temperature?",
|
||
"answer": "HRC = 47"
|
||
},
|
||
{
|
||
"question": "What is the microstructure of an eutectoid steel that has been heated to 800 degrees Celsius for 1 hour, quenched to 350 degrees Celsius and held for 750 seconds, and finally quenched to room temperature?",
|
||
"answer": "The microstructure is all bainite."
|
||
}
|
||
]
|
||
处理第 159/821 条数据...
|
||
It's a single issue.
|
||
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||
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|
||
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|
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||
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|
||
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|
||
[
|
||
{
|
||
"question": "Calculate % IC of the interatomic bonds for the intermetallic compound TiAl3.",
|
||
"answer": "The percent ionic character is a function of the electron negativities of the ions XA and XB according to Equation. The electronegativities for Al and Ti are both 1.5. Therefore, the percent ionic character is determined using Equation as follows: % IC = [1 - exp (-0.25)(1.5 - 1.5)^2] x 100 = 0%."
|
||
},
|
||
{
|
||
"question": "On the basis of this result what type of interatomic bonding would you expect to be found in TiAl3? (Options: van der Waals, ionic, metallic, covalent)",
|
||
"answer": "Because the percent ionic character is zero and this intermetallic compound is composed of two metals, the bonding is completely metallic."
|
||
}
|
||
]
|
||
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|
||
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|
||
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||
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|
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||
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|
||
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|
||
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|
||
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|
||
```json
|
||
[
|
||
{
|
||
"question": "Given rhodium's atomic radius of 0.1345 nm, density of 12.41 g/cm³, and atomic weight of 102.91 g/mol, calculate the number of atoms per unit cell using the formula n = (ρ * V_C * N_A) / A.",
|
||
"answer": "The number of atoms per unit cell is calculated as follows: n = (12.41 g/cm³ * V_C * 6.022 × 10²³ atoms/mol) / 102.91 g/mol = 7.26 × 10²² * V_C atoms/cm³."
|
||
},
|
||
{
|
||
"question": "Determine the unit cell volume V_C for a simple cubic (SC) crystal structure given the atomic radius R = 0.1345 nm.",
|
||
"answer": "For SC, V_C = a³ = (2R)³ = 8R³ = 8 * (1.345 × 10⁻⁸ cm)³ = 8 * 2.434 × 10⁻²⁴ cm³ = 1.947 × 10⁻²³ cm³."
|
||
},
|
||
{
|
||
"question": "Calculate the number of atoms per unit cell n for a simple cubic (SC) crystal structure using the derived V_C.",
|
||
"answer": "n = 7.26 × 10²² atoms/cm³ * 1.947 × 10⁻²³ cm³ = 1.41 atoms."
|
||
},
|
||
{
|
||
"question": "Determine the unit cell volume V_C for a body-centered cubic (BCC) crystal structure given the atomic radius R = 0.1345 nm.",
|
||
"answer": "For BCC, V_C = a³ = (4R/√3)³ = 64R³ / (3√3) = 64 * (1.345 × 10⁻⁸ cm)³ / (3√3) = 64 * 2.434 × 10⁻²⁴ cm³ / 5.196 = 2.999 × 10⁻²³ cm³."
|
||
},
|
||
{
|
||
"question": "Calculate the number of atoms per unit cell n for a body-centered cubic (BCC) crystal structure using the derived V_C.",
|
||
"answer": "n = 7.26 × 10²² atoms/cm³ * 2.999 × 10⁻²³ cm³ = 2.18 atoms."
|
||
},
|
||
{
|
||
"question": "Determine the unit cell volume V_C for a face-centered cubic (FCC) crystal structure given the atomic radius R = 0.1345 nm.",
|
||
"answer": "For FCC, V_C = a³ = (2R√2)³ = 16R³√2 = 16 * (1.345 × 10⁻⁸ cm)³ * √2 = 16 * 2.434 × 10⁻²⁴ cm³ * 1.414 = 5.508 × 10⁻²³ cm³."
|
||
},
|
||
{
|
||
"question": "Calculate the number of atoms per unit cell n for a face-centered cubic (FCC) crystal structure using the derived V_C.",
|
||
"answer": "n = 7.26 × 10²² atoms/cm³ * 5.508 × 10⁻²³ cm³ = 4.0 atoms."
|
||
},
|
||
{
|
||
"question": "Based on the calculated number of atoms per unit cell n for SC, BCC, and FCC crystal structures, determine rhodium's crystal structure.",
|
||
"answer": "The calculated n values are: SC = 1.41 atoms, BCC = 2.18 atoms, FCC = 4.0 atoms. Since only FCC matches the theoretical value of 4.0 atoms per unit cell, rhodium's crystal structure is FCC."
|
||
}
|
||
]
|
||
```
|
||
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|
||
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|
||
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|
||
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|
||
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||
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|
||
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||
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||
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||
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|
||
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|
||
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|
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||
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|
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||
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|
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||
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|
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||
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|
||
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|
||
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|
||
处理第 199/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Polyethylene may be fluorinated by inducing the random substitution of fluorine atoms for hydrogen. For this polymer, determine the concentration of F (in wt %) that must be added if this substitution occurs for 18.6% of all of the original hydrogen atoms. Atomic weights: Carbon 12.01 g/mol, Hydrogen 1.008 g/mol, Fluorine 19.00 g/mol.",
|
||
"answer": "Consider 50 carbon atoms; for polyethylene, there are 100 possible side-bonding sites. For 18.6% substitution, 18.6 sites are occupied by F and 81.4 by H. The mass of 50 carbon atoms, m_C, is (50)(12.01 g/mol) = 600.5 g. The mass of hydrogen, m_H, is (81.4)(1.008 g/mol) = 82.05 g. The mass of fluorine, m_F, is (18.6)(19.00 g/mol) = 353.4 g. The concentration of fluorine, C_F, is (353.4 g) / (600.5 g + 82.05 g + 353.4 g) × 100 = 34.1 wt %."
|
||
},
|
||
{
|
||
"question": "Polyethylene may be fluorinated by inducing the random substitution of fluorine atoms for hydrogen. For this polymer, determine the concentration of F (in wt %) that must be added to completely fluorinate the material, i.e., to produce polytetrafluoroethylene (PTFE). Atomic weights: Carbon 12.01 g/mol, Fluorine 19.00 g/mol.",
|
||
"answer": "For complete fluorination, consider 50 carbon atoms with 100 side-bonding sites, all occupied by F. The mass of 50 carbon atoms, m_C, is (50)(12.01 g/mol) = 600.5 g. The mass of fluorine, m_F, is (100)(19.00 g/mol) = 1900.00 g. The concentration of fluorine, C_F, is (1900.00 g) / (600.5 g + 1900.00 g) × 100 = 76.0 wt %."
|
||
}
|
||
]
|
||
```
|
||
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|
||
It's a single issue.
|
||
处理第 201/821 条数据...
|
||
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|
||
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|
||
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|
||
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|
||
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|
||
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|
||
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|
||
处理第 205/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is the equation to calculate the number of vacancies per cubic meter in a metal at a given temperature?",
|
||
"answer": "The number of vacancies per cubic meter may be calculated using the equation: N_v = N exp (-Q_v / kT), where N_v is the number of vacancies, N is the total number of atomic sites per cubic meter, Q_v is the energy for vacancy formation, k is the Boltzmann constant, and T is the temperature in Kelvin."
|
||
},
|
||
{
|
||
"question": "How can the total number of atomic sites per cubic meter (N) be determined if the number of vacancies (N_v) and other parameters are known?",
|
||
"answer": "The total number of atomic sites per cubic meter (N) can be determined by solving the equation N = N_v / exp (-Q_v / kT) = N_v exp (Q_v / kT)."
|
||
},
|
||
{
|
||
"question": "Given that the density of the metal is the same at two different temperatures, how are the total number of atomic sites (N) related at these temperatures?",
|
||
"answer": "If the densities are the same at both temperatures, the total number of atomic sites per cubic meter (N) will also be the same at both temperatures. This means N_v1 exp (Q_v / kT1) = N_v2 exp (Q_v / kT2), where subscripts 1 and 2 denote the lower and higher temperatures, respectively."
|
||
},
|
||
{
|
||
"question": "How can the number of vacancies at a lower temperature (N_v1) be calculated if the number of vacancies at a higher temperature (N_v2) and other parameters are known?",
|
||
"answer": "The number of vacancies at the lower temperature (N_v1) can be calculated using the equation: N_v1 = N_v2 exp (Q_v / k) (1/T2 - 1/T1), where T1 and T2 are the lower and higher temperatures in Kelvin, respectively."
|
||
},
|
||
{
|
||
"question": "Given N_v2 = 1.1 x 10^24 m^-3 at T2 = 864°C, Q_v = 1.25 eV/atom, and k = 8.62 x 10^-5 eV/atom-K, calculate the number of vacancies (N_v1) at T1 = 463°C.",
|
||
"answer": "First, convert temperatures to Kelvin: T2 = 864 + 273 = 1137 K, T1 = 463 + 273 = 736 K. Then, calculate N_v1 using the equation: N_v1 = (1.1 x 10^24 m^-3) exp (1.25 eV/atom / 8.62 x 10^-5 eV/atom-K) (1/1137 K - 1/736 K). The result is N_v1 = 5.45 x 10^22 m^-3."
|
||
}
|
||
]
|
||
```
|
||
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|
||
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|
||
处理第 207/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Which of the following oxides would you expect to form substitutional solid solutions that have complete (i.e., 100%) solubility with MnO? Explain your answer for MgO.",
|
||
"answer": "In order to have a high degree (or 100% solubility) of one ceramic compound in another ceramic compound, the following criteria must be met: the relative sizes of cations/anions must be similar (< ±15%); the valence (or charges) on the cations and anions of both compounds must be the same; and the crystals structures of the two compounds must the same or similar. For MgO, the ionic radii of the Mn2+ and Mg2+ are 0.067 nm and 0.072 nm, respectively. Therefore the percentage difference in ionic radii, Δr% is determined as follows: Δr% = (0.072 nm - 0.067 nm) / 0.072 nm × 100 = 6.9%, which value is within the acceptable range for a high degree of solubility."
|
||
},
|
||
{
|
||
"question": "Which of the following oxides would you expect to form substitutional solid solutions that have complete (i.e., 100%) solubility with MnO? Explain your answer for CaO.",
|
||
"answer": "In order to have a high degree (or 100% solubility) of one ceramic compound in another ceramic compound, the following criteria must be met: the relative sizes of cations/anions must be similar (< ±15%); the valence (or charges) on the cations and anions of both compounds must be the same; and the crystals structures of the two compounds must the same or similar. For CaO, the ionic radii of the Mn2+ and Ca2+ are 0.067 nm and 0.100 nm, respectively. Therefore, the percentage difference in ionic radii, Δr% is determined as follows: Δr% = (0.100 nm - 0.067 nm) / 0.100 nm × 100 = 33%. This Δr% value is much larger than the ±15% range, and, therefore, CaO is not expected to experience any appreciable solubility in MnO."
|
||
},
|
||
{
|
||
"question": "Which of the following oxides would you expect to form substitutional solid solutions that have complete (i.e., 100%) solubility with MnO? Explain your answer for BeO.",
|
||
"answer": "In order to have a high degree (or 100% solubility) of one ceramic compound in another ceramic compound, the following criteria must be met: the relative sizes of cations/anions must be similar (< ±15%); the valence (or charges) on the cations and anions of both compounds must be the same; and the crystals structures of the two compounds must the same or similar. For BeO, the ionic radii of the Mn2+ and Be2+ are 0.067 nm and 0.035 nm, respectively. Therefore, the percentage difference in ionic radii, Δr% is determined as follows: Δr% = (0.067 nm - 0.035 nm) / 0.067 nm × 100 = 48%. This Δr% value is much larger than the ±15% range, and, therefore, BeO is not expected to experience any appreciable solubility in MnO."
|
||
},
|
||
{
|
||
"question": "Which of the following oxides would you expect to form substitutional solid solutions that have complete (i.e., 100%) solubility with MnO? Explain your answer for NiO.",
|
||
"answer": "In order to have a high degree (or 100% solubility) of one ceramic compound in another ceramic compound, the following criteria must be met: the relative sizes of cations/anions must be similar (< ±15%); the valence (or charges) on the cations and anions of both compounds must be the same; and the crystals structures of the two compounds must the same or similar. For NiO, the ionic radii of the Mn2+ and Ni2+ are 0.067 nm and 0.069 nm, respectively. Therefore, the percentage difference in ionic radii Δr% is determined as follows: Δr% = (0.069 nm - 0.067 nm) / 0.069 nm × 100 = 3%, which value is, of course, within the acceptable range for a high degree of solubility."
|
||
}
|
||
]
|
||
```
|
||
处理第 208/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Which of the given elements would form a substitutional solid solution with copper having complete solubility, considering atomic radius difference less than ±15%, same crystal structure, similar electronegativity, and same or nearly same valence?",
|
||
"answer": "Ni, Pd, and Pt meet all of the criteria and thus form substitutional solid solutions having complete solubility. At elevated temperatures Co and Fe experience allotropic transformations to the FCC crystal structure, and thus display complete solid solubility at these temperatures."
|
||
},
|
||
{
|
||
"question": "Which of the given elements would form a substitutional solid solution with copper having incomplete solubility, considering differences in atomic radii, crystal structures, electronegativities, and valences?",
|
||
"answer": "Ag, Al, Co, Cr, Fe, and Zn form substitutional solid solutions of incomplete solubility. All these metals have either BCC or HCP crystal structures, and/or the difference between their atomic radii and that for Cu are greater than ±15%, and/or have a valence different than 2+."
|
||
},
|
||
{
|
||
"question": "Which of the given elements would form an interstitial solid solution with copper, considering significantly smaller atomic radii compared to copper?",
|
||
"answer": "C, H, and O form interstitial solid solutions. These elements have atomic radii that are significantly smaller than the atomic radius of Cu."
|
||
}
|
||
]
|
||
```
|
||
处理第 209/821 条数据...
|
||
It's a single issue.
|
||
处理第 210/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is the average density of the alloy composed of 12.5 wt% of metal A and 87.5 wt% of metal B, given the densities of metals A and B are 4.27 g/cm³ and 6.35 g/cm³, respectively?",
|
||
"answer": "The average density of the alloy is calculated using the formula: ρ_ave = 100 / ( (C_A/ρ_A) + (C_B/ρ_B) ), where C_A = 12.5 wt%, C_B = 87.5 wt%, ρ_A = 4.27 g/cm³, and ρ_B = 6.35 g/cm³. Substituting the values: ρ_ave = 100 / ( (12.5/4.27) + (87.5/6.35) ) = 5.96 g/cm³."
|
||
},
|
||
{
|
||
"question": "What is the average atomic weight of the alloy composed of 12.5 wt% of metal A and 87.5 wt% of metal B, given the atomic weights of metals A and B are 61.4 g/mol and 125.7 g/mol, respectively?",
|
||
"answer": "The average atomic weight of the alloy is calculated using the formula: A_ave = 100 / ( (C_A/A_A) + (C_B/A_B) ), where C_A = 12.5 wt%, C_B = 87.5 wt%, A_A = 61.4 g/mol, and A_B = 125.7 g/mol. Substituting the values: A_ave = 100 / ( (12.5/61.4) + (87.5/125.7) ) = 113.5 g/mol."
|
||
},
|
||
{
|
||
"question": "Given the average density of the alloy (ρ_ave = 5.96 g/cm³), the average atomic weight (A_ave = 113.5 g/mol), the unit cell edge length (a = 0.395 nm), and Avogadro's number (N_A = 6.022 × 10²³ atoms/mol), determine the number of atoms per unit cell (n) for the alloy.",
|
||
"answer": "The number of atoms per unit cell is calculated using the formula: n = (ρ_ave * a³ * N_A) / A_ave. First, convert the unit cell edge length to cm: a = 0.395 nm = 3.95 × 10⁻⁸ cm. Then, substitute the values: n = (5.96 g/cm³ * (3.95 × 10⁻⁸ cm)³ * 6.022 × 10²³ atoms/mol) / 113.5 g/mol = 2.00 atoms/unit cell."
|
||
},
|
||
{
|
||
"question": "Based on the number of atoms per unit cell (n = 2.00), determine whether the crystal structure for this alloy is simple cubic, face-centered cubic, or body-centered cubic.",
|
||
"answer": "For n = 1, the crystal structure is simple cubic; for n = 2, it is body-centered cubic (BCC); and for n = 4, it is face-centered cubic (FCC). Since n = 2.00, the crystal structure is body-centered cubic."
|
||
}
|
||
]
|
||
```
|
||
处理第 211/821 条数据...
|
||
It's a single issue.
|
||
处理第 212/821 条数据...
|
||
It's a single issue.
|
||
处理第 213/821 条数据...
|
||
It's a single issue.
|
||
处理第 214/821 条数据...
|
||
It's a single issue.
|
||
处理第 215/821 条数据...
|
||
It's a single issue.
|
||
处理第 216/821 条数据...
|
||
It's a single issue.
|
||
处理第 217/821 条数据...
|
||
It's a single issue.
|
||
处理第 218/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "If an increased steady-state flow rate of O2 (oxygen molecules per second) to the cornea is desired, is increasing the contact lens thickness likely to be useful? Note: the flow rate is equal to product of the diffusion flux and an area of interest through which diffusion occurs.",
|
||
"answer": "If the contact lens thickness is increased and all other factors are preserved, the concentration gradient decrease and the oxygen diffusion flux should decrease."
|
||
},
|
||
{
|
||
"question": "If an increased steady-state flow rate of O2 (oxygen molecules per second) to the cornea is desired, is increasing the diffusivity of oxygen gas by decreasing the contact lens porosity likely to be useful? Note: the flow rate is equal to product of the diffusion flux and an area of interest through which diffusion occurs.",
|
||
"answer": "If the lens features more voids, we expect the oxygen to permeate the membrane faster since oxygen should be able to transit voids faster than the bulk lens material."
|
||
},
|
||
{
|
||
"question": "If an increased steady-state flow rate of O2 (oxygen molecules per second) to the cornea is desired, is increasing the ambient temperature likely to be useful? Note: the flow rate is equal to product of the diffusion flux and an area of interest through which diffusion occurs.",
|
||
"answer": "Increasing the temperature should increase the diffusivity and therefore increase the diffusion flux and flow rate."
|
||
},
|
||
{
|
||
"question": "If an increased steady-state flow rate of O2 (oxygen molecules per second) to the cornea is desired, is increasing the ambient partial pressure of oxygen gas likely to be useful? Note: the flow rate is equal to product of the diffusion flux and an area of interest through which diffusion occurs.",
|
||
"answer": "Increasing the concentration of oxygen in the ambient environment will increase the concentration gradient, thereby increasing the diffusion flux of oxygen."
|
||
}
|
||
]
|
||
```
|
||
处理第 219/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For a steel alloy, given a carburizing heat treatment of 16 h duration at 757°C raises the carbon concentration to 0.5 wt% at a point 2.3 mm from the surface, estimate the time necessary to achieve the same concentration at an 8 mm position for an identical steel and at a carburizing temperature of 1130°C. Assume that D0 is 4.6×10−5 m2/s and Qd is 104 kJ/mol.",
|
||
"answer": "The solution involves using Fick's second law and the temperature dependence of the diffusion coefficient. The key equation is:\n\nt2 = (t1 * x2^2 / x1^2) * exp[-(Qd/R) * (1/T1 - 1/T2)]\n\nGiven:\nt1 = 16 h\nT1 = 757°C = 1030 K\nx1 = 2.3 mm = 2.3×10−3 m\nx2 = 8 mm = 8×10−3 m\nT2 = 1130°C = 1403 K\nQd = 104 kJ/mol = 104,000 J/mol\nR = 8.31 J/mol-K\n\nSubstituting these values:\n\nt2 = [(16 h) * (8×10−3 m)^2 / (2.3×10−3 m)^2] * exp[-(104,000 J/mol / 8.31 J/mol-K) * (1/1030 K - 1/1403 K)]\n\nCalculating the exponent:\n(1/1030 - 1/1403) = (9.7087×10−4 - 7.1276×10−4) = 2.5811×10−4\n\nExponent term:\nexp[-(104,000 / 8.31) * 2.5811×10−4] = exp[-3.233] ≈ 0.0394\n\nTime ratio:\n(8/2.3)^2 = (3.478)^2 ≈ 12.10\n\nFinal calculation:\nt2 = 16 h * 12.10 * 0.0394 ≈ 7.63 h"
|
||
}
|
||
]
|
||
```
|
||
处理第 220/821 条数据...
|
||
It's a single issue.
|
||
处理第 221/821 条数据...
|
||
It's a single issue.
|
||
处理第 222/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For a bronze alloy, the stress at which plastic deformation begins is 277 MPa and the modulus of elasticity is 117 GPa. What is the maximum load that may be applied to a specimen having a cross-sectional area of 327 mm2 without plastic deformation?",
|
||
"answer": "The maximum load that may be applied without yielding is calculated using the formula Fy = σy * A0. Given σy = 277 MPa = 277 × 10^6 N/m2 and A0 = 327 mm2 = 327 × 10^-6 m2, then Fy = (277 × 10^6 N/m2)(327 × 10^-6 m2) = 90,579 N."
|
||
},
|
||
{
|
||
"question": "For a bronze alloy, the stress at which plastic deformation begins is 277 MPa and the modulus of elasticity is 117 GPa. If the original specimen length is 148 mm, what is the maximum length to which it may be stretched without causing plastic deformation?",
|
||
"answer": "The maximum length to which the specimen may be stretched without plastic deformation is calculated using the formula li = l0(σy/E + 1). Given l0 = 148 mm = 148 × 10^-3 m, σy = 277 MPa = 277 × 10^6 N/m2, and E = 117 GPa = 117 × 10^9 N/m2, then li = (148 × 10^-3 m)((277 × 10^6 N/m2)/(117 × 10^9 N/m2) + 1) = 0.14835 m = 148.35 mm."
|
||
}
|
||
]
|
||
```
|
||
处理第 223/821 条数据...
|
||
It's a single issue.
|
||
处理第 224/821 条数据...
|
||
It's a single issue.
|
||
处理第 225/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Which of the four metals or alloys (Aluminum alloy, Brass alloy, Copper, Steel alloy) will not experience plastic deformation when a tensile load of 24,500 N is applied to a cylindrical rod with a diameter of 10.0 mm?",
|
||
"answer": "To determine which materials will not experience plastic deformation, compute the stress using the formula σ = F/A0. The cross-sectional area A0 is π*(10.0×10^-3 m/2)^2 = 7.854×10^-5 m^2. The stress σ = 24,500 N / 7.854×10^-5 m^2 = 312 MPa. The yield strengths of the materials are: Aluminum alloy (255 MPa), Brass alloy (345 MPa), Copper (250 MPa), Steel alloy (450 MPa). Only Brass alloy and Steel alloy have yield strengths greater than 312 MPa."
|
||
},
|
||
{
|
||
"question": "For the remaining candidates (Brass alloy and Steel alloy) that do not experience plastic deformation, which will not elongate more than 0.9 mm under the given load and rod dimensions (length = 380 mm, diameter = 10.0 mm)?",
|
||
"answer": "Compute the elongation for Brass alloy: Δl = (σ * l0)/E = (312 MPa * 380 mm)/(100×10^3 MPa) = 1.19 mm. Compute the elongation for Steel alloy: Δl = (312 MPa * 380 mm)/(207×10^3 MPa) = 0.57 mm. Only Steel alloy has an elongation less than 0.9 mm."
|
||
}
|
||
]
|
||
```
|
||
处理第 226/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Given a metal alloy with a true plastic strain of 0.12 at a true stress of 280 MPa and a strain hardening exponent of 0.3, what is the value of the strength coefficient K?",
|
||
"answer": "To find the strength coefficient K, we use the relationship between true stress and true strain: σ_T = K * (ε_T)^n. Rearranging to solve for K gives K = σ_T / (ε_T)^n. Substituting the given values: K = 280 MPa / (0.12)^0.3 = 280 MPa / 0.6813 = 410.9 MPa."
|
||
},
|
||
{
|
||
"question": "Using the strength coefficient K = 410.9 MPa and strain hardening exponent n = 0.3, what true plastic strain would be expected for a total true plastic stress of 330 MPa?",
|
||
"answer": "To find the true plastic strain at 330 MPa, we use the relationship ε_T = (σ_T / K)^(1/n). Substituting the given values: ε_T = (330 MPa / 410.9 MPa)^(1/0.3) = (0.803)^3.333 = 0.171."
|
||
}
|
||
]
|
||
```
|
||
处理第 227/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Given a cylindrical specimen of a metal alloy with initial length 48.8 mm and diameter 9.09 mm, calculate the true strain when the specimen is plastically elongated to a length of 55 mm.",
|
||
"answer": "The true strain is calculated using the formula: ε_T = ln(l_i / l_0). Substituting the given values: ε_T = ln(55 mm / 48.8 mm) = ln(1.127) ≈ 0.119."
|
||
},
|
||
{
|
||
"question": "Using the true stress of 327 MPa and the true strain of 0.119, along with the strain-hardening exponent of 0.3, calculate the strength coefficient K for the material.",
|
||
"answer": "The relationship between true stress and true strain is given by σ_T = K * (ε_T)^n. Solving for K: K = σ_T / (ε_T)^n = 327 MPa / (0.119)^0.3 ≈ 327 MPa / 0.668 ≈ 489.5 MPa."
|
||
},
|
||
{
|
||
"question": "Calculate the true strain when the specimen is elongated from the initial length of 48.8 mm to a final length of 57.6 mm.",
|
||
"answer": "The true strain is calculated as: ε_T = ln(l_i / l_0) = ln(57.6 mm / 48.8 mm) = ln(1.180) ≈ 0.166."
|
||
},
|
||
{
|
||
"question": "Using the strength coefficient K of 489.5 MPa and the strain-hardening exponent of 0.3, calculate the true stress required to plastically elongate the specimen from 48.8 mm to 57.6 mm.",
|
||
"answer": "The true stress is calculated using the formula: σ_T = K * (ε_T)^n = 489.5 MPa * (0.166)^0.3 ≈ 489.5 MPa * 0.715 ≈ 350.0 MPa."
|
||
}
|
||
]
|
||
```
|
||
处理第 228/821 条数据...
|
||
[
|
||
{
|
||
"question": "Predict whether or not a cylindrical specimen of aluminum oxide with a reported flexural strength of 390 MPa (56,600 psi), radius of 2.5 mm (0.10 in.), and support point separation distance of 30 mm (1.2 in.) would fracture when a load of 620 N (140 lbf) is applied in a three-point bending test.",
|
||
"answer": "Using the equation σ = FL / (πR^3), where F = 620 N, L = 30 × 10^-3 m, and R = 2.5 × 10^-3 m, the flexural strength is calculated as σ = (620 N)(30 × 10^-3 m) / (π)(2.5 × 10^-3 m)^3 = 379 × 10^6 N/m^2 = 379 MPa (53,500 psi). Since this value is less than the given flexural strength of 390 MPa, fracture is not predicted. However, there is some chance of fracture due to variability in ceramic materials."
|
||
},
|
||
{
|
||
"question": "Calculate the value of flexural strength for the three-point bending test on the cylindrical specimen of aluminum oxide with radius 2.5 mm (0.10 in.) and support point separation distance 30 mm (1.2 in.) when a load of 620 N (140 lbf) is applied.",
|
||
"answer": "The flexural strength is calculated using the equation σ = FL / (πR^3), where F = 620 N, L = 30 × 10^-3 m, and R = 2.5 × 10^-3 m. The calculation yields σ = (620 N)(30 × 10^-3 m) / (π)(2.5 × 10^-3 m)^3 = 379 × 10^6 N/m^2 = 379 MPa (53,500 psi)."
|
||
}
|
||
]
|
||
处理第 229/821 条数据...
|
||
[
|
||
{
|
||
"question": "Compute the flexural strength for a completely nonporous specimen of this material given the flexural strength and associated volume fraction porosity for two specimens: (100 MPa, 0.05) and (50 MPa, 0.20).",
|
||
"answer": "Given the flexural strengths at two different volume fraction porosities, we determine the flexural strength for a nonporous material using the equation ln σ_fs = ln σ_0 - nP. Using the provided data, two simultaneous equations are written as ln(100 MPa) = ln σ_0 - (0.05)n and ln(50 MPa) = ln σ_0 - (0.20)n. Solving for n and σ_0 leads to n = 4.62 and σ_0 = 126 MPa. For the nonporous material, P = 0, and σ_0 = σ_fs. Thus, σ_fs for P = 0 is 126 MPa."
|
||
},
|
||
{
|
||
"question": "Compute the flexural strength for a 0.10 volume fraction porosity given the flexural strength and associated volume fraction porosity for two specimens: (100 MPa, 0.05) and (50 MPa, 0.20).",
|
||
"answer": "We are asked for σ_fs at P = 0.10 for this material. Utilizing the equation σ_fs = σ_0 exp(-nP), we substitute σ_0 = 126 MPa and n = 4.62 to get σ_fs = (126 MPa) exp[-(4.62)(0.10)] = 79.4 MPa."
|
||
}
|
||
]
|
||
处理第 230/821 条数据...
|
||
It's a single issue.
|
||
处理第 231/821 条数据...
|
||
It's a single issue.
|
||
处理第 232/821 条数据...
|
||
It's a single issue.
|
||
处理第 233/821 条数据...
|
||
It's a single issue.
|
||
处理第 234/821 条数据...
|
||
It's a single issue.
|
||
处理第 235/821 条数据...
|
||
It's a single issue.
|
||
处理第 236/821 条数据...
|
||
It's a single issue.
|
||
处理第 237/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "During the recovery of a cold-worked material, is some of the internal strain energy relieved?",
|
||
"answer": "Some of the internal strain energy is relieved."
|
||
},
|
||
{
|
||
"question": "During the recovery of a cold-worked material, is all of the internal strain energy relieved?",
|
||
"answer": "All of the internal strain energy is not relieved."
|
||
},
|
||
{
|
||
"question": "During the recovery of a cold-worked material, is there some reduction in the number of dislocations?",
|
||
"answer": "There is some reduction in the number of dislocations."
|
||
},
|
||
{
|
||
"question": "During the recovery of a cold-worked material, is there a significant reduction in the number of dislocations, to approximately the number found in the precold-worked state?",
|
||
"answer": "There is not a significant reduction in the number of dislocations, to approximately the number found in the precold-worked state."
|
||
},
|
||
{
|
||
"question": "During the recovery of a cold-worked material, is the electrical conductivity recovered to its precold-worked state?",
|
||
"answer": "The electrical conductivity is recovered to its precold-worked state."
|
||
},
|
||
{
|
||
"question": "During the recovery of a cold-worked material, is the thermal conductivity recovered to its precold-worked state?",
|
||
"answer": "The thermal conductivity is recovered to its precold-worked state."
|
||
},
|
||
{
|
||
"question": "During the recovery of a cold-worked material, does the metal become more ductile, as in its precold-worked state?",
|
||
"answer": "The metal does not become more ductile, as in its precold-worked state."
|
||
},
|
||
{
|
||
"question": "During the recovery of a cold-worked material, are grains with high strains replaced with new, unstrained grains?",
|
||
"answer": "Grains with high strains are not replaced with new, unstrained grains."
|
||
}
|
||
]
|
||
```
|
||
处理第 238/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "During the recrystallization of a cold-worked material, is some of the internal strain energy relieved?",
|
||
"answer": "All of the internal strain energy is relieved."
|
||
},
|
||
{
|
||
"question": "During the recrystallization of a cold-worked material, is there some reduction in the number of dislocations?",
|
||
"answer": "There is significant reduction in the number of dislocations."
|
||
},
|
||
{
|
||
"question": "During the recrystallization of a cold-worked material, does the metal become more ductile, as in its precold-worked state?",
|
||
"answer": "The metal becomes more ductile, as in its precold-worked state."
|
||
},
|
||
{
|
||
"question": "During the recrystallization of a cold-worked material, are grains with high strains replaced with new, unstrained grains?",
|
||
"answer": "Grains with high strains are replaced with new, unstrained grains."
|
||
}
|
||
]
|
||
```
|
||
处理第 239/821 条数据...
|
||
It's a single issue.
|
||
处理第 240/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Given a metal alloy with initial grain diameter d0 = 2.4 × 10^-2 mm, final grain diameter d = 7.3 × 10^-2 mm after heat treatment at 575°C for t = 500 min, and grain diameter exponent n = 2.2, compute the parameter K in the grain growth equation.",
|
||
"answer": "K = (d^2.2 - d0^2.2) / t = ((7.3 × 10^-2 mm)^2.2 - (2.4 × 10^-2 mm)^2.2) / 500 min = 5.77 × 10^-6 mm^2.2 / min"
|
||
},
|
||
{
|
||
"question": "Using the computed K = 5.77 × 10^-6 mm^2.2 / min, initial grain diameter d0 = 2.4 × 10^-2 mm, grain diameter exponent n = 2.2, and target grain diameter d = 5.5 × 10^-2 mm, calculate the required heat treatment time at 575°C.",
|
||
"answer": "t = (d^2.2 - d0^2.2) / K = ((5.5 × 10^-2 mm)^2.2 - (2.4 × 10^-2 mm)^2.2) / (5.77 × 10^-6 mm^2.2 / min) = 246 min"
|
||
}
|
||
]
|
||
```
|
||
处理第 241/821 条数据...
|
||
It's a single issue.
|
||
处理第 242/821 条数据...
|
||
It's a single issue.
|
||
处理第 243/821 条数据...
|
||
It's a single issue.
|
||
处理第 244/821 条数据...
|
||
It's a single issue.
|
||
处理第 245/821 条数据...
|
||
It's a single issue.
|
||
处理第 246/821 条数据...
|
||
It's a single issue.
|
||
处理第 247/821 条数据...
|
||
It's a single issue.
|
||
处理第 248/821 条数据...
|
||
[
|
||
{
|
||
"question": "Which kind of fracture (ductile or brittle) is associated with intergranular crack propagation?",
|
||
"answer": "Intergranular fracture is brittle. Crack propagation occurs along grain boundaries."
|
||
},
|
||
{
|
||
"question": "Which kind of fracture (ductile or brittle) is associated with transgranular crack propagation?",
|
||
"answer": "Transgranular fracture is brittle. Crack propagation occurs through the grains."
|
||
}
|
||
]
|
||
处理第 249/821 条数据...
|
||
It's a single issue.
|
||
处理第 250/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Given a metal alloy with a yield strength of 535 MPa and a plane strain fracture toughness of 31.3 MPa·m^(-1/2), and a design stress of 0.5 times the yield strength, what is the design stress?",
|
||
"answer": "The design stress is calculated as 0.5 times the yield strength: 0.5 * 535 MPa = 267.5 MPa."
|
||
},
|
||
{
|
||
"question": "Using the design stress of 267.5 MPa, a plane strain fracture toughness of 31.3 MPa·m^(-1/2), and a geometry factor Y of 1.8, what is the critical length of a surface flaw (a_c)?",
|
||
"answer": "The critical length of a surface flaw is calculated using the equation: a_c = (1/π) * (K_Ic / (σ * Y))^2. Substituting the given values: a_c = (1/π) * (31.3 MPa·m^(-1/2) / (267.5 MPa * 1.8))^2 = 2.0 × 10^(-3) m = 2.0 mm."
|
||
}
|
||
]
|
||
```
|
||
处理第 251/821 条数据...
|
||
It's a single issue.
|
||
处理第 252/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Does increasing temperature favor brittle fracture in polymers?",
|
||
"answer": "No, increasing temperature does not favor brittle fracture in polymers."
|
||
},
|
||
{
|
||
"question": "Does increasing strain rate favor brittle fracture in polymers?",
|
||
"answer": "Yes, increasing strain rate favors brittle fracture in polymers."
|
||
},
|
||
{
|
||
"question": "Does the presence of a sharp notch favor brittle fracture in polymers?",
|
||
"answer": "Yes, the presence of a sharp notch favors brittle fracture in polymers."
|
||
},
|
||
{
|
||
"question": "Does decreasing specimen thickness favor brittle fracture in polymers?",
|
||
"answer": "The answer does not specify the effect of decreasing specimen thickness on brittle fracture in polymers."
|
||
}
|
||
]
|
||
```
|
||
处理第 253/821 条数据...
|
||
It's a single issue.
|
||
处理第 254/821 条数据...
|
||
It's a single issue.
|
||
处理第 255/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Given a structural component made from an alloy with a plane strain fracture toughness of 45 MPa√m, a failure stress of 300 MPa, and a maximum surface crack length of 0.95 mm, calculate the geometry factor Y.",
|
||
"answer": "The geometry factor Y can be calculated using the rearranged equation: Y = K_Ic / (σ √(π a_c)). Substituting the given values: K_Ic = 45 MPa√m, σ = 300 MPa, and a_c = 0.95 mm (9.5 × 10^-4 m), we get Y = 45 / (300 √(π × 9.5 × 10^-4)) = 2.75."
|
||
},
|
||
{
|
||
"question": "Using the calculated geometry factor Y = 2.75, determine the maximum allowable surface crack length without fracture for the same component made from another alloy with a plane strain fracture toughness of 100.0 MPa√m and exposed to a stress of 300 MPa.",
|
||
"answer": "The maximum allowable surface crack length a_c can be calculated using the equation: a_c = 1/π (K_Ic / (σ Y))^2. Substituting the given values: K_Ic = 100.0 MPa√m, σ = 300 MPa, and Y = 2.75, we get a_c = 1/π (100.0 / (300 × 2.75))^2 = 0.00468 m = 4.68 mm."
|
||
}
|
||
]
|
||
```
|
||
处理第 256/821 条数据...
|
||
It's a single issue.
|
||
处理第 257/821 条数据...
|
||
It's a single issue.
|
||
处理第 258/821 条数据...
|
||
It's a single issue.
|
||
处理第 259/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Is solid ductile cast iron (ferrite solid solution + embedded graphite spheres) a two-phase material system?",
|
||
"answer": "Yes, solid ductile cast iron is a two-phase system because it consists of a ferrite solid solution with embedded graphite spheres, which are chemically and structurally distinct phases separated by physical boundaries."
|
||
},
|
||
{
|
||
"question": "Is solid sodium chloride (salt, NaCl) a two-phase material system?",
|
||
"answer": "No, solid sodium chloride is a single-phase compound with a fixed stoichiometry and orderly arrangement of ions in the crystal, not a two-phase system."
|
||
},
|
||
{
|
||
"question": "Is liquid bronze (Cu + Sn liquid solution) a two-phase material system?",
|
||
"answer": "No, liquid bronze is a single-phase liquid solution, as solutions by definition are single-phase."
|
||
},
|
||
{
|
||
"question": "Is solid gray cast iron (ferrite solid solution + embedded graphite flakes) a two-phase material system?",
|
||
"answer": "Yes, solid gray cast iron is a two-phase system because it consists of a ferrite solid solution with embedded graphite flakes, which are chemically and structurally distinct phases separated by physical boundaries."
|
||
},
|
||
{
|
||
"question": "Is solid aluminum featuring dissolved silicon a two-phase material system?",
|
||
"answer": "No, solid aluminum with dissolved silicon is a single-phase solid solution, as solutions by definition are single-phase."
|
||
},
|
||
{
|
||
"question": "Is solid lead-tin solder (a mixture of Pb-rich and Sn-rich solid solutions) a two-phase material system?",
|
||
"answer": "Yes, solid lead-tin solder is a two-phase system because it consists of Pb-rich and Sn-rich solid solutions, which are chemically and structurally distinct phases separated by physical boundaries."
|
||
},
|
||
{
|
||
"question": "Is partially melted aluminum a two-phase material system?",
|
||
"answer": "No, partially melted aluminum is not a two-phase system in the context of chemically distinct phases; it merely has structurally distinct volumes (crystalline and liquid)."
|
||
},
|
||
{
|
||
"question": "Is frozen water with trapped air bubbles a two-phase material system?",
|
||
"answer": "Yes, frozen water with trapped air bubbles is a two-phase system because it consists of ice and air bubbles, which are chemically and structurally distinct phases separated by physical boundaries."
|
||
},
|
||
{
|
||
"question": "Is epoxy embedded with carbon fibers a two-phase material system?",
|
||
"answer": "Yes, epoxy embedded with carbon fibers is a two-phase system because it consists of epoxy and carbon fibers, which are chemically and structurally distinct phases separated by physical boundaries."
|
||
}
|
||
]
|
||
```
|
||
处理第 260/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Complete the following statement regarding conditions that must be satisfied in order for a solid solution to exhibit extensive solubility: The solute and host species must have a very [w] sizes. (w = similar, different)",
|
||
"answer": "The solute and host species must have a very similar sizes."
|
||
},
|
||
{
|
||
"question": "Complete the following statement regarding conditions that must be satisfied in order for a solid solution to exhibit extensive solubility: The solute and host species must attempt to pack with [x] crystal structure. (x = a similar (or the same), a different)",
|
||
"answer": "The solute and host species must attempt to pack with a similar (or the same) crystal structure."
|
||
},
|
||
{
|
||
"question": "Complete the following statement regarding conditions that must be satisfied in order for a solid solution to exhibit extensive solubility: The solute and host species must feature [y] valence electron configuration. (y = a similar (or the same), a different)",
|
||
"answer": "The solute and host species must feature a similar (or the same) valence electron configuration."
|
||
},
|
||
{
|
||
"question": "Complete the following statement regarding conditions that must be satisfied in order for a solid solution to exhibit extensive solubility: The solute and host species must feature [z] ability to attract electrons (electronegativity). (z = a similar, a different)",
|
||
"answer": "The solute and host species must feature a similar ability to attract electrons (electronegativity)."
|
||
}
|
||
]
|
||
```
|
||
处理第 261/821 条数据...
|
||
It's a single issue.
|
||
处理第 262/821 条数据...
|
||
It's a single issue.
|
||
处理第 263/821 条数据...
|
||
It's a single issue.
|
||
处理第 264/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Which of the following kinds of information may be determined with the aid of a phase diagram: The phase(s) present at a specified temperature and composition?",
|
||
"answer": "With the aid of a phase diagram the following may be determined: The phase(s) present at a specified temperature and composition."
|
||
},
|
||
{
|
||
"question": "Which of the following kinds of information may be determined with the aid of a phase diagram: The composition(s) of phase(s) present at a specified temperature and composition?",
|
||
"answer": "With the aid of a phase diagram the following may be determined: The composition(s) of phase(s) present at a specified temperature and composition."
|
||
},
|
||
{
|
||
"question": "Which of the following kinds of information may be determined with the aid of a phase diagram: The fraction(s) of phase(s) present at specified temperature and composition?",
|
||
"answer": "With the aid of a phase diagram the following may be determined: The fraction(s) of phase(s) present at a specified temperature and composition."
|
||
}
|
||
]
|
||
```
|
||
处理第 265/821 条数据...
|
||
It's a single issue.
|
||
处理第 266/821 条数据...
|
||
It's a single issue.
|
||
处理第 267/821 条数据...
|
||
It's a single issue.
|
||
处理第 268/821 条数据...
|
||
It's a single issue.
|
||
处理第 269/821 条数据...
|
||
It's a single issue.
|
||
处理第 270/821 条数据...
|
||
It's a single issue.
|
||
处理第 271/821 条数据...
|
||
It's a single issue.
|
||
处理第 272/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Compute the atom concentration of aluminum at room temperature given the mass density of aluminum is 2.7 g/cm3 and the atomic weight of aluminum is 27 g/mol.",
|
||
"answer": "The atom concentration of aluminum at room temperature is: ((2.7 g/cm3)(6.022 x 10^23 atom/mol))/(27 g/mol) = 6.022 x 10^22 atom/cm3"
|
||
},
|
||
{
|
||
"question": "Compute the free electron concentration of aluminum at room temperature given the resistivity of aluminum is 2.63 x 10^-8 Ω·m, the electron mobility of aluminum is 0.0012 m2/(V·s), and the elementary charge is 1.6 x 10^-19 C.",
|
||
"answer": "The free electron concentration at room temperature is: 1/((2.63 x 10^-8 Ω·m)(1.6 x 10^-19 C)(0.0012 m2/(V·s))) = 1.98 x 10^29 electron/m3"
|
||
},
|
||
{
|
||
"question": "Compute the number of free electrons donated by each aluminum atom, on average, given the atom concentration is 6.022 x 10^22 atom/cm3 and the free electron concentration is 1.98 x 10^29 electron/m3.",
|
||
"answer": "The number of free electrons donated by each atom, on average, is: (1.98 x 10^29 electron/m3)/(6.022 x 10^28 atom/m3) = 3.29 electron/atom"
|
||
}
|
||
]
|
||
```
|
||
处理第 273/821 条数据...
|
||
It's a single issue.
|
||
处理第 274/821 条数据...
|
||
It's a single issue.
|
||
处理第 275/821 条数据...
|
||
It's a single issue.
|
||
处理第 276/821 条数据...
|
||
It's a single issue.
|
||
处理第 277/821 条数据...
|
||
It's a single issue.
|
||
处理第 278/821 条数据...
|
||
It's a single issue.
|
||
处理第 279/821 条数据...
|
||
It's a single issue.
|
||
处理第 280/821 条数据...
|
||
It's a single issue.
|
||
处理第 281/821 条数据...
|
||
It's a single issue.
|
||
处理第 282/821 条数据...
|
||
It's a single issue.
|
||
处理第 283/821 条数据...
|
||
It's a single issue.
|
||
处理第 284/821 条数据...
|
||
It's a single issue.
|
||
处理第 285/821 条数据...
|
||
It's a single issue.
|
||
处理第 286/821 条数据...
|
||
It's a single issue.
|
||
处理第 287/821 条数据...
|
||
It's a single issue.
|
||
处理第 288/821 条数据...
|
||
It's a single issue.
|
||
处理第 289/821 条数据...
|
||
It's a single issue.
|
||
处理第 290/821 条数据...
|
||
It's a single issue.
|
||
处理第 291/821 条数据...
|
||
[
|
||
{
|
||
"question": "[a] Aluminum alloys are generally not viable as lightweight structural materials in humid environments because they are highly susceptible to corrosion by water vapor.",
|
||
"answer": "F"
|
||
},
|
||
{
|
||
"question": "[b] Aluminum alloys are generally superior to pure aluminum, in terms of yield strength, because their microstructures often contain precipitate phases that strain the lattice, thereby hardening the alloy relative to pure aluminum.",
|
||
"answer": "T"
|
||
},
|
||
{
|
||
"question": "[c] Aluminum is not very workable at high temperatures in air, in terms of extrusion and rolling, because a non-protective oxide grows and consumes the metal, converting it to a hard and brittle ceramic.",
|
||
"answer": "F"
|
||
},
|
||
{
|
||
"question": "[d] Compared to most other metals, like steel, pure aluminum is very resistant to creep deformation.",
|
||
"answer": "F"
|
||
},
|
||
{
|
||
"question": "[e] The relatively low melting point of aluminum is often considered a significant limitation for structural applications.",
|
||
"answer": "T"
|
||
}
|
||
]
|
||
处理第 292/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "[a] Aluminum alloys are generally viable as lightweight structural materials in humid environments because they are not very susceptible to corrosion by water vapor.",
|
||
"answer": "T"
|
||
},
|
||
{
|
||
"question": "[b] Aluminum is not very workable at high temperatures in air, in terms of extrusion and rolling, because a non-protective oxide grows and consumes the metal, converting it to a hard and brittle ceramic.",
|
||
"answer": "F"
|
||
},
|
||
{
|
||
"question": "[c] Aluminum alloys are generally superior to pure aluminum, in terms of yield strength, because their microstructures often contain precipitate phases that strain the lattice, thereby hardening the alloy relative to pure aluminum.",
|
||
"answer": "T"
|
||
},
|
||
{
|
||
"question": "[d] Compared to other metals, like steel, pure aluminum is very resistant to failure via fatigue.",
|
||
"answer": "F"
|
||
},
|
||
{
|
||
"question": "[e] Aluminum exhibits one of the highest melting points of all metals, which makes it difficult and expensive to cast.",
|
||
"answer": "F"
|
||
}
|
||
]
|
||
```
|
||
处理第 293/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "[a] Copper has a higher elastic modulus than aluminum.",
|
||
"answer": "T"
|
||
},
|
||
{
|
||
"question": "[b] The density of copper is closer to that of aluminum than it is to iron.",
|
||
"answer": "F"
|
||
},
|
||
{
|
||
"question": "[c] Bronze is an alloy of copper and zinc.",
|
||
"answer": "F"
|
||
},
|
||
{
|
||
"question": "[d] Copper and its alloys form a green tarnish over time, consisting of sulfides and carbonates.",
|
||
"answer": "T"
|
||
},
|
||
{
|
||
"question": "[e] Copper is relatively resistant to corrosion by neutral and even mildly basic water, making it useful for freshwater plumbing applications.",
|
||
"answer": "T"
|
||
}
|
||
]
|
||
```
|
||
处理第 294/821 条数据...
|
||
[
|
||
{
|
||
"question": "Select T / F for the following statement regarding copper & copper alloys: Copper is much more abundant in the earth's crust compared to iron or aluminum.",
|
||
"answer": "F"
|
||
},
|
||
{
|
||
"question": "Select T / F for the following statement regarding copper & copper alloys: Copper is one of just a few metals that can be found in metallic form in nature.",
|
||
"answer": "T"
|
||
},
|
||
{
|
||
"question": "Select T / F for the following statement regarding copper & copper alloys: Pure and/or annealed copper is more difficult to machine compared to its work-hardened form or its alloys.",
|
||
"answer": "T"
|
||
},
|
||
{
|
||
"question": "Select T / F for the following statement regarding copper & copper alloys: Copper is a minor component (by weight) of most brass & bronze alloys.",
|
||
"answer": "F"
|
||
},
|
||
{
|
||
"question": "Select T / F for the following statement regarding copper & copper alloys: Amongst metals and alloys copper is one of the best conductors of heat.",
|
||
"answer": "T"
|
||
}
|
||
]
|
||
处理第 295/821 条数据...
|
||
[
|
||
{
|
||
"question": "[a] Nickel is majority component (by mass) in certain superalloys such as Waspaloy TM.",
|
||
"answer": "T"
|
||
},
|
||
{
|
||
"question": "[b] Tungsten is the lowest density metal that has structural use.",
|
||
"answer": "F"
|
||
},
|
||
{
|
||
"question": "[c] Tantalum offers extremely good corrosion resistance, especially at low temperatures.",
|
||
"answer": "T"
|
||
},
|
||
{
|
||
"question": "[d] Magnesium metal is very similar to aluminum, in terms of its physical and mechanical properties.",
|
||
"answer": "T"
|
||
},
|
||
{
|
||
"question": "[e] Beryllium metal is commonly used as an alloying agent in copper metal.",
|
||
"answer": "T"
|
||
}
|
||
]
|
||
处理第 296/821 条数据...
|
||
It's a single issue.
|
||
处理第 297/821 条数据...
|
||
It's a single issue.
|
||
处理第 298/821 条数据...
|
||
[
|
||
{
|
||
"question": "As the porosity of refractory ceramic bricks increases, what happens to the strength?",
|
||
"answer": "Strength decreases."
|
||
},
|
||
{
|
||
"question": "As the porosity of refractory ceramic bricks increases, what happens to the chemical resistance?",
|
||
"answer": "Chemical resistance decreases."
|
||
},
|
||
{
|
||
"question": "As the porosity of refractory ceramic bricks increases, what happens to the thermal insulation?",
|
||
"answer": "Thermal insulation increases."
|
||
}
|
||
]
|
||
处理第 299/821 条数据...
|
||
It's a single issue.
|
||
处理第 300/821 条数据...
|
||
It's a single issue.
|
||
处理第 301/821 条数据...
|
||
It's a single issue.
|
||
处理第 302/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Given that the bond strength across the fiber-epoxy interface is [s] MPa, and the shear yield strength of the epoxy is [y] MPa, determine which strength value (bond strength or shear yield strength) should be used as the critical shear strength (τ_c) for calculating the minimum fiber length.",
|
||
"answer": "Since τ_fiber-matrix < τ_y, matrix, the critical shear strength τ_c should be equal to the fiber-matrix bond strength, i.e., τ_c = τ_fiber-matrix = [s] MPa."
|
||
},
|
||
{
|
||
"question": "Compute the minimum fiber length (l), in millimeters, to guarantee that the fibers are conveying an optimum fraction of force that is applied to the composite, given the tensile strength of the carbon fibers is [f] MPa and the fiber diameter is [d] mm.",
|
||
"answer": "The minimum fiber length (l) is calculated using the formula: l = (σ_f * d) / (2 * τ_c) = ([f] MPa * [d] mm) / (2 * [s] MPa) = [f][d] / (2[s] * 1000) mm."
|
||
}
|
||
]
|
||
```
|
||
处理第 303/821 条数据...
|
||
It's a single issue.
|
||
处理第 304/821 条数据...
|
||
It's a single issue.
|
||
处理第 305/821 条数据...
|
||
It's a single issue.
|
||
处理第 306/821 条数据...
|
||
It's a single issue.
|
||
处理第 307/821 条数据...
|
||
It's a single issue.
|
||
处理第 308/821 条数据...
|
||
It's a single issue.
|
||
处理第 309/821 条数据...
|
||
It's a single issue.
|
||
处理第 310/821 条数据...
|
||
It's a single issue.
|
||
处理第 311/821 条数据...
|
||
It's a single issue.
|
||
处理第 312/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "[a] Composites are single-phase materials by definition.",
|
||
"answer": "F"
|
||
},
|
||
{
|
||
"question": "[b] The term \"composite\" applies to materials that feature polymeric materials only.",
|
||
"answer": "F"
|
||
},
|
||
{
|
||
"question": "[c] Structural composites are, in general, highly regarded for their specific strengths.",
|
||
"answer": "T"
|
||
},
|
||
{
|
||
"question": "[d] Composites featuring continuous and aligned fibers for reinforcement generally offer properties that are highly isotropic compared to most metals (random polycrystals).",
|
||
"answer": "F"
|
||
}
|
||
]
|
||
```
|
||
API调用错误: Expecting ',' delimiter: line 7 column 36 (char 158)
|
||
处理第 313/821 条数据...
|
||
It's a single issue.
|
||
处理第 314/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Match the fiber type 'Whiskers' with its description.",
|
||
"answer": "Whiskers are single crystals with extremely large length-to-diameter ratios."
|
||
},
|
||
{
|
||
"question": "Match the fiber type 'Fibers' with its description.",
|
||
"answer": "Fibers are polycrystalline or amorphous materials with small diameters."
|
||
},
|
||
{
|
||
"question": "Match the fiber type 'Wires' with its description.",
|
||
"answer": "Wires are metals having relatively large diameters."
|
||
}
|
||
]
|
||
```
|
||
处理第 315/821 条数据...
|
||
It's a single issue.
|
||
处理第 316/821 条数据...
|
||
It's a single issue.
|
||
处理第 317/821 条数据...
|
||
It's a single issue.
|
||
处理第 318/821 条数据...
|
||
It's a single issue.
|
||
处理第 319/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Do carbon fiber-reinforced composites have relatively high strengths?",
|
||
"answer": "Carbon fiber-reinforced composites have relatively high strengths."
|
||
},
|
||
{
|
||
"question": "Do carbon fiber-reinforced composites have relatively high stiffnesses?",
|
||
"answer": "Carbon fiber-reinforced composites have relatively high stiffnesses."
|
||
},
|
||
{
|
||
"question": "Do carbon fiber-reinforced composites have high service temperatures (>200°C)?",
|
||
"answer": "Carbon fiber-reinforced composites have high service temperatures (>200°C)."
|
||
}
|
||
]
|
||
```
|
||
处理第 320/821 条数据...
|
||
It's a single issue.
|
||
处理第 321/821 条数据...
|
||
It's a single issue.
|
||
处理第 322/821 条数据...
|
||
It's a single issue.
|
||
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|
||
It's a single issue.
|
||
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|
||
```json
|
||
[
|
||
{
|
||
"question": "Do carbon-carbon composites exhibit high tensile moduli at elevated temperatures?",
|
||
"answer": "Carbon-carbon composites have high tensile moduli at elevated temperatures."
|
||
},
|
||
{
|
||
"question": "Do carbon-carbon composites exhibit high tensile strengths at elevated temperatures?",
|
||
"answer": "Carbon-carbon composites have high tensile strengths at elevated temperatures."
|
||
},
|
||
{
|
||
"question": "Are carbon-carbon composites resistant to creep?",
|
||
"answer": "Carbon-carbon composites are highly resistant to creep."
|
||
},
|
||
{
|
||
"question": "Do carbon-carbon composites have large fracture toughness values?",
|
||
"answer": "Carbon-carbon composites have large fracture toughness values."
|
||
},
|
||
{
|
||
"question": "Do carbon-carbon composites have high thermal conductivities?",
|
||
"answer": "Carbon-carbon composites have high thermal conductivities."
|
||
},
|
||
{
|
||
"question": "Do carbon-carbon composites have low coefficients of thermal expansion?",
|
||
"answer": "Carbon-carbon composites have low coefficients of thermal expansion."
|
||
},
|
||
{
|
||
"question": "Are carbon-carbon composites resistant to oxidation at elevated temperatures?",
|
||
"answer": "The answer does not specify resistance to oxidation at elevated temperatures."
|
||
},
|
||
{
|
||
"question": "Are carbon-carbon composites low cost?",
|
||
"answer": "The answer does not specify low cost."
|
||
}
|
||
]
|
||
```
|
||
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|
||
It's a single issue.
|
||
处理第 326/821 条数据...
|
||
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|
||
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|
||
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|
||
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|
||
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|
||
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|
||
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|
||
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|
||
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|
||
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|
||
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|
||
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|
||
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|
||
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|
||
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|
||
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|
||
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|
||
处理第 335/821 条数据...
|
||
It's a single issue.
|
||
处理第 336/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Calculate the number of Bohr magnetons per unit cell (n_B) given the saturation magnetization (M_s = 1.0 x 10^4 A/m), the unit cell edge length (a = 1.2376 x 10^-9 m), and the Bohr magneton (μ_B = 9.27 x 10^-24 A·m^2/BM).",
|
||
"answer": "n_B = (M_s * a^3) / μ_B = (1.0 x 10^4 A/m) * (1.2376 x 10^-9 m)^3 / (9.27 x 10^-24 A·m^2/BM) = 2.04 Bohr magnetons/unit cell."
|
||
},
|
||
{
|
||
"question": "Determine the number of Bohr magnetons per formula unit given that there are 8 formula units per unit cell and the calculated n_B = 2.04 Bohr magnetons/unit cell.",
|
||
"answer": "Bohr magnetons per formula unit = n_B / 8 = 2.04 / 8 = 0.255 Bohr magnetons/formula unit."
|
||
},
|
||
{
|
||
"question": "Calculate the net magnetic moment contribution per formula unit from the Fe^3+ ions, given that there are 2 Fe^3+ ions on a sites and 3 Fe^3+ ions on d sites per formula unit, each with 5 Bohr magnetons, and their magnetic moments are aligned antiparallel.",
|
||
"answer": "Net magnetic moment from Fe^3+ ions = (2 * 5 BM) - (3 * 5 BM) = 10 BM - 15 BM = -5 BM per formula unit."
|
||
},
|
||
{
|
||
"question": "Compute the number of Bohr magnetons associated with each Y^3+ ion, given the net magnetic moment per formula unit from Fe^3+ ions is -5 BM, the Bohr magnetons per formula unit is 0.255 BM, and there are 3 Y^3+ ions per formula unit.",
|
||
"answer": "Number of Bohr magnetons per Y^3+ ion = (0.255 BM + 5 BM) / 3 = 5.255 BM / 3 = 1.75 BM."
|
||
}
|
||
]
|
||
```
|
||
处理第 337/821 条数据...
|
||
It's a single issue.
|
||
处理第 338/821 条数据...
|
||
It's a single issue.
|
||
处理第 339/821 条数据...
|
||
It's a single issue.
|
||
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|
||
It's a single issue.
|
||
处理第 341/821 条数据...
|
||
It's a single issue.
|
||
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|
||
It's a single issue.
|
||
处理第 343/821 条数据...
|
||
It's a single issue.
|
||
处理第 344/821 条数据...
|
||
It's a single issue.
|
||
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|
||
It's a single issue.
|
||
处理第 346/821 条数据...
|
||
[
|
||
{
|
||
"question": "Match the type of light transmission with its description: Transmits light with relative little absorption.",
|
||
"answer": "A transparent material transmits light with relatively little absorption."
|
||
},
|
||
{
|
||
"question": "Match the type of light transmission with its description: Transmits light diffusely.",
|
||
"answer": "A translucent material transmits light diffusely."
|
||
},
|
||
{
|
||
"question": "Match the type of light transmission with its description: Is impervious to light transmission.",
|
||
"answer": "An opaque material is impervious to light transmission."
|
||
}
|
||
]
|
||
处理第 347/821 条数据...
|
||
It's a single issue.
|
||
处理第 348/821 条数据...
|
||
It's a single issue.
|
||
处理第 349/821 条数据...
|
||
It's a single issue.
|
||
处理第 350/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What are the light transmission characteristics of single crystal electrical insulators?",
|
||
"answer": "Single crystal electrical insulators are transparent."
|
||
},
|
||
{
|
||
"question": "What are the light transmission characteristics of polycrystalline and nonporous electrical insulators?",
|
||
"answer": "Polycrystalline and nonporous electrical insulators are translucent."
|
||
},
|
||
{
|
||
"question": "What are the light transmission characteristics of porous electrical insulators?",
|
||
"answer": "Porous electrical insulators are opaque to visible light."
|
||
}
|
||
]
|
||
```
|
||
处理第 351/821 条数据...
|
||
It's a single issue.
|
||
处理第 352/821 条数据...
|
||
It's a single issue.
|
||
处理第 353/821 条数据...
|
||
It's a single issue.
|
||
处理第 354/821 条数据...
|
||
It's a single issue.
|
||
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|
||
It's a single issue.
|
||
处理第 356/821 条数据...
|
||
It's a single issue.
|
||
处理第 357/821 条数据...
|
||
It's a single issue.
|
||
处理第 358/821 条数据...
|
||
It's a single issue.
|
||
处理第 359/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "How many grams are there in one amu of a material?",
|
||
"answer": "In order to determine the number of grams in one amu of material, appropriate manipulation of the amu/atom, g/mol, and atom/mol relationships is all that is necessary, as # g/amu = (1 mol/6.022 x 10^23 atoms)(1 g/mol/1 amu/atom) = 1.66 x 10^-24 g/amu"
|
||
},
|
||
{
|
||
"question": "Mole, in the context of this book, is taken in units of gram-mole. On this basis, how many atoms are there in a pound-mole of a substance?",
|
||
"answer": "Since there are 453.6 g/lb_m, 1 lb-mol = (453.6 g/lb_m)(6.022 x 10^23 atoms/g-mol) = 2.73 x 10^26 atoms/lb-mol"
|
||
}
|
||
]
|
||
```
|
||
处理第 360/821 条数据...
|
||
[
|
||
{
|
||
"question": "Cite two important quantum-mechanical concepts associated with the Bohr model of the atom.",
|
||
"answer": "Two important quantum-mechanical concepts associated with the Bohr model of the atom are (1) that electrons are particles moving in discrete orbitals, and (2) electron energy is quantized into shells."
|
||
},
|
||
{
|
||
"question": "Cite two important additional refinements that resulted from the wave-mechanical atomic model.",
|
||
"answer": "Two important refinements resulting from the wave-mechanical atomic model are (1) that electron position is described in terms of a probability distribution, and (2) electron energy is quantized into both shells and subshells--each electron is characterized by four quantum numbers."
|
||
}
|
||
]
|
||
处理第 361/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Relative to electrons and electron states, what does the n quantum number specify?",
|
||
"answer": "The n quantum number designates the electron shell."
|
||
},
|
||
{
|
||
"question": "Relative to electrons and electron states, what does the l quantum number specify?",
|
||
"answer": "The l quantum number designates the electron subshell."
|
||
},
|
||
{
|
||
"question": "Relative to electrons and electron states, what does the m_j quantum number specify?",
|
||
"answer": "The m_j quantum number designates the number of electron states in each electron subshell."
|
||
},
|
||
{
|
||
"question": "Relative to electrons and electron states, what does the m_s quantum number specify?",
|
||
"answer": "The m_s quantum number designates the spin moment on each electron."
|
||
}
|
||
]
|
||
```
|
||
处理第 362/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For the L shell (n=2), what are the four quantum numbers for all of the electrons, and which correspond to the s and p subshells?",
|
||
"answer": "For the L state, n=2, and eight electron states are possible. Possible l values are 0 and 1, while possible ml values are 0 and ±1; and possible ms values are ±1/2. Therefore, for the s states (l=0), the quantum numbers are 200(1/2) and 200(-1/2). For the p states (l=1), the quantum numbers are 210(1/2), 210(-1/2), 211(1/2), 211(-1/2), 21(-1)(1/2), and 21(-1)(-1/2)."
|
||
},
|
||
{
|
||
"question": "For the M shell (n=3), what are the four quantum numbers for all of the electrons, and which correspond to the s, p, and d subshells?",
|
||
"answer": "For the M state, n=3, and 18 states are possible. Possible l values are 0, 1, and 2; possible ml values are 0, ±1, and ±2; and possible ms values are ±1/2. Therefore, for the s states (l=0), the quantum numbers are 300(1/2) and 300(-1/2). For the p states (l=1), the quantum numbers are 310(1/2), 310(-1/2), 311(1/2), 311(-1/2), 31(-1)(1/2), and 31(-1)(-1/2). For the d states (l=2), the quantum numbers are 320(1/2), 320(-1/2), 321(1/2), 321(-1/2), 32(-1)(1/2), 32(-1)(-1/2), 322(1/2), 322(-1/2), 32(-2)(1/2), and 32(-2)(-1/2)."
|
||
}
|
||
]
|
||
```
|
||
处理第 363/821 条数据...
|
||
[
|
||
{
|
||
"question": "Give the electron configuration for the ion Fe2+.",
|
||
"answer": "From Table 2.2, the electron configuration for an atom of iron is 1s2 2s2 2p6 3s2 3p6 3d6 4s2. In order to become an ion with a plus two charge, it must lose two electrons-in this case the two 4s. Thus, the electron configuration for an Fe2+ ion is 1s2 2s2 2p6 3s2 3p6 3d6."
|
||
},
|
||
{
|
||
"question": "Give the electron configuration for the ion Al3+.",
|
||
"answer": "From Table 2.2, the electron configuration for an atom of aluminum is 1s2 2s2 2p6 3s2 3p1. In order to become an ion with a plus three charge, it must lose three electrons-in this case two 3s and the one 3p. Thus, the electron configuration for an Al3+ ion is 1s2 2s2 2p6."
|
||
},
|
||
{
|
||
"question": "Give the electron configuration for the ion Cu+.",
|
||
"answer": "From Table 2.2, the electron configuration for an atom of copper is 1s2 2s2 2p6 3s2 3p6 3d10 4s1. In order to become an ion with a plus one charge, it must lose one electron-in this case the 4s. Thus, the electron configuration for a Cu+ ion is 1s2 2s2 2p6 3s2 3p6 3d10."
|
||
},
|
||
{
|
||
"question": "Give the electron configuration for the ion Ba2+.",
|
||
"answer": "The atomic number for barium is 56 (Figure 2.6), and inasmuch as it is not a transition element the electron configuration for one of its atoms is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 6s2. In order to become an ion with a plus two charge, it must lose two electrons-in this case the two 6s. Thus, the electron configuration for a Ba2+ ion is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6."
|
||
},
|
||
{
|
||
"question": "Give the electron configuration for the ion Br-.",
|
||
"answer": "From Table 2.2, the electron configuration for an atom of bromine is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5. In order to become an ion with a minus one charge, it must acquire one electron-in this case another 4p. Thus, the electron configuration for a Br- ion is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6."
|
||
},
|
||
{
|
||
"question": "Give the electron configuration for the ion O2-.",
|
||
"answer": "From Table 2.2, the electron configuration for an atom of oxygen is 1s2 2s2 2p4. In order to become an ion with a minus two charge, it must acquire two electrons-in this case another two 2p. Thus, the electron configuration for an O2- ion is 1s2 2s2 2p6."
|
||
}
|
||
]
|
||
处理第 364/821 条数据...
|
||
It's a single issue.
|
||
处理第 365/821 条数据...
|
||
It's a single issue.
|
||
处理第 366/821 条数据...
|
||
It's a single issue.
|
||
处理第 367/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Determine whether the electron configuration \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7} 4 s^{2}\) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.",
|
||
"answer": "The \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7} 4 s^{2}\) electron configuration is that of a transition metal because of an incomplete \(d\) subshell."
|
||
},
|
||
{
|
||
"question": "Determine whether the electron configuration \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}\) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.",
|
||
"answer": "The \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}\) electron configuration is that of an inert gas because of filled \(3 s\) and \(3 p\) subshells."
|
||
},
|
||
{
|
||
"question": "Determine whether the electron configuration \(1 s^{2} 2 s^{2} 2 p^{5}\) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.",
|
||
"answer": "The \(1 s^{2} 2 s^{2} 2 p^{5}\) electron configuration is that of a halogen because it is one electron deficient from having a filled \(L\) shell."
|
||
},
|
||
{
|
||
"question": "Determine whether the electron configuration \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}\) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.",
|
||
"answer": "The \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}\) electron configuration is that of an alkaline earth metal because of two \(s\) electrons."
|
||
},
|
||
{
|
||
"question": "Determine whether the electron configuration \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{2} 4 s^{2}\) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.",
|
||
"answer": "The \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{2} 4 s^{2}\) electron configuration is that of a transition metal because of an incomplete \(d\) subshell."
|
||
},
|
||
{
|
||
"question": "Determine whether the electron configuration \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}\) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.",
|
||
"answer": "The \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}\) electron configuration is that of an alkali metal because of a single \(s\) electron."
|
||
}
|
||
]
|
||
```
|
||
API调用错误: Invalid \escape: line 3 column 67 (char 74)
|
||
处理第 368/821 条数据...
|
||
[
|
||
{
|
||
"question": "What electron subshell is being filled for the rare earth series of elements on the periodic table?",
|
||
"answer": "The 4 f subshell is being filled for the rare earth series of elements."
|
||
},
|
||
{
|
||
"question": "What electron subshell is being filled for the actinide series of elements on the periodic table?",
|
||
"answer": "The 5 f subshell is being filled for the actinide series of elements."
|
||
}
|
||
]
|
||
处理第 369/821 条数据...
|
||
It's a single issue.
|
||
处理第 370/821 条数据...
|
||
[
|
||
{
|
||
"question": "Briefly cite the main differences between ionic, covalent, and metallic bonding.",
|
||
"answer": "The main differences between the various forms of primary bonding are: Ionic--there is electrostatic attraction between oppositely charged ions. Covalent--there is electron sharing between two adjacent atoms such that each atom assumes a stable electron configuration. Metallic--the positively charged ion cores are shielded from one another, and also 'glued' together by the sea of valence electrons."
|
||
},
|
||
{
|
||
"question": "State the Pauli exclusion principle.",
|
||
"answer": "The Pauli exclusion principle states that each electron state can hold no more than two electrons, which must have opposite spins."
|
||
}
|
||
]
|
||
处理第 371/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What type(s) of bonding would be expected for brass (a copper-zinc alloy)?",
|
||
"answer": "For brass, the bonding is metallic since it is a metal alloy."
|
||
},
|
||
{
|
||
"question": "What type(s) of bonding would be expected for rubber?",
|
||
"answer": "For rubber, the bonding is covalent with some van der Waals. (Rubber is composed primarily of carbon and hydrogen atoms.)"
|
||
},
|
||
{
|
||
"question": "What type(s) of bonding would be expected for barium sulfide (BaS)?",
|
||
"answer": "For BaS, the bonding is predominantly ionic (but with some covalent character) on the basis of the relative positions of Ba and S in the periodic table."
|
||
},
|
||
{
|
||
"question": "What type(s) of bonding would be expected for solid xenon?",
|
||
"answer": "For solid xenon, the bonding is van der Waals since xenon is an inert gas."
|
||
},
|
||
{
|
||
"question": "What type(s) of bonding would be expected for bronze?",
|
||
"answer": "For bronze, the bonding is metallic since it is a metal alloy (composed of copper and tin)."
|
||
},
|
||
{
|
||
"question": "What type(s) of bonding would be expected for nylon?",
|
||
"answer": "For nylon, the bonding is covalent with perhaps some van der Waals. (Nylon is composed primarily of carbon and hydrogen.)"
|
||
},
|
||
{
|
||
"question": "What type(s) of bonding would be expected for aluminum phosphide (AlP)?",
|
||
"answer": "For AlP the bonding is predominantly covalent (but with some ionic character) on the basis of the relative positions of Al and P in the periodic table."
|
||
}
|
||
]
|
||
```
|
||
处理第 372/821 条数据...
|
||
It's a single issue.
|
||
处理第 373/821 条数据...
|
||
It's a single issue.
|
||
处理第 374/821 条数据...
|
||
It's a single issue.
|
||
处理第 375/821 条数据...
|
||
It's a single issue.
|
||
处理第 376/821 条数据...
|
||
It's a single issue.
|
||
处理第 377/821 条数据...
|
||
It's a single issue.
|
||
处理第 378/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Determine whether the crystal structure of alloy A (atomic weight = 77.4 g/mol, density = 8.22 g/cm³, atomic radius = 0.125 mm) is FCC, BCC, or simple cubic and justify your determination.",
|
||
"answer": "For alloy A, assuming a simple cubic crystal structure, the density is calculated as follows: ρ = (n * A_A) / ((2R)^3 * N_A) = (1 atom/unit cell * 77.4 g/mol) / ([(2)(1.25 × 10^-8 cm)]^3 / unit cell * 6.022 × 10^23 atoms/mol) = 8.22 g/cm³. Therefore, its crystal structure is simple cubic."
|
||
},
|
||
{
|
||
"question": "Determine whether the crystal structure of alloy B (atomic weight = 107.6 g/mol, density = 13.42 g/cm³, atomic radius = 0.133 mm) is FCC, BCC, or simple cubic and justify your determination.",
|
||
"answer": "For alloy B, assuming an FCC crystal structure, the density is calculated as follows: ρ = (n * A_B) / ((2R√2)^3 * N_A) = (4 atoms/unit cell * 107.6 g/mol) / ([(2√2)(1.33 × 10^-8 cm)]^3 / unit cell * 6.022 × 10^23 atoms/mol) = 13.42 g/cm³. Therefore, its crystal structure is FCC."
|
||
},
|
||
{
|
||
"question": "Determine whether the crystal structure of alloy C (atomic weight = 127.3 g/mol, density = 9.23 g/cm³, atomic radius = 0.142 mm) is FCC, BCC, or simple cubic and justify your determination.",
|
||
"answer": "For alloy C, assuming a simple cubic crystal structure, the density is calculated as follows: ρ = (n * A_C) / ((2R)^3 * N_A) = (1 atom/unit cell * 127.3 g/mol) / ([(2)(1.42 × 10^-8 cm)]^3 / unit cell * 6.022 × 10^23 atoms/mol) = 9.23 g/cm³. Therefore, its crystal structure is simple cubic."
|
||
}
|
||
]
|
||
```
|
||
处理第 379/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Given the tetragonal unit cell for tin with lattice parameters a = 0.583 nm and c = 0.318 nm, density = 7.30 g/cm³, atomic weight = 118.69 g/mol, and Avogadro's number = 6.022 × 10²³ atoms/mol, calculate the number of atoms per unit cell (n).",
|
||
"answer": "n = (ρ * V_C * N_A) / A_Sn = (7.30 g/cm³ * (5.83 × 10⁻⁸ cm)² * (3.18 × 10⁻⁸ cm) * 6.022 × 10²³ atoms/mol) / 118.69 g/mol = 4.00 atoms/unit cell"
|
||
},
|
||
{
|
||
"question": "Given the atomic radius R = 0.151 nm and the number of atoms per unit cell n = 4, calculate the total sphere volume (V_S) for the tin unit cell.",
|
||
"answer": "V_S = n * (4/3 * π * R³) = 4 * (4/3 * π * (1.51 × 10⁻⁸ cm)³) = 4 * (4/3 * π * 3.45 × 10⁻²⁴ cm³) = 1.44 × 10⁻²³ cm³"
|
||
},
|
||
{
|
||
"question": "Given the tetragonal unit cell parameters a = 0.583 nm and c = 0.318 nm, calculate the unit cell volume (V_C).",
|
||
"answer": "V_C = a² * c = (5.83 × 10⁻⁸ cm)² * (3.18 × 10⁻⁸ cm) = 3.40 × 10⁻²³ cm³ * 3.18 × 10⁻⁸ cm = 1.08 × 10⁻²² cm³"
|
||
},
|
||
{
|
||
"question": "Using the total sphere volume V_S = 1.44 × 10⁻²³ cm³ and unit cell volume V_C = 1.08 × 10⁻²² cm³, calculate the atomic packing factor (APF) for tin.",
|
||
"answer": "APF = V_S / V_C = (1.44 × 10⁻²³ cm³) / (1.08 × 10⁻²² cm³) = 0.534"
|
||
}
|
||
]
|
||
```
|
||
处理第 380/821 条数据...
|
||
[
|
||
{
|
||
"question": "List the point coordinates of the barium ions for a unit cell of the perovskite crystal structure.",
|
||
"answer": "The barium ions are situated at all corner positions. The point coordinates for these ions are as follows: 000, 100, 110, 010, 001, 101, 111, and 011."
|
||
},
|
||
{
|
||
"question": "List the point coordinates of the oxygen ions for a unit cell of the perovskite crystal structure.",
|
||
"answer": "The oxygen ions are located at all face-centered positions; therefore, their coordinates are 1/2 1/2 0, 1/2 1/2 1, 1/2 0 1/2, 0 1/2 1/2, 1/2 1 1/2, and 1/2 0 1/2."
|
||
},
|
||
{
|
||
"question": "List the point coordinates of the titanium ion for a unit cell of the perovskite crystal structure.",
|
||
"answer": "The titanium ion resides at the center of the cubic unit cell, with coordinates 1/2 1/2 1/2."
|
||
}
|
||
]
|
||
处理第 381/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "List the point coordinates of all corner position atoms in the diamond cubic unit cell.",
|
||
"answer": "The coordinates of the corner position atoms are: 000, 100, 110, 010, 001, 101, 111, and 011."
|
||
},
|
||
{
|
||
"question": "List the point coordinates of all face-centered position atoms in the diamond cubic unit cell.",
|
||
"answer": "The coordinates of the face-centered position atoms are: 1/2 1/2 0, 1/2 0 1/2, 0 1/2 1/2, 1/2 1/2 1, 1/2 1 1/2, and 1 1/2 1/2."
|
||
},
|
||
{
|
||
"question": "List the point coordinates of all interior position atoms in the diamond cubic unit cell.",
|
||
"answer": "The coordinates of the interior position atoms are: 1/4 1/4 1/4, 3/4 3/4 1/4, 3/4 1/4 3/4, and 1/4 3/4 3/4."
|
||
}
|
||
]
|
||
```
|
||
处理第 382/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "How to define atom types, colors, and sizes for Au and Cu in the Molecule Definition Utility for generating a unit cell of AuCu3?",
|
||
"answer": "In the 'Step 1' window of the Molecule Definition Utility, define 'Au' as the name for gold atoms and 'Cu' for copper atoms. Choose colors from the pull-down menu, e.g., 'Yellow' for Au and 'Red' for Cu. Enter atom sizes based on atomic radii: 0.288 nm for Au (2 × 0.144 nm) and 0.256 nm for Cu (2 × 0.128 nm). Click 'Register' and then 'Go to Step 2'."
|
||
},
|
||
{
|
||
"question": "How to specify the positions of gold atoms in the cubic unit cell of AuCu3 with an edge length of 0.374 nm?",
|
||
"answer": "In 'Step 2', select the yellow sphere for Au. Enter the following coordinates for the 8 corner positions: 0, 0, 0; 0.374, 0, 0; 0, 0.374, 0; 0, 0, 0.374; 0, 0.374, 0.374; 0.374, 0, 0.374; 0.374, 0.374, 0; 0.374, 0.374, 0.374. Click 'Register Atom Position' after each entry."
|
||
},
|
||
{
|
||
"question": "How to specify the positions of copper atoms in the cubic unit cell of AuCu3 with an edge length of 0.374 nm?",
|
||
"answer": "In 'Step 2', select the red sphere for Cu. Enter the following coordinates for the 6 face-centered positions: 0.187, 0.187, 0; 0.187, 0, 0.187; 0, 0.187, 0.187; 0.374, 0.187, 0.187; 0.187, 0.374, 0.187; 0.187, 0.187, 0.374. Click 'Register Atom Position' after each entry."
|
||
},
|
||
{
|
||
"question": "How to proceed after specifying all atom positions in the Molecule Definition Utility for AuCu3?",
|
||
"answer": "In 'Step 3', you may choose to represent unit cell edges as bonds or elect to not represent any bonds. The image can be rotated using mouse click-and-drag. Note that data or images cannot be saved in this version of the utility; use screen capture software to record the image."
|
||
}
|
||
]
|
||
```
|
||
处理第 383/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Convert the (010) plane into the four-index Miller-Bravais scheme for hexagonal unit cells.",
|
||
"answer": "For (010), h=0, k=1, and l=0, and, from Equation, the value of i is equal to i=-(h+k)=-(0+1)=-1. Therefore, the (010) plane becomes (01-10)."
|
||
},
|
||
{
|
||
"question": "Convert the (101) plane into the four-index Miller-Bravais scheme for hexagonal unit cells.",
|
||
"answer": "For the (101) plane, h=1, k=0, and l=1, and computation of i using Equation leads to i=-(h+k)=-[1+0]=-1 such that (101) becomes (10-11)."
|
||
}
|
||
]
|
||
```
|
||
处理第 384/821 条数据...
|
||
[
|
||
{
|
||
"question": "Derive the planar density expression for the HCP (0001) plane in terms of the atomic radius R.",
|
||
"answer": "A (0001) plane for an HCP unit cell is shown below. Each of the 6 perimeter atoms in this plane is shared with three other unit cells, whereas the center atom is shared with no other unit cells; this gives rise to three equivalent atoms belonging to this plane. In terms of the atomic radius R, the area of each of the 6 equilateral triangles that have been drawn is R^(2) sqrt(3), or the total area of the plane shown is 6 R^(2) sqrt(3). And the planar density for this (0001) plane is equal to PD_(0001) = (number of atoms centered on (0001) plane) / (area of (0001) plane) = (3 atoms) / (6 R^(2) sqrt(3)) = 1 / (2 R^(2) sqrt(3))"
|
||
},
|
||
{
|
||
"question": "Compute the planar density value for the HCP (0001) plane for magnesium, given the atomic radius for magnesium is 0.160 nm.",
|
||
"answer": "From the table inside the front cover, the atomic radius for magnesium is 0.160 nm. Therefore, the planar density for the (0001) plane is PD_(0001)(Mg) = 1 / (2 R^(2) sqrt(3)) = 1 / (2 (0.160 nm)^(2) sqrt(3)) = 11.28 nm^(-2) = 1.128 x 10^(19) m^(-2)"
|
||
}
|
||
]
|
||
处理第 385/821 条数据...
|
||
It's a single issue.
|
||
处理第 386/821 条数据...
|
||
[
|
||
{
|
||
"question": "The metal iridium has an FCC crystal structure. If the angle of diffraction for the (220) set of planes occurs at 69.22 degrees (first-order reflection) when monochromatic x-radiation having a wavelength of 0.1542 nm is used, compute the interplanar spacing for this set of planes.",
|
||
"answer": "From the data given in the problem, and realizing that 69.22 degrees = 2 theta, the interplanar spacing for the (220) set of planes for iridium may be computed using the equation d220 = (n * lambda) / (2 * sin(theta)) = (1 * 0.1542 nm) / (2 * sin(34.61 degrees)) = 0.1357 nm."
|
||
},
|
||
{
|
||
"question": "The metal iridium has an FCC crystal structure. Given the interplanar spacing for the (220) set of planes is 0.1357 nm, compute the atomic radius for an iridium atom.",
|
||
"answer": "In order to compute the atomic radius we must first determine the lattice parameter, a, using the equation a = d220 * sqrt(h^2 + k^2 + l^2) = (0.1357 nm) * sqrt(8) = 0.3838 nm. And, from the equation for FCC crystal structure, R = a / (2 * sqrt(2)) = (0.3838 nm) / (2 * sqrt(2)) = 0.1357 nm."
|
||
}
|
||
]
|
||
处理第 387/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Which of the elements listed would form a substitutional solid solution having complete solubility with copper (Cu)? Consider the criteria: atomic radius difference within ±15%, same crystal structure, similar electronegativity, and same or nearly same valence.",
|
||
"answer": "Ni, Pd, and Pt meet all of the criteria and thus form substitutional solid solutions having complete solubility. At elevated temperatures Co and Fe experience allotropic transformations to the FCC crystal structure, and thus display complete solid solubility at these temperatures."
|
||
},
|
||
{
|
||
"question": "Which of the elements listed would form a substitutional solid solution of incomplete solubility with copper (Cu)? Consider the criteria: atomic radius difference greater than ±15%, different crystal structure, dissimilar electronegativity, or different valence.",
|
||
"answer": "Ag, Al, Co, Cr, Fe, and Zn form substitutional solid solutions of incomplete solubility. All these metals have either BCC or HCP crystal structures, and/or the difference between their atomic radii and that for Cu are greater than ±15%, and/or have a valence different than 2+."
|
||
},
|
||
{
|
||
"question": "Which of the elements listed would form an interstitial solid solution with copper (Cu)? Consider the criteria: significantly smaller atomic radius compared to Cu.",
|
||
"answer": "C, H, and O form interstitial solid solutions. These elements have atomic radii that are significantly smaller than the atomic radius of Cu."
|
||
}
|
||
]
|
||
```
|
||
处理第 388/821 条数据...
|
||
It's a single issue.
|
||
处理第 389/821 条数据...
|
||
It's a single issue.
|
||
处理第 390/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is the mass of copper in grams, given 99.7 lbm of copper?",
|
||
"answer": "The mass of copper is (99.7 lbm)(453.6 g/lbm) = 45,224 g."
|
||
},
|
||
{
|
||
"question": "What is the mass of zinc in grams, given 102 lbm of zinc?",
|
||
"answer": "The mass of zinc is (102 lbm)(453.1 g/lbm) = 46,267 g."
|
||
},
|
||
{
|
||
"question": "What is the mass of lead in grams, given 2.1 lbm of lead?",
|
||
"answer": "The mass of lead is (2.1 lbm)(453.3 g/lbm) = 953 g."
|
||
},
|
||
{
|
||
"question": "How many moles of copper are there in 45,224 g of copper, given the atomic weight of copper is 63.55 g/mol?",
|
||
"answer": "The number of moles of copper is 45,224 g / 63.55 g/mol = 711.6 mol."
|
||
},
|
||
{
|
||
"question": "How many moles of zinc are there in 46,267 g of zinc, given the atomic weight of zinc is 65.41 g/mol?",
|
||
"answer": "The number of moles of zinc is 46,267 g / 65.41 g/mol = 707.3 mol."
|
||
},
|
||
{
|
||
"question": "How many moles of lead are there in 953 g of lead, given the atomic weight of lead is 207.2 g/mol?",
|
||
"answer": "The number of moles of lead is 953 g / 207.2 g/mol = 4.6 mol."
|
||
},
|
||
{
|
||
"question": "What is the atom percent of copper in the alloy, given 711.6 mol of copper, 707.3 mol of zinc, and 4.6 mol of lead?",
|
||
"answer": "The atom percent of copper is (711.6 mol / (711.6 mol + 707.3 mol + 4.6 mol)) × 100 = 50.0 at%."
|
||
},
|
||
{
|
||
"question": "What is the atom percent of zinc in the alloy, given 711.6 mol of copper, 707.3 mol of zinc, and 4.6 mol of lead?",
|
||
"answer": "The atom percent of zinc is (707.3 mol / (711.6 mol + 707.3 mol + 4.6 mol)) × 100 = 49.7 at%."
|
||
},
|
||
{
|
||
"question": "What is the atom percent of lead in the alloy, given 711.6 mol of copper, 707.3 mol of zinc, and 4.6 mol of lead?",
|
||
"answer": "The atom percent of lead is (4.6 mol / (711.6 mol + 707.3 mol + 4.6 mol)) × 100 = 0.3 at%."
|
||
}
|
||
]
|
||
```
|
||
处理第 391/821 条数据...
|
||
It's a single issue.
|
||
处理第 392/821 条数据...
|
||
It's a single issue.
|
||
处理第 393/821 条数据...
|
||
It's a single issue.
|
||
处理第 394/821 条数据...
|
||
It's a single issue.
|
||
处理第 395/821 条数据...
|
||
It's a single issue.
|
||
处理第 396/821 条数据...
|
||
It's a single issue.
|
||
处理第 397/821 条数据...
|
||
[
|
||
{
|
||
"question": "For a given material, would you expect the surface energy to be greater than, the same as, or less than the grain boundary energy? Why?",
|
||
"answer": "The surface energy will be greater than the grain boundary energy. For grain boundaries, some atoms on one side of a boundary will bond to atoms on the other side; such is not the case for surface atoms. Therefore, there will be fewer unsatisfied bonds along a grain boundary."
|
||
},
|
||
{
|
||
"question": "The grain boundary energy of a small-angle grain boundary is less than for a high-angle one. Why is this so?",
|
||
"answer": "The small-angle grain boundary energy is lower than for a high-angle one because more atoms bond across the boundary for the small-angle, and, thus, there are fewer unsatisfied bonds."
|
||
}
|
||
]
|
||
处理第 398/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Briefly describe a twin and a twin boundary.",
|
||
"answer": "A twin boundary is an interface such that atoms on one side are located at mirror image positions of those atoms situated on the other boundary side. The region on one side of this boundary is called a twin."
|
||
},
|
||
{
|
||
"question": "Cite the difference between mechanical and annealing twins.",
|
||
"answer": "Mechanical twins are produced as a result of mechanical deformation and generally occur in BCC and HCP metals. Annealing twins form during annealing heat treatments, most often in FCC metals."
|
||
}
|
||
]
|
||
```
|
||
处理第 399/821 条数据...
|
||
It's a single issue.
|
||
处理第 400/821 条数据...
|
||
[
|
||
{
|
||
"question": "Briefly explain the concept of a driving force.",
|
||
"answer": "The driving force is that which compels a reaction to occur."
|
||
},
|
||
{
|
||
"question": "What is the driving force for steady-state diffusion?",
|
||
"answer": "The driving force for steady-state diffusion is the concentration gradient."
|
||
}
|
||
]
|
||
处理第 401/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For a steel alloy, given that a carburizing heat treatment of 10-h duration raises the carbon concentration to 0.45 wt% at a point 2.5 mm from the surface, estimate the time necessary to achieve the same concentration at a 5.0-mm position for an identical steel and at the same carburizing temperature.",
|
||
"answer": "This problem is solved using the relation x²/Dt = constant. Since D is constant at the same temperature, x²/t = constant. Thus, (2.5 mm)²/10 h = (5.0 mm)²/t₂. Solving for t₂ gives t₂ = 40 h."
|
||
}
|
||
]
|
||
```
|
||
处理第 402/821 条数据...
|
||
It's a single issue.
|
||
处理第 403/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Given the activation energy for the diffusion of carbon in chromium is 111,000 J/mol and the diffusion coefficient D at 1400 K is 6.25 × 10^-11 m²/s, calculate the pre-exponential factor D₀.",
|
||
"answer": "D₀ = D exp (Q_d / (R T)) = (6.25 × 10^-11 m²/s) exp [111,000 J/mol / (8.31 J/mol-K × 1400 K)] = 8.7 × 10^-7 m²/s"
|
||
},
|
||
{
|
||
"question": "Using the pre-exponential factor D₀ = 8.7 × 10^-7 m²/s and activation energy Q_d = 111,000 J/mol, calculate the diffusion coefficient D at 1100 K.",
|
||
"answer": "D = D₀ exp (-Q_d / (R T)) = (8.7 × 10^-7 m²/s) exp [-111,000 J/mol / (8.31 J/mol-K × 1100 K)] = 4.6 × 10^-12 m²/s"
|
||
}
|
||
]
|
||
```
|
||
处理第 404/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "The diffusion coefficients for iron in nickel are given at two temperatures: T1 = 1273 K, D1 = 9.4 × 10^-16 m²/s and T2 = 1473 K, D2 = 2.4 × 10^-14 m²/s. Determine the values of D0 and the activation energy Qd.",
|
||
"answer": "Using the Arrhenius equation, we set up two simultaneous equations with Qd and D0 as unknowns: ln D1 = ln D0 - (Qd / R)(1 / T1) and ln D2 = ln D0 - (Qd / R)(1 / T2). Solving for Qd: Qd = -R (ln D1 - ln D2) / (1 / T1 - 1 / T2) = -(8.31 J/mol-K) [ln(9.4 × 10^-16) - ln(2.4 × 10^-14)] / (1 / 1273 K - 1 / 1473 K) = 252,400 J/mol. Solving for D0: D0 = D1 exp(Qd / R T1) = (9.4 × 10^-16 m²/s) exp[252,400 J/mol / (8.31 J/mol-K)(1273 K)] = 2.2 × 10^-5 m²/s."
|
||
},
|
||
{
|
||
"question": "Using the values of D0 = 2.2 × 10^-5 m²/s and Qd = 252,400 J/mol determined from the diffusion coefficients for iron in nickel at T1 = 1273 K and T2 = 1473 K, what is the magnitude of D at 1100°C (1373 K)?",
|
||
"answer": "Using the Arrhenius equation with the determined values of D0 and Qd, D at 1373 K is calculated as: D = D0 exp(-Qd / R T) = (2.2 × 10^-5 m²/s) exp[-252,400 J/mol / (8.31 J/mol-K)(1373 K)] = 5.4 × 10^-15 m²/s."
|
||
}
|
||
]
|
||
```
|
||
处理第 405/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Determine the values of D0 and Qd given the diffusion coefficients for silver in copper at two temperatures: T1 = 650°C (923 K) with D1 = 5.5 × 10^-16 m²/s and T2 = 900°C (1173 K) with D2 = 1.3 × 10^-13 m²/s.",
|
||
"answer": "Using the Arrhenius equation, set up two simultaneous equations: ln D1 = ln D0 - (Qd / R)(1 / T1) and ln D2 = ln D0 - (Qd / R)(1 / T2). Solving for Qd: Qd = -R (ln D1 - ln D2) / (1 / T1 - 1 / T2) = -(8.31 J/mol-K) [ln(5.5 × 10^-16) - ln(1.3 × 10^-13)] / (1/923 K - 1/1173 K) = 196,700 J/mol. Solving for D0: D0 = D1 exp(Qd / R T1) = (5.5 × 10^-16 m²/s) exp[196,700 J/mol / (8.31 J/mol-K)(923 K)] = 7.5 × 10^-5 m²/s."
|
||
},
|
||
{
|
||
"question": "What is the magnitude of D at 875°C (1148 K) using the values of D0 = 7.5 × 10^-5 m²/s and Qd = 196,700 J/mol?",
|
||
"answer": "Using the Arrhenius equation: D = D0 exp(-Qd / R T) = (7.5 × 10^-5 m²/s) exp[-196,700 J/mol / (8.31 J/mol-K)(1148 K)] = 8.3 × 10^-14 m²/s."
|
||
}
|
||
]
|
||
```
|
||
处理第 406/821 条数据...
|
||
It's a single issue.
|
||
处理第 407/821 条数据...
|
||
[
|
||
{
|
||
"question": "Calculate the value of D0 from the given data at 727°C (1000 K) with a diffusion flux of 5.4×10−10 kg/m2-s and a concentration gradient of −350 kg/m4, assuming an activation energy for diffusion of 125,000 J/mol.",
|
||
"answer": "To calculate D0, use the equation J = -D0 (ΔC/Δx) exp(-Qd/RT). Solving for D0 gives D0 = - (J / (ΔC/Δx)) exp(Qd/RT). Plugging in the values: D0 = - (5.4×10−10 kg/m2-s / -350 kg/m4) exp(125,000 J/mol / (8.31 J/mol-K)(1000 K)) = 5.26×10−6 m2/s."
|
||
},
|
||
{
|
||
"question": "Calculate the diffusion flux at 1027°C (1300 K) for the same concentration gradient of −350 kg/m4 and activation energy for diffusion of 125,000 J/mol, using the previously calculated D0 value of 5.26×10−6 m2/s.",
|
||
"answer": "Using the equation J = -D0 (ΔC/Δx) exp(-Qd/RT), the diffusion flux at 1300 K is calculated as: J = - (5.26×10−6 m2/s)(-350 kg/m4) exp(-125,000 J/mol / (8.31 J/mol-K)(1300 K)) = 1.74×10−8 kg/m2-s."
|
||
}
|
||
]
|
||
处理第 408/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For a bronze alloy, the stress at which plastic deformation begins is 275 MPa (40,000 psi), and the modulus of elasticity is 115 GPa (16.7 × 10^6 psi). What is the maximum load that may be applied to a specimen with a cross-sectional area of 325 mm^2 (0.5 in.^2) without plastic deformation?",
|
||
"answer": "This portion of the problem calls for a determination of the maximum load that can be applied without plastic deformation (F_y). Taking the yield strength to be 275 MPa, and employment of Equation leads to F_y = σ_y A_0 = (275 × 10^6 N/m^2)(325 × 10^-6 m^2) = 89,375 N (20,000 lb_f)."
|
||
},
|
||
{
|
||
"question": "For a bronze alloy, the stress at which plastic deformation begins is 275 MPa (40,000 psi), and the modulus of elasticity is 115 GPa (16.7 × 10^6 psi). If the original specimen length is 115 mm (4.5 in.), what is the maximum length to which it may be stretched without causing plastic deformation?",
|
||
"answer": "The maximum length to which the sample may be deformed without plastic deformation is determined from Equations as l_i = l_0 (1 + σ/E) = (115 mm)[1 + (275 MPa)/(115 × 10^3 MPa)] = 115.28 mm (4.51 in.)."
|
||
}
|
||
]
|
||
```
|
||
处理第 409/821 条数据...
|
||
It's a single issue.
|
||
处理第 410/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Compute the elastic modulus for titanium alloy using the stress-strain behavior observed in the 'Tensile Tests' module of Virtual Materials Science and Engineering (VMSE). How does this value compare with that presented in Table 6.1 for titanium?",
|
||
"answer": "The elastic modulus is calculated as E = (σ2 - σ1)/(ε2 - ε1). For titanium, σ2 = 404.2 MPa and ε2 = 0.0038. Therefore, E = (404.2 MPa - 0 MPa)/(0.0038 - 0) = 106,400 MPa = 106.4 GPa. The elastic modulus for titanium given in Table 6.1 is 107 GPa, which is in very good agreement with this value."
|
||
},
|
||
{
|
||
"question": "Compute the elastic modulus for tempered steel using the stress-strain behavior observed in the 'Tensile Tests' module of Virtual Materials Science and Engineering (VMSE). How does this value compare with that presented in Table 6.1 for tempered steel?",
|
||
"answer": "The elastic modulus is calculated as E = (σ2 - σ1)/(ε2 - ε1). For tempered steel, σ2 = 962.2 MPa and ε2 = 0.0047. Therefore, E = (962.2 MPa - 0 MPa)/(0.0047 - 0) = 204,700 MPa = 204.7 GPa. The elastic modulus for steel given in Table 6.1 is 207 GPa, which is in reasonably good agreement with this value."
|
||
},
|
||
{
|
||
"question": "Compute the elastic modulus for aluminum using the stress-strain behavior observed in the 'Tensile Tests' module of Virtual Materials Science and Engineering (VMSE). How does this value compare with that presented in Table 6.1 for aluminum?",
|
||
"answer": "The elastic modulus is calculated as E = (σ2 - σ1)/(ε2 - ε1). For aluminum, σ2 = 145.1 MPa and ε2 = 0.0021. Therefore, E = (145.1 MPa - 0 MPa)/(0.0021 - 0) = 69,100 MPa = 69.1 GPa. The elastic modulus for aluminum given in Table 6.1 is 69 GPa, which is in excellent agreement with this value."
|
||
},
|
||
{
|
||
"question": "Compute the elastic modulus for carbon steel using the stress-strain behavior observed in the 'Tensile Tests' module of Virtual Materials Science and Engineering (VMSE). How does this value compare with that presented in Table 6.1 for carbon steel?",
|
||
"answer": "The elastic modulus is calculated as E = (σ2 - σ1)/(ε2 - ε1). For carbon steel, σ2 = 129 MPa and ε2 = 0.0006. Therefore, E = (129 MPa - 0 MPa)/(0.0006 - 0) = 215,000 MPa = 215 GPa. The elastic modulus for steel given in Table 6.1 is 207 GPa, which is in reasonable agreement with this value."
|
||
}
|
||
]
|
||
```
|
||
处理第 411/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is the modulus of elasticity (E) for a cylindrical specimen of a hypothetical metal alloy with a diameter of 8.0 mm (0.31 in.), subjected to a tensile force of 1000 N (225 lbf), given that the elastic reduction in diameter is 2.8 x 10^-4 mm (1.10 x 10^-5 in.) and Poisson's ratio is 0.30?",
|
||
"answer": "The modulus of elasticity (E) is calculated using the formula: E = (4 F v) / (π d0 Δd), where F is the tensile force, v is Poisson's ratio, d0 is the original diameter, and Δd is the reduction in diameter. Substituting the given values: F = 1000 N, v = 0.30, d0 = 8 x 10^-3 m, Δd = 2.8 x 10^-7 m, we get E = (4 x 1000 N x 0.30) / (π x 8 x 10^-3 m x 2.8 x 10^-7 m) = 1.705 x 10^11 Pa = 170.5 GPa (24.7 x 10^6 psi)."
|
||
}
|
||
]
|
||
```
|
||
处理第 412/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For a cylindrical rod 100 mm long with a diameter of 10.0 mm subjected to a tensile load of 27,500 N, which of the listed materials (Aluminum alloy, Brass alloy, Steel alloy, Titanium alloy) will not experience plastic deformation? The yield strengths are: Aluminum alloy - 200 MPa, Brass alloy - 300 MPa, Steel alloy - 400 MPa, Titanium alloy - 650 MPa.",
|
||
"answer": "First, compute the stress: sigma = F / A0 = 27,500 N / (pi * (10.0 mm / 2)^2) = 350 MPa. The materials with yield strengths greater than 350 MPa are Steel alloy (400 MPa) and Titanium alloy (650 MPa). Thus, Steel and Titanium alloys will not experience plastic deformation."
|
||
},
|
||
{
|
||
"question": "For the Steel alloy (Modulus of Elasticity = 207 GPa, Poisson's Ratio = 0.30) under the same loading conditions, will the diameter reduction be less than 7.5 x 10^-3 mm?",
|
||
"answer": "Calculate the diameter reduction: Delta d = - (v * sigma * d0) / E = - (0.30 * 350 MPa * 10 mm) / 207,000 MPa = -5.1 x 10^-3 mm. Since |Delta d| = 5.1 x 10^-3 mm < 7.5 x 10^-3 mm, the Steel alloy meets the diameter reduction criterion."
|
||
},
|
||
{
|
||
"question": "For the Titanium alloy (Modulus of Elasticity = 107 GPa, Poisson's Ratio = 0.34) under the same loading conditions, will the diameter reduction be less than 7.5 x 10^-3 mm?",
|
||
"answer": "Calculate the diameter reduction: Delta d = - (v * sigma * d0) / E = - (0.34 * 350 MPa * 10 mm) / 107,000 MPa = -11.1 x 10^-3 mm. Since |Delta d| = 11.1 x 10^-3 mm > 7.5 x 10^-3 mm, the Titanium alloy does not meet the diameter reduction criterion."
|
||
}
|
||
]
|
||
```
|
||
处理第 413/821 条数据...
|
||
[
|
||
{
|
||
"question": "A cylindrical metal specimen having an original diameter of 12.8 mm (0.505 in.) and gauge length of 50.80 mm (2.000 in.) is pulled in tension until fracture occurs. The diameter at the point of fracture is 6.60 mm (0.260 in.). Calculate the ductility in terms of percent reduction in area.",
|
||
"answer": "Percent reduction in area is computed using the equation: %RA = (π(d0/2)^2 - π(df/2)^2) / π(d0/2)^2 × 100, where d0 and df are the original and fracture diameters. Thus, %RA = (π(12.8 mm/2)^2 - π(6.60 mm/2)^2) / π(12.8 mm/2)^2 × 100 = 73.4%."
|
||
},
|
||
{
|
||
"question": "A cylindrical metal specimen having an original diameter of 12.8 mm (0.505 in.) and gauge length of 50.80 mm (2.000 in.) is pulled in tension until fracture occurs. The fractured gauge length is 72.14 mm (2.840 in.). Calculate the ductility in terms of percent elongation.",
|
||
"answer": "Percent elongation is computed using the equation: %EL = (lf - l0) / l0 × 100, where l0 and lf are the original and fractured gauge lengths. Thus, %EL = (72.14 mm - 50.80 mm) / 50.80 mm × 100 = 42%."
|
||
}
|
||
]
|
||
处理第 414/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For a metal alloy with a strain-hardening exponent n of 0.25, a true stress of 415 MPa produces a plastic true strain of 0.475. Calculate the strength coefficient K.",
|
||
"answer": "The strength coefficient K can be calculated using the equation K = σ_T / (ε_T)^n. Substituting the given values: K = 415 MPa / (0.475)^0.25 = 500 MPa (72,500 psi)."
|
||
},
|
||
{
|
||
"question": "Using the calculated strength coefficient K of 500 MPa and strain-hardening exponent n of 0.25, determine the true strain ε_T when a true stress of 325 MPa is applied.",
|
||
"answer": "The true strain ε_T can be calculated using the equation ε_T = (σ_T / K)^(1/n). Substituting the given values: ε_T = (325 MPa / 500 MPa)^(1/0.25) = 0.179."
|
||
},
|
||
{
|
||
"question": "Given the original length l0 of 300 mm and the calculated true strain ε_T of 0.179, find the final length li after elongation.",
|
||
"answer": "The final length li can be calculated using the equation li = l0 * e^(ε_T). Substituting the given values: li = 300 mm * e^(0.179) = 358.8 mm (14.11 in)."
|
||
},
|
||
{
|
||
"question": "Calculate the elongation Δl of the specimen using the original length l0 of 300 mm and the final length li of 358.8 mm.",
|
||
"answer": "The elongation Δl is calculated as Δl = li - l0 = 358.8 mm - 300 mm = 58.8 mm (2.31 in)."
|
||
}
|
||
]
|
||
```
|
||
处理第 415/821 条数据...
|
||
It's a single issue.
|
||
处理第 416/821 条数据...
|
||
[
|
||
{
|
||
"question": "Consider a metal specimen that has a dislocation density of 10^4 mm^-2. Suppose that all the dislocations in 1000 mm^3 (1 cm^3) were somehow removed and linked end to end. How far (in miles) would this chain extend?",
|
||
"answer": "The total length in 1000 mm^3 of material having a density of 10^4 mm^-2 is (10^4 mm^-2)(1000 mm^3) = 10^7 mm = 10^4 m = 6.2 mi."
|
||
},
|
||
{
|
||
"question": "Suppose that the dislocation density is increased to 10^10 mm^-2 by cold working. What would be the chain length of dislocations in 1000 mm^3 of material?",
|
||
"answer": "For a dislocation density of 10^10 mm^-2, the total length is (10^10 mm^-2)(1000 mm^3) = 10^13 mm = 10^10 m = 6.2 x 10^6 mi."
|
||
}
|
||
]
|
||
处理第 417/821 条数据...
|
||
[
|
||
{
|
||
"question": "For an edge dislocation, cite the relationship between the direction of the applied shear stress and the direction of dislocation line motion.",
|
||
"answer": "edge dislocation--parallel"
|
||
},
|
||
{
|
||
"question": "For a screw dislocation, cite the relationship between the direction of the applied shear stress and the direction of dislocation line motion.",
|
||
"answer": "screw dislocation--perpendicular"
|
||
},
|
||
{
|
||
"question": "For a mixed dislocation, cite the relationship between the direction of the applied shear stress and the direction of dislocation line motion.",
|
||
"answer": "mixed dislocation--neither parallel nor perpendicular"
|
||
}
|
||
]
|
||
处理第 418/821 条数据...
|
||
[
|
||
{
|
||
"question": "Define a slip system.",
|
||
"answer": "A slip system is a crystallographic plane, and, within that plane, a direction along which dislocation motion (or slip) occurs."
|
||
},
|
||
{
|
||
"question": "Do all metals have the same slip system? Why or why not?",
|
||
"answer": "All metals do not have the same slip system. The reason for this is that for most metals, the slip system will consist of the most densely packed crystallographic plane, and within that plane the most closely packed direction. This plane and direction will vary from crystal structure to crystal structure."
|
||
}
|
||
]
|
||
处理第 419/821 条数据...
|
||
It's a single issue.
|
||
处理第 420/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "A single crystal of aluminum is oriented for a tensile test such that its slip plane normal makes an angle of 28.1 degrees with the tensile axis. Three possible slip directions make angles of 62.4 degrees, 72.0 degrees, and 81.1 degrees with the same tensile axis. Which of these three slip directions is most favored?",
|
||
"answer": "Slip will occur along that direction for which (cos φ cos λ) is a maximum, or, in this case, for the largest cos λ. Cosines for the possible λ values are given below. cos(62.4 degrees) = 0.46, cos(72.0 degrees) = 0.31, cos(81.1 degrees) = 0.15. Thus, the slip direction is at an angle of 62.4 degrees with the tensile axis."
|
||
},
|
||
{
|
||
"question": "If plastic deformation begins at a tensile stress of 1.95 MPa (280 psi) for the aluminum crystal oriented as described, determine the critical resolved shear stress for aluminum.",
|
||
"answer": "The critical resolved shear stress is given by τ_crss = σ_y (cos φ cos λ)_max. Using φ = 28.1 degrees and λ = 62.4 degrees, τ_crss = (1.95 MPa)[cos(28.1 degrees) cos(62.4 degrees)] = 0.80 MPa (114 psi)."
|
||
}
|
||
]
|
||
```
|
||
处理第 421/821 条数据...
|
||
It's a single issue.
|
||
处理第 422/821 条数据...
|
||
[
|
||
{
|
||
"question": "What is the difference between deformation by twinning and deformation by slip relative to crystallographic reorientation?",
|
||
"answer": "With slip deformation there is no crystallographic reorientation, whereas with twinning there is a reorientation."
|
||
},
|
||
{
|
||
"question": "What is the difference between deformation by twinning and deformation by slip relative to atomic displacements?",
|
||
"answer": "For slip, the atomic displacements occur in atomic spacing multiples, whereas for twinning, these displacements may be other than by atomic spacing multiples."
|
||
},
|
||
{
|
||
"question": "What is the difference between deformation by twinning and deformation by slip relative to the conditions of occurrence in metals?",
|
||
"answer": "Slip occurs in metals having many slip systems, whereas twinning occurs in metals having relatively few slip systems."
|
||
},
|
||
{
|
||
"question": "What is the difference between deformation by twinning and deformation by slip relative to the final deformation result?",
|
||
"answer": "Normally slip results in relatively large deformations, whereas only small deformations result for twinning."
|
||
}
|
||
]
|
||
处理第 423/821 条数据...
|
||
It's a single issue.
|
||
处理第 424/821 条数据...
|
||
It's a single issue.
|
||
处理第 425/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For the first cylindrical specimen with initial radius of 16 mm and deformed radius of 11 mm, calculate the percent cold work (%CW).",
|
||
"answer": "The percent cold work (%CW) for the first specimen is calculated as follows: %CW = (A0 - Ad)/A0 × 100 = (πr0² - πrd²)/πr0² × 100 = (π(16 mm)² - π(11 mm)²)/π(16 mm)² × 100 = 52.7% CW."
|
||
},
|
||
{
|
||
"question": "For the second cylindrical specimen with initial radius of 12 mm to achieve the same deformed hardness as the first specimen (52.7% CW), compute its deformed radius (rd).",
|
||
"answer": "The deformed radius (rd) for the second specimen is computed using the equation: rd = r0 √(1 - %CW/100) = (12 mm) √(1 - 52.7% CW/100) = 8.25 mm."
|
||
}
|
||
]
|
||
```
|
||
处理第 426/821 条数据...
|
||
It's a single issue.
|
||
处理第 427/821 条数据...
|
||
It's a single issue.
|
||
处理第 428/821 条数据...
|
||
It's a single issue.
|
||
处理第 429/821 条数据...
|
||
It's a single issue.
|
||
处理第 430/821 条数据...
|
||
It's a single issue.
|
||
处理第 431/821 条数据...
|
||
It's a single issue.
|
||
处理第 432/821 条数据...
|
||
It's a single issue.
|
||
处理第 433/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Given a 4340 steel alloy specimen with a plane strain fracture toughness of 45 MPa sqrt(m) (41 ksi sqrt(in)), a stress of 1000 MPa (145,000 psi), and the largest surface crack of 0.75 mm (0.03 in) long with parameter Y = 1.0, calculate the critical stress (σ_c) required for fracture.",
|
||
"answer": "The critical stress (σ_c) is calculated using the formula σ_c = K_IC / (Y * sqrt(π * a)). Substituting the given values: σ_c = 45 MPa sqrt(m) / (1.0 * sqrt(π * 0.75 × 10^-3 m)) = 927 MPa (133,500 psi)."
|
||
},
|
||
{
|
||
"question": "Will the 4340 steel alloy specimen fracture under the given stress of 1000 MPa (145,000 psi) when the critical stress (σ_c) is 927 MPa (133,500 psi)? Why or why not?",
|
||
"answer": "Fracture will most likely occur because the applied stress of 1000 MPa (145,000 psi) exceeds the critical stress of 927 MPa (133,500 psi) that the specimen can tolerate before fracture."
|
||
}
|
||
]
|
||
```
|
||
处理第 434/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For an aluminum alloy with a plane strain fracture toughness of 35 MPa sqrt(m), given that fracture occurs at a stress of 250 MPa when the maximum internal crack length is 2.0 mm, calculate the parameter Y.",
|
||
"answer": "Y is calculated using the formula Y = K_Ic / (σ sqrt(πa)), where K_Ic = 35 MPa sqrt(m), σ = 250 MPa, and a = 1.0 mm (half of 2.0 mm). Thus, Y = 35 MPa sqrt(m) / (250 MPa sqrt(π × 1.0 × 10^-3 m)) = 2.50."
|
||
},
|
||
{
|
||
"question": "Using the calculated Y value of 2.50, determine if fracture will occur for the same alloy under a stress level of 325 MPa and a maximum internal crack length of 1.0 mm.",
|
||
"answer": "Calculate Yσ sqrt(πa) with Y = 2.50, σ = 325 MPa, and a = 0.5 mm (half of 1.0 mm). Thus, Yσ sqrt(πa) = 2.50 × 325 MPa × sqrt(π × 0.5 × 10^-3 m) = 32.2 MPa sqrt(m). Since 32.2 MPa sqrt(m) is less than the K_Ic of 35 MPa sqrt(m), fracture will not occur."
|
||
}
|
||
]
|
||
```
|
||
处理第 435/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For an aluminum alloy with a plane strain fracture toughness of 40 MPa sqrt(m), given that fracture occurs at a stress of 365 MPa when the maximum internal crack length is 2.5 mm, determine the parameter Y for these conditions.",
|
||
"answer": "Y = (K_IC) / (σ sqrt(πa)) = (40 MPa sqrt(m)) / (365 MPa sqrt(π (2.5 × 10^-3 m / 2))) = 1.75"
|
||
},
|
||
{
|
||
"question": "Using the determined parameter Y (1.75) and the same fracture toughness (40 MPa sqrt(m)), compute the stress level at which fracture will occur for a critical internal crack length of 4.0 mm.",
|
||
"answer": "σ_c = (K_IC) / (Y sqrt(πa)) = (40 MPa sqrt(m)) / (1.75 sqrt(π (4 × 10^-3 m / 2))) = 288 MPa (41,500 psi)"
|
||
}
|
||
]
|
||
```
|
||
处理第 436/821 条数据...
|
||
It's a single issue.
|
||
处理第 437/821 条数据...
|
||
It's a single issue.
|
||
处理第 438/821 条数据...
|
||
It's a single issue.
|
||
处理第 439/821 条数据...
|
||
It's a single issue.
|
||
处理第 440/821 条数据...
|
||
[
|
||
{
|
||
"question": "Briefly explain the difference between fatigue striations and beachmarks in terms of size.",
|
||
"answer": "With regard to size, beachmarks are normally of macroscopic dimensions and may be observed with the naked eye; fatigue striations are of microscopic size and it is necessary to observe them using electron microscopy."
|
||
},
|
||
{
|
||
"question": "Briefly explain the difference between fatigue striations and beachmarks in terms of origin.",
|
||
"answer": "With regard to origin, beachmarks result from interruptions in the stress cycles; each fatigue striation corresponds to the advance of a fatigue crack during a single load cycle."
|
||
}
|
||
]
|
||
处理第 441/821 条数据...
|
||
It's a single issue.
|
||
处理第 442/821 条数据...
|
||
[
|
||
{
|
||
"question": "What is the approximate temperature at which creep deformation becomes an important consideration for nickel, given that creep becomes important at about 0.4 Tm, where Tm is the absolute melting temperature of the metal?",
|
||
"answer": "For Ni, 0.4 Tm = (0.4)(1455 + 273) = 691 K or 418 degrees Celsius (785 degrees Fahrenheit)"
|
||
},
|
||
{
|
||
"question": "What is the approximate temperature at which creep deformation becomes an important consideration for copper, given that creep becomes important at about 0.4 Tm, where Tm is the absolute melting temperature of the metal?",
|
||
"answer": "For Cu, 0.4 Tm = (0.4)(1085 + 273) = 543 K or 270 degrees Celsius (518 degrees Fahrenheit)"
|
||
},
|
||
{
|
||
"question": "What is the approximate temperature at which creep deformation becomes an important consideration for iron, given that creep becomes important at about 0.4 Tm, where Tm is the absolute melting temperature of the metal?",
|
||
"answer": "For Fe, 0.4 Tm = (0.4)(1538 + 273) = 725 K or 450 degrees Celsius (845 degrees Fahrenheit)"
|
||
},
|
||
{
|
||
"question": "What is the approximate temperature at which creep deformation becomes an important consideration for tungsten, given that creep becomes important at about 0.4 Tm, where Tm is the absolute melting temperature of the metal?",
|
||
"answer": "For W, 0.4 Tm = (0.4)(3410 + 273) = 1473 K or 1200 degrees Celsius (2190 degrees Fahrenheit)"
|
||
},
|
||
{
|
||
"question": "What is the approximate temperature at which creep deformation becomes an important consideration for lead, given that creep becomes important at about 0.4 Tm, where Tm is the absolute melting temperature of the metal?",
|
||
"answer": "For Pb, 0.4 Tm = (0.4)(327 + 273) = 240 K or -33 degrees Celsius (-27 degrees Fahrenheit)"
|
||
},
|
||
{
|
||
"question": "What is the approximate temperature at which creep deformation becomes an important consideration for aluminum, given that creep becomes important at about 0.4 Tm, where Tm is the absolute melting temperature of the metal?",
|
||
"answer": "For Al, 0.4 Tm = (0.4)(660 + 273) = 373 K or 100 degrees Celsius (212 degrees Fahrenheit)"
|
||
}
|
||
]
|
||
处理第 443/821 条数据...
|
||
It's a single issue.
|
||
处理第 444/821 条数据...
|
||
It's a single issue.
|
||
处理第 445/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For the alloy composition 60 wt% A-40 wt% B with fraction α phase 0.57 and fraction β phase 0.43, determine the composition of the phase boundary (or solubility limit) for both α and β phases at this temperature.",
|
||
"answer": "Using the lever rule expression for this alloy composition: W_α1 = 0.57 = (C_β - C_01) / (C_β - C_α) = (C_β - 60) / (C_β - C_α). This equation relates the phase boundary compositions C_α and C_β for the given alloy."
|
||
},
|
||
{
|
||
"question": "For the alloy composition 30 wt% A-70 wt% B with fraction α phase 0.14 and fraction β phase 0.86, determine the composition of the phase boundary (or solubility limit) for both α and β phases at this temperature.",
|
||
"answer": "Using the lever rule expression for this alloy composition: W_α2 = 0.14 = (C_β - C_02) / (C_β - C_α) = (C_β - 30) / (C_β - C_α). This equation relates the phase boundary compositions C_α and C_β for the given alloy."
|
||
},
|
||
{
|
||
"question": "Solve the system of equations derived from the lever rule expressions for both alloy compositions to determine the phase boundary compositions C_α and C_β.",
|
||
"answer": "Solving the system of equations: 0.57 = (C_β - 60) / (C_β - C_α) and 0.14 = (C_β - 30) / (C_β - C_α), yields the phase boundary compositions: C_α = 90 wt% A-10 wt% B and C_β = 20.2 wt% A-79.8 wt% B."
|
||
}
|
||
]
|
||
```
|
||
处理第 446/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Given an A-B alloy with composition 55 wt% B-45 wt% A at a certain temperature, mass fractions of 0.5 for both α and β phases, and the composition of the β phase as 90 wt% B-10 wt% A, what is the composition of the α phase?",
|
||
"answer": "Using the lever rule for Wα: Wα = 0.5 = (Cβ - C0) / (Cβ - Cα) = (90 - 55) / (90 - Cα). Solving for Cα: Cα = 20 (or 20 wt% B-80 wt% A)."
|
||
}
|
||
]
|
||
```
|
||
处理第 447/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For a 11.20 kg magnesium-lead alloy of composition 30 wt% Pb-70 wt% Mg, is it possible, at equilibrium, to have alpha and Mg2Pb phases with masses of 7.39 kg and 3.81 kg respectively?",
|
||
"answer": "Yes, it is possible to have a 30 wt% Pb-70 wt% Mg alloy with masses of 7.39 kg and 3.81 kg for the alpha and Mg2Pb phases, respectively. The mass fractions are calculated as follows: W_alpha = m_alpha / (m_alpha + m_Mg2Pb) = 7.39 kg / (7.39 kg + 3.81 kg) = 0.66, W_Mg2Pb = 1.00 - 0.66 = 0.34."
|
||
},
|
||
{
|
||
"question": "What is the composition of the alpha phase (C_alpha) in the magnesium-lead alloy, given the mass fractions and overall composition?",
|
||
"answer": "The composition of the alpha phase (C_alpha) is determined using the lever rule: W_alpha = (C_Mg2Pb - C_0) / (C_Mg2Pb - C_alpha). Given C_Mg2Pb = 81 wt% Pb and C_0 = 30 wt% Pb, solving for C_alpha yields C_alpha = 3.7 wt% Pb."
|
||
},
|
||
{
|
||
"question": "What is the approximate temperature of the alloy given the composition of the alpha phase (C_alpha = 3.7 wt% Pb)?",
|
||
"answer": "The approximate temperature of the alloy, corresponding to the alpha phase composition of 3.7 wt% Pb along the alpha-(alpha + Mg2Pb) phase boundary, is approximately 190 degrees Celsius."
|
||
}
|
||
]
|
||
```
|
||
处理第 448/821 条数据...
|
||
[
|
||
{
|
||
"question": "(a) Briefly describe the phenomenon of coring and why it occurs.",
|
||
"answer": "Coring is the phenomenon whereby concentration gradients exist across grains in polycrystalline alloys, with higher concentrations of the component having the lower melting temperature at the grain boundaries. It occurs, during solidification, as a consequence of cooling rates that are too rapid to allow for the maintenance of the equilibrium composition of the solid phase."
|
||
},
|
||
{
|
||
"question": "(b) Cite one undesirable consequence of coring.",
|
||
"answer": "One undesirable consequence of a cored structure is that, upon heating, the grain boundary regions will melt first and at a temperature below the equilibrium phase boundary from the phase diagram; this melting results in a loss in mechanical integrity of the alloy."
|
||
}
|
||
]
|
||
处理第 449/821 条数据...
|
||
It's a single issue.
|
||
处理第 450/821 条数据...
|
||
It's a single issue.
|
||
处理第 451/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Given the intermetallic compound AB with composition 34.3 wt% A-65.7 wt% B, and knowing that element A is potassium (atomic weight 39.10 g/mol), determine the atomic weight of element B.",
|
||
"answer": "First, consider the AB intermetallic compound, which has equal numbers of A and B atoms, resulting in 50 at% A-50 at% B. Using the equation for atomic percent composition: CB = (CB * AA) / (CA * AB + CB * AA) * 100, where AA and AB are the atomic weights of A and B, and CA and CB are their weight percentages. Substituting the given values: 50 at% B = (65.7 wt% B * 39.10 g/mol) / (34.3 wt% A * AB + 65.7 wt% B * 39.10 g/mol) * 100. Solving this expression yields AB = 1.916 * 39.10 g/mol = 74.92 g/mol."
|
||
},
|
||
{
|
||
"question": "Identify element B from its atomic weight of 74.92 g/mol.",
|
||
"answer": "Upon consulting the periodic table, the element with an atomic weight closest to 74.92 g/mol is arsenic (As). Therefore, element B is arsenic, and the intermetallic compounds are KAs and KAs2."
|
||
}
|
||
]
|
||
```
|
||
处理第 452/821 条数据...
|
||
It's a single issue.
|
||
处理第 453/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Compute the mass fraction of α ferrite in pearlite, given C_Fe3C = 6.70 wt% C, C_0 = 0.76 wt% C, and C_α = 0.022 wt% C.",
|
||
"answer": "The lever-rule expression for ferrite is W_α = (C_Fe3C - C_0) / (C_Fe3C - C_α). Substituting the given values: W_α = (6.70 - 0.76) / (6.70 - 0.022) = 0.89."
|
||
},
|
||
{
|
||
"question": "Compute the mass fraction of cementite in pearlite, given C_Fe3C = 6.70 wt% C, C_0 = 0.76 wt% C, and C_α = 0.022 wt% C.",
|
||
"answer": "The lever-rule expression for cementite is W_Fe3C = (C_0 - C_α) / (C_Fe3C - C_α). Substituting the given values: W_Fe3C = (0.76 - 0.022) / (6.70 - 0.022) = 0.11."
|
||
}
|
||
]
|
||
```
|
||
处理第 454/821 条数据...
|
||
[
|
||
{
|
||
"question": "What is the distinction between hypoeutectoid and hypereutectoid steels?",
|
||
"answer": "A hypoeutectoid steel has a carbon concentration less than the eutectoid; on the other hand, a hypereutectoid steel has a carbon content greater than the eutectoid."
|
||
},
|
||
{
|
||
"question": "In a hypoeutectoid steel, both eutectoid and proeutectoid ferrite exist. Explain the difference between them. What will be the carbon concentration in each?",
|
||
"answer": "For a hypoeutectoid steel, the proeutectoid ferrite is a microconstituent that formed above the eutectoid temperature. The eutectoid ferrite is one of the constituents of pearlite that formed at a temperature below the eutectoid. The carbon concentration for both ferrites is 0.022 wt% C."
|
||
}
|
||
]
|
||
处理第 455/821 条数据...
|
||
It's a single issue.
|
||
处理第 456/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is the proeutectoid phase for an iron-carbon alloy in which the mass fractions of total ferrite and total cementite are 0.92 and 0.08, respectively?",
|
||
"answer": "In this problem we are given values of W_α and W_Fe3C (0.92 and 0.08, respectively) for an iron-carbon alloy and then are asked to specify the proeutectoid phase. Employment of the lever rule for total α leads to W_α=0.92=(C_Fe3C-C_0)/(C_Fe3C-C_α)=(6.70-C_0)/(6.70-0.022). Now, solving for C_0, the alloy composition, leads to C_0=0.56 wt% C. Therefore, the proeutectoid phase is α ferrite since C_0 is less than 0.76 wt% C."
|
||
},
|
||
{
|
||
"question": "Why is the proeutectoid phase α ferrite for an iron-carbon alloy with mass fractions of total ferrite and total cementite of 0.92 and 0.08, respectively?",
|
||
"answer": "The proeutectoid phase is α ferrite because the calculated alloy composition C_0=0.56 wt% C is less than the eutectoid composition of 0.76 wt% C, which is the threshold below which the proeutectoid phase is α ferrite."
|
||
}
|
||
]
|
||
```
|
||
处理第 457/821 条数据...
|
||
[
|
||
{
|
||
"question": "Consider 1.0 kg of austenite containing 1.15 wt% C, cooled to below 727 C (1341 F). What is the proeutectoid phase?",
|
||
"answer": "The proeutectoid phase will be Fe3C since 1.15 wt% C is greater than the eutectoid composition (0.76 wt% C)."
|
||
},
|
||
{
|
||
"question": "Consider 1.0 kg of austenite containing 1.15 wt% C, cooled to below 727 C (1341 F). How many kilograms each of total ferrite and cementite form?",
|
||
"answer": "For total ferrite: Wα = (CFe3C - C0) / (CFe3C - Cα) = (6.70 - 1.15) / (6.70 - 0.022) = 0.83, which gives 0.83 kg of total ferrite. For total cementite: WFe3C = (C0 - Cα) / (CFe3C - Cα) = (1.15 - 0.022) / (6.70 - 0.022) = 0.17, which gives 0.17 kg of total cementite."
|
||
},
|
||
{
|
||
"question": "Consider 1.0 kg of austenite containing 1.15 wt% C, cooled to below 727 C (1341 F). How many kilograms each of pearlite and the proeutectoid phase form?",
|
||
"answer": "For pearlite: Wp = (6.70 - C1') / (6.70 - 0.76) = (6.70 - 1.15) / (6.70 - 0.76) = 0.93, which gives 0.93 kg of pearlite. For the proeutectoid phase (cementite): WFe3C = (C1' - 0.76) / 5.94 = (1.15 - 0.76) / 5.94 = 0.07, which gives 0.07 kg of the proeutectoid phase."
|
||
}
|
||
]
|
||
处理第 458/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Consider 2.5 kg of austenite containing 0.65 wt% C, cooled to below 727°C (1341°F). What is the proeutectoid phase?",
|
||
"answer": "Ferrite is the proeutectoid phase since 0.65 wt% C is less than 0.76 wt% C."
|
||
},
|
||
{
|
||
"question": "Consider 2.5 kg of austenite containing 0.65 wt% C, cooled to below 727°C (1341°F). How many kilograms each of total ferrite and cementite form?",
|
||
"answer": "For ferrite, application of the appropriate lever rule expression yields W_α = (6.70 - 0.65)/(6.70 - 0.022) = 0.91, which corresponds to (0.91)(2.5 kg) = 2.27 kg of total ferrite. Similarly, for total cementite, W_Fe3C = (0.65 - 0.022)/(6.70 - 0.022) = 0.09, or (0.09)(2.5 kg) = 0.23 kg of total cementite form."
|
||
},
|
||
{
|
||
"question": "Consider 2.5 kg of austenite containing 0.65 wt% C, cooled to below 727°C (1341°F). How many kilograms each of pearlite and the proeutectoid phase form?",
|
||
"answer": "Using the equation W_p = (0.65 - 0.022)/0.74 = 0.85, this corresponds to (0.85)(2.5 kg) = 2.12 kg of pearlite. Also, from the equation W_α' = (0.76 - 0.65)/0.74 = 0.15, or there are (0.15)(2.5 kg) = 0.38 kg of proeutectoid ferrite."
|
||
}
|
||
]
|
||
```
|
||
处理第 459/821 条数据...
|
||
It's a single issue.
|
||
处理第 460/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Given the mass fraction of pearlite (Wp) is 0.714 in an iron-carbon alloy, determine the carbon concentration (C0) in the alloy using the equation Wp = (C0 - 0.022) / 0.74.",
|
||
"answer": "Using the equation Wp = (C0 - 0.022) / 0.74, where Wp = 0.714, we solve for C0: 0.714 = (C0 - 0.022) / 0.74. Rearranging gives C0 - 0.022 = 0.714 * 0.74. Calculating the right side: 0.714 * 0.74 = 0.52836. Thus, C0 = 0.52836 + 0.022 = 0.55036 wt% C. The carbon concentration in the alloy is approximately 0.55 wt% C."
|
||
}
|
||
]
|
||
```
|
||
处理第 461/821 条数据...
|
||
It's a single issue.
|
||
处理第 462/821 条数据...
|
||
It's a single issue.
|
||
处理第 463/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Consider 2.0 kg of a 99.6 wt % Fe-0.4 wt % C alloy that is cooled to a temperature just below the eutectoid. How many kilograms of proeutectoid ferrite form?",
|
||
"answer": "To compute the amount of proeutectoid ferrite that forms, the following equation is used: W_alpha' = (0.76 - C0') / 0.74 = (0.76 - 0.40) / 0.74 = 0.49. Or, (0.49)(2.0 kg) = 0.98 kg of proeutectoid ferrite forms."
|
||
},
|
||
{
|
||
"question": "Consider 2.0 kg of a 99.6 wt % Fe-0.4 wt % C alloy that is cooled to a temperature just below the eutectoid. How many kilograms of eutectoid ferrite form?",
|
||
"answer": "To determine the amount of eutectoid ferrite, first compute the amount of total ferrite using the lever rule applied across the alpha + Fe3C phase field: W_alpha = (C_Fe3C - C0') / (C_Fe3C - C_alpha) = (6.70 - 0.40) / (6.70 - 0.022) = 0.94, which corresponds to (0.94)(2.0 kg) = 1.88 kg. The amount of eutectoid ferrite is the difference between total and proeutectoid ferrites: 1.88 kg - 0.98 kg = 0.90 kg."
|
||
},
|
||
{
|
||
"question": "Consider 2.0 kg of a 99.6 wt % Fe-0.4 wt % C alloy that is cooled to a temperature just below the eutectoid. How many kilograms of cementite form?",
|
||
"answer": "To determine the amount of cementite that forms, apply the lever rule across the alpha + Fe3C phase field: W_Fe3C = (C0' - C_alpha) / (C_Fe3C - C_alpha) = (0.40 - 0.022) / (6.70 - 0.022) = 0.057, which amounts to (0.057)(2.0 kg) = 0.114 kg cementite in the alloy."
|
||
}
|
||
]
|
||
```
|
||
处理第 464/821 条数据...
|
||
It's a single issue.
|
||
处理第 465/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Is it possible to have an iron-carbon alloy for which the mass fraction of total ferrite (W_α) is 0.846? Why or why not?",
|
||
"answer": "To determine if an iron-carbon alloy with W_α=0.846 is possible, we use the lever rule expression for the mass fraction of total ferrite: W_α=(C_Fe3C - C_0)/(C_Fe3C - C_α)=(6.70 - C_0)/(6.70 - 0.022)=0.846. Solving for C_0 yields C_0=1.05 wt% C. This composition is within the possible range for iron-carbon alloys, so such an alloy is possible."
|
||
},
|
||
{
|
||
"question": "Is it possible to have an iron-carbon alloy for which the mass fraction of proeutectoid cementite (W_Fe3C') is 0.049? Why or why not?",
|
||
"answer": "To determine if an iron-carbon alloy with W_Fe3C'=0.049 is possible, we use the lever rule expression for the mass fraction of proeutectoid cementite: W_Fe3C'=(C_1' - 0.76)/5.94=0.049. Solving for C_1' yields C_1'=1.05 wt% C. This composition is within the possible range for iron-carbon alloys, so such an alloy is possible."
|
||
},
|
||
{
|
||
"question": "Given that both W_α=0.846 and W_Fe3C'=0.049 yield an alloy composition of 1.05 wt% C, is such an iron-carbon alloy possible? Why or why not?",
|
||
"answer": "Since both the mass fraction of total ferrite (W_α=0.846) and the mass fraction of proeutectoid cementite (W_Fe3C'=0.049) yield the same alloy composition (C_0=C_1'=1.05 wt% C), and this composition is within the possible range for iron-carbon alloys, such an alloy is possible."
|
||
}
|
||
]
|
||
```
|
||
处理第 466/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Is it possible to have an iron-carbon alloy with a mass fraction of total cementite (W_Fe3C) of 0.039? Why or why not?",
|
||
"answer": "To determine if an iron-carbon alloy with W_Fe3C = 0.039 is possible, we use the lever rule expression: W_Fe3C = (C_0 - C_a)/(C_Fe3C - C_a) = (C_0 - 0.022)/(6.70 - 0.022) = 0.039. Solving for C_0 yields C_0 = 0.28 wt% C. This value is valid as it falls within the possible composition range for iron-carbon alloys."
|
||
},
|
||
{
|
||
"question": "Is it possible to have an iron-carbon alloy with a mass fraction of pearlite (W_p) of 0.417? Why or why not?",
|
||
"answer": "To determine if an iron-carbon alloy with W_p = 0.417 is possible, we use the lever rule expression for hypoeutectoid alloys: W_p = (C_0' - 0.022)/0.74 = 0.417. Solving for C_0' yields C_0' = 0.33 wt% C. This value is valid as it falls within the possible composition range for iron-carbon alloys."
|
||
},
|
||
{
|
||
"question": "Can an iron-carbon alloy simultaneously have mass fractions of total cementite (W_Fe3C) = 0.039 and pearlite (W_p) = 0.417? Why or why not?",
|
||
"answer": "For an alloy to simultaneously have W_Fe3C = 0.039 and W_p = 0.417, the calculated compositions C_0 and C_0' must be equal. However, C_0 = 0.28 wt% C and C_0' = 0.33 wt% C are different. Therefore, such an alloy is not possible."
|
||
}
|
||
]
|
||
```
|
||
处理第 467/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Compute the mass fraction of total ferrite (W_α) in an iron-carbon alloy that contains 0.43 wt% C using the lever rule and a tie line that extends across the entire α + Fe3C phase field.",
|
||
"answer": "The mass fraction of total ferrite is calculated using the lever rule as W_α = (C_Fe3C - C_0) / (C_Fe3C - C_α) = (6.70 - 0.43) / (6.70 - 0.022) = 0.939."
|
||
},
|
||
{
|
||
"question": "Compute the mass fraction of proeutectoid ferrite (W_α') in an iron-carbon alloy that contains 0.43 wt% C.",
|
||
"answer": "The mass fraction of proeutectoid ferrite is calculated as W_α' = (0.76 - C_0) / 0.74 = (0.76 - 0.43) / 0.74 = 0.446."
|
||
},
|
||
{
|
||
"question": "Compute the mass fraction of eutectoid ferrite (W_α'') in an iron-carbon alloy that contains 0.43 wt% C by subtracting the mass fraction of proeutectoid ferrite from the mass fraction of total ferrite.",
|
||
"answer": "The mass fraction of eutectoid ferrite is calculated as W_α'' = W_α - W_α' = 0.939 - 0.446 = 0.493."
|
||
}
|
||
]
|
||
```
|
||
处理第 468/821 条数据...
|
||
[
|
||
{
|
||
"question": "Is it possible to determine the composition of an iron-carbon alloy if the mass fraction of eutectoid cementite is 0.104?",
|
||
"answer": "Yes, it is possible to determine the alloy composition; and, in fact, there are two possible answers."
|
||
},
|
||
{
|
||
"question": "What is the first possible composition of the alloy where the eutectoid cementite exists in addition to proeutectoid cementite?",
|
||
"answer": "For the first case, the mass fraction of eutectoid cementite is the difference between total cementite and proeutectoid cementite mass fractions. The expression is W_Fe3C* = (C0 - C_α)/(C_Fe3C - C_α) - (C0 - 0.76)/5.94 = 0.104. Solving for C0 yields C0 = 1.11 wt% C."
|
||
},
|
||
{
|
||
"question": "What is the second possible composition of the alloy where all of the cementite is eutectoid cementite in a hypoeutectoid alloy?",
|
||
"answer": "For the second possibility, the mass fraction of total cementite is 0.104. The lever rule expression is W_Fe3C = (C0 - C_α)/(C_Fe3C - C_α) = (C0 - 0.022)/(6.70 - 0.022). Solving for C0 in this case would complete the calculation, but the exact value is not provided in the given answer."
|
||
}
|
||
]
|
||
处理第 469/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Is it possible to determine the composition of an iron-carbon alloy for which the mass fraction of eutectoid ferrite is 0.82?",
|
||
"answer": "Yes, it is possible to determine the alloy composition; and, in fact, there are two possible answers."
|
||
},
|
||
{
|
||
"question": "What is the composition of the alloy if the eutectoid ferrite exists in addition to proeutectoid ferrite?",
|
||
"answer": "For this case, the mass fraction of eutectoid ferrite (W_α') is the difference between total ferrite and proeutectoid ferrite mass fractions. The expression is W_α' = W_α - W_α'. Solving for C_0 yields C_0 = 0.70 wt% C."
|
||
},
|
||
{
|
||
"question": "What is the composition of the alloy if it is a hypereutectoid alloy wherein all of the ferrite is eutectoid ferrite?",
|
||
"answer": "For this case, the mass fraction of total ferrite is 0.82. The lever rule expression is W_α = (C_Fe3C - C_0) / (C_Fe3C - C_α) = (6.70 - C_0) / (6.70 - 0.022) = 0.82. Solving for C_0 yields C_0 = 1.22 wt% C."
|
||
}
|
||
]
|
||
```
|
||
处理第 470/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is the first stage involved in the formation of particles of a new phase? Briefly describe it.",
|
||
"answer": "The first stage is nucleation. The nucleation process involves the formation of normally very small particles of the new phase(s) which are stable and capable of continued growth."
|
||
},
|
||
{
|
||
"question": "What is the second stage involved in the formation of particles of a new phase? Briefly describe it.",
|
||
"answer": "The second stage is growth. The growth stage is simply the increase in size of the new phase particles."
|
||
}
|
||
]
|
||
```
|
||
处理第 471/821 条数据...
|
||
It's a single issue.
|
||
处理第 472/821 条数据...
|
||
It's a single issue.
|
||
处理第 473/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Given the fraction recrystallized-time data for the recrystallization at 600°C of a previously deformed steel, determine the values of n and k in the Avrami relationship using the data points (0.20, 13.1 min) and (0.70, 29.1 min).",
|
||
"answer": "To determine the values of n and k, we use the Avrami relationship in the form ln[ln(1/(1-y))] = ln k + n ln t. For the first data point (0.20, 13.1 min): ln[ln(1/(1-0.20))] = ln k + n ln(13.1). For the second data point (0.70, 29.1 min): ln[ln(1/(1-0.70))] = ln k + n ln(29.1). Solving these two equations simultaneously yields n = 2.112 and k = 9.75 × 10^-4."
|
||
},
|
||
{
|
||
"question": "Using the Avrami relationship with n = 2.112 and k = 9.75 × 10^-4, determine the fraction recrystallized after a total time of 22.8 min at 600°C.",
|
||
"answer": "Using the Avrami relationship y = 1 - exp(-k t^n), we substitute n = 2.112, k = 9.75 × 10^-4, and t = 22.8 min: y = 1 - exp[-(9.75 × 10^-4)(22.8)^2.112] = 0.51."
|
||
}
|
||
]
|
||
```
|
||
处理第 474/821 条数据...
|
||
[
|
||
{
|
||
"question": "In terms of heat treatment and the development of microstructure, what is one major limitation of the iron-iron carbide phase diagram related to nonequilibrium phases?",
|
||
"answer": "The nonequilibrium martensite does not appear on the iron-iron carbide phase diagram."
|
||
},
|
||
{
|
||
"question": "In terms of heat treatment and the development of microstructure, what is another major limitation of the iron-iron carbide phase diagram related to time-temperature relationships?",
|
||
"answer": "The iron-iron carbide phase diagram provides no indication as to the time-temperature relationships for the formation of pearlite, bainite, and spheroidite, all of which are composed of the equilibrium ferrite and cementite phases."
|
||
}
|
||
]
|
||
处理第 475/821 条数据...
|
||
[
|
||
{
|
||
"question": "Briefly describe the phenomena of superheating and supercooling.",
|
||
"answer": "Superheating and supercooling correspond, respectively, to heating or cooling above or below a phase transition temperature without the occurrence of the transformation."
|
||
},
|
||
{
|
||
"question": "Why do superheating and supercooling phenomena occur?",
|
||
"answer": "These phenomena occur because right at the phase transition temperature, the driving force is not sufficient to cause the transformation to occur. The driving force is enhanced during superheating or supercooling."
|
||
}
|
||
]
|
||
处理第 476/821 条数据...
|
||
[
|
||
{
|
||
"question": "Briefly cite the differences between pearlite, bainite, and spheroidite relative to microstructure.",
|
||
"answer": "The microstructures of pearlite, bainite, and spheroidite all consist of α-ferrite and cementite phases. For pearlite, the two phases exist as layers which alternate with one another. Bainite consists of very fine and parallel needle-shaped particles of cementite that are surrounded by an α-ferrite matrix. For spheroidite, the matrix is ferrite, and the cementite phase is in the shape of sphere-shaped particles."
|
||
},
|
||
{
|
||
"question": "Briefly cite the differences between pearlite, bainite, and spheroidite relative to mechanical properties.",
|
||
"answer": "Bainite is harder and stronger than pearlite, which, in turn, is harder and stronger than spheroidite."
|
||
}
|
||
]
|
||
处理第 477/821 条数据...
|
||
It's a single issue.
|
||
处理第 478/821 条数据...
|
||
[
|
||
{
|
||
"question": "Name the microstructural products of eutectoid iron-carbon alloy (0.76 wt% C) specimens that are first completely transformed to austenite, then cooled to room temperature at a rate of 200°C/s.",
|
||
"answer": "At a rate of 200°C/s, only martensite forms."
|
||
},
|
||
{
|
||
"question": "Name the microstructural products of eutectoid iron-carbon alloy (0.76 wt% C) specimens that are first completely transformed to austenite, then cooled to room temperature at a rate of 100°C/s.",
|
||
"answer": "At a rate of 100°C/s, both martensite and pearlite form."
|
||
},
|
||
{
|
||
"question": "Name the microstructural products of eutectoid iron-carbon alloy (0.76 wt% C) specimens that are first completely transformed to austenite, then cooled to room temperature at a rate of 20°C/s.",
|
||
"answer": "At a rate of 20°C/s, only fine pearlite forms."
|
||
}
|
||
]
|
||
处理第 479/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is one important difference between continuous cooling transformation diagrams for plain carbon and alloy steels regarding the presence of a bainite nose?",
|
||
"answer": "For an alloy steel, a bainite nose will be present, which nose will be absent for plain carbon alloys."
|
||
},
|
||
{
|
||
"question": "What is one important difference between continuous cooling transformation diagrams for plain carbon and alloy steels regarding the position of the pearlite-proeutectoid noses?",
|
||
"answer": "The pearlite-proeutectoid noses for plain carbon steel alloys are positioned at shorter times than for the alloy steels."
|
||
}
|
||
]
|
||
```
|
||
处理第 480/821 条数据...
|
||
It's a single issue.
|
||
处理第 481/821 条数据...
|
||
[
|
||
{
|
||
"question": "Name the microstructural products of 4340 alloy steel specimens that are first completely transformed to austenite, then cooled to room temperature at a rate of 10 degrees Celsius per second.",
|
||
"answer": "At a cooling rate of 10 degrees Celsius per second, only martensite forms."
|
||
},
|
||
{
|
||
"question": "Name the microstructural products of 4340 alloy steel specimens that are first completely transformed to austenite, then cooled to room temperature at a rate of 1 degree Celsius per second.",
|
||
"answer": "At a cooling rate of 1 degree Celsius per second, both martensite and bainite form."
|
||
},
|
||
{
|
||
"question": "Name the microstructural products of 4340 alloy steel specimens that are first completely transformed to austenite, then cooled to room temperature at a rate of 0.1 degrees Celsius per second.",
|
||
"answer": "At a cooling rate of 0.1 degrees Celsius per second, martensite, proeutectoid ferrite, and bainite form."
|
||
},
|
||
{
|
||
"question": "Name the microstructural products of 4340 alloy steel specimens that are first completely transformed to austenite, then cooled to room temperature at a rate of 0.01 degrees Celsius per second.",
|
||
"answer": "At a cooling rate of 0.01 degrees Celsius per second, martensite, proeutectoid ferrite, pearlite, and bainite form."
|
||
}
|
||
]
|
||
处理第 482/821 条数据...
|
||
It's a single issue.
|
||
处理第 483/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Why is fine pearlite harder and stronger than coarse pearlite in iron-carbon alloys with α-ferrite and cementite phases?",
|
||
"answer": "Fine pearlite is harder and stronger than coarse pearlite because the alternating ferrite-cementite layers are thinner in fine pearlite, resulting in more phase boundary area. The greater phase boundary area increases hardness and strength by impeding dislocation motion and restricting deformation of the ferrite phase near the boundaries."
|
||
},
|
||
{
|
||
"question": "Why is coarse pearlite harder and stronger than spheroidite in iron-carbon alloys with α-ferrite and cementite phases?",
|
||
"answer": "Coarse pearlite is harder and stronger than spheroidite because the phase boundary area between the sphere-like cementite particles and the ferrite matrix in spheroidite is less than the alternating layered microstructure in coarse pearlite. The reduced boundary area in spheroidite decreases its ability to impede dislocations and restrict deformation."
|
||
}
|
||
]
|
||
```
|
||
处理第 484/821 条数据...
|
||
It's a single issue.
|
||
处理第 485/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Rank the following iron-carbon alloys and associated microstructures from the highest to the lowest tensile strength: (a) 0.25 wt% C with spheroidite, (b) 0.25 wt% C with coarse pearlite, (c) 0.60 wt% C with fine pearlite, and (d) 0.60 wt% C with coarse pearlite. Justify this ranking.",
|
||
"answer": "The ranking from highest to lowest tensile strength is as follows: 0.60 wt% C, fine pearlite; 0.60 wt% C, coarse pearlite; 0.25 wt% C, coarse pearlite; 0.25 wt% C, spheroidite. The 0.25 wt% C, coarse pearlite is stronger than the 0.25 wt% C, spheroidite since coarse pearlite is stronger than spheroidite; the composition of the alloys is the same. The 0.60 wt% C, coarse pearlite is stronger than the 0.25 wt% C, coarse pearlite, since increasing the carbon content increases the strength. Finally, the 0.60 wt% C, fine pearlite is stronger than the 0.60 wt% C, coarse pearlite inasmuch as the strength of fine pearlite is greater than coarse pearlite because of the many more ferrite-cementite phase boundaries in fine pearlite."
|
||
}
|
||
]
|
||
```
|
||
处理第 486/821 条数据...
|
||
[
|
||
{
|
||
"question": "Why does the hardness of tempered martensite diminish with tempering time at constant temperature?",
|
||
"answer": "The hardness of tempered martensite diminishes with tempering time at constant temperature because the microstructure consists of small sphere-like particles of cementite embedded within a ferrite matrix. As tempering time increases, the cementite particles grow, which reduces the ferrite-cementite phase boundary area. Since these phase boundaries act as barriers to dislocation motion, the reduction in phase boundary area leads to a decrease in hardness."
|
||
},
|
||
{
|
||
"question": "Why does the hardness of tempered martensite diminish with increasing temperature at constant tempering time?",
|
||
"answer": "The hardness of tempered martensite diminishes with increasing temperature at constant tempering time because higher temperatures accelerate the rate of cementite particle growth. This growth reduces the ferrite-cementite phase boundary area, which acts as barriers to dislocation motion. Consequently, the decrease in phase boundary area results in a reduction of hardness."
|
||
}
|
||
]
|
||
处理第 487/821 条数据...
|
||
[
|
||
{
|
||
"question": "Briefly describe the microstructural difference between spheroidite and tempered martensite.",
|
||
"answer": "Both tempered martensite and spheroidite have sphere-like cementite particles within a ferrite matrix; however, these particles are much larger for spheroidite."
|
||
},
|
||
{
|
||
"question": "Explain why tempered martensite is much harder and stronger.",
|
||
"answer": "Tempered martensite is harder and stronger inasmuch as there is much more ferrite-cementite phase boundary area for the smaller particles; thus, there is greater reinforcement of the ferrite phase, and more phase boundary barriers to dislocation motion."
|
||
}
|
||
]
|
||
处理第 488/821 条数据...
|
||
[
|
||
{
|
||
"question": "List the four classifications of steels.",
|
||
"answer": "Low Carbon Steels, Medium Carbon Steels, High Carbon Steels, High Alloy Steels (Stainless and Tool)."
|
||
},
|
||
{
|
||
"question": "For Low Carbon Steels, briefly describe the properties and typical applications.",
|
||
"answer": "Properties: nonresponsive to heat treatments; relatively soft and weak; machinable and weldable. Typical applications: automobile bodies, structural shapes, pipelines, buildings, bridges, and tin cans."
|
||
},
|
||
{
|
||
"question": "For Medium Carbon Steels, briefly describe the properties and typical applications.",
|
||
"answer": "Properties: heat treatable, relatively large combinations of mechanical characteristics. Typical applications: railway wheels and tracks, gears, crankshafts, and machine parts."
|
||
},
|
||
{
|
||
"question": "For High Carbon Steels, briefly describe the properties and typical applications.",
|
||
"answer": "Properties: hard, strong, and relatively brittle. Typical applications: chisels, hammers, knives, and hacksaw blades."
|
||
},
|
||
{
|
||
"question": "For High Alloy Steels (Stainless and Tool), briefly describe the properties and typical applications.",
|
||
"answer": "Properties: hard and wear resistant; resistant to corrosion in a large variety of environments. Typical applications: cutting tools, drills, cutlery, food processing, and surgical tools."
|
||
}
|
||
]
|
||
处理第 489/821 条数据...
|
||
[
|
||
{
|
||
"question": "Cite three reasons why ferrous alloys are used so extensively.",
|
||
"answer": "Ferrous alloys are used extensively because: (1) Iron ores exist in abundant quantities. (2) Economical extraction, refining, and fabrication techniques are available. (3) The alloys may be tailored to have a wide range of properties."
|
||
},
|
||
{
|
||
"question": "Cite three characteristics of ferrous alloys that limit their utilization.",
|
||
"answer": "Disadvantages of ferrous alloys are: (1) They are susceptible to corrosion. (2) They have a relatively high density. (3) They have relatively low electrical conductivities."
|
||
}
|
||
]
|
||
处理第 490/821 条数据...
|
||
It's a single issue.
|
||
处理第 491/821 条数据...
|
||
It's a single issue.
|
||
处理第 492/821 条数据...
|
||
[
|
||
{
|
||
"question": "Compare white and nodular cast irons with respect to composition and heat treatment.",
|
||
"answer": "White iron--2.5 to 4.0 wt% C and less than 1.0 wt% Si. No heat treatment; however, cooling is rapid during solidification. Nodular cast iron--2.5 to 4.0 wt% C, 1.0 to 3.0 wt% Si, and a small amount of Mg or Ce. A heat treatment at about 700°C may be necessary to produce a ferritic matrix."
|
||
},
|
||
{
|
||
"question": "Compare white and nodular cast irons with respect to microstructure.",
|
||
"answer": "White iron--There are regions of cementite interspersed within pearlite. Nodular cast iron--Nodules of graphite are embedded in a ferrite or pearlite matrix."
|
||
},
|
||
{
|
||
"question": "Compare white and nodular cast irons with respect to mechanical characteristics.",
|
||
"answer": "White iron--Extremely hard and brittle. Nodular cast iron--Moderate strength and ductility."
|
||
}
|
||
]
|
||
处理第 493/821 条数据...
|
||
It's a single issue.
|
||
处理第 494/821 条数据...
|
||
It's a single issue.
|
||
处理第 495/821 条数据...
|
||
It's a single issue.
|
||
处理第 496/821 条数据...
|
||
It's a single issue.
|
||
处理第 497/821 条数据...
|
||
[
|
||
{
|
||
"question": "Cite advantages of cold working",
|
||
"answer": "The advantages of cold working are: (1) A high quality surface finish. (2) The mechanical properties may be varied. (3) Close dimensional tolerances."
|
||
},
|
||
{
|
||
"question": "Cite disadvantages of cold working",
|
||
"answer": "The disadvantages of cold working are: (1) High deformation energy requirements. (2) Large deformations must be accomplished in steps, which may be expensive. (3) A loss of ductility."
|
||
},
|
||
{
|
||
"question": "Cite advantages of hot working",
|
||
"answer": "The advantages of hot working are: (1) Large deformations are possible, which may be repeated. (2) Deformation energy requirements are relatively low."
|
||
},
|
||
{
|
||
"question": "Cite disadvantages of hot working",
|
||
"answer": "The disadvantages of hot working are: (1) A poor surface finish. (2) A variety of mechanical properties is not possible."
|
||
}
|
||
]
|
||
处理第 498/821 条数据...
|
||
[
|
||
{
|
||
"question": "Cite advantages of forming metals by extrusion as opposed to rolling.",
|
||
"answer": "The advantages of extrusion as opposed to rolling are as follows: (1) Pieces having more complicated cross-sectional geometries may be formed. (2) Seamless tubing may be produced."
|
||
},
|
||
{
|
||
"question": "Cite some disadvantages of forming metals by extrusion as opposed to rolling.",
|
||
"answer": "The disadvantages of extrusion over rolling are as follows: (1) Nonuniform deformation over the cross-section. (2) A variation in properties may result over a cross-section of an extruded piece."
|
||
}
|
||
]
|
||
处理第 499/821 条数据...
|
||
It's a single issue.
|
||
处理第 500/821 条数据...
|
||
[
|
||
{
|
||
"question": "What are the characteristics of sand casting technique?",
|
||
"answer": "For sand casting, sand is the mold material, a two-piece mold is used, ordinarily the surface finish is not an important consideration, the sand may be reused (but the mold may not), casting rates are low, and large pieces are usually cast."
|
||
},
|
||
{
|
||
"question": "What are the characteristics of die casting technique?",
|
||
"answer": "For die casting, a permanent mold is used, casting rates are high, the molten metal is forced into the mold under pressure, a two-piece mold is used, and small pieces are normally cast."
|
||
},
|
||
{
|
||
"question": "What are the characteristics of investment casting technique?",
|
||
"answer": "For investment casting, a single-piece mold is used, which is not reusable; it results in high dimensional accuracy, good reproduction of detail, and a fine surface finish; and casting rates are low."
|
||
},
|
||
{
|
||
"question": "What are the characteristics of lost foam casting technique?",
|
||
"answer": "For lost foam casting, the pattern is polystyrene foam, whereas the mold material is sand. Complex geometries and tight tolerances are possible. Casting rates are higher than for investment, and there are few environmental wastes."
|
||
},
|
||
{
|
||
"question": "What are the characteristics of continuous casting technique?",
|
||
"answer": "For continuous casting, at the conclusion of the extraction process, the molten metal is cast into a continuous strand having either a rectangular or circular cross-section; these shapes are desirable for subsequent secondary metal-forming operations. The chemical composition and mechanical properties are relatively uniform throughout the cross-section."
|
||
}
|
||
]
|
||
处理第 501/821 条数据...
|
||
It's a single issue.
|
||
处理第 502/821 条数据...
|
||
[
|
||
{
|
||
"question": "Cite three sources of internal residual stresses in metal components.",
|
||
"answer": "Three sources of residual stresses in metal components are plastic deformation processes, nonuniform cooling of a piece that was cooled from an elevated temperature, and a phase transformation in which parent and product phases have different densities."
|
||
},
|
||
{
|
||
"question": "What are two possible adverse consequences of internal residual stresses in metal components?",
|
||
"answer": "Two adverse consequences of these stresses are distortion (or warpage) and fracture."
|
||
}
|
||
]
|
||
处理第 503/821 条数据...
|
||
[
|
||
{
|
||
"question": "Give the approximate minimum temperature at which it is possible to austenitize a 0.20 wt% C iron-carbon alloy during a normalizing heat treatment.",
|
||
"answer": "For 0.20 wt% C, heat to at least 905°C (1660°F) since the A3 temperature is 850°C (1560°F)."
|
||
},
|
||
{
|
||
"question": "Give the approximate minimum temperature at which it is possible to austenitize a 0.76 wt% C iron-carbon alloy during a normalizing heat treatment.",
|
||
"answer": "For 0.76 wt% C, heat to at least 782°C (1440°F) since the A3' temperature is 727°C (1340°F)."
|
||
},
|
||
{
|
||
"question": "Give the approximate minimum temperature at which it is possible to austenitize a 0.95 wt% C iron-carbon alloy during a normalizing heat treatment.",
|
||
"answer": "For 0.95 wt% C, heat to at least 840°C (1545°F) since the Acm temperature is 785°C (1445°F)."
|
||
}
|
||
]
|
||
处理第 504/821 条数据...
|
||
[
|
||
{
|
||
"question": "Give the approximate temperature at which it is desirable to heat a 0.25 wt% C iron-carbon alloy during a full anneal heat treatment.",
|
||
"answer": "For 0.25 wt% C, heat to about 880°C (1510°F) since the A3 temperature is 830°C (1420°F)."
|
||
},
|
||
{
|
||
"question": "Give the approximate temperature at which it is desirable to heat a 0.45 wt% C iron-carbon alloy during a full anneal heat treatment.",
|
||
"answer": "For 0.45 wt% C, heat to about 830°C (1525°F) since the A3 temperature is 780°C (1435°F)."
|
||
},
|
||
{
|
||
"question": "Give the approximate temperature at which it is desirable to heat a 0.85 wt% C iron-carbon alloy during a full anneal heat treatment.",
|
||
"answer": "For 0.85 wt% C, heat to about 777°C (1430°F) since the A1 temperature is 727°C (1340°F)."
|
||
},
|
||
{
|
||
"question": "Give the approximate temperature at which it is desirable to heat a 1.10 wt% C iron-carbon alloy during a full anneal heat treatment.",
|
||
"answer": "For 1.10 wt% C, heat to about 777°C (1430°F) since the A1 temperature is 727°C (1340°F)."
|
||
}
|
||
]
|
||
处理第 505/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is the purpose of a spheroidizing heat treatment?",
|
||
"answer": "The purpose of a spheroidizing heat treatment is to produce a very soft and ductile steel alloy having a spheroiditic microstructure."
|
||
},
|
||
{
|
||
"question": "On what classes of alloys is spheroidizing heat treatment normally used?",
|
||
"answer": "It is normally used on medium- and high-carbon steels, which, by virtue of carbon content, are relatively hard and strong."
|
||
}
|
||
]
|
||
```
|
||
处理第 506/821 条数据...
|
||
It's a single issue.
|
||
处理第 507/821 条数据...
|
||
It's a single issue.
|
||
处理第 508/821 条数据...
|
||
It's a single issue.
|
||
处理第 509/821 条数据...
|
||
It's a single issue.
|
||
处理第 510/821 条数据...
|
||
[
|
||
{
|
||
"question": "The zinc blende crystal structure is one that may be generated from close-packed planes of anions. Will the stacking sequence for this structure be FCC or HCP? Why?",
|
||
"answer": "The stacking sequence of close-packed planes of anions for the zinc blende crystal structure will be the same as FCC (and not HCP) because the anion packing is FCC."
|
||
},
|
||
{
|
||
"question": "The zinc blende crystal structure is one that may be generated from close-packed planes of anions. Will cations fill tetrahedral or octahedral positions? Why?",
|
||
"answer": "The cations will fill tetrahedral positions since the coordination number for cations is four."
|
||
},
|
||
{
|
||
"question": "The zinc blende crystal structure is one that may be generated from close-packed planes of anions. What fraction of the positions will be occupied?",
|
||
"answer": "Only one-half of the tetrahedral positions will be occupied because there are two tetrahedral sites per anion, and yet only one cation per anion."
|
||
}
|
||
]
|
||
处理第 511/821 条数据...
|
||
It's a single issue.
|
||
处理第 512/821 条数据...
|
||
It's a single issue.
|
||
处理第 513/821 条数据...
|
||
It's a single issue.
|
||
处理第 514/821 条数据...
|
||
It's a single issue.
|
||
处理第 515/821 条数据...
|
||
It's a single issue.
|
||
处理第 516/821 条数据...
|
||
[
|
||
{
|
||
"question": "Suppose that Li2O is added as an impurity to CaO. If the Li+ substitutes for Ca2+, what kind of vacancies would you expect to form? How many of these vacancies are created for every Li+ added?",
|
||
"answer": "For Li+ substituting for Ca2+ in CaO, oxygen vacancies would be created. For each Li+ substituting for Ca2+, one positive charge is removed; in order to maintain charge neutrality, a single negative charge may be removed. Negative charges are eliminated by creating oxygen vacancies, and for every two Li+ ions added, a single oxygen vacancy is formed."
|
||
},
|
||
{
|
||
"question": "Suppose that CaCl2 is added as an impurity to CaO. If the Cl substitutes for O2-, what kind of vacancies would you expect to form? How many of the vacancies are created for every Cl- added?",
|
||
"answer": "For Cl- substituting for O2- in CaO, calcium vacancies would be created. For each Cl- substituting for an O2-, one negative charge is removed; in order to maintain charge neutrality, two Cl- ions will lead to the formation of one calcium vacancy."
|
||
}
|
||
]
|
||
处理第 517/821 条数据...
|
||
[
|
||
{
|
||
"question": "What point defects are possible for Al2O3 as an impurity in MgO?",
|
||
"answer": "The possible point defects are Al3+ substitution for Mg2+ and the formation of Mg2+ vacancies to maintain charge neutrality."
|
||
},
|
||
{
|
||
"question": "How many Al3+ ions must be added to form each of these defects in MgO?",
|
||
"answer": "For every Al3+ ion that substitutes for Mg2+, a single positive charge is added. To maintain charge neutrality, one Mg2+ vacancy is formed for every two Al3+ ions added."
|
||
}
|
||
]
|
||
处理第 518/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "When kaolinite clay [Al2(Si2O5)(OH)4] is heated to a sufficiently high temperature, chemical water is driven off. Under these circumstances, what is the composition of the remaining product (in weight percent Al2O3)?",
|
||
"answer": "The chemical formula for kaolinite clay may also be written as Al2O3·2SiO2·2H2O. Thus, if we remove the chemical water, the formula becomes Al2O3·2SiO2. The formula weight for Al2O3 is just (2)(26.98 g/mol)+(3)(16.00 g/mol)=101.96 g/mol; and for SiO2 the formula weight is 28.09 g/mol+(2)(16.00 g/mol)=60.09 g/mol. Thus, the composition of this product, in terms of the concentration of Al2O3, CAl2O3, in weight percent is just CAl2O3=(101.96 g/mol)/(101.96 g/mol+(2)(60.09 g/mol))×100=45.9 wt%."
|
||
},
|
||
{
|
||
"question": "What are the liquidus and solidus temperatures of the material remaining after heating kaolinite clay [Al2(Si2O5)(OH)4] to drive off chemical water?",
|
||
"answer": "The liquidus and solidus temperatures for this material as determined from the SiO2-Al2O3 phase diagram, Figure 12.27, are 1825°C and 1587°C, respectively."
|
||
}
|
||
]
|
||
```
|
||
处理第 519/821 条数据...
|
||
[
|
||
{
|
||
"question": "Why may there be significant scatter in the fracture strength for some given ceramic material?",
|
||
"answer": "There may be significant scatter in the fracture strength for some given ceramic material because the fracture strength depends on the probability of the existence of a flaw that is capable of initiating a crack; this probability varies from specimen to specimen of the same material."
|
||
},
|
||
{
|
||
"question": "Why does fracture strength increase with decreasing specimen size?",
|
||
"answer": "The fracture strength increases with decreasing specimen size because as specimen size decreases, the probability of the existence of a flaw that is capable of initiating a crack diminishes."
|
||
}
|
||
]
|
||
处理第 520/821 条数据...
|
||
It's a single issue.
|
||
处理第 521/821 条数据...
|
||
It's a single issue.
|
||
处理第 522/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "A three-point transverse bending test is conducted on a cylindrical specimen of aluminum oxide having a reported flexural strength of 390 MPa (56,600 psi). If the specimen radius is 2.5 mm (0.10 in.) and the support point separation distance is 30 mm (1.2 in.), predict whether or not you would expect the specimen to fracture when a load of 620 N (140 lbf) is applied. Justify your prediction.",
|
||
"answer": "This portion of the problem asks that we determine whether or not a cylindrical specimen of aluminum oxide having a flexural strength of 390 MPa (56,600 psi) and a radius of 2.5 mm will fracture when subjected to a load of 620 N in a three-point bending test; the support point separation is given as 30 mm. Using Equation we will calculate the value of σ; if this value is greater than σ_fs (390 MPa), then fracture is expected to occur. Employment of Equation yields σ = (FL)/(πR^3) = (620 N)(30 × 10^-3 m)/(π(2.5 × 10^-3 m)^3) = 379 × 10^6 N/m^2 = 379 MPa (53,500 psi). Since this value is less than the given value of σ_fs (390 MPa), then fracture is not predicted."
|
||
},
|
||
{
|
||
"question": "Would you be 100% certain of the prediction in part (a)? Why or why not?",
|
||
"answer": "The certainty of this prediction is not 100% because there is always some variability in the flexural strength for ceramic materials, and since this value of σ (379 MPa) is relatively close to σ_fs (390 MPa) then there is some chance that fracture will occur."
|
||
}
|
||
]
|
||
```
|
||
处理第 523/821 条数据...
|
||
It's a single issue.
|
||
处理第 524/821 条数据...
|
||
[
|
||
{
|
||
"question": "Compute the modulus of elasticity for nonporous beryllium oxide (BeO) given that the modulus of elasticity for BeO with 5 vol% porosity is 310 GPa (45 x 10^6 psi).",
|
||
"answer": "To compute the modulus of elasticity for nonporous BeO (E0), we use the equation E0 = E / (1 - 1.9P + 0.9P^2), where E = 310 GPa and P = 0.05. Substituting these values, we get E0 = 310 GPa / (1 - (1.9)(0.05) + (0.9)(0.05)^2) = 342 GPa (49.6 x 10^6 psi)."
|
||
},
|
||
{
|
||
"question": "Compute the modulus of elasticity for beryllium oxide (BeO) with 10 vol% porosity, given that the modulus of elasticity for nonporous BeO is 342 GPa (49.6 x 10^6 psi).",
|
||
"answer": "To compute the modulus of elasticity for BeO with 10 vol% porosity (E), we use the equation E = E0 (1 - 1.9P + 0.9P^2), where E0 = 342 GPa and P = 0.10. Substituting these values, we get E = 342 GPa [1 - (1.9)(0.10) + (0.9)(0.10)^2] = 280 GPa (40.6 x 10^6 psi)."
|
||
}
|
||
]
|
||
处理第 525/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Compute the modulus of elasticity for the nonporous material given that the modulus of elasticity for boron carbide (B4C) having 5 vol% porosity is 290 GPa (42 x 10^6 psi).",
|
||
"answer": "To compute the modulus of elasticity for nonporous B4C, we solve the equation E0 = E / (1 - 1.9P + 0.9P^2) using P = 0.05. This gives E0 = 290 GPa / (1 - (1.9)(0.05) + (0.9)(0.05)^2) = 320 GPa (46.3 x 10^6 psi)."
|
||
},
|
||
{
|
||
"question": "At what volume percent porosity will the modulus of elasticity be 235 GPa (34 x 10^6 psi) for boron carbide (B4C), given that the modulus of elasticity for the nonporous material is 320 GPa?",
|
||
"answer": "To find the volume percent porosity at which the elastic modulus is 235 GPa, we use the equation E/E0 = 1 - 1.9P + 0.9P^2, where E0 = 320 GPa. This gives 235 GPa / 320 GPa = 0.734 = 1 - 1.9P + 0.9P^2. Rearranging, we get 0.9P^2 - 1.9P + 0.266 = 0. Solving the quadratic equation yields P = (1.9 ± sqrt((-1.9)^2 - (4)(0.9)(0.266))) / (2)(0.9). The physically meaningful root is P = 0.151, or 15.1 vol%."
|
||
}
|
||
]
|
||
```
|
||
处理第 526/821 条数据...
|
||
It's a single issue.
|
||
处理第 527/821 条数据...
|
||
[
|
||
{
|
||
"question": "What is crystallization?",
|
||
"answer": "Crystallization is the process whereby a glass material is caused to transform to a crystalline solid, usually as a result of a heat treatment."
|
||
},
|
||
{
|
||
"question": "Cite two properties that may be improved by crystallization.",
|
||
"answer": "Two properties that may be improved by crystallization are (1) a lower coefficient of thermal expansion, and (2) higher strength."
|
||
}
|
||
]
|
||
处理第 528/821 条数据...
|
||
[
|
||
{
|
||
"question": "For refractory ceramic materials, what are three characteristics that improve with increasing porosity?",
|
||
"answer": "Three characteristics that improve with increasing porosity are (1) decreased thermal expansion and contraction upon thermal cycling, (2) improved thermal insulation, and (3) improved resistance to thermal shock."
|
||
},
|
||
{
|
||
"question": "For refractory ceramic materials, what are two characteristics that are adversely affected by increasing porosity?",
|
||
"answer": "Two characteristics that are adversely affected are (1) load-bearing capacity and (2) resistance to attack by corrosive materials."
|
||
}
|
||
]
|
||
处理第 529/821 条数据...
|
||
[
|
||
{
|
||
"question": "How do aggregate particles become bonded together in clay-based mixtures during firing?",
|
||
"answer": "For clay-based aggregates, a liquid phase forms during firing, which infiltrates the pores between the unmelted particles; upon cooling, this liquid becomes a glass, that serves as the bonding phase."
|
||
},
|
||
{
|
||
"question": "How do aggregate particles become bonded together in cements during setting?",
|
||
"answer": "With cements, the bonding process is a chemical, hydration reaction between the water that has been added and the various cement constituents. The cement particles are bonded together by reactions that occur at the particle surfaces."
|
||
}
|
||
]
|
||
处理第 530/821 条数据...
|
||
It's a single issue.
|
||
处理第 531/821 条数据...
|
||
It's a single issue.
|
||
处理第 532/821 条数据...
|
||
It's a single issue.
|
||
处理第 533/821 条数据...
|
||
[
|
||
{
|
||
"question": "Explain why residual thermal stresses are introduced into a glass piece when it is cooled.",
|
||
"answer": "Residual thermal stresses are introduced into a glass piece when it is cooled because surface and interior regions cool at different rates, and, therefore, contract different amounts; since the material will experience very little, if any deformation, stresses are established."
|
||
},
|
||
{
|
||
"question": "Are thermal stresses introduced upon heating? Why or why not?",
|
||
"answer": "Yes, thermal stresses will be introduced because of thermal expansion upon heating for the same reason as for thermal contraction upon cooling."
|
||
}
|
||
]
|
||
处理第 534/821 条数据...
|
||
It's a single issue.
|
||
处理第 535/821 条数据...
|
||
It's a single issue.
|
||
处理第 536/821 条数据...
|
||
It's a single issue.
|
||
处理第 537/821 条数据...
|
||
It's a single issue.
|
||
处理第 538/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What are the three main components of a whiteware ceramic such as porcelain?",
|
||
"answer": "The three components of a whiteware ceramic are clay, quartz, and a flux."
|
||
},
|
||
{
|
||
"question": "What role does each component play in the forming and firing procedures of a whiteware ceramic such as porcelain?",
|
||
"answer": "Quartz acts as a filler material. Clay facilitates the forming operation since, when mixed with water, the mass may be made to become either hydroplastic or form a slip. Also, since clays melt over a range of temperatures, the shape of the piece being fired will be maintained. The flux facilitates the formation of a glass having a relatively low melting temperature."
|
||
}
|
||
]
|
||
```
|
||
处理第 539/821 条数据...
|
||
[
|
||
{
|
||
"question": "Why is it so important to control the rate of drying of a ceramic body that has been hydroplastically formed or slip cast?",
|
||
"answer": "It is important to control the rate of drying inasmuch as if the rate of drying is too rapid, there will be nonuniform shrinkage between surface and interior regions, such that warping and/or cracking of the ceramic ware may result."
|
||
},
|
||
{
|
||
"question": "Cite three factors that influence the rate of drying of a ceramic body, and explain how each affects the rate.",
|
||
"answer": "Three factors that affect the rate of drying are temperature, humidity, and rate of air flow. The rate of drying is enhanced by increasing both the temperature and rate of air flow, and by decreasing the humidity of the air."
|
||
}
|
||
]
|
||
处理第 540/821 条数据...
|
||
It's a single issue.
|
||
处理第 541/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "(a) Name three factors that influence the degree to which vitrification occurs in clay-based ceramic wares.",
|
||
"answer": "Three factors that influence the degree to which vitrification occurs in clay-based ceramic wares are: (1) composition (especially the concentration of flux present); (2) the temperature of firing; and (3) the time at the firing temperature."
|
||
},
|
||
{
|
||
"question": "(b) Explain how density, firing distortion, strength, corrosion resistance, and thermal conductivity are affected by the extent of vitrification.",
|
||
"answer": "Density will increase with degree of vitrification since the total remaining pore volume decreases. Firing distortion will increase with degree of vitrification since more liquid phase will be present at the firing temperature. Strength will also increase with degree of vitrification inasmuch as more of the liquid phase forms, which fills in a greater fraction of pore volume. Upon cooling, the liquid forms a glass matrix of relatively high strength. Corrosion resistance normally increases also, especially at service temperatures below that at which the glass phase begins to soften. The rate of corrosion is dependent on the amount of surface area exposed to the corrosive medium; hence, decreasing the total surface area by filling in some of the surface pores, diminishes the corrosion rate. Thermal conductivity will increase with degree of vitrification. The glass phase has a higher conductivity than the pores that it has filled."
|
||
}
|
||
]
|
||
```
|
||
处理第 542/821 条数据...
|
||
It's a single issue.
|
||
处理第 543/821 条数据...
|
||
[
|
||
{
|
||
"question": "Compute repeat unit molecular weight for poly(vinyl chloride), where each repeat unit consists of two carbons, three hydrogens, and one chlorine. The atomic weights are: carbon (12.01 g/mol), hydrogen (1.008 g/mol), and chlorine (35.45 g/mol).",
|
||
"answer": "For poly(vinyl chloride), the molecular weight is calculated as: m = 2(12.01 g/mol) + 3(1.008 g/mol) + 35.45 g/mol = 62.49 g/mol."
|
||
},
|
||
{
|
||
"question": "Compute repeat unit molecular weight for poly(ethylene terephthalate), where each repeat unit has ten carbons, eight hydrogens, and four oxygens. The atomic weights are: carbon (12.01 g/mol), hydrogen (1.008 g/mol), and oxygen (16.00 g/mol).",
|
||
"answer": "For poly(ethylene terephthalate), the molecular weight is calculated as: m = 10(12.01 g/mol) + 8(1.008 g/mol) + 4(16.00 g/mol) = 192.16 g/mol."
|
||
},
|
||
{
|
||
"question": "Compute repeat unit molecular weight for polycarbonate, where each repeat unit has sixteen carbons, fourteen hydrogens, and three oxygens. The atomic weights are: carbon (12.01 g/mol), hydrogen (1.008 g/mol), and oxygen (16.00 g/mol).",
|
||
"answer": "For polycarbonate, the molecular weight is calculated as: m = 16(12.01 g/mol) + 14(1.008 g/mol) + 3(16.00 g/mol) = 254.27 g/mol."
|
||
},
|
||
{
|
||
"question": "Compute repeat unit molecular weight for polydimethylsiloxane, where each repeat unit has two carbons, six hydrogens, one silicon, and one oxygen. The atomic weights are: carbon (12.01 g/mol), hydrogen (1.008 g/mol), silicon (28.09 g/mol), and oxygen (16.00 g/mol).",
|
||
"answer": "For polydimethylsiloxane, the molecular weight is calculated as: m = 2(12.01 g/mol) + 6(1.008 g/mol) + 28.09 g/mol + 16.00 g/mol = 74.16 g/mol."
|
||
}
|
||
]
|
||
处理第 544/821 条数据...
|
||
It's a single issue.
|
||
处理第 545/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Compute the repeat unit molecular weight of polystyrene.",
|
||
"answer": "The repeat unit molecular weight of polystyrene is called for in this portion of the problem. For polystyrene, each repeat unit has eight carbons and eight hydrogens. Thus, m = 8(A_C) + 8(A_H) = (8)(12.01 g/mol) + (8)(1.008 g/mol) = 104.14 g/mol"
|
||
},
|
||
{
|
||
"question": "Compute the number-average molecular weight for a polystyrene for which the degree of polymerization is 25,000.",
|
||
"answer": "We are now asked to compute the number-average molecular weight. Since the degree of polymerization is 25,000, using the equation M_n = (DP)m = (25,000)(104.14 g/mol) = 2.60 × 10^6 g/mol"
|
||
}
|
||
]
|
||
```
|
||
处理第 546/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "High-density polyethylene may be chlorinated by inducing the random substitution of chlorine atoms for hydrogen. Determine the concentration of Cl (in wt%) that must be added if this substitution occurs for 5% of all the original hydrogen atoms.",
|
||
"answer": "For chlorinated polyethylene, we are asked to determine the weight percent of chlorine added for 5% Cl substitution of all original hydrogen atoms. Consider 50 carbon atoms; there are 100 possible side-bonding sites. Ninety-five are occupied by hydrogen and five are occupied by Cl. Thus, the mass of these 50 carbon atoms, m_C, is just m_C=50(A_C)=(50)(12.01 g/mol)=600.5 g. Likewise, for hydrogen and chlorine, m_H=95(A_H)=(95)(1.008 g/mol)=95.76 g and m_Cl=5(A_Cl)=(5)(35.45 g/mol)=177.25 g. Thus, the concentration of chlorine, C_Cl, is determined as C_Cl = (m_Cl)/(m_C + m_H + m_Cl) × 100 = (177.25 g)/(600.5 g + 95.76 g + 177.25 g) × 100 = 20.3 wt%."
|
||
},
|
||
{
|
||
"question": "In what ways does chlorinated polyethylene differ from poly(vinyl chloride)?",
|
||
"answer": "Chlorinated polyethylene differs from poly(vinyl chloride), in that, for PVC, (1) 25% of the side-bonding sites are substituted with Cl, and (2) the substitution is probably much less random."
|
||
}
|
||
]
|
||
```
|
||
处理第 547/821 条数据...
|
||
[
|
||
{
|
||
"question": "Make comparisons of thermoplastic and thermosetting polymers on the basis of mechanical characteristics upon heating.",
|
||
"answer": "Thermoplastic polymers soften when heated and harden when cooled, whereas thermosetting polymers harden upon heating, while further heating will not lead to softening."
|
||
},
|
||
{
|
||
"question": "Make comparisons of thermoplastic and thermosetting polymers according to possible molecular structures.",
|
||
"answer": "Thermoplastic polymers have linear and branched structures, while for thermosetting polymers, the structures will normally be network or crosslinked."
|
||
}
|
||
]
|
||
处理第 548/821 条数据...
|
||
[
|
||
{
|
||
"question": "Is it possible to grind up and reuse phenol-formaldehyde? Why or why not?",
|
||
"answer": "It is not possible to grind up and reuse phenol-formaldehyde because it is a network thermoset polymer and, therefore, is not amenable to remolding."
|
||
},
|
||
{
|
||
"question": "Is it possible to grind up and reuse polypropylene? Why or why not?",
|
||
"answer": "Yes, it is possible to grind up and reuse polypropylene since it is a thermoplastic polymer, will soften when reheated, and, thus, may be remolded."
|
||
}
|
||
]
|
||
处理第 549/821 条数据...
|
||
It's a single issue.
|
||
处理第 550/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Calculate the repeat unit molecular weight of acrylonitrile in poly(acrylonitrile-butadiene) copolymer, given that it consists of three carbon, one nitrogen, and three hydrogen atoms.",
|
||
"answer": "The repeat unit molecular weight of acrylonitrile is calculated as: (3)(12.01 g/mol) + 14.01 g/mol + (3)(1.008 g/mol) = 53.06 g/mol."
|
||
},
|
||
{
|
||
"question": "Calculate the repeat unit molecular weight of butadiene in poly(acrylonitrile-butadiene) copolymer, given that it consists of four carbon and six hydrogen atoms.",
|
||
"answer": "The repeat unit molecular weight of butadiene is calculated as: (4)(12.01 g/mol) + (6)(1.008 g/mol) = 54.09 g/mol."
|
||
},
|
||
{
|
||
"question": "Calculate the average repeat unit molecular weight of the poly(acrylonitrile-butadiene) copolymer, given that the fraction of butadiene repeat units is 0.30 and the fraction of acrylonitrile repeat units is 0.70.",
|
||
"answer": "The average repeat unit molecular weight is calculated as: (0.70)(53.06 g/mol) + (0.30)(54.09 g/mol) = 53.37 g/mol."
|
||
},
|
||
{
|
||
"question": "Calculate the number-average molecular weight of the poly(acrylonitrile-butadiene) copolymer, given that the degree of polymerization is 2000 and the average repeat unit molecular weight is 53.37 g/mol.",
|
||
"answer": "The number-average molecular weight is calculated as: (53.37 g/mol)(2000) = 106740 g/mol."
|
||
}
|
||
]
|
||
```
|
||
处理第 551/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is the average molecular weight per repeat unit (m̄) for an alternating copolymer with a number-average molecular weight of 250,000 g/mol and a degree of polymerization of 3420?",
|
||
"answer": "The average molecular weight per repeat unit (m̄) is calculated as m̄ = M̄n / DP = 250,000 g/mol / 3420 = 73.10 g/mol."
|
||
},
|
||
{
|
||
"question": "What is the molecular weight of the styrene repeat unit (ms) in the alternating copolymer?",
|
||
"answer": "The molecular weight of the styrene repeat unit (ms) is calculated as ms = 8(AC) + 8(AH) = 8(12.01 g/mol) + 8(1.008 g/mol) = 104.14 g/mol."
|
||
},
|
||
{
|
||
"question": "What is the molecular weight of the unknown repeat unit (mx) in the alternating copolymer, given that the chain fractions of styrene and the unknown repeat unit are both 0.5?",
|
||
"answer": "The molecular weight of the unknown repeat unit (mx) is calculated as mx = (m̄ - fs * ms) / fx = (73.10 g/mol - (0.5)(104.14 g/mol)) / 0.5 = 42.06 g/mol."
|
||
},
|
||
{
|
||
"question": "What are the molecular weights of the possible repeat units: ethylene, propylene, tetrafluoroethylene, and vinyl chloride?",
|
||
"answer": "The molecular weights are calculated as follows: m_ethylene = 2(AC) + 4(AH) = 2(12.01 g/mol) + 4(1.008 g/mol) = 28.05 g/mol; m_propylene = 3(AC) + 6(AH) = 3(12.01 g/mol) + 6(1.008 g/mol) = 42.08 g/mol; m_TFE = 2(AC) + 4(AF) = 2(12.01 g/mol) + 4(19.00 g/mol) = 100.02 g/mol; m_VC = 2(AC) + 3(AH) + (ACl) = 2(12.01 g/mol) + 3(1.008 g/mol) + 35.45 g/mol = 62.49 g/mol."
|
||
},
|
||
{
|
||
"question": "Which of the possible repeat units (ethylene, propylene, tetrafluoroethylene, or vinyl chloride) matches the calculated molecular weight of the unknown repeat unit (42.06 g/mol)?",
|
||
"answer": "Propylene is the other repeat unit type since its molecular weight (42.08 g/mol) is almost the same as the calculated mx (42.06 g/mol)."
|
||
}
|
||
]
|
||
```
|
||
处理第 552/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Determine the ratio of butadiene to styrene repeat units in a copolymer having a number-average molecular weight of 350,000 g/mol and degree of polymerization of 4425.",
|
||
"answer": "To determine the ratio of butadiene to styrene repeat units, first calculate the average repeat unit molecular weight of the copolymer, m̄, using the formula: m̄ = M̄_n / DP = 350,000 g/mol / 4425 = 79.10 g/mol. Let f_b be the chain fraction of butadiene repeat units, then the chain fraction of styrene repeat units f_s is 1 - f_b. The repeat unit molecular weights are: m_b = 4(12.01 g/mol) + 6(1.008 g/mol) = 54.09 g/mol for butadiene and m_s = 8(12.01 g/mol) + 8(1.008 g/mol) = 104.14 g/mol for styrene. Solving for f_b: f_b = (m̄ - m_s) / (m_b - m_s) = (79.10 g/mol - 104.14 g/mol) / (54.09 g/mol - 104.14 g/mol) = 0.50. Thus, f_s = 1 - 0.50 = 0.50, and the ratio is f_b / f_s = 0.50 / 0.50 = 1.0."
|
||
},
|
||
{
|
||
"question": "Which type(s) of copolymer(s) will this copolymer be, considering the following possibilities: random, alternating, graft, and block? Why?",
|
||
"answer": "The copolymer can be alternating, random, graft, or block. The alternating copolymer requires a 1:1 ratio of repeat units, which matches the result from part (a). However, random, graft, and block copolymers can also have a 1:1 ratio, so all four types are possible."
|
||
}
|
||
]
|
||
```
|
||
处理第 553/821 条数据...
|
||
It's a single issue.
|
||
处理第 554/821 条数据...
|
||
It's a single issue.
|
||
处理第 555/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For the pair of linear and syndiotactic poly(vinyl chloride) and linear and isotactic polystyrene, (1) state whether or not it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why.",
|
||
"answer": "Yes, for these two polymers it is possible to decide. The linear and syndiotactic poly(vinyl chloride) is more likely to crystallize; the phenyl side-group for polystyrene is bulkier than the Cl side-group for poly(vinyl chloride). Syndiotactic and isotactic isomers are equally likely to crystallize."
|
||
},
|
||
{
|
||
"question": "For the pair of network phenol-formaldehyde and linear and heavily crosslinked cis-isoprene, (1) state whether or not it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why.",
|
||
"answer": "No, it is not possible to decide for these two polymers. Both heavily crosslinked and network polymers are not likely to crystallize."
|
||
},
|
||
{
|
||
"question": "For the pair of linear polyethylene and lightly branched isotactic polypropylene, (1) state whether or not it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why.",
|
||
"answer": "Yes, it is possible to decide for these two polymers. The linear polyethylene is more likely to crystallize. The repeat unit structure for polypropylene is chemically more complicated than is the repeat unit structure for polyethylene. Furthermore, branched structures are less likely to crystallize than are linear structures."
|
||
},
|
||
{
|
||
"question": "For the pair of alternating poly(styrene-ethylene) copolymer and random poly(vinyl chloride-tetrafluoroethylene) copolymer, (1) state whether or not it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why.",
|
||
"answer": "Yes, it is possible to decide for these two copolymers. The alternating poly(styrene-ethylene) copolymer is more likely to crystallize. Alternating copolymers crystallize more easily than do random copolymers."
|
||
}
|
||
]
|
||
```
|
||
处理第 556/821 条数据...
|
||
It's a single issue.
|
||
处理第 557/821 条数据...
|
||
It's a single issue.
|
||
处理第 558/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Compute the permeability coefficient (P_M) at 350 K for hydrogen in poly(dimethyl siloxane) (PDMSO) using the given equation P_M = P_M0 exp(-Q_p / R T), where P_M0 = 1.45 × 10^-8 (cm³ STP)(cm) / (cm²-s-Pa), Q_p = 13.7 kJ/mol, and R = 8.31 J/mol-K.",
|
||
"answer": "P_M = [1.45 × 10^-8 (cm³ STP)(cm) / (cm²-s-Pa)] exp[-13,700 J/mol / (8.31 J/mol-K)(350 K)] = 1.31 × 10^-10 (cm³ STP)(cm) / (cm²-s-Pa)"
|
||
},
|
||
{
|
||
"question": "Compute the diffusion flux (J) for hydrogen through a PDMSO sheet 20 mm thick with hydrogen pressures of 10 kPa and 1 kPa at the two faces, using the permeability coefficient P_M = 1.31 × 10^-10 (cm³ STP)(cm) / (cm²-s-Pa) and the equation J = P_M (ΔP / Δx), where ΔP = P2 - P1 and Δx is the thickness.",
|
||
"answer": "J = 1.31 × 10^-10 (cm³ STP)(cm) / (cm²-s-Pa) × (10,000 Pa - 1,000 Pa) / 2.0 cm = 5.90 × 10^-7 (cm³ STP) / (cm²-s)"
|
||
}
|
||
]
|
||
```
|
||
处理第 559/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Determine the relaxation time (τ) for the viscoelastic polymer given the initial stress σ(0) = 2.76 MPa, stress after 60 s σ(60) = 1.72 MPa, and elapsed time t = 60 s.",
|
||
"answer": "The relaxation time τ is calculated using the formula τ = -t / ln[σ(t)/σ(0)]. Substituting the given values, τ = -60 s / ln[1.72 MPa / 2.76 MPa] = 127 s."
|
||
},
|
||
{
|
||
"question": "Calculate the stress σ(10) at t = 10 s for the viscoelastic polymer using the relaxation time τ = 127 s and initial stress σ(0) = 2.76 MPa.",
|
||
"answer": "The stress at t = 10 s is calculated using σ(t) = σ(0) exp(-t/τ). Substituting the values, σ(10) = 2.76 MPa exp(-10 s / 127 s) = 2.55 MPa."
|
||
},
|
||
{
|
||
"question": "Determine the relaxation modulus E_r(10) for the viscoelastic polymer given the stress σ(10) = 2.55 MPa and the constant strain ε0 = 0.6.",
|
||
"answer": "The relaxation modulus is calculated using E_r(t) = σ(t)/ε0. Substituting the values, E_r(10) = 2.55 MPa / 0.6 = 4.25 MPa (616 psi)."
|
||
}
|
||
]
|
||
```
|
||
处理第 560/821 条数据...
|
||
[
|
||
{
|
||
"question": "Contrast the manner in which stress relaxation and viscoelastic creep tests are conducted.",
|
||
"answer": "Stress relaxation tests are conducted by rapidly straining the material elastically in tension, holding the strain level constant, and then measuring the stress as a function of time. For viscoelastic creep tests, a stress (usually tensile) is applied instantaneously and maintained constant while strain is measured as a function of time."
|
||
},
|
||
{
|
||
"question": "For each of these tests, cite the experimental parameter of interest and how it is determined.",
|
||
"answer": "The experimental parameters of interest from the stress relaxation and viscoelastic creep tests are the relaxation modulus and creep modulus (or creep compliance), respectively. The relaxation modulus is the ratio of stress measured after 10 s and strain; creep modulus is the ratio of stress and strain taken at a specific time."
|
||
}
|
||
]
|
||
处理第 561/821 条数据...
|
||
It's a single issue.
|
||
处理第 562/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For the pair of polymers: random acrylonitrile-butadiene copolymer with 10% of possible sites crosslinked; alternating acrylonitrile-butadiene copolymer with 5% of possible sites crosslinked, (1) state whether or not it is possible to decide whether one polymer has a higher tensile modulus than the other; (2) if this is possible, note which has the higher tensile modulus and then cite the reason(s) for your choice; and (3) if it is not possible to decide, then state why.",
|
||
"answer": "No, it is not possible. The random acrylonitrile-butadiene copolymer will tend to a lower degree of crystallinity than the alternating acrylonitrile-butadiene copolymer inasmuch as random copolymers don't normally crystallize. On this basis only, the alternating material would have a higher modulus inasmuch as tensile modulus increases with degree of crystallinity. On the other hand, the random copolymer has a higher degree of crosslinking (10% versus 5% for the alternating copolymer), and, on this basis only would have the higher tensile modulus-an increase in crosslinking leads to an increase in E. Thus, this determination is not possible; with regard to degree of crystallinity the alternating material has the higher E, whereas the random copolymer would have a higher E value on the basis of degree of crosslinking."
|
||
},
|
||
{
|
||
"question": "For the pair of polymers: branched and syndiotactic polypropylene with a degree of polymerization of 5000; linear and isotactic polypropylene with a degree of polymerization of 3000, (1) state whether or not it is possible to decide whether one polymer has a higher tensile modulus than the other; (2) if this is possible, note which has the higher tensile modulus and then cite the reason(s) for your choice; and (3) if it is not possible to decide, then state why.",
|
||
"answer": "Yes, it is possible. The linear and isotactic polypropylene will have a greater tensile modulus. Linear polymers are more likely to crystallize that branched ones. The likelihood of crystallization for both syndiotactic and isotactic polypropylene is about the same, and, therefore, degree is crystallization is not a factor. Furthermore, tensile modulus is relatively insensitive to degree of polymerization (i.e., molecular weight)-the fact that branched PP has the higher molecular weight is not important."
|
||
},
|
||
{
|
||
"question": "For the pair of polymers: branched polyethylene with a number-average molecular weight of 250,000 g/mol; linear and isotactic poly(vinyl chloride) with a number-average molecular weight of 200,000 g/mol, (1) state whether or not it is possible to decide whether one polymer has a higher tensile modulus than the other; (2) if this is possible, note which has the higher tensile modulus and then cite the reason(s) for your choice; and (3) if it is not possible to decide, then state why.",
|
||
"answer": "No, it is not possible. Linear polymers have higher degrees of crystallization (and higher tensile moduli) than branched polymers - on this basis, the PVC material should have the higher value of E. On the other hand, PVC has a more complex repeat unit structure than does polyethylene, which means that, on this basis, the PE would have a higher degree of crystallinity and also a greater tensile modulus. Also, tensile modulus is relatively independent of number-average molecular weight. Therefore, this determination is not possible since it is not possible to determine which of the two materials has the greater degree of crystallinity."
|
||
}
|
||
]
|
||
```
|
||
处理第 563/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For syndiotactic polystyrene having a number-average molecular weight of 600,000 g/mol and atactic polystyrene having a number-average molecular weight of 500,000 g/mol, state whether or not it is possible to decide whether one polymer has a higher tensile strength than the other; if possible, note which has the higher tensile strength and cite the reason(s) for your choice; if not possible, state why.",
|
||
"answer": "Yes it is possible. The syndiotactic polystyrene has the higher tensile strength. Syndiotactic polymers are more likely to crystallize than atactic ones; the greater the crystallinity, the higher the tensile strength. Furthermore, the syndiotactic also has a higher molecular weight; increasing molecular weight also enhances the strength."
|
||
},
|
||
{
|
||
"question": "For random acrylonitrile-butadiene copolymer with 10% of possible sites crosslinked and block acrylonitrile-butadiene copolymer with 5% of possible sites crosslinked, state whether or not it is possible to decide whether one polymer has a higher tensile strength than the other; if possible, note which has the higher tensile strength and cite the reason(s) for your choice; if not possible, state why.",
|
||
"answer": "No it is not possible. The random acrylonitrile-butadiene copolymer has more crosslinking; increased crosslinking leads to an increase in strength. However, the block copolymeric material will most likely have a higher degree of crystallinity; and increasing crystallinity improves the strength."
|
||
},
|
||
{
|
||
"question": "For network polyester and lightly branched polypropylene, state whether or not it is possible to decide whether one polymer has a higher tensile strength than the other; if possible, note which has the higher tensile strength and cite the reason(s) for your choice; if not possible, state why.",
|
||
"answer": "Yes it is possible. The network polyester will display a greater tensile strength. Relative chain motion is much more restricted than for the lightly branched polypropylene since there are many more of the strong covalent bonds for the network structure."
|
||
}
|
||
]
|
||
```
|
||
处理第 564/821 条数据...
|
||
It's a single issue.
|
||
处理第 565/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Name the polymer(s) from the following list that would be suitable for the fabrication of cups to contain hot coffee: polyethylene, polypropylene, poly(vinyl chloride), PET polyester, and polycarbonate.",
|
||
"answer": "Of the polymers listed, polycarbonate would be suitable for the fabrication of cups to contain hot coffee."
|
||
},
|
||
{
|
||
"question": "Why is polycarbonate suitable for the fabrication of cups to contain hot coffee?",
|
||
"answer": "Polycarbonate has a glass transition temperature of 100 degrees Celsius or above, which means it does not soften at the maximum temperature of hot coffee (slightly below 100 degrees Celsius)."
|
||
}
|
||
]
|
||
```
|
||
处理第 566/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For the pair of polymers: isotactic polystyrene that has a density of 1.12 g/cm3 and a weight-average molecular weight of 150,000 g/mol; syndiotactic polystyrene that has a density of 1.10 g/cm3 and a weight-average molecular weight of 125,000 g/mol, (1) state whether or not it is possible to determine whether one polymer has a higher melting temperature than the other; (2) if it is possible, note which has the higher melting temperature and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why.",
|
||
"answer": "Yes, it is possible to determine which of the two polystyrenes has the higher Tm. The isotactic polystyrene will have the higher melting temperature because it has a higher density (i.e., less branching) and also the greater weight-average molecular weight."
|
||
},
|
||
{
|
||
"question": "For the pair of polymers: linear polyethylene that has a degree of polymerization of 5,000; linear and isotactic polypropylene that has a degree of polymerization of 6,500, (1) state whether or not it is possible to determine whether one polymer has a higher melting temperature than the other; (2) if it is possible, note which has the higher melting temperature and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why.",
|
||
"answer": "Yes, it is possible to determine which polymer has the higher melting temperature. The polypropylene will have the higher Tm because it has a bulky phenyl side group in its repeat unit structure, which is absent in the polyethylene. Furthermore, the polypropylene has a higher degree of polymerization."
|
||
},
|
||
{
|
||
"question": "For the pair of polymers: branched and isotactic polystyrene that has a degree of polymerization of 4,000; linear and isotactic polypropylene that has a degree of polymerization of 7,500, (1) state whether or not it is possible to determine whether one polymer has a higher melting temperature than the other; (2) if it is possible, note which has the higher melting temperature and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why.",
|
||
"answer": "No, it is not possible to determine which of the two polymers has the higher melting temperature. The polystyrene has a bulkier side group than the polypropylene; on the basis of this effect alone, the polystyrene should have the greater Tm. However, the polystyrene has more branching and a lower degree of polymerization; both of these factors lead to a lowering of the melting temperature."
|
||
}
|
||
]
|
||
```
|
||
处理第 567/821 条数据...
|
||
It's a single issue.
|
||
处理第 568/821 条数据...
|
||
It's a single issue.
|
||
处理第 569/821 条数据...
|
||
It's a single issue.
|
||
处理第 570/821 条数据...
|
||
It's a single issue.
|
||
处理第 571/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Estimate the maximum thermal conductivity value for a cermet that contains 85 vol % titanium carbide (TiC) particles in a cobalt matrix. Assume thermal conductivities of 27 and 69 W/m-K for TiC and Co, respectively.",
|
||
"answer": "The maximum thermal conductivity k_max is calculated as k_max = k_m V_m + k_p V_p = k_Co V_Co + k_TiC V_TiC = (69 W/m-K)(0.15) + (27 W/m-K)(0.85) = 33.3 W/m-K."
|
||
},
|
||
{
|
||
"question": "Estimate the minimum thermal conductivity value for a cermet that contains 85 vol % titanium carbide (TiC) particles in a cobalt matrix. Assume thermal conductivities of 27 and 69 W/m-K for TiC and Co, respectively.",
|
||
"answer": "The minimum thermal conductivity k_min is calculated as k_min = (k_Co k_TiC) / (V_Co k_TiC + V_TiC k_Co) = (69 W/m-K)(27 W/m-K) / [(0.15)(27 W/m-K) + (0.85)(69 W/m-K)] = 29.7 W/m-K."
|
||
}
|
||
]
|
||
```
|
||
处理第 572/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is the upper limit for the elastic modulus of the composite given the volume fractions of tungsten (0.60) and copper (0.40), and the moduli of elasticity for copper (110 GPa) and tungsten (407 GPa)?",
|
||
"answer": "The upper limit for the elastic modulus of the composite is calculated as follows: E_c(u) = E_Cu V_Cu + E_W V_W = (110 GPa)(0.40) + (407 GPa)(0.60) = 288 GPa."
|
||
},
|
||
{
|
||
"question": "What is the specific gravity of the composite given the volume fractions of tungsten (0.60) and copper (0.40), and the specific gravities for copper (8.9) and tungsten (19.3)?",
|
||
"answer": "The specific gravity of the composite is calculated as follows: ρ_c = ρ_Cu V_Cu + ρ_W V_W = (8.9)(0.40) + (19.3)(0.60) = 15.14."
|
||
},
|
||
{
|
||
"question": "What is the specific stiffness of the composite using the upper limit of elastic modulus (288 GPa) and the specific gravity (15.14)?",
|
||
"answer": "The specific stiffness of the composite is calculated as follows: Specific Stiffness = E_c(u) / ρ_c = 288 GPa / 15.14 = 19.0 GPa."
|
||
},
|
||
{
|
||
"question": "What is the specific stiffness of the composite using the specific stiffness-volume fraction product for both copper and tungsten, given the volume fractions of tungsten (0.60) and copper (0.40), the moduli of elasticity for copper (110 GPa) and tungsten (407 GPa), and the specific gravities for copper (8.9) and tungsten (19.3)?",
|
||
"answer": "The specific stiffness of the composite is calculated as follows: Specific Stiffness = (E_Cu / ρ_Cu) V_Cu + (E_W / ρ_W) V_W = (110 GPa / 8.9)(0.40) + (407 GPa / 19.3)(0.60) = 17.6 GPa."
|
||
}
|
||
]
|
||
```
|
||
处理第 573/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is the distinction between cement and concrete?",
|
||
"answer": "Concrete consists of an aggregate of particles that are bonded together by a cement."
|
||
},
|
||
{
|
||
"question": "Cite three important limitations that restrict the use of concrete as a structural material.",
|
||
"answer": "Three limitations of concrete are: (1) it is a relatively weak and brittle material; (2) it experiences relatively large thermal expansions (contractions) with changes in temperature; and (3) it may crack when exposed to freeze-thaw cycles."
|
||
},
|
||
{
|
||
"question": "Briefly explain three techniques that are utilized to strengthen concrete by reinforcement.",
|
||
"answer": "Three reinforcement strengthening techniques are: (1) reinforcement with steel wires, rods, etc.; (2) reinforcement with fine fibers of a high modulus material; and (3) introduction of residual compressive stresses by prestressing or postensioning."
|
||
}
|
||
]
|
||
```
|
||
处理第 574/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is one similarity between precipitation hardening and dispersion strengthening?",
|
||
"answer": "The similarity between precipitation hardening and dispersion strengthening is the strengthening mechanism--i.e., the precipitates/particles effectively hinder dislocation motion."
|
||
},
|
||
{
|
||
"question": "What is one difference between precipitation hardening and dispersion strengthening?",
|
||
"answer": "The hardening/strengthening effect is not retained at elevated temperatures for precipitation hardening--however, it is retained for dispersion strengthening."
|
||
},
|
||
{
|
||
"question": "What is another difference between precipitation hardening and dispersion strengthening?",
|
||
"answer": "The strength is developed by a heat treatment for precipitation hardening--such is not the case for dispersion strengthening."
|
||
}
|
||
]
|
||
```
|
||
处理第 575/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For a continuous and aligned fiber-reinforced composite consisting of 30 vol% aramid fibers and 70 vol% polycarbonate matrix, with the following mechanical characteristics: Aramid fiber modulus of elasticity = 131 GPa (19x10^6 psi), tensile strength = 3600 MPa (520,000 psi); Polycarbonate modulus of elasticity = 2.4 GPa (3.5x10^5 psi), tensile strength = 65 MPa (9425 psi); and stress on the polycarbonate matrix when aramid fibers fail = 45 MPa (6500 psi), compute the longitudinal tensile strength.",
|
||
"answer": "The longitudinal tensile strength is determined as: sigma_cl* = sigma_m*(1-V_f) + sigma_f* V_f = (45 MPa)(0.70) + (3600 MPa)(0.30) = 1100 MPa (160,000 psi)."
|
||
},
|
||
{
|
||
"question": "For a continuous and aligned fiber-reinforced composite consisting of 30 vol% aramid fibers and 70 vol% polycarbonate matrix, with the following mechanical characteristics: Aramid fiber modulus of elasticity = 131 GPa (19x10^6 psi), tensile strength = 3600 MPa (520,000 psi); Polycarbonate modulus of elasticity = 2.4 GPa (3.5x10^5 psi), tensile strength = 65 MPa (9425 psi); and stress on the polycarbonate matrix when aramid fibers fail = 45 MPa (6500 psi), compute the longitudinal modulus of elasticity.",
|
||
"answer": "The longitudinal elastic modulus is computed as: E_cl = E_m V_m + E_f V_f = (2.4 GPa)(0.70) + (131 GPa)(0.30) = 41 GPa (5.95x10^6 psi)."
|
||
}
|
||
]
|
||
```
|
||
处理第 576/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For a continuous and oriented fiber-reinforced composite with a volume fraction of fibers of 0.25, determine the modulus of elasticity of the matrix phase given the modulus of elasticity in the longitudinal direction is 19.7 GPa (2.8×10^5 psi).",
|
||
"answer": "Using the equation for the longitudinal elastic modulus: E_cl = E_m(1 - V_f) + E_f V_f, where E_cl = 19.7 GPa and V_f = 0.25, we can write: 19.7 GPa = E_m(1 - 0.25) + E_f(0.25). Solving this equation simultaneously with the transverse elastic modulus equation leads to E_m = 2.79 GPa (4.04×10^5 psi)."
|
||
},
|
||
{
|
||
"question": "For a continuous and oriented fiber-reinforced composite with a volume fraction of fibers of 0.25, determine the modulus of elasticity of the fiber phase given the modulus of elasticity in the transverse direction is 3.66 GPa (5.3×10^5 psi).",
|
||
"answer": "Using the equation for the transverse elastic modulus: E_ct = (E_m E_f) / [(1 - V_f) E_f + V_f E_m], where E_ct = 3.66 GPa and V_f = 0.25, we can write: 3.66 GPa = (E_m E_f) / [(1 - 0.25) E_f + 0.25 E_m]. Solving this equation simultaneously with the longitudinal elastic modulus equation leads to E_f = 70.4 GPa (10.2×10^6 psi)."
|
||
}
|
||
]
|
||
```
|
||
处理第 577/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "In an aligned and continuous glass fiber-reinforced nylon 6,6 composite, the fibers are to carry 94% of a load applied in the longitudinal direction. Using the data provided, determine the volume fraction of fibers that will be required. The modulus of elasticity for glass fiber is 72.5 GPa (10.5 x 10^6 psi) and for nylon 6,6 is 3.0 GPa (4.35 x 10^6 psi).",
|
||
"answer": "Given that the fibers carry 94% of the load, the ratio of the load carried by the fibers to that carried by the matrix is F_f/F_m = 0.94/0.06 = 15.67. Using the equation F_f/F_m = (E_f V_f)/(E_m (1-V_f)), where E_f is the modulus of elasticity of the fibers (72.5 GPa), E_m is the modulus of elasticity of the matrix (3.0 GPa), and V_f is the volume fraction of fibers, we substitute the known values to get 15.67 = (72.5 GPa V_f)/(3.0 GPa (1-V_f)). Solving for V_f yields V_f = 0.393."
|
||
},
|
||
{
|
||
"question": "What will be the tensile strength of this composite? Assume that the matrix stress at fiber failure is 30 MPa (4350 psi). The tensile strength of glass fiber is 3400 MPa (490,000 psi) and of nylon 6,6 is 76 MPa (11,000 psi). The volume fraction of fibers is 0.393.",
|
||
"answer": "The tensile strength of the composite is calculated using the equation σ_cl* = σ_m* (1-V_f) + σ_f* V_f, where σ_m* is the matrix stress at fiber failure (30 MPa), σ_f* is the tensile strength of the fibers (3400 MPa), and V_f is the volume fraction of fibers (0.393). Substituting the values gives σ_cl* = (30 MPa)(1-0.393) + (3400 MPa)(0.393) = 1354 MPa (196,400 psi)."
|
||
}
|
||
]
|
||
```
|
||
处理第 578/821 条数据...
|
||
[
|
||
{
|
||
"question": "For a polymer-matrix fiber-reinforced composite, list three functions of the matrix phase.",
|
||
"answer": "For polymer-matrix fiber-reinforced composites, three functions of the polymer-matrix phase are: (1) to bind the fibers together so that the applied stress is distributed among the fibers; (2) to protect the surface of the fibers from being damaged; and (3) to separate the fibers and inhibit crack propagation."
|
||
},
|
||
{
|
||
"question": "For a polymer-matrix fiber-reinforced composite, compare the desired mechanical characteristics of matrix and fiber phases.",
|
||
"answer": "The matrix phase must be ductile and is usually relatively soft, whereas the fiber phase must be stiff and strong."
|
||
},
|
||
{
|
||
"question": "For a polymer-matrix fiber-reinforced composite, cite two reasons why there must be a strong bond between fiber and matrix at their interface.",
|
||
"answer": "There must be a strong interfacial bond between fiber and matrix in order to: (1) maximize the stress transmittance between matrix and fiber phases; and (2) minimize fiber pull-out, and the probability of failure."
|
||
}
|
||
]
|
||
处理第 579/821 条数据...
|
||
[
|
||
{
|
||
"question": "What is the distinction between matrix and dispersed phases in a composite material?",
|
||
"answer": "The matrix phase is a continuous phase that surrounds the noncontinuous dispersed phase."
|
||
},
|
||
{
|
||
"question": "Contrast the mechanical characteristics of matrix and dispersed phases for fiber-reinforced composites.",
|
||
"answer": "In general, the matrix phase is relatively weak, has a low elastic modulus, but is quite ductile. On the other hand, the fiber phase is normally quite strong, stiff, and brittle."
|
||
}
|
||
]
|
||
处理第 580/821 条数据...
|
||
[
|
||
{
|
||
"question": "Briefly explain the difference between oxidation and reduction electrochemical reactions.",
|
||
"answer": "Oxidation is the process by which an atom gives up an electron (or electrons) to become a cation. Reduction is the process by which an atom acquires an extra electron (or electrons) and becomes an anion."
|
||
},
|
||
{
|
||
"question": "Which reaction occurs at the anode and which at the cathode?",
|
||
"answer": "Oxidation occurs at the anode; reduction at the cathode."
|
||
}
|
||
]
|
||
处理第 581/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "(a)(i) Write the possible oxidation and reduction half-reactions that occur when magnesium is immersed in HCl solution.",
|
||
"answer": "In HCl, possible reactions are\nMg -> Mg2+ + 2 e- (oxidation)\n2 H+ + 2 e- -> H2 (reduction)"
|
||
},
|
||
{
|
||
"question": "(a)(ii) Write the possible oxidation and reduction half-reactions that occur when magnesium is immersed in an HCl solution containing dissolved oxygen.",
|
||
"answer": "In an HCl solution containing dissolved oxygen, possible reactions are\nMg -> Mg2+ + 2 e- (oxidation)\n4 H+ + O2 + 4 e- -> 2 H2O (reduction)"
|
||
},
|
||
{
|
||
"question": "(a)(iii) Write the possible oxidation and reduction half-reactions that occur when magnesium is immersed in an HCl solution containing dissolved oxygen and Fe2+ ions.",
|
||
"answer": "In an HCl solution containing dissolved oxygen and Fe2+ ions, possible reactions are\nMg -> Mg2+ + 2 e- (oxidation)\n4 H+ + O2 + 4 e- -> 2 H2O (reduction)\nFe2+ + 2 e- -> Fe (reduction)"
|
||
},
|
||
{
|
||
"question": "(b) In which of these solutions would you expect the magnesium to oxidize most rapidly? Why?",
|
||
"answer": "The magnesium would probably oxidize most rapidly in the HCl solution containing dissolved oxygen and Fe2+ ions because there are two reduction reactions that will consume electrons from the oxidation of magnesium."
|
||
}
|
||
]
|
||
```
|
||
处理第 582/821 条数据...
|
||
[
|
||
{
|
||
"question": "Demonstrate that the value of F in Equation 17.19 is 96,500 C/mol.",
|
||
"answer": "The Faraday constant F is just the product of the charge per electron and Avogadro's number; that is F = e N_A = (1.602 x 10^-19 C/electron)(6.022 x 10^23 electrons/mol) = 96,472 C/mol."
|
||
},
|
||
{
|
||
"question": "Demonstrate that at 25°C (298 K), (RT/nF) ln x = (0.0592/n) log x.",
|
||
"answer": "At 25°C (298 K), (RT/nF) ln x = [(8.31 J/mol-K)(298 K)]/[(n)(96,472 C/mol)] (2.303) log x = (0.0592/n) log x. This gives units in volts since a volt is a J/C."
|
||
}
|
||
]
|
||
处理第 583/821 条数据...
|
||
[
|
||
{
|
||
"question": "Is a voltage generated between the two cell halves of a Zn/Zn2+ concentration cell where one half has a Zn2+ concentration of 1.0 M and the other has 10^-2 M?",
|
||
"answer": "Yes, a voltage is generated between the two cell halves."
|
||
},
|
||
{
|
||
"question": "What is the magnitude of the voltage generated in the Zn/Zn2+ concentration cell with Zn2+ concentrations of 1.0 M and 10^-2 M?",
|
||
"answer": "The magnitude of the voltage generated is 0.0592 V."
|
||
},
|
||
{
|
||
"question": "Which electrode will be oxidized in the Zn/Zn2+ concentration cell with Zn2+ concentrations of 1.0 M and 10^-2 M?",
|
||
"answer": "The electrode in the cell half with the Zn2+ concentration of 10^-2 M will be oxidized."
|
||
}
|
||
]
|
||
处理第 584/821 条数据...
|
||
It's a single issue.
|
||
处理第 585/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "A thick steel sheet of area 400 cm² is exposed to air near the ocean. After a one-year period it was found to experience a weight loss of 375 g due to corrosion. To what rate of corrosion, in mm/yr, does this correspond?",
|
||
"answer": "This problem asks for us to calculate the CPR in mm/yr for a thick steel sheet of area 400 cm² which experiences a weight loss of 375 g after one year. Employment of Equation leads to CPR(mm/yr) = (K W)/(ρ A t) = (87.6)(375 g)(10³ mg/g)/(7.9 g/cm³)(400 cm²)(24 h/day)(365 day/yr)(1 yr) = 1.2 mm/yr"
|
||
},
|
||
{
|
||
"question": "A thick steel sheet of area 400 cm² is exposed to air near the ocean. After a one-year period it was found to experience a weight loss of 375 g due to corrosion. To what rate of corrosion, in mpy, does this correspond?",
|
||
"answer": "This problem asks for us to calculate the CPR in mpy for a thick steel sheet of area 400 cm² which experiences a weight loss of 375 g after one year. Employment of Equation leads to CPR(mpy) = (534)(375 g)(10³ mg/g)/(7.9 g/cm³)(400 in.²)(1 in/2.54 cm)²(24 h/day)(365 day/yr)(1 yr) = 46.7 mpy"
|
||
}
|
||
]
|
||
```
|
||
处理第 586/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Cite the major differences between activation and concentration polarizations.",
|
||
"answer": "Activation polarization is the condition wherein a reaction rate is controlled by one step in a series of steps that takes place at the slowest rate. For corrosion, activation polarization is possible for both oxidation and reduction reactions. Concentration polarization occurs when a reaction rate is limited by diffusion in a solution. For corrosion, concentration polarization is possible only for reduction reactions."
|
||
},
|
||
{
|
||
"question": "Under what conditions is activation polarization rate controlling?",
|
||
"answer": "Activation polarization is rate controlling when the reaction rate is low and/or the concentration of active species in the liquid solution is high."
|
||
},
|
||
{
|
||
"question": "Under what conditions is concentration polarization rate controlling?",
|
||
"answer": "Concentration polarization is rate controlling when the reaction rate is high and/or the concentration of active species in the liquid solution is low."
|
||
}
|
||
]
|
||
```
|
||
处理第 587/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Assuming that activation polarization controls both oxidation and reduction reactions, determine the rate of corrosion of metal M (in mol/cm2-s) given the following data: V(M/M2+) = -0.47 V, V(H+/H2) = 0 V, i0 for M = 5 × 10^-10 A/cm2, i0 for H = 2 × 10^-9 A/cm2, β for M = +0.15, β for H = -0.12.",
|
||
"answer": "To compute the rate of oxidation for metal M, we establish potential expressions for both oxidation and reduction reactions. For hydrogen reduction: VH = V(H+/H2) + βH log(i/i0H). For M oxidation: VM = V(M/M2+) + βM log(i/i0M). Setting VH = VM and solving for log(iC) gives: log(iC) = [1/(βM - βH)][V(H+/H2) - V(M/M2+) - βH log(i0H) + βM log(i0M)]. Substituting the given values: log(iC) = [1/(0.15 - (-0.12))][0 - (-0.47) - (-0.12)log(2 × 10^-9) + 0.15log(5 × 10^-10)] = -7.293. Thus, iC = 10^-7.293 = 5.09 × 10^-8 A/cm2. The corrosion rate is then: r = iC/(nF) = (5.09 × 10^-8 C/s-cm2)/(2 × 96,500 C/mol) = 2.64 × 10^-13 mol/cm2-s."
|
||
},
|
||
{
|
||
"question": "Compute the corrosion potential for the reaction involving metal M given the following data: V(H+/H2) = 0 V, i0 for H = 2 × 10^-9 A/cm2, β for H = -0.12, and the corrosion current density iC = 5.09 × 10^-8 A/cm2.",
|
||
"answer": "The corrosion potential VC can be calculated using the potential expression for hydrogen reduction: VC = V(H+/H2) + βH log(iC/i0H). Substituting the given values: VC = 0 + (-0.12 V) log[(5.09 × 10^-8 A/cm2)/(2 × 10^-9 A/cm2)] = -0.169 V."
|
||
}
|
||
]
|
||
```
|
||
处理第 588/821 条数据...
|
||
[
|
||
{
|
||
"question": "Briefly describe the phenomenon of passivity.",
|
||
"answer": "Passivity is the loss of chemical reactivity, under particular environmental conditions, of normally active metals and alloys."
|
||
},
|
||
{
|
||
"question": "Name two common types of alloy that passivate.",
|
||
"answer": "Stainless steels and aluminum alloys often passivate."
|
||
}
|
||
]
|
||
处理第 589/821 条数据...
|
||
It's a single issue.
|
||
处理第 590/821 条数据...
|
||
It's a single issue.
|
||
处理第 591/821 条数据...
|
||
It's a single issue.
|
||
处理第 592/821 条数据...
|
||
[
|
||
{
|
||
"question": "What are inhibitors?",
|
||
"answer": "Inhibitors are substances that, when added to a corrosive environment in relatively low concentrations, decrease the environment's corrosiveness."
|
||
},
|
||
{
|
||
"question": "What possible mechanisms account for their effectiveness?",
|
||
"answer": "Possible mechanisms that account for the effectiveness of inhibitors are: (1) elimination of a chemically active species in the solution; (2) attachment of inhibitor molecules to the corroding surface so as to interfere with either the oxidation or reduction reaction; and (3) the formation of a very thin and protective coating on the corroding surface."
|
||
}
|
||
]
|
||
处理第 593/821 条数据...
|
||
[
|
||
{
|
||
"question": "What is the sacrificial anode technique used for galvanic protection?",
|
||
"answer": "A sacrificial anode is electrically coupled to the metal piece to be protected, which anode is also situated in the corrosion environment. The sacrificial anode is a metal or alloy that is chemically more reactive in the particular environment. It (the anode) preferentially oxidizes, and, upon giving up electrons to the other metal, protects it from electrochemical corrosion."
|
||
},
|
||
{
|
||
"question": "What is the impressed current technique used for galvanic protection?",
|
||
"answer": "An impressed current from an external dc power source provides excess electrons to the metallic structure to be protected."
|
||
}
|
||
]
|
||
处理第 594/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Compute the electrical conductivity of a 5.1-mm (0.2-in.) diameter cylindrical silicon specimen 51 mm (2 in.) long in which a current of 0.1 A passes in an axial direction. A voltage of 12.5 V is measured across two probes that are separated by 38 mm (1.5 in.).",
|
||
"answer": "We use Equations for the conductivity, as sigma = 1/rho = (I l)/(V A) = (I l)/(V pi (d/2)^2). And, incorporating values for the several parameters provided in the problem statement, leads to sigma = (0.1 A)(38 × 10^-3 m)/(12.5 V)(pi)((5.1 × 10^-3 m)/2)^2 = 14.9 (Ω·m)^-1."
|
||
},
|
||
{
|
||
"question": "Compute the resistance over the entire 51 mm (2 in.) of the specimen.",
|
||
"answer": "The resistance, R, may be computed using Equations, as R = rho l/A = l/(sigma A) = l/(sigma pi (d/2)^2) = (51 × 10^-3 m)/([14.9 (Ω·m)^-1](pi)((5.1 × 10^-3 m)/2)^2) = 168 Ω."
|
||
}
|
||
]
|
||
```
|
||
处理第 595/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Calculate the drift velocity of electrons in germanium at room temperature and when the magnitude of the electric field is 1000 V/m.",
|
||
"answer": "The drift velocity of electrons in Ge may be determined using Equation. Since the room temperature mobility of electrons is 0.38 m2/V-s, and the electric field is 1000 V/m (as stipulated in the problem statement), vd = μe E = (0.38 m2/V-s)(1000 V/m) = 380 m/s."
|
||
},
|
||
{
|
||
"question": "Under these circumstances, how long does it take an electron to traverse a 25-mm (1-in.) length of crystal?",
|
||
"answer": "The time, t, required to traverse a given length, l(=25 mm), is just t = l / vd = (25 x 10-3 m) / (380 m/s) = 6.6 x 10-5 s."
|
||
}
|
||
]
|
||
```
|
||
处理第 596/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "At room temperature the electrical conductivity and the electron mobility for copper are 6.0 x 10^7 (Ω·m)^-1 and 0.0030 m^2/V·s, respectively. Compute the number of free electrons per cubic meter for copper at room temperature.",
|
||
"answer": "The number of free electrons per cubic meter for copper at room temperature may be computed using the equation n = σ / (|e| μ_e), where σ is the electrical conductivity, |e| is the electron charge, and μ_e is the electron mobility. Substituting the given values: n = (6.0 x 10^7 (Ω·m)^-1) / ((1.602 x 10^-19 C)(0.0030 m^2/V·s)) = 1.25 x 10^29 m^-3."
|
||
},
|
||
{
|
||
"question": "At room temperature the electrical conductivity and the electron mobility for copper are 6.0 x 10^7 (Ω·m)^-1 and 0.0030 m^2/V·s, respectively, and the density of copper is 8.9 g/cm^3. What is the number of free electrons per copper atom?",
|
||
"answer": "To calculate the number of free electrons per copper atom, first determine the number of copper atoms per cubic meter, N_Cu, using the equation N_Cu = (N_A ρ') / A_Cu, where N_A is Avogadro's number, ρ' is the density, and A_Cu is the atomic weight of copper. Substituting the given values: N_Cu = ((6.022 x 10^23 atoms/mol)(8.9 g/cm^3)(10^6 cm^3/m^3)) / 63.55 g/mol = 8.43 x 10^28 m^-3. Then, the number of free electrons per copper atom is n / N_Cu = (1.25 x 10^29 m^-3) / (8.43 x 10^28 m^-3) = 1.48."
|
||
}
|
||
]
|
||
```
|
||
处理第 597/821 条数据...
|
||
It's a single issue.
|
||
处理第 598/821 条数据...
|
||
[
|
||
{
|
||
"question": "Define the term 'intrinsic' as it pertains to semiconducting materials and provide an example.",
|
||
"answer": "Intrinsic refers to pure, undoped semiconducting materials. Example: high purity (undoped) Si, GaAs, CdS, etc."
|
||
},
|
||
{
|
||
"question": "Define the term 'extrinsic' as it pertains to semiconducting materials and provide an example.",
|
||
"answer": "Extrinsic refers to doped semiconducting materials. Example: P-doped Ge, B-doped Si, S-doped GaP, etc."
|
||
},
|
||
{
|
||
"question": "Define the term 'compound' as it pertains to semiconducting materials and provide an example.",
|
||
"answer": "Compound refers to semiconducting materials composed of two or more elements. Example: GaAs, InP, CdS, etc."
|
||
},
|
||
{
|
||
"question": "Define the term 'elemental' as it pertains to semiconducting materials and provide an example.",
|
||
"answer": "Elemental refers to semiconducting materials composed of a single element. Example: Ge and Si."
|
||
}
|
||
]
|
||
处理第 599/821 条数据...
|
||
[
|
||
{
|
||
"question": "Explain why no hole is generated by the electron excitation involving a donor impurity atom.",
|
||
"answer": "No hole is generated by an electron excitation involving a donor impurity atom because the excitation comes from a level within the band gap, and thus, no missing electron is created within the normally filled valence band."
|
||
},
|
||
{
|
||
"question": "Explain why no free electron is generated by the electron excitation involving an acceptor impurity atom.",
|
||
"answer": "No free electron is generated by an electron excitation involving an acceptor impurity atom because the electron is excited from the valence band into the impurity level within the band gap; no free electron is introduced into the conduction band."
|
||
}
|
||
]
|
||
处理第 600/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Germanium to which 5 × 10^22 m^-3 Sb atoms have been added is an extrinsic semiconductor at room temperature, and virtually all the Sb atoms may be thought of as being ionized (i.e., one charge carrier exists for each Sb atom). Is this material n-type or p-type?",
|
||
"answer": "This germanium material to which has been added 5 × 10^22 m^-3 Sb atoms is n-type since Sb is a donor in Ge. (Antimony is from group VA of the periodic table--Ge is from group IVA.)"
|
||
},
|
||
{
|
||
"question": "Germanium to which 5 × 10^22 m^-3 Sb atoms have been added is an extrinsic semiconductor at room temperature, and virtually all the Sb atoms may be thought of as being ionized (i.e., one charge carrier exists for each Sb atom). Calculate the electrical conductivity of this material, assuming electron and hole mobilities of 0.1 and 0.05 m^2/V-s, respectively.",
|
||
"answer": "Since this material is n-type extrinsic, each Sb will donate a single electron, or the electron concentration is equal to the Sb concentration since all of the Sb atoms are ionized at room temperature; that is n = 5 × 10^22 m^-3, and, as given in the problem statement, μe = 0.1 m^2/V-s. Thus σ = n|e|μe = (5 × 10^22 m^-3)(1.602 × 10^-19 C)(0.1 m^2/V-s) = 800 (Ω·m)^-1."
|
||
}
|
||
]
|
||
```
|
||
处理第 601/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Calculate the electron and hole mobilities for intrinsic InP given the following electrical characteristics at room temperature: conductivity (sigma) = 2.5 x 10^-6 (Omega.m)^-1, electron concentration (n) = 3.0 x 10^13 m^-3, hole concentration (p) = 3.0 x 10^13 m^-3.",
|
||
"answer": "For intrinsic InP, the conductivity expression is: sigma = n|e| mu_e + p|e| mu_h. Substituting the given values: 2.5 x 10^-6 (Omega.m)^-1 = (3.0 x 10^13 m^-3)(1.602 x 10^-19 C) mu_e + (3.0 x 10^13 m^-3)(1.602 x 10^-19 C) mu_h. This simplifies to 0.52 = mu_e + mu_h."
|
||
},
|
||
{
|
||
"question": "Calculate the electron and hole mobilities for extrinsic n-type InP given the following electrical characteristics at room temperature: conductivity (sigma) = 3.6 x 10^-5 (Omega.m)^-1, electron concentration (n) = 4.5 x 10^14 m^-3, hole concentration (p) = 2.0 x 10^12 m^-3.",
|
||
"answer": "For extrinsic n-type InP, the conductivity expression is: sigma = n|e| mu_e + p|e| mu_h. Substituting the given values: 3.6 x 10^-5 (Omega.m)^-1 = (4.5 x 10^14 m^-3)(1.602 x 10^-19 C) mu_e + (2.0 x 10^12 m^-3)(1.602 x 10^-19 C) mu_h. This simplifies to 112.4 = 225 mu_e + mu_h."
|
||
},
|
||
{
|
||
"question": "Solve the simultaneous equations 0.52 = mu_e + mu_h and 112.4 = 225 mu_e + mu_h to find the electron mobility (mu_e) and hole mobility (mu_h).",
|
||
"answer": "Solving the simultaneous equations: 0.52 = mu_e + mu_h and 112.4 = 225 mu_e + mu_h, we subtract the first equation from the second to get 111.88 = 224 mu_e. Solving for mu_e gives mu_e = 0.50 m^2/V-s. Substituting mu_e back into the first equation gives mu_h = 0.02 m^2/V-s."
|
||
}
|
||
]
|
||
```
|
||
处理第 602/821 条数据...
|
||
It's a single issue.
|
||
处理第 603/821 条数据...
|
||
It's a single issue.
|
||
处理第 604/821 条数据...
|
||
It's a single issue.
|
||
处理第 605/821 条数据...
|
||
It's a single issue.
|
||
处理第 606/821 条数据...
|
||
[
|
||
{
|
||
"question": "For each of the three types of polarization, briefly describe the mechanism by which dipoles are induced and/or oriented by the action of an applied electric field.",
|
||
"answer": "For electronic polarization, the electric field causes a net displacement of the center of the negatively charged electron cloud relative to the positive nucleus. With ionic polarization, the cations and anions are displaced in opposite directions as a result of the application of an electric field. Orientation polarization is found in substances that possess permanent dipole moments; these dipole moments become aligned in the direction of the electric field."
|
||
},
|
||
{
|
||
"question": "For solid lead titanate (PbTiO3), gaseous neon, diamond, solid KCl, and liquid NH3, what kind(s) of polarization is (are) possible? Why?",
|
||
"answer": "Electronic, ionic, and orientation polarizations would be observed in lead titanate. The lead, titanium, and oxygen would undoubtedly be largely ionic in character. Furthermore, orientation polarization is also possible inasmuch as permanent dipole moments may be induced in the same manner as for BaTiO3. Only electronic polarization is to be found in gaseous neon; being an inert gas, its atoms will not be ionized nor possess permanent dipole moments. Only electronic polarization is to be found in solid diamond; this material does not have molecules with permanent dipole moments, nor is it an ionic material. Both electronic and ionic polarizations will be found in solid KCl, since it is strongly ionic. In all probability, no permanent dipole moments will be found in this material. Both electronic and orientation polarizations are found in liquid NH3. The NH3 molecules have permanent dipole moments that are easily oriented in the liquid state."
|
||
}
|
||
]
|
||
处理第 607/821 条数据...
|
||
[
|
||
{
|
||
"question": "For aluminum, the heat capacity at constant volume Cv at 30 K is 0.81 J/mol-K, and the Debye temperature is 375 K. Estimate the specific heat at 50 K.",
|
||
"answer": "For aluminum, Cv at 50 K may be approximated by Equation, since this temperature is significantly below the Debye temperature (375 K). The value of Cv at 30 K is used to determine the constant A as A = Cv/T^3 = 0.81 J/mol-K / (30 K)^3 = 3.00 × 10^-5 J/mol-K^4. Therefore, at 50 K, Cv = A T^3 = (3.00 × 10^-5 J/mol-K^4)(50 K)^3 = 3.75 J/mol-K. The specific heat cv is then calculated as cv = (3.75 J/mol-K)(1 mol/26.98 g)(1000 g/kg) = 139 J/kg-K."
|
||
},
|
||
{
|
||
"question": "For aluminum, the heat capacity at constant volume Cv at 30 K is 0.81 J/mol-K, and the Debye temperature is 375 K. Estimate the specific heat at 425 K.",
|
||
"answer": "Since 425 K is above the Debye temperature, a good approximation for Cv is Cv = 3 R = (3)(8.31 J/mol-K) = 24.9 J/mol-K. The specific heat cv is then calculated as cv = (24.9 J/mol-K)(1 mol/26.98 g)(1000 g/kg) = 923 J/kg-K."
|
||
}
|
||
]
|
||
处理第 608/821 条数据...
|
||
[
|
||
{
|
||
"question": "Briefly explain why Cv rises with increasing temperature at temperatures near 0 K.",
|
||
"answer": "The reason that Cv rises with increasing temperature at temperatures near 0 K is because, in this temperature range, the allowed vibrational energy levels of the lattice waves are far apart relative to the available thermal energy, and only a portion of the lattice waves may be excited. As temperature increases, more of the lattice waves may be excited by the available thermal energy, and, hence, the ability of the solid to absorb energy (i.e., the magnitude of the heat capacity) increases."
|
||
},
|
||
{
|
||
"question": "Briefly explain why Cv becomes virtually independent of temperature at temperatures far removed from 0 K.",
|
||
"answer": "At temperatures far removed from 0 K, Cv becomes independent of temperature because all of the lattice waves have been excited and the energy required to produce an incremental temperature change is nearly constant."
|
||
}
|
||
]
|
||
处理第 609/821 条数据...
|
||
It's a single issue.
|
||
处理第 610/821 条数据...
|
||
It's a single issue.
|
||
处理第 611/821 条数据...
|
||
It's a single issue.
|
||
处理第 612/821 条数据...
|
||
[
|
||
{
|
||
"question": "Briefly explain why porosity decreases the thermal conductivity of ceramic and polymeric materials, rendering them more thermally insulative.",
|
||
"answer": "Porosity decreases the thermal conductivity of ceramic and polymeric materials because the thermal conductivity of a gas phase that occupies pore space is extremely small relative to that of the solid material. Furthermore, contributions from gaseous convection are generally insignificant."
|
||
},
|
||
{
|
||
"question": "Briefly explain how the degree of crystallinity affects the thermal conductivity of polymeric materials and why.",
|
||
"answer": "Increasing the degree of crystallinity of a semicrystalline polymer enhances its thermal conductivity; the vibrations, rotations, etc. of the molecular chains are more effective modes of thermal transport when a crystalline structure prevails."
|
||
}
|
||
]
|
||
处理第 613/821 条数据...
|
||
It's a single issue.
|
||
处理第 614/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For the pair of materials pure copper and aluminum bronze (95 wt% Cu-5 wt% Al), which has the larger thermal conductivity? Justify your choice.",
|
||
"answer": "Pure copper will have a larger conductivity than aluminum bronze because the impurity atoms in the latter will lead to a greater degree of free electron scattering."
|
||
},
|
||
{
|
||
"question": "For the pair of materials fused silica and quartz, which has the larger thermal conductivity? Justify your choice.",
|
||
"answer": "Quartz will have a larger conductivity than fused silica because fused silica is noncrystalline (whereas quartz is crystalline) and lattice vibrations are more effectively scattered in noncrystalline materials."
|
||
},
|
||
{
|
||
"question": "For the pair of materials linear polyethylene and branched polyethylene, which has the larger thermal conductivity? Justify your choice.",
|
||
"answer": "The linear polyethylene will have the larger conductivity than the branched polyethylene because the former will have the higher degree of crystallinity. Linear polymers have higher degrees of crystallinity than branched polymers. Since heat transfer is accomplished by molecular chain vibrations, and the coordination of these vibrations increases with percent crystallinity, the higher the crystallinity, the greater the thermal conductivity."
|
||
},
|
||
{
|
||
"question": "For the pair of materials random poly(styrene-butadiene) copolymer and alternating poly(styrene-butadiene) copolymer, which has the larger thermal conductivity? Justify your choice.",
|
||
"answer": "The alternating poly(styrene-butadiene) copolymer will have a higher crystallinity than the random copolymer; alternating copolymers crystallize more easily than random ones. The influence of crystallinity on conductivity is explained in part (c)."
|
||
}
|
||
]
|
||
```
|
||
处理第 615/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Explain the two sources of magnetic moments for electrons.",
|
||
"answer": "The two sources of magnetic moments for electrons are the electron's orbital motion around the nucleus, and also, its spin."
|
||
},
|
||
{
|
||
"question": "Do all electrons have a net magnetic moment? Why or why not?",
|
||
"answer": "Each electron will have a net magnetic moment from spin, and possibly, orbital contributions, which do not cancel for an isolated atom."
|
||
},
|
||
{
|
||
"question": "Do all atoms have a net magnetic moment? Why or why not?",
|
||
"answer": "All atoms do not have a net magnetic moment. If an atom has completely filled electron shells or subshells, there will be a cancellation of both orbital and spin magnetic moments."
|
||
}
|
||
]
|
||
```
|
||
处理第 616/821 条数据...
|
||
It's a single issue.
|
||
处理第 617/821 条数据...
|
||
It's a single issue.
|
||
处理第 618/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Calculate the number of Bohr magnetons per unit cell (n_B) for yttrium iron garnet given the saturation magnetization (1.0 x 10^4 A/m), cubic unit cell edge length (1.2376 nm), and Bohr magneton value (9.27 x 10^-24 A·m^2/BM).",
|
||
"answer": "n_B = (M_s * a^3) / μ_B = (1.0 x 10^4 A/m) * (1.2376 x 10^-9 m)^3 / (9.27 x 10^-24 A·m^2/BM) = 2.04 Bohr magnetons/unit cell."
|
||
},
|
||
{
|
||
"question": "Determine the number of Bohr magnetons per formula unit for yttrium iron garnet, given there are 8 formula units per unit cell and the previously calculated n_B value (2.04 Bohr magnetons/unit cell).",
|
||
"answer": "Bohr magnetons per formula unit = n_B / 8 = 2.04 / 8 = 0.255 Bohr magnetons/formula unit."
|
||
},
|
||
{
|
||
"question": "Calculate the net magnetic moment contribution per formula unit from the Fe^3+ ions in yttrium iron garnet, given there are 2 Fe^3+ ions on a sites and 3 Fe^3+ ions on d sites per formula unit, each with 5 Bohr magnetons, and their magnetic moments are aligned antiparallel.",
|
||
"answer": "Net magnetic moment from Fe^3+ ions = (2 * 5 BM) - (3 * 5 BM) = 10 BM - 15 BM = -5 Bohr magnetons/formula unit (antiparallel to Y^3+ contribution)."
|
||
},
|
||
{
|
||
"question": "Compute the number of Bohr magnetons associated with each Y^3+ ion in yttrium iron garnet, given the Bohr magnetons per formula unit (0.255 BM), the net magnetic moment from Fe^3+ ions (-5 BM), and there are 3 Y^3+ ions per formula unit.",
|
||
"answer": "Number of Bohr magnetons/Y^3+ = (0.255 BM + 5 BM) / 3 = 5.255 BM / 3 = 1.75 Bohr magnetons/Y^3+ ion."
|
||
}
|
||
]
|
||
```
|
||
处理第 619/821 条数据...
|
||
[
|
||
{
|
||
"question": "Why does the magnitude of the saturation magnetization decrease with increasing temperature for ferromagnetic materials?",
|
||
"answer": "The saturation magnetization decreases with increasing temperature because the atomic thermal vibrational motions counteract the coupling forces between the adjacent atomic dipole moments, causing some magnetic dipole misalignment."
|
||
},
|
||
{
|
||
"question": "Why does ferromagnetic behavior cease above the Curie temperature for ferromagnetic materials?",
|
||
"answer": "Ferromagnetic behavior ceases above the Curie temperature because the atomic thermal vibrations are sufficiently violent so as to completely destroy the mutual spin coupling forces."
|
||
}
|
||
]
|
||
处理第 620/821 条数据...
|
||
It's a single issue.
|
||
处理第 621/821 条数据...
|
||
[
|
||
{
|
||
"question": "Visible light having a wavelength of 6 × 10^-7 m appears orange. Compute the frequency of a photon of this light.",
|
||
"answer": "v = c / λ = (3 × 10^8 m/s) / (6 × 10^-7 m) = 5 × 10^14 s^-1"
|
||
},
|
||
{
|
||
"question": "Visible light having a wavelength of 6 × 10^-7 m appears orange. Compute the energy of a photon of this light.",
|
||
"answer": "E = hc / λ = (6.63 × 10^-34 J·s × 3 × 10^8 m/s) / (6 × 10^-7 m) = 3.31 × 10^-19 J (2.07 eV)"
|
||
}
|
||
]
|
||
处理第 622/821 条数据...
|
||
[
|
||
{
|
||
"question": "What are the characteristics of opaque materials in terms of their appearance and light transmittance?",
|
||
"answer": "Opaque materials are impervious to light transmission; it is not possible to see through them."
|
||
},
|
||
{
|
||
"question": "What are the characteristics of translucent materials in terms of their appearance and light transmittance?",
|
||
"answer": "Light is transmitted diffusely through translucent materials (there is some internal light scattering). Objects are not clearly distinguishable when viewed through a translucent material."
|
||
},
|
||
{
|
||
"question": "What are the characteristics of transparent materials in terms of their appearance and light transmittance?",
|
||
"answer": "Virtually all of the incident light is transmitted through transparent materials, and one can see clearly through them."
|
||
}
|
||
]
|
||
处理第 623/821 条数据...
|
||
It's a single issue.
|
||
处理第 624/821 条数据...
|
||
It's a single issue.
|
||
处理第 625/821 条数据...
|
||
It's a single issue.
|
||
处理第 626/821 条数据...
|
||
It's a single issue.
|
||
处理第 627/821 条数据...
|
||
It's a single issue.
|
||
处理第 628/821 条数据...
|
||
[
|
||
{
|
||
"question": "What determines the characteristic color of a metal?",
|
||
"answer": "The characteristic color of a metal is determined by the distribution of wavelengths of the nonabsorbed light radiation that is reflected."
|
||
},
|
||
{
|
||
"question": "What determines the characteristic color of a transparent nonmetal?",
|
||
"answer": "The characteristic color of a transparent nonmetal is determined by the distribution of wavelengths of the nonabsorbed light radiation that is transmitted through the material."
|
||
}
|
||
]
|
||
处理第 629/821 条数据...
|
||
It's a single issue.
|
||
处理第 630/821 条数据...
|
||
It's a single issue.
|
||
处理第 631/821 条数据...
|
||
It's a single issue.
|
||
处理第 632/821 条数据...
|
||
It's a single issue.
|
||
处理第 633/821 条数据...
|
||
[
|
||
{
|
||
"question": "The K+ ion has an electron structure that is identical to which inert gas?",
|
||
"answer": "The K+ ion has an electron configuration the same as argon."
|
||
},
|
||
{
|
||
"question": "The I- ion has an electron structure that is identical to which inert gas?",
|
||
"answer": "The I- ion has an electron configuration the same as xenon."
|
||
}
|
||
]
|
||
处理第 634/821 条数据...
|
||
It's a single issue.
|
||
处理第 635/821 条数据...
|
||
It's a single issue.
|
||
处理第 636/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Determine whether the electron configuration \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5}\) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.",
|
||
"answer": "The \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5}\) electron configuration is that of a halogen because it is one electron deficient from having a filled \(p\) subshell."
|
||
},
|
||
{
|
||
"question": "Determine whether the electron configuration \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7} 4 s^{2}\) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.",
|
||
"answer": "The \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7} 4 s^{2}\) electron configuration is that of a transition metal because of an incomplete \(d\) subshell."
|
||
},
|
||
{
|
||
"question": "Determine whether the electron configuration \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6}\) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.",
|
||
"answer": "The \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6}\) electron configuration is that of an inert gas because of filled \(4 s\) and \(4 p\) subshells."
|
||
},
|
||
{
|
||
"question": "Determine whether the electron configuration \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}\) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.",
|
||
"answer": "The \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}\) electron configuration is that of an alkali metal because of a single \(s\) electron."
|
||
},
|
||
{
|
||
"question": "Determine whether the electron configuration \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{5} 5 s^{2}\) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.",
|
||
"answer": "The \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{5} 5 s^{2}\) electron configuration is that of a transition metal because of an incomplete \(d\) subshell."
|
||
},
|
||
{
|
||
"question": "Determine whether the electron configuration \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}\) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.",
|
||
"answer": "The \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}\) electron configuration is that of an alkaline earth metal because of two \(s\) electrons."
|
||
}
|
||
]
|
||
```
|
||
API调用错误: Invalid \escape: line 3 column 67 (char 74)
|
||
处理第 637/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Differentiate the net potential energy equation \(E_{N}=\frac{A}{r}+\frac{B}{r^{n}}\) with respect to \(r\) and set the resulting expression equal to zero to find the condition for the minimum energy.",
|
||
"answer": "Differentiation of the equation yields \(\frac{d E_{N}}{d r} = \frac{A}{r^{2}} - \frac{n B}{r^{(n+1)}} = 0\)."
|
||
},
|
||
{
|
||
"question": "Solve the differentiated equation \(\frac{A}{r_{0}^{2}} = \frac{n B}{r_{0}^{(n+1)}}\) for \(r\) in terms of \(A, B\), and \(n\) to find the equilibrium interionic spacing \(r_{0}\).",
|
||
"answer": "Solving for \(r_{0}\) gives \(r_{0} = \left(\frac{A}{n B}\right)^{1/(1-n)}\)."
|
||
},
|
||
{
|
||
"question": "Substitute \(r_{0}\) back into the original net potential energy equation \(E_{N}=\frac{A}{r}+\frac{B}{r^{n}}\) to determine the expression for the bonding energy \(E_{0}\) in terms of \(A, B\), and \(n\).",
|
||
"answer": "Substituting \(r_{0}\) into the equation yields \(E_{0} = -\frac{A}{\left(\frac{A}{n B}\right)^{1/(1-n)}} + \frac{B}{\left(\frac{A}{n B}\right)^{n/(1-n)}}\)."
|
||
}
|
||
]
|
||
```
|
||
API调用错误: Invalid \escape: line 3 column 70 (char 77)
|
||
处理第 638/821 条数据...
|
||
It's a single issue.
|
||
处理第 639/821 条数据...
|
||
[
|
||
{
|
||
"question": "What type(s) of bonding would be expected for solid xenon?",
|
||
"answer": "For solid xenon, the bonding is van der Waals since xenon is an inert gas."
|
||
},
|
||
{
|
||
"question": "What type(s) of bonding would be expected for calcium fluoride (CaF2)?",
|
||
"answer": "For CaF2, the bonding is predominantly ionic (but with some slight covalent character) on the basis of the relative positions of Ca and F in the periodic table."
|
||
},
|
||
{
|
||
"question": "What type(s) of bonding would be expected for bronze?",
|
||
"answer": "For bronze, the bonding is metallic since it is a metal alloy (composed of copper and tin)."
|
||
},
|
||
{
|
||
"question": "What type(s) of bonding would be expected for cadmium telluride (CdTe)?",
|
||
"answer": "For CdTe, the bonding is predominantly covalent (with some slight ionic character) on the basis of the relative positions of Cd and Te in the periodic table."
|
||
},
|
||
{
|
||
"question": "What type(s) of bonding would be expected for rubber?",
|
||
"answer": "For rubber, the bonding is covalent with some van der Waals. (Rubber is composed primarily of carbon and hydrogen atoms.)"
|
||
},
|
||
{
|
||
"question": "What type(s) of bonding would be expected for tungsten?",
|
||
"answer": "For tungsten, the bonding is metallic since it is a metallic element from the periodic table."
|
||
}
|
||
]
|
||
处理第 640/821 条数据...
|
||
It's a single issue.
|
||
处理第 641/821 条数据...
|
||
It's a single issue.
|
||
处理第 642/821 条数据...
|
||
It's a single issue.
|
||
处理第 643/821 条数据...
|
||
It's a single issue.
|
||
处理第 644/821 条数据...
|
||
It's a single issue.
|
||
处理第 645/821 条数据...
|
||
It's a single issue.
|
||
处理第 646/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Titanium (Ti) has an HCP crystal structure and a density of 4.51 g/cm³. What is the volume of its unit cell in cubic meters?",
|
||
"answer": "The volume of the Ti unit cell may be computed using a rearranged form of Equation as V_C = (n A_Ti) / (ρ N_A). For HCP, n = 6 atoms/unit cell, and the atomic weight for Ti, A_Ti = 47.87 g/mol. Thus, V_C = (6 atoms/unit cell)(47.87 g/mol) / (4.51 g/cm³)(6.022 × 10^23 atoms/mol) = 1.058 × 10^-22 cm³/unit cell = 1.058 × 10^-28 m³/unit cell."
|
||
},
|
||
{
|
||
"question": "Titanium (Ti) has an HCP crystal structure and a density of 4.51 g/cm³. If the c/a ratio is 1.58, compute the values of c and a.",
|
||
"answer": "From Equation, for HCP the unit cell volume is V_C = (3 a² c √3) / 2. Since, for Ti, c = 1.58 a, then V_C = (3(1.58) a³ √3) / 2 = 1.058 × 10^-22 cm³/unit cell. Now, solving for a: a = [(2)(1.058 × 10^-22 cm³) / (3)(1.58)(√3)]^(1/3) = 2.96 × 10^-8 cm = 0.296 nm. And finally, the value of c is c = 1.58 a = (1.58)(0.296 nm) = 0.468 nm."
|
||
}
|
||
]
|
||
```
|
||
处理第 647/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Magnesium (Mg) has an HCP crystal structure and a density of 1.74 g/cm³. What is the volume of its unit cell in cubic centimeters?",
|
||
"answer": "The volume of the Mg unit cell may be computed using a rearranged form of the equation as V_C = (n A_Mg) / (ρ N_A). For HCP, n = 6 atoms/unit cell, and the atomic weight for Mg, A_Mg = 24.31 g/mol. Thus, V_C = (6 atoms/unit cell)(24.31 g/mol) / (1.74 g/cm³)(6.022 × 10²³ atoms/mol) = 1.39 × 10⁻²² cm³/unit cell."
|
||
},
|
||
{
|
||
"question": "Magnesium (Mg) has an HCP crystal structure, a density of 1.74 g/cm³, and a c/a ratio of 1.624. Compute the values of c and a.",
|
||
"answer": "From the equation for HCP, the unit cell volume is V_C = (3 a² c √3) / 2. Given c = 1.624 a and V_C = 1.39 × 10⁻²² cm³/unit cell, solving for a gives a = [(2)(1.39 × 10⁻²² cm³) / (3)(1.624)(√3)]^(1/3) = 3.21 × 10⁻⁸ cm = 0.321 nm. The value of c is c = 1.624 a = (1.624)(0.321 nm) = 0.521 nm."
|
||
}
|
||
]
|
||
```
|
||
处理第 648/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Given the orthorhombic unit cell parameters for uranium (U) as a = 0.286 nm, b = 0.587 nm, and c = 0.495 nm, calculate the unit cell volume (V_C).",
|
||
"answer": "The unit cell volume (V_C) is calculated as the product of the three lattice parameters: V_C = a * b * c = (0.286 nm) * (0.587 nm) * (0.495 nm) = 0.286 * 0.587 * 0.495 * 10^-21 cm^3 = 8.31 * 10^-23 cm^3."
|
||
},
|
||
{
|
||
"question": "Using the density (19.05 g/cm^3), atomic weight (238.03 g/mol), and the calculated unit cell volume (V_C = 8.31 * 10^-23 cm^3) for uranium (U), determine the number of atoms per unit cell (n).",
|
||
"answer": "The number of atoms per unit cell (n) is calculated using the formula: n = (density * V_C * N_A) / atomic weight = (19.05 g/cm^3 * 8.31 * 10^-23 cm^3 * 6.022 * 10^23 atoms/mol) / 238.03 g/mol = 4.01 atoms/unit cell."
|
||
},
|
||
{
|
||
"question": "Given the atomic radius of uranium (U) as 0.1385 nm, calculate the volume of a single uranium atom (V_atom).",
|
||
"answer": "The volume of a single uranium atom (V_atom) is calculated using the formula for the volume of a sphere: V_atom = (4/3) * π * R^3 = (4/3) * π * (0.1385 nm)^3 = (4/3) * π * (1.385 * 10^-8 cm)^3 = 1.11 * 10^-23 cm^3."
|
||
},
|
||
{
|
||
"question": "Using the number of atoms per unit cell (n = 4.01) and the volume of a single uranium atom (V_atom = 1.11 * 10^-23 cm^3), calculate the total sphere volume (V_S) in the unit cell.",
|
||
"answer": "The total sphere volume (V_S) is calculated as the product of the number of atoms per unit cell and the volume of a single atom: V_S = n * V_atom = 4.01 * 1.11 * 10^-23 cm^3 = 4.45 * 10^-23 cm^3."
|
||
},
|
||
{
|
||
"question": "Using the total sphere volume (V_S = 4.45 * 10^-23 cm^3) and the unit cell volume (V_C = 8.31 * 10^-23 cm^3), compute the atomic packing factor (APF) for uranium (U).",
|
||
"answer": "The atomic packing factor (APF) is calculated as the ratio of the total sphere volume to the unit cell volume: APF = V_S / V_C = (4.45 * 10^-23 cm^3) / (8.31 * 10^-23 cm^3) = 0.536."
|
||
}
|
||
]
|
||
```
|
||
处理第 649/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Indium (In) has a tetragonal unit cell for which the a and c lattice parameters are 0.459 and 0.495 nm, respectively. If the atomic packing factor and atomic radius are 0.693 and 0.1625 nm, respectively, determine the number of atoms in each unit cell.",
|
||
"answer": "For indium, and from the definition of the APF is APF = VS / VC = n(4/3 πR3) / a2c we may solve for the number of atoms per unit cell, n, as n = (APF) a2 c / (4/3 πR3) = (0.693)(4.59)2(4.95)(10^-24 cm3) / (4/3 π(1.625 × 10^-8 cm)3) = 4.0 atoms/unit cell"
|
||
},
|
||
{
|
||
"question": "The atomic weight of indium is 114.82 g/mol; compute its theoretical density using the tetragonal unit cell parameters a = 0.459 nm and c = 0.495 nm, and the number of atoms per unit cell determined to be 4.0.",
|
||
"answer": "In order to compute the density, we just employ the equation ρ = n AIn / (a2 c NA) = (4 atoms/unit cell)(114.82 g/mol) / [(4.59 × 10^-8 cm)2(4.95 × 10^-8 cm)/unit cell](6.022 × 10^23 atoms/mol) = 7.31 g/cm3"
|
||
}
|
||
]
|
||
```
|
||
处理第 650/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Beryllium (Be) has an HCP unit cell for which the ratio of the lattice parameters c/a is 1.568. If the radius of the Be atom is 0.1143 nm, determine the unit cell volume.",
|
||
"answer": "The volume of an HCP unit cell is given by V_C = 6 R^2 c sqrt(3). Given c = 1.568 a and a = 2 R, then c = (1.568)(2 R) = 3.136 R. Substituting this expression for c into the volume equation leads to V_C = (6)(3.14) R^3 sqrt(3) = (6)(3.14)(sqrt(3))[0.1143 × 10^-7 cm]^3 = 4.87 × 10^-23 cm^3/unit cell."
|
||
},
|
||
{
|
||
"question": "Using the unit cell volume calculated previously (4.87 × 10^-23 cm^3/unit cell) and given that Be has an atomic weight of 9.01 g/mol, calculate the theoretical density of Be and compare it with the literature value.",
|
||
"answer": "The theoretical density of Be is determined using the formula rho = (n A_Be)/(V_C N_A). For HCP, n = 6 atoms/unit cell, and for Be, A_Be = 9.01 g/mol. Using the value of V_C = 4.87 × 10^-23 cm^3/unit cell, the theoretical density of Be is rho = (6 atoms/unit cell)(9.01 g/mol)/(4.87 × 10^-23 cm^3/unit cell)(6.022 × 10^23 atoms/mol) = 1.84 g/cm^3. The literature value is 1.85 g/cm^3."
|
||
}
|
||
]
|
||
```
|
||
处理第 651/821 条数据...
|
||
It's a single issue.
|
||
处理第 652/821 条数据...
|
||
It's a single issue.
|
||
处理第 653/821 条数据...
|
||
It's a single issue.
|
||
处理第 654/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Given a hypothetical AX type of ceramic material with a density of 2.10 g/cm³, a cubic unit cell edge length of 0.57 nm, and atomic weights of A and X elements as 28.5 g/mol and 30.0 g/mol respectively, calculate the number of formula units per unit cell (n').",
|
||
"answer": "The number of formula units per unit cell (n') is calculated as follows: n' = (ρ * V_C * N_A) / (ΣA_C + ΣA_A) = (2.10 g/cm³ * (5.70 × 10⁻⁸ cm)³/unit cell * 6.022 × 10²³ formula units/mol) / (30.0 g/mol + 28.5 g/mol) = 4.00 formula units/unit cell."
|
||
},
|
||
{
|
||
"question": "Based on the calculated number of formula units per unit cell (n' = 4.00), which of the following crystal structures are possible for the AX type of ceramic material: sodium chloride, cesium chloride, or zinc blende?",
|
||
"answer": "For n' = 4.00 formula units/unit cell, the possible crystal structures are sodium chloride and zinc blende, as cesium chloride has n' = 1."
|
||
}
|
||
]
|
||
```
|
||
处理第 655/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Iron titanate, FeTiO3, forms in the ilmenite crystal structure that consists of an HCP arrangement of O2- ions. Which type of interstitial site will the Fe2+ ions occupy? Why?",
|
||
"answer": "From Table, the cation-anion radius ratio is r_Fe2+/r_O2- = 0.077 nm / 0.140 nm = 0.550. Since this ratio is between 0.414 and 0.732, the Fe2+ ions will occupy octahedral sites."
|
||
},
|
||
{
|
||
"question": "Iron titanate, FeTiO3, forms in the ilmenite crystal structure that consists of an HCP arrangement of O2- ions. Which type of interstitial site will the Ti4+ ions occupy? Why?",
|
||
"answer": "The titanium-oxygen ionic radius ratio is r_Ti4+/r_O2- = 0.0601 nm / 0.140 nm = 0.436. Since this ratio is between 0.414 and 0.723, the Ti4+ ions will also occupy octahedral sites."
|
||
},
|
||
{
|
||
"question": "Iron titanate, FeTiO3, forms in the ilmenite crystal structure that consists of an HCP arrangement of O2- ions. What fraction of the total tetrahedral sites will be occupied?",
|
||
"answer": "Since both Fe2+ and Ti4+ ions occupy octahedral sites, no tetrahedral sites will be occupied."
|
||
},
|
||
{
|
||
"question": "Iron titanate, FeTiO3, forms in the ilmenite crystal structure that consists of an HCP arrangement of O2- ions. What fraction of the total octahedral sites will be occupied?",
|
||
"answer": "For every FeTiO3 formula unit, there are three O2- ions, and therefore three octahedral sites; since there is one ion each of Fe2+ and Ti4+ per formula unit, two-thirds of these octahedral sites will be occupied."
|
||
}
|
||
]
|
||
```
|
||
处理第 656/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "To what equation does the expression for interplanar spacing \(d_{\text {hkl }}\) reduce for crystals having cubic symmetry, given the general orthorhombic formula \(\frac{1}{d_{h k l}^{2}}=\frac{h^{2}}{a^{2}}+\frac{k^{2}}{b^{2}}+\frac{l^{2}}{c^{2}}\) where \(a, b\), and \(c\) are the lattice parameters?",
|
||
"answer": "For the crystals having cubic symmetry, \(a=b=c\). Making this substitution into the above equation leads to \(\frac{1}{d_{h k l}^{2}}=\frac{h^{2}}{a^{2}}+\frac{k^{2}}{a^{2}}+\frac{l^{2}}{a^{2}}\) which simplifies to \(\frac{h^{2}+k^{2}+l^{2}}{a^{2}}\)."
|
||
},
|
||
{
|
||
"question": "To what equation does the expression for interplanar spacing \(d_{\text {hkl }}\) reduce for crystals having tetragonal symmetry, given the general orthorhombic formula \(\frac{1}{d_{h k l}^{2}}=\frac{h^{2}}{a^{2}}+\frac{k^{2}}{b^{2}}+\frac{l^{2}}{c^{2}}\) where \(a, b\), and \(c\) are the lattice parameters?",
|
||
"answer": "For crystals having tetragonal symmetry, \(a=b \neq c\). Replacing \(b\) with \(a\) in the equation found in the problem statement leads to \(\frac{1}{d_{h k l}^{2}}=\frac{h^{2}}{a^{2}}+\frac{k^{2}}{a^{2}}+\frac{l^{2}}{c^{2}}\) which simplifies to \(\frac{h^{2}+k^{2}}{a^{2}}+\frac{l^{2}}{c^{2}}\)."
|
||
}
|
||
]
|
||
```
|
||
API调用错误: Invalid \escape: line 3 column 83 (char 90)
|
||
处理第 657/821 条数据...
|
||
It's a single issue.
|
||
处理第 658/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "High-density polyethylene may be chlorinated by inducing the random substitution of chlorine atoms for hydrogen. Determine the concentration of Cl (in wt%) that must be added if this substitution occurs for 8% of all the original hydrogen atoms.",
|
||
"answer": "For chlorinated polyethylene, we are asked to determine the weight percent of chlorine added for 8% Cl substitution of all original hydrogen atoms. Consider 50 carbon atoms; for polyethylene there are 100 possible side-bonding sites. Ninety-two are occupied by hydrogen and eight are occupied by Cl. Thus, the mass of these 50 carbon atoms, m_C, is just m_C=50(A_C)=(50)(12.01 g/mol)=600.5 g. Likewise, for hydrogen and chlorine, m_H=92(A_H)=(92)(1.008 g/mol)=92.74 g, m_Cl=8(A_Cl)=(8)(35.45 g/mol)=283.60 g. Thus, the concentration of chlorine, C_Cl, is determined as follows: C_Cl=(m_Cl)/(m_C+m_H+m_Cl)×100=(283.60 g)/(600.5 g+92.74 g+283.60 g)×100=29.0 wt%."
|
||
},
|
||
{
|
||
"question": "In what ways does chlorinated polyethylene differ from poly(vinyl chloride)?",
|
||
"answer": "Chlorinated polyethylene differs from poly(vinyl chloride), in that, for PVC, (1) 25% of the side-bonding sites are substituted with Cl, and (2) the substitution is probably much less random."
|
||
}
|
||
]
|
||
```
|
||
处理第 659/821 条数据...
|
||
[
|
||
{
|
||
"question": "Compare thermoplastic and thermosetting polymers on the basis of mechanical characteristics upon heating.",
|
||
"answer": "Thermoplastic polymers soften when heated and harden when cooled, whereas thermosetting polymers harden upon heating, while further heating will not lead to softening."
|
||
},
|
||
{
|
||
"question": "Compare thermoplastic and thermosetting polymers according to possible molecular structures.",
|
||
"answer": "Thermoplastic polymers have linear and branched structures, while for thermosetting polymers, the structures will normally be network or crosslinked."
|
||
}
|
||
]
|
||
处理第 660/821 条数据...
|
||
It's a single issue.
|
||
处理第 661/821 条数据...
|
||
It's a single issue.
|
||
处理第 662/821 条数据...
|
||
It's a single issue.
|
||
处理第 663/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Determine the ratio of butadiene to acrylonitrile repeat units in a copolymer having a number-average molecular weight of 250,000 g/mol and a degree of polymerization of 4640.",
|
||
"answer": "To determine the ratio of butadiene to acrylonitrile repeat units, first calculate the average repeat unit molecular weight of the copolymer, m_bar, using the formula: m_bar = M_n / DP = 250,000 g/mol / 4640 = 53.88 g/mol. Designate f_b as the chain fraction of butadiene repeat units, so the chain fraction of acrylonitrile repeat units f_a is 1 - f_b. The repeat unit molecular weights are: m_b = 4(12.01 g/mol) + 6(1.008 g/mol) = 54.09 g/mol for butadiene and m_a = 3(12.01 g/mol) + 3(1.008 g/mol) + 14.01 g/mol = 53.06 g/mol for acrylonitrile. Solve for f_b using the equation: f_b = (m_bar - m_a) / (m_b - m_a) = (53.88 g/mol - 53.06 g/mol) / (54.09 g/mol - 53.06 g/mol) = 0.80. Then, f_a = 1 - 0.80 = 0.20. The ratio is f_b / f_a = 0.80 / 0.20 = 4.0."
|
||
},
|
||
{
|
||
"question": "Which type(s) of copolymer(s) will this copolymer be, considering the following possibilities: random, alternating, graft, and block? Why?",
|
||
"answer": "The possible copolymers are random, graft, and block. The alternating copolymer is excluded because it requires a 1:1 ratio of repeat unit types, which is not the case here (the ratio is 4:1)."
|
||
}
|
||
]
|
||
```
|
||
处理第 664/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For a copolymer consisting of 35 wt% ethylene and 65 wt% propylene, determine the molecular weight of the ethylene repeat unit (C2H4).",
|
||
"answer": "The molecular weight of ethylene (C2H4) is calculated as follows: m(ethylene) = 2(12.01 g/mol) + 4(1.008 g/mol) = 28.05 g/mol."
|
||
},
|
||
{
|
||
"question": "For a copolymer consisting of 35 wt% ethylene and 65 wt% propylene, determine the molecular weight of the propylene repeat unit (C3H6).",
|
||
"answer": "The molecular weight of propylene (C3H6) is calculated as follows: m(propylene) = 3(12.01 g/mol) + 6(1.008 g/mol) = 42.08 g/mol."
|
||
},
|
||
{
|
||
"question": "For a copolymer consisting of 35 wt% ethylene and 65 wt% propylene, determine the number of moles of ethylene repeat units in 100 g of the material.",
|
||
"answer": "The number of moles of ethylene in 100 g of the material is calculated as follows: n_m(ethylene) = 35 g / 28.05 g/mol = 1.25 mol."
|
||
},
|
||
{
|
||
"question": "For a copolymer consisting of 35 wt% ethylene and 65 wt% propylene, determine the number of moles of propylene repeat units in 100 g of the material.",
|
||
"answer": "The number of moles of propylene in 100 g of the material is calculated as follows: n_m(propylene) = 65 g / 42.08 g/mol = 1.54 mol."
|
||
},
|
||
{
|
||
"question": "For a copolymer consisting of 35 wt% ethylene and 65 wt% propylene, determine the fraction of ethylene repeat units.",
|
||
"answer": "The fraction of ethylene repeat units is calculated as follows: f(ethylene) = 1.25 mol / (1.25 mol + 1.54 mol) = 0.45."
|
||
},
|
||
{
|
||
"question": "For a copolymer consisting of 35 wt% ethylene and 65 wt% propylene, determine the fraction of propylene repeat units.",
|
||
"answer": "The fraction of propylene repeat units is calculated as follows: f(propylene) = 1.54 mol / (1.25 mol + 1.54 mol) = 0.55."
|
||
}
|
||
]
|
||
```
|
||
处理第 665/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For the pair of polymers: linear and atactic poly(vinyl chloride); linear and isotactic polypropylene, (1) State whether it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why.",
|
||
"answer": "No, it is not possible to decide for these two polymers. On the basis of tacticity, the isotactic PP is more likely to crystallize than the atactic PVC. On the other hand, with regard to side-group bulkiness, the PVC is more likely to crystallize."
|
||
},
|
||
{
|
||
"question": "For the pair of polymers: linear and syndiotactic polypropylene; crosslinked cis-polyisoprene, (1) State whether it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why.",
|
||
"answer": "Yes, it is possible to decide for these two copolymers. The linear and syndiotactic polypropylene is more likely to crystallize than crosslinked cis-isoprene because linear polymers are more likely to crystallize than crosslinked ones."
|
||
},
|
||
{
|
||
"question": "For the pair of polymers: network phenol-formaldehyde; linear and isotactic polystyrene, (1) State whether it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why.",
|
||
"answer": "Yes, it is possible to decide for these two polymers. The linear and isotactic polystyrene is more likely to crystallize than network phenol-formaldehyde; network polymers rarely crystallize, whereas isotactic ones crystallize relatively easily."
|
||
},
|
||
{
|
||
"question": "For the pair of polymers: block poly(acrylonitrile-isoprene) copolymer; graft poly(chloroprene-isobutylene) copolymer, (1) State whether it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why.",
|
||
"answer": "Yes, it is possible to decide for these two copolymers. The block poly(acrylonitrileisoprene) copolymer is more likely to crystallize than the graft poly(chloroprene-isobutylene) copolymer. Block copolymers crystallize more easily than graft ones."
|
||
}
|
||
]
|
||
```
|
||
处理第 666/821 条数据...
|
||
It's a single issue.
|
||
处理第 667/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Calculate the fraction of atom sites that are vacant for copper (Cu) at its melting temperature of 1084 degrees Celsius (1357 K). Assume an energy for vacancy formation of 0.90 eV/atom.",
|
||
"answer": "To compute the fraction of atom sites that are vacant in copper at 1357 K, we use the equation: NV/N = exp(-QV/kT). Given QV = 0.90 eV/atom, the calculation is as follows: NV/N = exp[-0.90 eV/atom / (8.62 × 10^-5 eV/atom-K × 1357 K)] = 4.56 × 10^-4."
|
||
},
|
||
{
|
||
"question": "Calculate the fraction of atom sites that are vacant for copper (Cu) at room temperature (298 K). Assume an energy for vacancy formation of 0.90 eV/atom.",
|
||
"answer": "To compute the fraction of atom sites that are vacant in copper at 298 K, we use the equation: NV/N = exp(-QV/kT). Given QV = 0.90 eV/atom, the calculation is as follows: NV/N = exp[-0.90 eV/atom / (8.62 × 10^-5 eV/atom-K × 298 K)] = 6.08 × 10^-16."
|
||
},
|
||
{
|
||
"question": "What is the ratio of NV/N(1357 K) and NV/N(298 K) for copper (Cu)?",
|
||
"answer": "The ratio of NV/N(1357 K) and NV/N(298 K) is calculated as follows: (4.56 × 10^-4) / (6.08 × 10^-16) = 7.5 × 10^11."
|
||
}
|
||
]
|
||
```
|
||
处理第 668/821 条数据...
|
||
It's a single issue.
|
||
处理第 669/821 条数据...
|
||
It's a single issue.
|
||
处理第 670/821 条数据...
|
||
It's a single issue.
|
||
处理第 671/821 条数据...
|
||
[
|
||
{
|
||
"question": "If cupric oxide (CuO) is exposed to reducing atmospheres at elevated temperatures, some of the Cu2+ ions will become Cu+. Under these conditions, name one crystalline defect that you would expect to form in order to maintain charge neutrality.",
|
||
"answer": "For a Cu2+ O2- compound in which a small fraction of the copper ions exist as Cu+, for each Cu+ formed there is one less positive charge introduced (or one more negative charge). In order to maintain charge neutrality, we must either add an additional positive charge or subtract a negative charge. This may be accomplished be either creating Cu2+ interstitials or O2- vacancies."
|
||
},
|
||
{
|
||
"question": "If cupric oxide (CuO) is exposed to reducing atmospheres at elevated temperatures, some of the Cu2+ ions will become Cu+. How many Cu+ ions are required for the creation of each defect?",
|
||
"answer": "There will be two Cu+ ions required for each of these defects."
|
||
},
|
||
{
|
||
"question": "If cupric oxide (CuO) is exposed to reducing atmospheres at elevated temperatures, some of the Cu2+ ions will become Cu+. How would you express the chemical formula for this nonstoichiometric material?",
|
||
"answer": "The chemical formula for this nonstoichiometric material is Cu1+ O or CuO1-x, where x is some small fraction."
|
||
}
|
||
]
|
||
处理第 672/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Which of the following oxides would you expect to form substitutional solid solutions that have complete (i.e., 100%) solubility with MgO? Explain your answer for FeO.",
|
||
"answer": "In order to have a high degree (or 100% solubility) of one ceramic compound in another ceramic compound, the following criteria must be met: the relative sizes of cations/anions must be similar (< ±15%); the valence (or charges) on the cations and anions of both compounds must be the same; and the crystal structures of the two compounds must be the same or similar. For FeO, the ionic radii of the Mg2+ and Fe2+ are 0.072 nm and 0.077 nm, respectively. The percentage difference in ionic radii, Δr%, is calculated as follows: Δr% = (0.077 nm - 0.072 nm) / 0.077 nm × 100 = 6.5%, which is within the acceptable range for a high degree of solubility. Additionally, the crystal structures of both MgO and FeO are rock salt. Therefore, FeO and MgO are expected to form high degrees of solid solubility, and experimentally, these two ceramics exhibit 100% solubility."
|
||
},
|
||
{
|
||
"question": "Which of the following oxides would you expect to form substitutional solid solutions that have complete (i.e., 100%) solubility with MgO? Explain your answer for BaO.",
|
||
"answer": "For BaO, the ionic radii of the Mg2+ and Ba2+ are 0.072 nm and 0.136 nm, respectively. The percentage difference in ionic radii, Δr%, is calculated as follows: Δr% = (0.136 nm - 0.072 nm) / 0.136 nm × 100 = 47%. This Δr% value is much larger than the ±15% range, and therefore, BaO is not expected to experience any appreciable solubility in MgO. Experimentally, the solubility of BaO in MgO is very small; this is also the case for the solubility of MgO in BaO."
|
||
},
|
||
{
|
||
"question": "Which of the following oxides would you expect to form substitutional solid solutions that have complete (i.e., 100%) solubility with MgO? Explain your answer for PbO.",
|
||
"answer": "For PbO, the ionic radii of the Mg2+ and Pb2+ are 0.072 nm and 0.120 nm, respectively. The percentage difference in ionic radii, Δr%, is calculated as follows: Δr% = (0.120 nm - 0.072 nm) / 0.120 nm × 100 = 40%. This Δr% value is much larger than the ±15% range, and therefore, PbO is not expected to experience any appreciable solubility in MgO."
|
||
},
|
||
{
|
||
"question": "Which of the following oxides would you expect to form substitutional solid solutions that have complete (i.e., 100%) solubility with MgO? Explain your answer for CoO.",
|
||
"answer": "For CoO, the ionic radii of the Mg2+ and Co2+ are 0.072 nm and 0.072 nm, respectively. The percentage difference in ionic radii, Δr%, is calculated as follows: Δr% = (0.072 nm - 0.072 nm) / 0.072 nm × 100 = 0%, which is within the acceptable range for a high degree (probably 100%) of solubility."
|
||
}
|
||
]
|
||
```
|
||
处理第 673/821 条数据...
|
||
[
|
||
{
|
||
"question": "Suppose that CaO is added as an impurity to Li2O. If the Ca2+ substitutes for Li+, what kind of vacancies would you expect to form? How many of these vacancies are created for every Ca2+ added?",
|
||
"answer": "For Ca2+ substituting for Li+ in Li2O, lithium vacancies would be created. For each Ca2+ substituting for Li+, one positive charge is added; in order to maintain charge neutrality, a single positive charge may be removed. Positive charges are eliminated by creating lithium vacancies, and for every Ca2+ ion added, a single lithium vacancy is formed."
|
||
},
|
||
{
|
||
"question": "Suppose that CaO is added as an impurity to CaCl2. If the O2- substitutes for Cl-, what kind of vacancies would you expect to form? How many of these vacancies are created for every O2- added?",
|
||
"answer": "For O2- substituting for Cl- in CaCl2, chlorine vacancies would be created. For each O2- substituting for a Cl-, one negative charge is added; negative charges are eliminated by creating chlorine vacancies. In order to maintain charge neutrality, one O2- ion will lead to the formation of one chlorine vacancy."
|
||
}
|
||
]
|
||
处理第 674/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Which of these elements would you expect to form a substitutional solid solution having complete solubility with nickel? The Hume-Rothery rules must be satisfied: (1) the difference in atomic radii between Ni and the other element must be less than ±15%; (2) the crystal structures must be the same; (3) the electronegativities must be similar; and (4) the valences should be the same. The elements and their properties are given in the table.",
|
||
"answer": "Pt is the only element that meets all of the criteria and thus forms a substitutional solid solution having complete solubility. At elevated temperatures Co and Fe experience allotropic transformations to the FCC crystal structure, and thus display complete solid solubility at these temperatures."
|
||
},
|
||
{
|
||
"question": "Which of these elements would you expect to form a substitutional solid solution of incomplete solubility with nickel? The Hume-Rothery rules must be considered: (1) the difference in atomic radii between Ni and the other element must be less than ±15%; (2) the crystal structures must be the same; (3) the electronegativities must be similar; and (4) the valences should be the same. The elements and their properties are given in the table.",
|
||
"answer": "Ag, Al, Co, Cr, Fe, and Zn form substitutional solid solutions of incomplete solubility. All these metals have either BCC or HCP crystal structures, and/or the difference between their atomic radii and that for Ni are greater than ±15%, and/or have a valence different than 2+."
|
||
},
|
||
{
|
||
"question": "Which of these elements would you expect to form an interstitial solid solution with nickel? The elements and their atomic radii are given in the table.",
|
||
"answer": "C, H, and O form interstitial solid solutions. These elements have atomic radii that are significantly smaller than the atomic radius of Ni."
|
||
}
|
||
]
|
||
```
|
||
处理第 675/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is the composition, in atom percent, of silver (Ag) in an alloy that consists of 92.5 wt% Ag and 7.5 wt% Cu, given the atomic weights of Ag (107.87 g/mol) and Cu (63.55 g/mol)?",
|
||
"answer": "The composition of silver in atom percent is calculated as follows: C_Ag = (C_Ag * A_Cu) / (C_Ag * A_Cu + C_Cu * A_Ag) * 100 = (92.5 * 63.55 g/mol) / (92.5 * 63.55 g/mol + 7.5 * 107.87 g/mol) * 100 = 87.9 at%."
|
||
},
|
||
{
|
||
"question": "What is the composition, in atom percent, of copper (Cu) in an alloy that consists of 92.5 wt% Ag and 7.5 wt% Cu, given the atomic weights of Ag (107.87 g/mol) and Cu (63.55 g/mol)?",
|
||
"answer": "The composition of copper in atom percent is calculated as follows: C_Cu = (C_Cu * A_Ag) / (C_Ag * A_Cu + C_Cu * A_Ag) * 100 = (7.5 * 107.87 g/mol) / (92.5 * 63.55 g/mol + 7.5 * 107.87 g/mol) * 100 = 12.1 at%."
|
||
}
|
||
]
|
||
```
|
||
处理第 676/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is the atomic percent of Pb in an alloy that consists of 5.5 wt% Pb and 94.5 wt% Sn, given the atomic weights of Pb (207.2 g/mol) and Sn (118.71 g/mol)?",
|
||
"answer": "The atomic percent of Pb is calculated as follows: C_Pb = (5.5 * 118.71) / (5.5 * 118.71 + 94.5 * 207.2) * 100 = 3.23 at%."
|
||
},
|
||
{
|
||
"question": "What is the atomic percent of Sn in an alloy that consists of 5.5 wt% Pb and 94.5 wt% Sn, given the atomic weights of Pb (207.2 g/mol) and Sn (118.71 g/mol)?",
|
||
"answer": "The atomic percent of Sn is calculated as follows: C_Sn = (94.5 * 207.2) / (5.5 * 118.71 + 94.5 * 207.2) * 100 = 96.77 at%."
|
||
}
|
||
]
|
||
```
|
||
处理第 677/821 条数据...
|
||
It's a single issue.
|
||
处理第 678/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Calculate the weight percent of iron in an alloy that contains 105 kg of iron, 0.2 kg of carbon, and 1.0 kg of chromium.",
|
||
"answer": "The concentration of iron (C_Fe) is calculated as follows: C_Fe = (m_Fe / (m_Fe + m_C + m_Cr)) * 100 = (105 kg / (105 kg + 0.2 kg + 1.0 kg)) * 100 = 98.87 wt%."
|
||
},
|
||
{
|
||
"question": "Calculate the weight percent of carbon in an alloy that contains 105 kg of iron, 0.2 kg of carbon, and 1.0 kg of chromium.",
|
||
"answer": "The concentration of carbon (C_C) is calculated as follows: C_C = (m_C / (m_Fe + m_C + m_Cr)) * 100 = (0.2 kg / (105 kg + 0.2 kg + 1.0 kg)) * 100 = 0.19 wt%."
|
||
},
|
||
{
|
||
"question": "Calculate the weight percent of chromium in an alloy that contains 105 kg of iron, 0.2 kg of carbon, and 1.0 kg of chromium.",
|
||
"answer": "The concentration of chromium (C_Cr) is calculated as follows: C_Cr = (m_Cr / (m_Fe + m_C + m_Cr)) * 100 = (1.0 kg / (105 kg + 0.2 kg + 1.0 kg)) * 100 = 0.94 wt%."
|
||
}
|
||
]
|
||
```
|
||
处理第 679/821 条数据...
|
||
It's a single issue.
|
||
处理第 680/821 条数据...
|
||
It's a single issue.
|
||
处理第 681/821 条数据...
|
||
It's a single issue.
|
||
处理第 682/821 条数据...
|
||
It's a single issue.
|
||
处理第 683/821 条数据...
|
||
It's a single issue.
|
||
处理第 684/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is the average atomic weight (A_ave) for an 80 wt% Ag-20 wt% Pd alloy, given the atomic weights of Ag (107.87 g/mol) and Pd (106.4 g/mol)?",
|
||
"answer": "The average atomic weight (A_ave) is calculated using the formula: A_ave = 100 / (C_Ag/A_Ag + C_Pd/A_Pd), where C_Ag = 80 wt%, C_Pd = 20 wt%, A_Ag = 107.87 g/mol, and A_Pd = 106.4 g/mol. Substituting these values gives A_ave = 100 / (80/107.87 + 20/106.4) = 107.44 g/mol."
|
||
},
|
||
{
|
||
"question": "What is the average density (ρ_ave) for an 80 wt% Ag-20 wt% Pd alloy, given the densities of Ag (10.49 g/cm³) and Pd (12.02 g/cm³)?",
|
||
"answer": "The average density (ρ_ave) is calculated using the formula: ρ_ave = 100 / (C_Ag/ρ_Ag + C_Pd/ρ_Pd), where C_Ag = 80 wt%, C_Pd = 20 wt%, ρ_Ag = 10.49 g/cm³, and ρ_Pd = 12.02 g/cm³. Substituting these values gives ρ_ave = 100 / (80/10.49 + 20/12.02) = 10.76 g/cm³."
|
||
},
|
||
{
|
||
"question": "Calculate the unit cell edge length (a) for an FCC 80 wt% Ag-20 wt% Pd alloy, given the average atomic weight (A_ave = 107.44 g/mol), average density (ρ_ave = 10.76 g/cm³), and Avogadro's number (N_A = 6.022 × 10²³ atoms/mol).",
|
||
"answer": "The unit cell edge length (a) for an FCC structure is calculated using the formula: a = ( (n × A_ave) / (ρ_ave × N_A) )^(1/3), where n = 4 atoms/unit cell for FCC, A_ave = 107.44 g/mol, ρ_ave = 10.76 g/cm³, and N_A = 6.022 × 10²³ atoms/mol. Substituting these values gives a = ( (4 × 107.44) / (10.76 × 6.022 × 10²³) )^(1/3) = 4.050 × 10⁻⁸ cm = 0.4050 nm."
|
||
}
|
||
]
|
||
```
|
||
处理第 685/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Given an alloy composed of 25 wt% of metal A and 75 wt% of metal B, with densities of 6.17 g/cm³ and 8.00 g/cm³, and atomic weights of 171.3 g/mol and 162.0 g/mol, respectively, calculate the average density of the alloy.",
|
||
"answer": "The average density (ρ_ave) of the alloy is calculated using the formula for the reciprocal of the average density based on weight percentages: ρ_ave = 100 / ( (C_A / ρ_A) + (C_B / ρ_B) ), where C_A = 25 wt%, C_B = 75 wt%, ρ_A = 6.17 g/cm³, and ρ_B = 8.00 g/cm³. Substituting these values gives: ρ_ave = 100 / ( (25 / 6.17) + (75 / 8.00) ) = 7.33 g/cm³."
|
||
},
|
||
{
|
||
"question": "Given the same alloy composition, calculate the average atomic weight of the alloy.",
|
||
"answer": "The average atomic weight (A_ave) of the alloy is calculated using the formula: A_ave = 100 / ( (C_A / A_A) + (C_B / A_B) ), where C_A = 25 wt%, C_B = 75 wt%, A_A = 171.3 g/mol, and A_B = 162.0 g/mol. Substituting these values gives: A_ave = 100 / ( (25 / 171.3) + (75 / 162.0) ) = 164.1 g/mol."
|
||
},
|
||
{
|
||
"question": "Using the average density (7.33 g/cm³), average atomic weight (164.1 g/mol), and given unit cell edge length (0.332 nm), determine the number of atoms per unit cell (n) for the alloy.",
|
||
"answer": "The number of atoms per unit cell (n) is calculated using the formula: n = (ρ_ave * a³ * N_A) / A_ave, where ρ_ave = 7.33 g/cm³, a = 0.332 nm = 3.32 × 10⁻⁸ cm, N_A = 6.022 × 10²³ atoms/mol, and A_ave = 164.1 g/mol. Substituting these values gives: n = (7.33 * (3.32 × 10⁻⁸)³ * 6.022 × 10²³) / 164.1 = 1.00 atom/unit cell."
|
||
},
|
||
{
|
||
"question": "Based on the calculated number of atoms per unit cell (n = 1.00), determine the crystal structure of the alloy (simple cubic, face-centered cubic, or body-centered cubic).",
|
||
"answer": "For n = 1.00 atom/unit cell, the crystal structure is simple cubic."
|
||
}
|
||
]
|
||
```
|
||
处理第 686/821 条数据...
|
||
It's a single issue.
|
||
处理第 687/821 条数据...
|
||
It's a single issue.
|
||
处理第 688/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For a BCC iron-carbon alloy that contains 0.1 wt% C, calculate the number of carbon atoms per cubic centimeter of alloy.",
|
||
"answer": "To compute the number of carbon atoms per cubic centimeter of alloy, we use the following parameters: A_C = 12.011 g/mol, C_C = 0.1 wt%, ρ_C = 2.25 g/cm³, ρ_Fe = 7.87 g/cm³. The calculation is as follows: N_C = (6.022 × 10²³ atoms/mol)(0.1 wt%) / [(0.1 wt%)(12.011 g/mol) / 2.25 g/cm³ + (12.011 g/mol / 7.87 g/cm³)(100 - 0.1 wt%)] = 3.94 × 10²⁰ C atoms/cm³."
|
||
},
|
||
{
|
||
"question": "For a BCC iron-carbon alloy that contains 0.1 wt% C, calculate the number of iron atoms per cubic centimeter of alloy.",
|
||
"answer": "To compute the number of iron atoms per cubic centimeter of alloy, we use the following parameters: A_Fe = 55.85 g/mol, C_Fe = 99.9 wt%, ρ_Fe = 7.87 g/cm³, ρ_C = 2.25 g/cm³. The calculation is as follows: N_Fe = (6.022 × 10²³ atoms/mol)(99.9 wt%) / [(99.9 wt%)(55.85 g/mol) / 7.87 g/cm³ + (55.85 g/mol / 2.25 g/cm³)(100 - 99.9 wt%)] = 8.46 × 10²² Fe atoms/cm³."
|
||
},
|
||
{
|
||
"question": "For a BCC iron-carbon alloy that contains 0.1 wt% C, calculate the number of unit cells per cubic centimeter of alloy.",
|
||
"answer": "From Section 3.4, there are two Fe atoms associated with each unit cell. Therefore, the number of unit cells per cubic centimeter is: n_i = (8.46 × 10²² atoms/cm³) / (2 atoms/unit cell) = 4.23 × 10²² unit cells/cm³."
|
||
},
|
||
{
|
||
"question": "For a BCC iron-carbon alloy that contains 0.1 wt% C, calculate the fraction of unit cells that contain carbon atoms.",
|
||
"answer": "The fraction of unit cells that contain carbon atoms, n_f, is the number of carbon atoms per cubic centimeter divided by the number of unit cells per cubic centimeter: n_f = (3.94 × 10²⁰ C atoms/cm³) / (4.23 × 10²² unit cells/cm³) = 9.31 × 10⁻³ atoms/unit cell. The reciprocal of n_f is 107.5 unit cells/atom, meaning there is one carbon atom per 107.5 unit cells."
|
||
}
|
||
]
|
||
```
|
||
处理第 689/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is the weight percent of gold (Au) that must be added to silver (Ag) to yield an alloy containing 5.5 × 10^21 Au atoms per cubic centimeter, given the densities of pure Au and Ag are 19.32 g/cm³ and 10.49 g/cm³, respectively, and the atomic weights of Au and Ag are 196.97 g/mol and 107.87 g/mol?",
|
||
"answer": "To compute the weight percent of gold (Au) in the alloy, use the following equation with the given values: N_Au = 5.5 × 10^21 atoms/cm³, ρ_Au = 19.32 g/cm³, ρ_Ag = 10.49 g/cm³, A_Au = 196.97 g/mol, A_Ag = 107.87 g/mol, and Avogadro's number N_A = 6.022 × 10^23 atoms/mol. The calculation is as follows: C_Au = 100 / [1 + (N_A × ρ_Ag) / (N_Au × A_Au) - (ρ_Ag / ρ_Au)] = 100 / [1 + (6.022 × 10^23 × 10.49) / (5.5 × 10^21 × 196.97) - (10.49 / 19.32)] = 15.9 wt%."
|
||
}
|
||
]
|
||
```
|
||
处理第 690/821 条数据...
|
||
It's a single issue.
|
||
处理第 691/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For an ASTM grain size of 6, approximately how many grains would there be per square inch at a magnification of 100x?",
|
||
"answer": "This part of problem asks that we compute the number of grains per square inch for an ASTM grain size of 6 at a magnification of 100x. All we need do is solve for the parameter n in Equation, inasmuch as G=6. Thus n = 2^(G-1) = 2^(6-1) = 32 grains/in.^2"
|
||
},
|
||
{
|
||
"question": "For an ASTM grain size of 6, approximately how many grains would there be per square inch without any magnification?",
|
||
"answer": "Now it is necessary to compute the value of n for no magnification. In order to solve this problem it is necessary to use Equation: n_M (M/100)^2 = 2^(G-1) where n_M = the number of grains per square inch at magnification M, and G is the ASTM grain size number. Without any magnification, M in the above equation is 1, and therefore, n_1 (1/100)^2 = 2^(6-1) = 32. And, solving for n_1, n_1 = 320,000 grains/in.^2."
|
||
}
|
||
]
|
||
```
|
||
处理第 692/821 条数据...
|
||
It's a single issue.
|
||
处理第 693/821 条数据...
|
||
It's a single issue.
|
||
处理第 694/821 条数据...
|
||
It's a single issue.
|
||
处理第 695/821 条数据...
|
||
It's a single issue.
|
||
处理第 696/821 条数据...
|
||
[
|
||
{
|
||
"question": "Compare interstitial and vacancy atomic mechanisms for diffusion.",
|
||
"answer": "With vacancy diffusion, atomic motion is from one lattice site to an adjacent vacancy. Self-diffusion and the diffusion of substitutional impurities proceed via this mechanism. On the other hand, atomic motion is from interstitial site to adjacent interstitial site for the interstitial diffusion mechanism."
|
||
},
|
||
{
|
||
"question": "Cite two reasons why interstitial diffusion is normally more rapid than vacancy diffusion.",
|
||
"answer": "Interstitial diffusion is normally more rapid than vacancy diffusion because: (1) interstitial atoms, being smaller, are more mobile; and (2) the probability of an empty adjacent interstitial site is greater than for a vacancy adjacent to a host (or substitutional impurity) atom."
|
||
}
|
||
]
|
||
处理第 697/821 条数据...
|
||
It's a single issue.
|
||
处理第 698/821 条数据...
|
||
It's a single issue.
|
||
处理第 699/821 条数据...
|
||
It's a single issue.
|
||
处理第 700/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Given the activation energy for the diffusion of copper in silver is 193,000 J/mol and the diffusion coefficient D at 1000 K is 1.0 × 10^-14 m²/s, calculate the pre-exponential factor D₀.",
|
||
"answer": "To solve for D₀, use the rearranged equation: D₀ = D exp(Qd / (R T)). Substituting the given values: D₀ = (1.0 × 10^-14 m²/s) exp[193,000 J/mol / (8.31 J/mol-K × 1000 K)] = 1.22 × 10^-4 m²/s."
|
||
},
|
||
{
|
||
"question": "Using the pre-exponential factor D₀ = 1.22 × 10^-4 m²/s and the activation energy Qd = 193,000 J/mol, calculate the diffusion coefficient D at 1200 K.",
|
||
"answer": "To solve for D at 1200 K, use the equation: D = D₀ exp(-Qd / (R T)). Substituting the values: D = (1.22 × 10^-4 m²/s) exp[-193,000 J/mol / (8.31 J/mol-K × 1200 K)] = 4.8 × 10^-13 m²/s."
|
||
}
|
||
]
|
||
```
|
||
处理第 701/821 条数据...
|
||
It's a single issue.
|
||
处理第 702/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Given the steady-state diffusion flux through a metal plate is 7.8 × 10^-8 kg/m²·s at a temperature of 1200°C (1473 K) and a concentration gradient of -500 kg/m⁴, calculate the value of D₀ assuming an activation energy for diffusion of 145,000 J/mol.",
|
||
"answer": "To compute D₀, use the expression: D₀ = - (J / (ΔC/Δx)) * exp(Qd / (R T)). Substituting the given values: D₀ = - (7.8 × 10^-8 kg/m²·s / -500 kg/m⁴) * exp(145,000 J/mol / (8.31 J/mol·K * 1473 K)) = 2.18 × 10^-5 m²/s."
|
||
},
|
||
{
|
||
"question": "Using the calculated D₀ value of 2.18 × 10^-5 m²/s and the same concentration gradient of -500 kg/m⁴, compute the diffusion flux at 1000°C (1273 K) with an activation energy for diffusion of 145,000 J/mol.",
|
||
"answer": "The diffusion flux at 1273 K is calculated using: J = -D₀ (ΔC/Δx) exp(-Qd / (R T)). Substituting the values: J = - (2.18 × 10^-5 m²/s) (-500 kg/m⁴) exp(-145,000 J/mol / (8.31 J/mol·K * 1273 K)) = 1.21 × 10^-8 kg/m²·s."
|
||
}
|
||
]
|
||
```
|
||
处理第 703/821 条数据...
|
||
It's a single issue.
|
||
处理第 704/821 条数据...
|
||
It's a single issue.
|
||
处理第 705/821 条数据...
|
||
It's a single issue.
|
||
处理第 706/821 条数据...
|
||
It's a single issue.
|
||
处理第 707/821 条数据...
|
||
It's a single issue.
|
||
处理第 708/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For a brass alloy, the stress at which plastic deformation begins is 345 MPa (50,000 psi), and the modulus of elasticity is 103 GPa (15.0 x 10^6 psi). What is the maximum load that can be applied to a specimen with a cross-sectional area of 130 mm^2 (0.2 in.^2) without plastic deformation?",
|
||
"answer": "This portion of the problem calls for a determination of the maximum load that can be applied without plastic deformation (F_y). Taking the yield strength to be 345 MPa, and employment of Equation leads to F_y = sigma_y A_0 = (345 x 10^6 N/m^2)(133 mm^2)(1 m/10^3 mm)^2 = 44,850 N (10,000 lbf)."
|
||
},
|
||
{
|
||
"question": "For a brass alloy, the stress at which plastic deformation begins is 345 MPa (50,000 psi), and the modulus of elasticity is 103 GPa (15.0 x 10^6 psi). If the original specimen length is 76 mm (3.0 in.), what is the maximum length to which it can be stretched without causing plastic deformation?",
|
||
"answer": "The maximum length to which the sample may be deformed without plastic deformation is determined by combining Equations as l_i = z l_0 + l_0 = l_0(z + 1) l_i = l_0(sigma/E + 1). Incorporating values of l_0, sigma, and E provided in the problem statement leads to the computation of l_i as follows: l_i = (76 mm)[(345 MPa)/(103 x 10^3 MPa) + 1] = 76.25 mm (3.01 in.)."
|
||
}
|
||
]
|
||
```
|
||
处理第 709/821 条数据...
|
||
It's a single issue.
|
||
处理第 710/821 条数据...
|
||
It's a single issue.
|
||
处理第 711/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is the formula to calculate the modulus of elasticity (E) for a cylindrical specimen under tensile force, given the force (F), initial diameter (d0), transverse strain (Δd/d0), and Poisson's ratio (ν)?",
|
||
"answer": "The modulus of elasticity (E) can be calculated using the formula: E = (4 * F * ν) / (π * d0 * Δd), where F is the tensile force, ν is Poisson's ratio, d0 is the initial diameter, and Δd is the change in diameter."
|
||
},
|
||
{
|
||
"question": "Given a cylindrical specimen with an initial diameter of 10.0 mm (0.39 in.), a tensile force of 1500 N (340 lb), a reduction in diameter of 6.7 × 10^-4 mm (2.64 × 10^-5 in.), and Poisson's ratio of 0.35, compute the modulus of elasticity (E).",
|
||
"answer": "Using the formula E = (4 * F * ν) / (π * d0 * Δd), substitute the given values: F = 1500 N, ν = 0.35, d0 = 10.0 × 10^-3 m, Δd = -6.7 × 10^-7 m. The calculation is E = (4 * 1500 * 0.35) / (π * 10.0 × 10^-3 * -6.7 × 10^-7) = 10^11 Pa = 100 GPa (14.7 × 10^6 psi)."
|
||
}
|
||
]
|
||
```
|
||
处理第 712/821 条数据...
|
||
[
|
||
{
|
||
"question": "For a cylindrical rod 500 mm long with a diameter of 12.7 mm subjected to a tensile load of 29,000 N, which of the four metals or alloys (aluminum alloy, brass alloy, copper, steel alloy) will not experience plastic deformation? Use the given yield strength values from the table.",
|
||
"answer": "First, compute the stress: sigma = F / A0 = 29000 N / (pi * (12.7e-3 m / 2)^2) = 230 MPa. Materials with yield strength greater than 230 MPa are: aluminum alloy (255 MPa), brass alloy (345 MPa), and steel alloy (450 MPa). Copper (210 MPa) is excluded."
|
||
},
|
||
{
|
||
"question": "For the same cylindrical rod under the same load, which of the candidate materials (aluminum alloy, brass alloy, steel alloy) will not elongate more than 1.3 mm? Use the given modulus of elasticity values from the table.",
|
||
"answer": "Calculate elongation for each candidate: For aluminum alloy (E = 70 GPa): delta_l = (230 MPa * 500 mm) / 70e3 MPa = 1.64 mm (exceeds limit). For brass alloy (E = 100 GPa): delta_l = (230 MPa * 500 mm) / 100e3 MPa = 1.15 mm (acceptable). For steel alloy (E = 207 GPa): delta_l = (230 MPa * 500 mm) / 207e3 MPa = 0.56 mm (acceptable). Final candidates: brass alloy and steel alloy."
|
||
}
|
||
]
|
||
处理第 713/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "A cylindrical metal specimen having an original diameter of 12.8 mm and gauge length of 50.80 mm is pulled in tension until fracture occurs. The diameter at the point of fracture is 8.13 mm. Calculate the ductility in terms of percent reduction in area.",
|
||
"answer": "Percent reduction in area is computed using the formula: %RA = ((A0 - Af)/A0) * 100 = ((π*(d0/2)^2 - π*(df/2)^2)/(π*(d0/2)^2)) * 100. Substituting the given values: %RA = ((π*(12.8 mm/2)^2 - π*(8.13 mm/2)^2)/(π*(12.8 mm/2)^2)) * 100 = 60%."
|
||
},
|
||
{
|
||
"question": "A cylindrical metal specimen having an original gauge length of 50.80 mm is pulled in tension until fracture occurs. The fractured gauge length is 74.17 mm. Calculate the ductility in terms of percent elongation.",
|
||
"answer": "Percent elongation is computed using the formula: %EL = ((lf - l0)/l0) * 100. Substituting the given values: %EL = ((74.17 mm - 50.80 mm)/50.80 mm) * 100 = 46%."
|
||
}
|
||
]
|
||
```
|
||
处理第 714/821 条数据...
|
||
It's a single issue.
|
||
处理第 715/821 条数据...
|
||
It's a single issue.
|
||
处理第 716/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is the flexural strength of the aluminum oxide specimen with a circular cross section of radius 5.0 mm (0.20 in.) that fractured at a load of 3000 N (675 lbf) when the distance between the support points was 40 mm (1.6 in.)?",
|
||
"answer": "The flexural strength is calculated as follows: sigma_fs = (F_f * L) / (pi * R^3) = (3000 N * 40 x 10^-3 m) / (pi * (5.0 x 10^-3 m)^3) = 306 x 10^6 N/m^2 = 306 MPa (42,970 psi)."
|
||
},
|
||
{
|
||
"question": "At what load would a specimen of the same material with a square cross section of 15 mm (0.6 in.) length on each edge fracture if the support point separation is maintained at 40 mm (1.6 in.), given the flexural strength is 306 MPa (42,970 psi)?",
|
||
"answer": "The fracture load is calculated as follows: F_f = (2 * sigma_fs * d^3) / (3 * L) = (2 * 306 x 10^6 N/m^2 * (15 x 10^-3 m)^3) / (3 * 40 x 10^-3 m) = 17,200 N (3870 lbf)."
|
||
}
|
||
]
|
||
```
|
||
处理第 717/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "A three-point transverse bending test is conducted on a cylindrical specimen of aluminum oxide having a reported flexural strength of 300 MPa (43,500 psi). If the specimen radius is 5.0 mm (0.20 in.) and the support point separation distance is 15.0 mm (0.61 in.), would you expect the specimen to fracture when a load of 7500 N (1690 lbf) is applied? Justify your answer.",
|
||
"answer": "This portion of the problem asks that we determine whether or not a cylindrical specimen of aluminum oxide having a flexural strength of 300 MPa and a radius of 5.0 mm will fracture when subjected to a load of 7500 N in a three-point bending test; the support point separation is given as 15.0 mm. Using Equation we will calculate the value of σ, in this case the flexural stress; if this value is greater than the flexural strength, σ_fs (300 MPa), then fracture is expected to occur. Employment of Equation to compute the flexural stress yields σ = (FL)/(πR^3) = (7500 N)(15 x 10^-3 m)/(π)(5 x 10^-3 m)^3 = 286.5 x 10^6 N/m^2 = 286.5 MPa (40,300 psi). Since this value (286.5 MPa) is less than the given value of σ_fs (300 MPa), then fracture is not predicted."
|
||
},
|
||
{
|
||
"question": "Would you be 100% certain of the answer in part (a)? Why or why not?",
|
||
"answer": "However, the certainty of this prediction is not 100% because there is always some variability in the flexural strength for ceramic materials, and since this value of the flexural stress, σ, is relatively close to σ_fs there is some chance that fracture will occur."
|
||
}
|
||
]
|
||
```
|
||
处理第 718/821 条数据...
|
||
[
|
||
{
|
||
"question": "Compute the modulus of elasticity for nonporous spinel (MgAl2O4) given that the modulus of elasticity for spinel having 5 vol% porosity is 240 GPa (35 x 10^6 psi).",
|
||
"answer": "To compute the modulus of elasticity for the nonporous material (E0), we use the equation: E0 = E / (1 - 1.9P + 0.9P^2), where E = 240 GPa and P = 0.05. Substituting the values: E0 = 240 GPa / (1 - (1.9)(0.05) + (0.9)(0.05)^2) = 265 GPa (38.6 x 10^6 psi)."
|
||
},
|
||
{
|
||
"question": "Compute the modulus of elasticity for spinel (MgAl2O4) having 15 vol% porosity, given that the modulus of elasticity for nonporous spinel is 265 GPa (38.6 x 10^6 psi).",
|
||
"answer": "To compute the modulus of elasticity for the material with 15 vol% porosity (E), we use the equation: E = E0 (1 - 1.9P + 0.9P^2), where E0 = 265 GPa and P = 0.15. Substituting the values: E = (265 GPa) [1 - (1.9)(0.15) + (0.9)(0.15)^2] = 195 GPa (28.4 x 10^6 psi)."
|
||
}
|
||
]
|
||
处理第 719/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Compute the modulus of elasticity for nonporous titanium carbide (TiC) given that the modulus of elasticity for TiC with 5 vol% porosity is 310 GPa (45 x 10^6 psi).",
|
||
"answer": "To compute the modulus of elasticity for nonporous TiC (E0), we use the given modulus of elasticity (E) of 310 GPa for a material with 5 vol% porosity (P = 0.05). The equation is as follows: E0 = E / (1 - 1.9P + 0.9P^2). Substituting the values, we get E0 = 310 GPa / (1 - (1.9)(0.05) + (0.9)(0.05)^2) = 342 GPa (49.6 x 10^6 psi)."
|
||
},
|
||
{
|
||
"question": "Determine the volume percent porosity at which the modulus of elasticity for titanium carbide (TiC) becomes 240 GPa (35 x 10^6 psi), given that the modulus of elasticity for nonporous TiC is 342 GPa.",
|
||
"answer": "To find the volume percent porosity (P) that results in a modulus of elasticity (E) of 240 GPa, we use the equation E/E0 = 1 - 1.9P + 0.9P^2, where E0 is 342 GPa. Substituting the values, we get 240 GPa / 342 GPa = 0.702 = 1 - 1.9P + 0.9P^2. Rearranging gives 0.9P^2 - 1.9P + 0.298 = 0. Solving the quadratic equation yields roots P+ = 1.94 and P- = 0.171. Only the negative root (0.171) is physically meaningful, corresponding to 17.1 vol% porosity."
|
||
}
|
||
]
|
||
```
|
||
处理第 720/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Compute the flexural strength for a completely nonporous specimen of this material given the flexural strength and associated volume fraction porosity for two specimens: (70 MPa, 0.10) and (60 MPa, 0.15).",
|
||
"answer": "Given the flexural strengths at two different volume fraction porosities, we determine the flexural strength for a nonporous material. Taking the natural logarithm of both sides of Equation 7.22 yields the following expression: ln σ_fs = ln σ_0 - nP. Using the data provided, two simultaneous equations may be written as: ln(70 MPa) = ln σ_0 - (0.10)n and ln(60 MPa) = ln σ_0 - (0.15)n. Solving for n and σ_0 leads to the values: n = 3.08 and σ_0 = 95.3 MPa. For the nonporous material, P = 0, and from Equation 7.22, σ_0 = σ_fs. Thus, σ_fs for P = 0 is 95.3 MPa."
|
||
},
|
||
{
|
||
"question": "Compute the flexural strength for a 0.20 volume fraction porosity given the flexural strength and associated volume fraction porosity for two specimens: (70 MPa, 0.10) and (60 MPa, 0.15), and the previously determined values n = 3.08 and σ_0 = 95.3 MPa.",
|
||
"answer": "We are asked for σ_fs at P = 0.20 for this material. From Equation 7.22: σ_fs = σ_0 exp(-nP) = (95.3 MPa) exp[-(3.08)(0.20)] = 51.5 MPa."
|
||
}
|
||
]
|
||
```
|
||
处理第 721/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Determine the relaxation time (τ) for the viscoelastic polymer given the initial stress σ(0) = 3.5 MPa, the stress at t = 30 s σ(30) = 0.5 MPa, and the elapsed time t = 30 s.",
|
||
"answer": "To determine τ, we use the equation: τ = -t / ln[σ(t)/σ(0)]. Substituting the given values: τ = -30 s / ln[0.5 MPa / 3.5 MPa] = 15.4 s."
|
||
},
|
||
{
|
||
"question": "Calculate the stress σ(10) at t = 10 s for the viscoelastic polymer using the relaxation time τ = 15.4 s and the initial stress σ(0) = 3.5 MPa.",
|
||
"answer": "Using the equation σ(t) = σ(0) exp(-t/τ), we find: σ(10) = 3.5 MPa exp(-10 s / 15.4 s) = 1.83 MPa."
|
||
},
|
||
{
|
||
"question": "Determine the flexural modulus E_f(10) for the viscoelastic polymer at t = 10 s given the strain ε_0 = 0.5 and the stress σ(10) = 1.83 MPa.",
|
||
"answer": "The flexural modulus is calculated as: E_f(10) = σ(10) / ε_0 = 1.83 MPa / 0.5 = 3.66 MPa (522 psi)."
|
||
}
|
||
]
|
||
```
|
||
处理第 722/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Compute the average hardness value from the given Rockwell G hardness measurements: 47.3, 48.7, 47.1, 52.1, 50.0, 50.4, 45.6, 46.2, 45.9, 49.9, 48.3, 46.4, 47.6, 51.1, 48.5, 50.4, 46.7, 49.7",
|
||
"answer": "The average hardness value is calculated by summing all the hardness values and dividing by the number of measurements. Sum = 47.3 + 48.7 + 47.1 + 52.1 + 50.0 + 50.4 + 45.6 + 46.2 + 45.9 + 49.9 + 48.3 + 46.4 + 47.6 + 51.1 + 48.5 + 50.4 + 46.7 + 49.7 = 871.9. Number of measurements = 18. Average = 871.9 / 18 = 48.4."
|
||
},
|
||
{
|
||
"question": "Compute the standard deviation of hardness values from the given Rockwell G hardness measurements: 47.3, 48.7, 47.1, 52.1, 50.0, 50.4, 45.6, 46.2, 45.9, 49.9, 48.3, 46.4, 47.6, 51.1, 48.5, 50.4, 46.7, 49.7, with an average hardness of 48.4",
|
||
"answer": "The standard deviation is calculated by taking the square root of the average of the squared differences from the mean. Squared differences: (47.3-48.4)^2 = 1.21, (48.7-48.4)^2 = 0.09, (47.1-48.4)^2 = 1.69, (52.1-48.4)^2 = 13.69, (50.0-48.4)^2 = 2.56, (50.4-48.4)^2 = 4.00, (45.6-48.4)^2 = 7.84, (46.2-48.4)^2 = 4.84, (45.9-48.4)^2 = 6.25, (49.9-48.4)^2 = 2.25, (48.3-48.4)^2 = 0.01, (46.4-48.4)^2 = 4.00, (47.6-48.4)^2 = 0.64, (51.1-48.4)^2 = 7.29, (48.5-48.4)^2 = 0.01, (50.4-48.4)^2 = 4.00, (46.7-48.4)^2 = 2.89, (49.7-48.4)^2 = 1.69. Sum of squared differences = 64.95. Variance = 64.95 / 17 = 3.82. Standard deviation = sqrt(3.82) = 1.95."
|
||
}
|
||
]
|
||
```
|
||
处理第 723/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Compute the average yield strength (in MPa) from the given values: 274.3, 277.1, 263.8, 267.5, 258.6, 271.2, 255.4, 266.9, 257.6, 270.8, 260.1, 264.3, 261.7, 279.4, 260.5",
|
||
"answer": "The average yield strength is calculated as (274.3 + 277.1 + 263.8 + 267.5 + 258.6 + 271.2 + 255.4 + 266.9 + 257.6 + 270.8 + 260.1 + 264.3 + 261.7 + 279.4 + 260.5) / 15 = 3988.2 / 15 = 265.9 MPa"
|
||
},
|
||
{
|
||
"question": "Compute the standard deviation of yield strength (in MPa) from the given values: 274.3, 277.1, 263.8, 267.5, 258.6, 271.2, 255.4, 266.9, 257.6, 270.8, 260.1, 264.3, 261.7, 279.4, 260.5, with an average of 265.9 MPa",
|
||
"answer": "The standard deviation is calculated as sqrt([(274.3 - 265.9)^2 + (277.1 - 265.9)^2 + (263.8 - 265.9)^2 + (267.5 - 265.9)^2 + (258.6 - 265.9)^2 + (271.2 - 265.9)^2 + (255.4 - 265.9)^2 + (266.9 - 265.9)^2 + (257.6 - 265.9)^2 + (270.8 - 265.9)^2 + (260.1 - 265.9)^2 + (264.3 - 265.9)^2 + (261.7 - 265.9)^2 + (279.4 - 265.9)^2 + (260.5 - 265.9)^2] / 14) = sqrt(753.75 / 14) = 7.34 MPa"
|
||
}
|
||
]
|
||
```
|
||
处理第 724/821 条数据...
|
||
It's a single issue.
|
||
处理第 725/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 60 degrees and 35 degrees, respectively, with the tensile axis. If the critical resolved shear stress is 6.2 MPa (900 psi), will an applied stress of 12 MPa (1750 psi) cause the single crystal to yield?",
|
||
"answer": "The resolved shear stress is calculated using the equation: tau_R = sigma * cos(phi) * cos(lambda) = (12 MPa) * (cos 60 degrees) * (cos 35 degrees) = 4.91 MPa (717 psi). Since the resolved shear stress (4.91 MPa) is less than the critical resolved shear stress (6.2 MPa), the single crystal will not yield."
|
||
},
|
||
{
|
||
"question": "What stress will be necessary to cause the metal single crystal to yield, given the normal to the slip plane and the slip direction are at angles of 60 degrees and 35 degrees, respectively, with the tensile axis, and the critical resolved shear stress is 6.2 MPa (900 psi)?",
|
||
"answer": "The stress at which yielding occurs is calculated using the equation: sigma_y = tau_crss / (cos(phi) * cos(lambda)) = 6.2 MPa / (cos 60 degrees * cos 35 degrees) = 15.1 MPa (2200 psi)."
|
||
}
|
||
]
|
||
```
|
||
处理第 726/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "A single crystal of zinc is oriented for a tensile test such that its slip plane normal makes an angle of 65 degrees with the tensile axis. Three possible slip directions make angles of 30 degrees, 48 degrees, and 78 degrees with the same tensile axis. Which of these three slip directions is most favored?",
|
||
"answer": "Slip will occur along that direction for which (cos φ cos λ) is a maximum, or, in this case, for the largest cos λ. Cosines for the possible λ values are given below. cos(30 degrees) = 0.87, cos(48 degrees) = 0.67, cos(78 degrees) = 0.21. Thus, the slip direction is at an angle of 30 degrees with the tensile axis."
|
||
},
|
||
{
|
||
"question": "A single crystal of zinc is oriented for a tensile test such that its slip plane normal makes an angle of 65 degrees with the tensile axis. The most favored slip direction makes an angle of 30 degrees with the tensile axis. If plastic deformation begins at a tensile stress of 2.5 MPa, determine the critical resolved shear stress for zinc.",
|
||
"answer": "The critical resolved shear stress is just crss = σy(cos φ cos λ)max = (2.5 MPa)[cos(65 degrees) cos(30 degrees)] = 0.91 MPa (130 psi)."
|
||
}
|
||
]
|
||
```
|
||
处理第 727/821 条数据...
|
||
It's a single issue.
|
||
处理第 728/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What are the values of σ0 and ky for an iron with given yield points and grain diameters? (Given: σy = 230 MPa at d = 1 × 10−2 mm, σy = 275 MPa at d = 6 × 10−3 mm)",
|
||
"answer": "The values are σ0 = 75.4 MPa and ky = 15.46 MPa(mm)1/2. These are obtained by solving the simultaneous equations: 230 MPa = σ0 + (10.0) ky and 275 MPa = σ0 + (12.91) ky, where d−1/2 values are 10.0 mm−1/2 and 12.91 mm−1/2 respectively."
|
||
},
|
||
{
|
||
"question": "What grain diameter (d) will result in a lower yield point of 310 MPa for the same iron? (Given: σ0 = 75.4 MPa, ky = 15.46 MPa(mm)1/2)",
|
||
"answer": "The grain diameter will be 4.34 × 10−3 mm. This is calculated using the equation 310 MPa = 75.4 MPa + [15.46 MPa(mm)1/2] d−1/2, solving for d−1/2 = (310 MPa − 75.4 MPa) / 15.46 MPa(mm)1/2, then d = [15.46 MPa(mm)1/2 / (310 MPa − 75.4 MPa)]2."
|
||
}
|
||
]
|
||
```
|
||
处理第 729/821 条数据...
|
||
It's a single issue.
|
||
处理第 730/821 条数据...
|
||
It's a single issue.
|
||
处理第 731/821 条数据...
|
||
It's a single issue.
|
||
处理第 732/821 条数据...
|
||
[
|
||
{
|
||
"question": "What is the driving force for recrystallization?",
|
||
"answer": "The driving force for recrystallization is the difference in internal energy between the strained and unstrained material."
|
||
},
|
||
{
|
||
"question": "What is the driving force for grain growth?",
|
||
"answer": "The driving force for grain growth is the reduction in grain boundary energy as the total grain boundary area decreases."
|
||
}
|
||
]
|
||
处理第 733/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What was the original grain diameter for a brass material measured as a function of time at 650 degrees Celsius, given the following data: Time (min) = 40, Grain Diameter (mm) = 5.6 x 10^-2; Time (min) = 100, Grain Diameter (mm) = 8.0 x 10^-2?",
|
||
"answer": "Using the data given and Equation (for n=2)—that is d^2 - d0^2 = K t, we may set up two simultaneous equations with d0 and K as unknowns, as follows: (5.6 x 10^-2 mm)^2 - d0^2 = (40 min) K; (8.0 x 10^-2 mm)^2 - d0^2 = (100 min) K. Solution of these expressions yields a value for d0, the original grain diameter, of d0 = 0.031 mm."
|
||
},
|
||
{
|
||
"question": "What grain diameter would you predict after 200 min at 650 degrees Celsius for the brass material, given the original grain diameter d0 = 0.031 mm and K = 5.44 x 10^-5 mm^2/min?",
|
||
"answer": "At 200 min, the diameter d is computed using a rearranged form of Equation (incorporating values of d0 and K that were just determined) as follows: d = sqrt(d0^2 + K t) = sqrt((0.031 mm)^2 + (5.44 x 10^-5 mm^2/min)(200 min)) = 0.109 mm."
|
||
}
|
||
]
|
||
```
|
||
处理第 734/821 条数据...
|
||
It's a single issue.
|
||
处理第 735/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Calculate the constant \(k_{y}\) in the equation for a brass specimen with yield strength \(150 \mathrm{MPa}\), \(\sigma_{0}\) of \(25 \mathrm{MPa}\), and grain size \(0.01 \mathrm{~mm}\).",
|
||
"answer": "The constant \(k_{y}\) is calculated as follows: \(k_{y} = \frac{\sigma_{y} - \sigma_{0}}{d^{-1 / 2}} = \frac{150 \mathrm{MPa} - 25 \mathrm{MPa}}{(0.01 \mathrm{~mm})^{-1 / 2}} = 12.5 \mathrm{MPa} \cdot \mathrm{mm}^{1 / 2}\)."
|
||
},
|
||
{
|
||
"question": "Determine the average grain size of the brass specimen after being heated to \(500^{\circ} \mathrm{C}\) for \(1000 \mathrm{~s}\).",
|
||
"answer": "From Figure 8.25, the average grain size after heating to \(500^{\circ} \mathrm{C}\) for \(1000 \mathrm{~s}\) is about \(0.016 \mathrm{~mm}\)."
|
||
},
|
||
{
|
||
"question": "Estimate the yield strength of the brass specimen after heating to \(500^{\circ} \mathrm{C}\) for \(1000 \mathrm{~s}\), given \(\sigma_{0} = 25 \mathrm{MPa}\) and \(k_{y} = 12.5 \mathrm{MPa} \cdot \mathrm{mm}^{1 / 2}\), with the new grain size of \(0.016 \mathrm{~mm}\).",
|
||
"answer": "The yield strength is calculated as follows: \(\sigma_{y} = \sigma_{0} + k_{y} d^{-1 / 2} = 25 \mathrm{MPa} + (12.5 \mathrm{MPa} \cdot \mathrm{mm}^{1/2})(0.016 \mathrm{~mm})^{-1 / 2} = 124 \mathrm{MPa}\)."
|
||
}
|
||
]
|
||
```
|
||
API调用错误: Invalid \escape: line 3 column 45 (char 52)
|
||
处理第 736/821 条数据...
|
||
It's a single issue.
|
||
处理第 737/821 条数据...
|
||
It's a single issue.
|
||
处理第 738/821 条数据...
|
||
It's a single issue.
|
||
处理第 739/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For the pair of polymers: Branched and atactic poly(vinyl chloride) with a weight-average molecular weight of 100,000 g/mol; linear and isotactic poly(vinyl chloride) having a weight-average molecular weight of 75,000 g/mol, (1) state whether it is possible to decide whether one polymer has a higher tensile modulus than the other; (2) if this is possible, note which has the higher tensile modulus and cite the reason(s) for your choice; and (3) if it is not possible to decide, state why.",
|
||
"answer": "Yes, it is possible. The linear and isotactic poly(vinyl chloride) will display a greater tensile modulus. Linear polymers are more likely to crystallize that branched ones. In addition, polymers having isotactic structures will normally have a higher degree of crystallinity that those having atactic structures. Increasing a polymer's crystallinity leads to an increase in its tensile modulus. In addition, tensile modulus is independent of molecular weight - the atactic/branched material has the higher molecular weight."
|
||
},
|
||
{
|
||
"question": "For the pair of polymers: Random styrene-butadiene copolymer with 5% of possible sites crosslinked; block styrene-butadiene copolymer with 10% of possible sites crosslinked, (1) state whether it is possible to decide whether one polymer has a higher tensile modulus than the other; (2) if this is possible, note which has the higher tensile modulus and cite the reason(s) for your choice; and (3) if it is not possible to decide, state why.",
|
||
"answer": "Yes, it is possible. The block styrene-butadiene copolymer with 10% of possible sites crosslinked will have the higher modulus. Block copolymers normally have higher degrees of crystallinity than random copolymers of the same material. A higher degree of crystallinity favors larger moduli. In addition, the block copolymer also has a higher degree of crosslinking; increasing the amount of crosslinking also enhances the tensile modulus."
|
||
},
|
||
{
|
||
"question": "For the pair of polymers: Branched polyethylene with a number-average molecular weight of 100,000 g/mol; atactic polypropylene with a number-average molecular weight of 150,000 g/mol, (1) state whether it is possible to decide whether one polymer has a higher tensile modulus than the other; (2) if this is possible, note which has the higher tensile modulus and cite the reason(s) for your choice; and (3) if it is not possible to decide, state why.",
|
||
"answer": "No, it is not possible. Branched polyethylene will tend to have a low degree of crystallinity since branched polymers don't normally crystallize. The atactic polypropylene probably also has a relatively low degree of crystallinity; atactic structures also don't tend to crystallize, and polypropylene has a more complex repeat unit structure than does polyethylene. Tensile modulus increases with degree of crystallinity, and it is not possible to determine which polymer is more crystalline. Furthermore, tensile modulus is independent of molecular weight."
|
||
}
|
||
]
|
||
```
|
||
处理第 740/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For the pair of linear and isotactic poly(vinyl chloride) with a weight-average molecular weight of 100,000 g/mol and branched and atactic poly(vinyl chloride) having a weight-average molecular weight of 75,000 g/mol, (1) state whether it is possible to decide whether one polymer has a higher tensile strength than the other; (2) if this is possible, note which has the higher tensile strength and cite the reason(s) for your choice; and (3) if it is not possible to decide, state why.",
|
||
"answer": "Yes, it is possible. The linear and isotactic material will have the higher tensile strength. Both linearity and isotacticity favor a higher degree of crystallinity than do branching and atacticity; and tensile strength increases with increasing degree of crystallinity. Furthermore, the molecular weight of the linear/isotactic material is higher (100,000 g/mol versus 75,000 g/mol), and tensile strength increases with increasing molecular weight."
|
||
},
|
||
{
|
||
"question": "For the pair of graft acrylonitrile-butadiene copolymer with 10% of possible sites crosslinked and alternating acrylonitrile-butadiene copolymer with 5% of possible sites crosslinked, (1) state whether it is possible to decide whether one polymer has a higher tensile strength than the other; (2) if this is possible, note which has the higher tensile strength and cite the reason(s) for your choice; and (3) if it is not possible to decide, state why.",
|
||
"answer": "No, it is not possible. Alternating copolymers tend to be more crystalline than graft copolymers, and tensile strength increases with degree of crystallinity. However, the graft material has a higher degree of crosslinking, and tensile strength increases with the percentage of crosslinks."
|
||
},
|
||
{
|
||
"question": "For the pair of network polyester and lightly branched polytetrafluoroethylene, (1) state whether it is possible to decide whether one polymer has a higher tensile strength than the other; (2) if this is possible, note which has the higher tensile strength and cite the reason(s) for your choice; and (3) if it is not possible to decide, state why.",
|
||
"answer": "Yes, it is possible. The network polyester will display a greater tensile strength. Relative chain motion is much more restricted than for the lightly branched polytetrafluoroethylene since there are many more of the strong covalent bonds for the network structure."
|
||
}
|
||
]
|
||
```
|
||
处理第 741/821 条数据...
|
||
It's a single issue.
|
||
处理第 742/821 条数据...
|
||
It's a single issue.
|
||
处理第 743/821 条数据...
|
||
It's a single issue.
|
||
处理第 744/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For an aluminum alloy with a plane strain fracture toughness of 40 MPa sqrt(m), determine the parameter Y when fracture occurs at a stress of 300 MPa and a maximum internal crack length of 4.0 mm.",
|
||
"answer": "Y is calculated using the formula Y = K_IC / (σ sqrt(π a)), where K_IC = 40 MPa sqrt(m), σ = 300 MPa, and a = 2.0 mm (half of the maximum internal crack length). Substituting these values, Y = 40 MPa sqrt(m) / (300 MPa sqrt(π × 2.0 × 10^-3 m)) = 1.68."
|
||
},
|
||
{
|
||
"question": "For the same component and alloy, calculate the product Y σ sqrt(π a) when the stress level is 260 MPa and the maximum internal crack length is 6.0 mm, using the previously determined Y value of 1.68.",
|
||
"answer": "The product Y σ sqrt(π a) is calculated as (1.68)(260 MPa) sqrt(π × 3.0 × 10^-3 m) = 42.4 MPa sqrt(m)."
|
||
},
|
||
{
|
||
"question": "Determine if fracture will occur at a stress level of 260 MPa and a maximum internal crack length of 6.0 mm, given that the plane strain fracture toughness (K_IC) of the alloy is 40 MPa sqrt(m).",
|
||
"answer": "Fracture will occur because the calculated Y σ sqrt(π a) value of 42.4 MPa sqrt(m) is greater than the K_IC of the material, which is 40 MPa sqrt(m)."
|
||
}
|
||
]
|
||
```
|
||
处理第 745/821 条数据...
|
||
It's a single issue.
|
||
处理第 746/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is the value of the design parameter Y for a structural component made from an alloy with a plane-strain fracture toughness of 62 MPa sqrt(m), failing at a stress of 250 MPa when the maximum length of a surface crack is 1.6 mm?",
|
||
"answer": "The design parameter Y is calculated using the formula Y = K_IC / (σ sqrt(π a_C)), where K_IC = 62 MPa sqrt(m), σ = 250 MPa, and a_C = 1.6 x 10^-3 m. Substituting these values, Y = 62 / (250 sqrt(π x 1.6 x 10^-3)) = 3.50."
|
||
},
|
||
{
|
||
"question": "What is the maximum allowable surface crack length (in mm) without fracture for the same component exposed to a stress of 250 MPa and made from another alloy that has a plane strain fracture toughness of 51 MPa sqrt(m), using the previously calculated Y value of 3.50?",
|
||
"answer": "The maximum crack length a_C is calculated using the formula a_C = 1/π (K_IC / (σ Y))^2, where K_IC = 51 MPa sqrt(m), σ = 250 MPa, and Y = 3.50. Substituting these values, a_C = 1/π (51 / (250 x 3.50))^2 = 0.00108 m = 1.08 mm."
|
||
}
|
||
]
|
||
```
|
||
处理第 747/821 条数据...
|
||
It's a single issue.
|
||
处理第 748/821 条数据...
|
||
It's a single issue.
|
||
处理第 749/821 条数据...
|
||
[
|
||
{
|
||
"question": "Why may there be significant scatter in the fracture strength for some given ceramic material?",
|
||
"answer": "There may be significant scatter in the fracture strength for some given ceramic material because the fracture strength depends on the probability of the existence of a flaw that is capable of initiating a crack; this probability varies from specimen to specimen of the same material."
|
||
},
|
||
{
|
||
"question": "Why does the fracture strength increase with decreasing specimen size?",
|
||
"answer": "The fracture strength increases with decreasing specimen size because as specimen size decreases, the probability of the existence of a flaw that is capable of initiating a crack diminishes."
|
||
}
|
||
]
|
||
处理第 750/821 条数据...
|
||
It's a single issue.
|
||
处理第 751/821 条数据...
|
||
It's a single issue.
|
||
处理第 752/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Give the approximate temperature at which creep deformation becomes an important consideration for tin (Sn), where creep becomes important at about 0.4 Tm, Tm being the absolute melting temperature of the metal.",
|
||
"answer": "For Sn, 0.4 Tm = (0.4)(232 + 273) = 202 K or -71 C (-96 F)"
|
||
},
|
||
{
|
||
"question": "Give the approximate temperature at which creep deformation becomes an important consideration for molybdenum (Mo), where creep becomes important at about 0.4 Tm, Tm being the absolute melting temperature of the metal.",
|
||
"answer": "For Mo, 0.4 Tm = (0.4)(2617 + 273) = 1049 K or 776 C (1429 F)"
|
||
},
|
||
{
|
||
"question": "Give the approximate temperature at which creep deformation becomes an important consideration for iron (Fe), where creep becomes important at about 0.4 Tm, Tm being the absolute melting temperature of the metal.",
|
||
"answer": "For Fe, 0.4 Tm = (0.4)(1538 + 273) = 724 K or 451 C (845 F)"
|
||
},
|
||
{
|
||
"question": "Give the approximate temperature at which creep deformation becomes an important consideration for gold (Au), where creep becomes important at about 0.4 Tm, Tm being the absolute melting temperature of the metal.",
|
||
"answer": "For Au, 0.4 Tm = (0.4)(1064 + 273) = 535 K or 262 C (504 F)"
|
||
},
|
||
{
|
||
"question": "Give the approximate temperature at which creep deformation becomes an important consideration for zinc (Zn), where creep becomes important at about 0.4 Tm, Tm being the absolute melting temperature of the metal.",
|
||
"answer": "For Zn, 0.4 Tm = (0.4)(420 + 273) = 277 K or 4 C (39 F)"
|
||
},
|
||
{
|
||
"question": "Give the approximate temperature at which creep deformation becomes an important consideration for chromium (Cr), where creep becomes important at about 0.4 Tm, Tm being the absolute melting temperature of the metal.",
|
||
"answer": "For Cr, 0.4 Tm = (0.4)(1875 + 273) = 859 K or 586 C (1087 F)"
|
||
}
|
||
]
|
||
```
|
||
处理第 753/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is the stress imposed on a cylindrical specimen of S-590 alloy with a diameter of 13.2 mm when exposed to a tensile load of 27,000 N?",
|
||
"answer": "The stress is calculated using the formula sigma = F / A0, where F is the force and A0 is the cross-sectional area. Given d0 = 13.2 mm (13.2 x 10^-3 m) and F = 27,000 N, the stress is equal to sigma = 27,000 N / (pi * (13.2 x 10^-3 m / 2)^2) = 197 x 10^6 N/m^2 = 197 MPa, or approximately 200 MPa."
|
||
},
|
||
{
|
||
"question": "At approximately what temperature will the steady-state creep be 10^-3 h^-1 for the S-590 alloy under a stress of 200 MPa?",
|
||
"answer": "From Figure 9.40, the point corresponding to 10^-3 h^-1 and 200 MPa lies approximately midway between 730°C and 815°C lines; therefore, the temperature would be approximately 775°C."
|
||
}
|
||
]
|
||
```
|
||
处理第 754/821 条数据...
|
||
It's a single issue.
|
||
处理第 755/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Given the steady-state creep rate data at 200°C (473 K) for an alloy with activation energy for creep of 140,000 J/mol, determine the stress exponent (n) and constant (K2) using the following data: 2.5×10−3 h−1 at 55 MPa and 2.4×10−2 h−1 at 69 MPa.",
|
||
"answer": "Using the equation ln(˙εs) = ln(K2) + n ln(σ) − (Qc/RT), set up two equations with the given data: ln(2.5×10−3 h−1) = ln(K2) + n ln(55 MPa) − (140,000 J/mol)/(8.31 J/mol·K)(473 K) and ln(2.4×10−2 h−1) = ln(K2) + n ln(69 MPa) − (140,000 J/mol)/(8.31 J/mol·K)(473 K). Solving these simultaneously yields n = 9.97 and K2 = 3.27×10−5 h−1."
|
||
},
|
||
{
|
||
"question": "Using the determined stress exponent (n = 9.97) and constant (K2 = 3.27×10−5 h−1), compute the steady-state creep rate at 250°C (523 K) and a stress level of 48 MPa (7000 psi).",
|
||
"answer": "Using the equation ˙εs = K2 σ^n exp(−Qc/RT), substitute the values: ˙εs = (3.27×10−5 h−1)(48 MPa)^9.97 exp[−(140,000 J/mol)/(8.31 J/mol·K)(523 K)]. The calculation results in a steady-state creep rate of 1.94×10−2 h−1."
|
||
}
|
||
]
|
||
```
|
||
处理第 756/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Given the steady-state creep data for an iron at a stress level of 140 MPa (20,000 psi) with strain rates of 6.6×10⁻⁴ h⁻¹ at 1090 K and 8.8×10⁻² h⁻¹ at 1200 K, and a stress exponent n=8.5, determine the values of K₂ and Qₐ.",
|
||
"answer": "Using the equation ln(ε̇ₛ) = ln(K₂) + n ln(σ) - (Qₐ)/(RT), we set up two equations:\n1. ln(6.6×10⁻⁴ h⁻¹) = ln(K₂) + 8.5 ln(140 MPa) - (Qₐ)/(8.31 J/mol·K × 1090 K)\n2. ln(8.8×10⁻² h⁻¹) = ln(K₂) + 8.5 ln(140 MPa) - (Qₐ)/(8.31 J/mol·K × 1200 K)\nSolving these simultaneously yields K₂ = 57.5 h⁻¹ and Qₐ = 483,500 J/mol."
|
||
},
|
||
{
|
||
"question": "Using the determined values of K₂=57.5 h⁻¹ and Qₐ=483,500 J/mol, compute the steady-state creep rate at a stress level of 83 MPa (12,000 psi) and temperature of 1300 K.",
|
||
"answer": "Applying the equation ε̇ₛ = K₂ σⁿ exp(-Qₐ/(RT)):\nε̇ₛ = (57.5 h⁻¹)(83 MPa)^8.5 exp[-483,500 J/mol / (8.31 J/mol·K × 1300 K)]\nThe calculated steady-state creep rate is 4.31×10⁻² h⁻¹."
|
||
}
|
||
]
|
||
```
|
||
处理第 757/821 条数据...
|
||
It's a single issue.
|
||
处理第 758/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Given the Avrami equation for recrystallization kinetics with n=5.0, and a fraction recrystallized y=0.30 after t=100 min, calculate the value of the rate constant k.",
|
||
"answer": "To calculate k, rearrange the Avrami equation: exp(-k t^n)=1-y. Take natural logarithms: -k t^n=ln(1-y). Solve for k: k=-ln(1-y)/t^n. Substituting y=0.30, t=100 min, n=5.0: k=-ln(1-0.30)/(100 min)^5=3.57×10^-11."
|
||
},
|
||
{
|
||
"question": "Using the calculated k=3.57×10^-11 and n=5.0, determine the time t_0.5 for the fraction recrystallized to reach y=0.5.",
|
||
"answer": "Rearrange the Avrami equation to solve for t: t^n=-ln(1-y)/k. For t_0.5: t_0.5=[-ln(1-0.5)/k]^(1/n). Substituting k=3.57×10^-11 and n=5.0: t_0.5=[-ln(1-0.5)/3.57×10^-11]^(1/5)=114.2 min."
|
||
},
|
||
{
|
||
"question": "Calculate the rate of recrystallization using the determined t_0.5=114.2 min.",
|
||
"answer": "The rate of recrystallization is the reciprocal of t_0.5: rate=1/t_0.5=1/114.2 min=8.76×10^-3 min^-1."
|
||
}
|
||
]
|
||
```
|
||
处理第 759/821 条数据...
|
||
It's a single issue.
|
||
处理第 760/821 条数据...
|
||
It's a single issue.
|
||
处理第 761/821 条数据...
|
||
[
|
||
{
|
||
"question": "Briefly cite the differences among pearlite, bainite, and spheroidite relative to microstructure.",
|
||
"answer": "The microstructures of pearlite, bainite, and spheroidite all consist of α-ferrite and cementite phases. For pearlite, the two phases exist as layers, which alternate with one another. Bainite consists of very fine and parallel needle-shaped particles of cementite that are surrounded by an α-ferrite matrix. For spheroidite, the matrix is ferrite, and the cementite phase is in the shape of sphere-shaped particles."
|
||
},
|
||
{
|
||
"question": "Briefly cite the differences among pearlite, bainite, and spheroidite relative to mechanical properties.",
|
||
"answer": "Bainite is harder and stronger than pearlite, which, in turn, is harder and stronger than spheroidite."
|
||
}
|
||
]
|
||
处理第 762/821 条数据...
|
||
It's a single issue.
|
||
处理第 763/821 条数据...
|
||
It's a single issue.
|
||
处理第 764/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For the pair of polymers: branched polyethylene having a number-average molecular weight of 850,000 g/mol and linear polyethylene having a number-average molecular weight of 850,000 g/mol, (1) state whether it is possible to determine whether one polymer has a higher melting temperature than the other; (2) if it is possible, note which has the higher melting temperature and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why.",
|
||
"answer": "Yes, it is possible to determine which polymer has the higher melting temperature. The linear polyethylene will most likely have a higher percent crystallinity, and, therefore, a higher melting temperature than the branched polyethylene. The molecular weights of both materials are the same and, thus, molecular weight is not a consideration."
|
||
},
|
||
{
|
||
"question": "For the pair of polymers: polytetrafluoroethylene having a density of 2.14 g/cm³ and a weight-average molecular weight of 600,000 g/mol and PTFE having a density of 2.20 g/cm³ and a weight-average molecular weight of 600,000 g/mol, (1) state whether it is possible to determine whether one polymer has a higher melting temperature than the other; (2) if it is possible, note which has the higher melting temperature and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why.",
|
||
"answer": "Yes, it is possible to determine which polymer has the higher melting temperature. Of these two polytetrafluoroethylene polymers, the PTFE with the higher density (2.20 g/cm³) will have the higher percent crystallinity, and, therefore, a higher melting temperature than the lower density PTFE. The molecular weights of both materials are the same and, thus, molecular weight is not a consideration."
|
||
},
|
||
{
|
||
"question": "For the pair of polymers: linear and syndiotactic poly(vinyl chloride) having a number-average molecular weight of 500,000 g/mol and linear polyethylene having a number-average molecular weight of 225,000 g/mol, (1) state whether it is possible to determine whether one polymer has a higher melting temperature than the other; (2) if it is possible, note which has the higher melting temperature and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why.",
|
||
"answer": "Yes, it is possible to determine which polymer has the higher melting temperature. The linear polyethylene will have the greater melting temperature inasmuch as it will have a higher degree of crystallinity; polymers having a syndiotactic structure do not crystallize as easily as those polymers having identical single-atom side groups. With regard to molecular weight, or rather, degree of polymerization, it is about the same for both materials (8000), and therefore, is not a consideration."
|
||
},
|
||
{
|
||
"question": "For the pair of polymers: linear and syndiotactic polypropylene having a weight-average molecular weight of 500,000 g/mol and linear and atactic polypropylene having a weight-average molecular weight of 750,000 g/mol, (1) state whether it is possible to determine whether one polymer has a higher melting temperature than the other; (2) if it is possible, note which has the higher melting temperature and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why.",
|
||
"answer": "No, it is not possible to determine which of the two polymers has the higher melting temperature. The syndiotactic polypropylene will have a higher degree of crystallinity than the atactic material. On the basis of this effect alone, the syndiotactic PP should have the greater Tm, since melting temperature increases with degree of crystallinity. However, the molecular weight for the syndiotactic polypropylene (500,000 g/mol) is less than for the atactic material (750,000 g/mol); a lowering of molecular weight generally results in a reduction of melting temperature."
|
||
}
|
||
]
|
||
```
|
||
处理第 765/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Compute the electrical conductivity of a cylindrical silicon specimen 7.0 mm (0.28 in.) diameter and 57 mm (2.25 in.) in length in which a current of 0.25 A passes in an axial direction. A voltage of 24 V is measured across two probes that are separated by 45 mm (1.75 in.).",
|
||
"answer": "This problem calls for us to compute the electrical conductivity of a silicon specimen. We use Equations for the conductivity, as follows: sigma = 1/rho = (I l)/(V A) = (I l)/(V pi (d/2)^2). Since the cross-sectional area of a cylinder having a diameter d is A = pi (d/2)^2. Now, incorporation of values for the several parameters provided in the problem statement, leads to the following: sigma = (I l)/(V pi (d/2)^2) = (0.25 A)(45 x 10^-3 m)/(24 V)(pi)((7.0 x 10^-3 m)/2)^2 = 12.2 (Ohm m)^-1."
|
||
},
|
||
{
|
||
"question": "Compute the resistance over the entire 57 mm (2.25 in.) of the specimen with diameter 7.0 mm (0.28 in.), given the electrical conductivity sigma = 12.2 (Ohm m)^-1.",
|
||
"answer": "The resistance, R, over the entire specimen length may be computed by combining and rearranging Equations as follows: R = (l rho)/A = (l (1/sigma))/A = l/(sigma A) = l/(sigma pi (d/2)^2). The value of R may be computed upon entering values given in the problem statement into the preceding expression: R = l/(sigma pi (d/2)^2) = 57 x 10^-3 m/[12.2 (Ohm m)^-1](pi)((7.0 x 10^-3 m)/2)^2 = 121.4 Ohm."
|
||
}
|
||
]
|
||
```
|
||
处理第 766/821 条数据...
|
||
It's a single issue.
|
||
处理第 767/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "In terms of electron energy band structure, why do metals have high electrical conductivity?",
|
||
"answer": "For metallic materials, there are vacant electron energy states adjacent to the highest filled state; thus, very little energy is required to excite large numbers of electrons into conducting states. These electrons are those that participate in the conduction process, and, because there are so many of them, metals are good electrical conductors."
|
||
},
|
||
{
|
||
"question": "In terms of electron energy band structure, why do semiconductors and insulators have lower electrical conductivity compared to metals?",
|
||
"answer": "There are no empty electron states adjacent to and above filled states for semiconductors and insulators, but rather, an energy band gap across which electrons must be excited in order to participate in the conduction process. Thermal excitation of electrons occurs, and the number of electrons excited will be less than for metals, and will depend on the band gap energy."
|
||
},
|
||
{
|
||
"question": "In terms of electron energy band structure, why do semiconductors have higher electrical conductivity than insulators?",
|
||
"answer": "For semiconductors, the band gap is narrower than for insulators; consequently, at a specific temperature more electrons will be excited for semiconductors, giving rise to higher conductivities."
|
||
}
|
||
]
|
||
```
|
||
处理第 768/821 条数据...
|
||
It's a single issue.
|
||
处理第 769/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Briefly state what is meant by the drift velocity of a free electron.",
|
||
"answer": "The drift velocity of a free electron is the average electron velocity in the direction of the force imposed by an electric field."
|
||
},
|
||
{
|
||
"question": "Briefly state what is meant by the mobility of a free electron.",
|
||
"answer": "The mobility is the proportionality constant between the drift velocity and the electric field. It is also a measure of the frequency of scattering events (and is inversely proportional to the frequency of scattering)."
|
||
}
|
||
]
|
||
```
|
||
处理第 770/821 条数据...
|
||
[
|
||
{
|
||
"question": "Calculate the drift velocity of electrons in silicon at room temperature when the magnitude of the electric field is 500 V/m. The room temperature mobility of electrons is 0.145 m2/V-s.",
|
||
"answer": "The drift velocity of electrons in Si may be determined using the equation vd = μeE. Since the room temperature mobility of electrons is 0.145 m2/V-s, and the electric field is 500 V/m, the drift velocity is equal to vd = (0.145 m2/V-s)(500 V/m) = 72.5 m/s."
|
||
},
|
||
{
|
||
"question": "Under the circumstances where the drift velocity of electrons in silicon is 72.5 m/s, how long does it take an electron to traverse a 25-mm (1 in.) length of crystal?",
|
||
"answer": "The time, t, required to traverse a given length, l (=25 mm), is just the length divided by the drift velocity, as follows: t = l/vd = (25 x 10-3 m)/(72.5 m/s) = 3.45 x 10-4 s."
|
||
}
|
||
]
|
||
处理第 771/821 条数据...
|
||
[
|
||
{
|
||
"question": "At room temperature the electrical conductivity and the electron mobility for aluminum are 3.8 × 10^7 (Ω·m)^-1 and 0.0012 m^2/V·s, respectively. Compute the number of free electrons per cubic meter for aluminum at room temperature.",
|
||
"answer": "The number of free electrons per cubic meter for aluminum at room temperature may be computed using rearranged form of Equation as follows: n = σ / |e| μ_e = 3.8 × 10^7 (Ω·m)^-1 / (1.602 × 10^-19 C)(0.0012 m^2/V·s) = 1.98 × 10^29 m^-3"
|
||
},
|
||
{
|
||
"question": "At room temperature the electrical conductivity and the electron mobility for aluminum are 3.8 × 10^7 (Ω·m)^-1 and 0.0012 m^2/V·s, respectively. What is the number of free electrons per aluminum atom? Assume a density of 2.7 g/cm^3.",
|
||
"answer": "In order to calculate the number of free electrons per aluminum atom, we must first determine the number of aluminum atoms per cubic meter, N_Al. From Equation (and using the atomic weight for Al found inside the front cover-viz. 26.98 g/mol) then the values of N_Al is determined as follows: N_Al = (N_A ρ') / A_Al = (6.022 × 10^23 atoms/mol)(2.7 g/cm^3)(10^6 cm^3/m^3) / 26.98 g/mol = 6.03 × 10^28 m^-3. And, finally, the number of free electrons per aluminum atom (n/N_Al) is equal to the following: n/N_Al = 1.98 × 10^29 m^-3 / 6.03 × 10^28 m^-3 = 3.28 electrons/Al atom"
|
||
}
|
||
]
|
||
处理第 772/821 条数据...
|
||
It's a single issue.
|
||
处理第 773/821 条数据...
|
||
It's a single issue.
|
||
处理第 774/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Predict whether nitrogen (N) will act as a donor or an acceptor when added to silicon (Si) as a substitutional impurity.",
|
||
"answer": "Nitrogen will act as a donor in Si. Since N is from group VA of the periodic table, and an N atom has one more valence electron than an Si atom."
|
||
},
|
||
{
|
||
"question": "Predict whether boron (B) will act as a donor or an acceptor when added to germanium (Ge) as a substitutional impurity.",
|
||
"answer": "Boron will act as an acceptor in Ge. Since B is from group IIIA of the periodic table, a B atom has one less valence electron than a Ge atom."
|
||
},
|
||
{
|
||
"question": "Predict whether sulfur (S) will act as a donor or an acceptor when added to indium antimonide (InSb) as a substitutional impurity.",
|
||
"answer": "Sulfur will act as a donor in InSb. Since S is from group VIA of the periodic table, it will substitute for Sb; an S atom has one more valence electron than an Sb atom."
|
||
},
|
||
{
|
||
"question": "Predict whether indium (In) will act as a donor or an acceptor when added to cadmium sulfide (CdS) as a substitutional impurity.",
|
||
"answer": "Indium will act as a donor in CdS. Since In is from group IIIA of the periodic table, it will substitute for Cd; an In atom has one more valence electron than a Cd atom."
|
||
},
|
||
{
|
||
"question": "Predict whether arsenic (As) will act as a donor or an acceptor when added to zinc telluride (ZnTe) as a substitutional impurity.",
|
||
"answer": "Arsenic will act as an acceptor in ZnTe. Since As is from group VA of the periodic table, it will substitute for Te; furthermore, an As atom has one less valence electron than a Te atom."
|
||
}
|
||
]
|
||
```
|
||
处理第 775/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Germanium to which 10^24 m^-3 As atoms have been added is an extrinsic semiconductor at room temperature, and virtually all the As atoms may be thought of as being ionized (i.e., one charge carrier exists for each As atom). Is this material n-type or p-type?",
|
||
"answer": "This germanium material to which has been added 10^24 m^-3 As atoms is n-type since As is a donor in Ge. (Arsenic is from group VA of the periodic table--Ge is from group IVA.)"
|
||
},
|
||
{
|
||
"question": "Germanium to which 10^24 m^-3 As atoms have been added is an extrinsic semiconductor at room temperature, and virtually all the As atoms may be thought of as being ionized (i.e., one charge carrier exists for each As atom). Calculate the electrical conductivity of this material, assuming electron and hole mobilities of 0.1 and 0.05 m^2/V-s, respectively.",
|
||
"answer": "Since this material is n-type extrinsic, Equation is valid. Furthermore, each As atom will donate a single electron, or the electron concentration is equal to the As concentration since all of the As atoms are ionized at room temperature; that is n=10^24 m^-3, and, as given in the problem statement, μe=0.1 m^2/V-s. Thus, the conductivity is equal to σ=n|e| μe =(10^24 m^-3)(1.602×10^-19 C)(0.1 m^2/V-s) =1.60×10^4 (Ω·m)^-1"
|
||
}
|
||
]
|
||
```
|
||
处理第 776/821 条数据...
|
||
It's a single issue.
|
||
处理第 777/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Given the activation energy (173,000 J/mol) and preexponential factor (4.0 x 10^-4 m^2/s) for the diffusion coefficient of Na+ in NaCl, compute the diffusion coefficient D_Na+ at 873 K.",
|
||
"answer": "D_Na+ = D0 exp(-Qd / RT) = (4.0 x 10^-4 m^2/s) exp[-173,000 J/mol / (8.31 J/mol-K)(873 K)] = 1.76 x 10^-14 m^2/s"
|
||
},
|
||
{
|
||
"question": "Using the computed diffusion coefficient D_Na+ (1.76 x 10^-14 m^2/s) at 873 K, calculate the mobility μ_Na+ for a Na+ ion, assuming the valence n_Na+ = 1.",
|
||
"answer": "μ_Na+ = (n_Na+ * e * D_Na+) / (k T) = (1)(1.602 x 10^-19 C/atom)(1.76 x 10^-14 m^2/s) / (1.38 x 10^-23 J/atom-K)(873 K) = 2.34 x 10^-13 m^2/V-s"
|
||
}
|
||
]
|
||
```
|
||
处理第 778/821 条数据...
|
||
It's a single issue.
|
||
处理第 779/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Consider a parallel-plate capacitor having an area of 3225 mm2 (5 in.2), a plate separation of 1 mm (0.04 in.), and a material having a dielectric constant of 3.5 positioned between the plates. What is the capacitance of this capacitor?",
|
||
"answer": "We are first asked to compute the capacitance. Combining Equations leads to the following expression: C = ε A / l = εr ε0 A / l. Incorporation of values given in the problem statement for parameters on the right-hand side of this expression, and solving for C yields C = (3.5)(8.85 × 10-12 F/m)(3225 mm2)(1 m2 / 106 mm2) / 10-3 m = 10-10 F = 100 pF."
|
||
},
|
||
{
|
||
"question": "Consider a parallel-plate capacitor having an area of 3225 mm2 (5 in.2), a plate separation of 1 mm (0.04 in.), and a material having a dielectric constant of 3.5 positioned between the plates. Compute the electric field that must be applied for 2 × 10-8 C to be stored on each plate.",
|
||
"answer": "Now we are asked to compute the electric field that must be applied in order for 2 × 10-8 C to be stored on each plate. First we need to solve for V in Equation as V = Q / C = (2 × 10-8 C) / (10-10 F) = 200 V. The electric field E may now be determined using Equation, as follows: E = V / l = (200 V) / (10-3 m) = 2.0 × 105 V/m."
|
||
}
|
||
]
|
||
```
|
||
处理第 780/821 条数据...
|
||
[
|
||
{
|
||
"question": "The polarization P of a dielectric material positioned within a parallel-plate capacitor is to be 4.0 × 10^-6 C/m^2. What must be the dielectric constant if an electric field of 10^5 V/m is applied?",
|
||
"answer": "We solve this portion of the problem by computing the value of εr using Equation. Rearranging this equation such that εr is the dependent parameter gives the following: εr = P / (ε0 E) + 1. When values of the parameters on the right-hand side of this equation are inserted the value of εr equal to εr = (4.0 × 10^-6 C/m^2) / ((8.85 × 10^-12 F/m)(1 × 10^5 V/m)) + 1 = 5.52"
|
||
},
|
||
{
|
||
"question": "The polarization P of a dielectric material positioned within a parallel-plate capacitor is to be 4.0 × 10^-6 C/m^2. What will be the dielectric displacement D when an electric field of 10^5 V/m is applied?",
|
||
"answer": "The dielectric displacement D may be determined using Equation, as follows: D = ε0 E + P = (8.85 × 10^-12 F/m)(1 × 10^5 V/m) + 4.0 × 10^-6 C/m^2 = 4.89 × 10^-6 C/m^2"
|
||
}
|
||
]
|
||
处理第 781/821 条数据...
|
||
[
|
||
{
|
||
"question": "(a) For each of the three types of polarization, briefly describe the mechanism by which dipoles are induced and/or oriented by the action of an applied electric field.",
|
||
"answer": "For electronic polarization, the electric field causes a net displacement of the center of the negatively charged electron cloud relative to the positive nucleus. With ionic polarization, the cations and anions are displaced in opposite directions as a result of the application of an electric field. Orientation polarization is found in substances that possess permanent dipole moments; these dipole moments become aligned in the direction of the electric field."
|
||
},
|
||
{
|
||
"question": "(b) For gaseous argon, solid LiF, liquid H2O, and solid Si, what kind(s) of polarization is (are) possible? Why?",
|
||
"answer": "Only electronic polarization is found in gaseous argon; being an inert gas, its atoms will not be ionized nor possess permanent dipole moments. Both electronic and ionic polarizations are found in solid LiF, since it is strongly ionic. In all probability, no permanent dipole moments will be found in this material. Both electronic and orientation polarizations are found in liquid H2O. The H2O molecules have permanent dipole moments that are easily oriented in the liquid state. Only electronic polarization is to be found in solid Si; this material does not have molecules with permanent dipole moments, nor is it an ionic material."
|
||
}
|
||
]
|
||
处理第 782/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "List the four classifications of steels.",
|
||
"answer": "The four classifications of steels are: Low Carbon Steels, Medium Carbon Steels, High Carbon Steels, and High Alloy Steels (Stainless and Tool)."
|
||
},
|
||
{
|
||
"question": "For Low Carbon Steels, briefly describe the properties and typical applications.",
|
||
"answer": "Properties: nonresponsive to heat treatments; relatively soft and weak; machinable and weldable. Typical applications: automobile bodies, structural shapes, pipelines, buildings, bridges, and tin cans."
|
||
},
|
||
{
|
||
"question": "For Medium Carbon Steels, briefly describe the properties and typical applications.",
|
||
"answer": "Properties: heat treatable, relatively large combinations of mechanical characteristics. Typical applications: railway wheels and tracks, gears, crankshafts, and machine parts."
|
||
},
|
||
{
|
||
"question": "For High Carbon Steels, briefly describe the properties and typical applications.",
|
||
"answer": "Properties: hard, strong, and relatively brittle. Typical applications: chisels, hammers, knives, and hacksaw blades."
|
||
},
|
||
{
|
||
"question": "For High Alloy Steels (Stainless and Tool), briefly describe the properties and typical applications.",
|
||
"answer": "Properties: hard and wear resistant; resistant to corrosion in a large variety of environments. Typical applications: cutting tools, drills, cutlery, food processing, and surgical tools."
|
||
}
|
||
]
|
||
```
|
||
处理第 783/821 条数据...
|
||
[
|
||
{
|
||
"question": "Compare gray and malleable cast irons with respect to composition and heat treatment",
|
||
"answer": "Gray iron--2.5 to 4.0 wt% C and 1.0 to 3.0 wt% Si. For most gray irons there is no heat treatment after solidification. Malleable iron--2.5 to 4.0 wt% C and less than 1.0 wt% Si. White iron is heated in a nonoxidizing atmosphere and at a temperature between 800 and 900 degrees C for an extended time period."
|
||
},
|
||
{
|
||
"question": "Compare gray and malleable cast irons with respect to microstructure",
|
||
"answer": "Gray iron--Graphite flakes are embedded in a ferrite or pearlite matrix. Malleable iron--Graphite clusters are embedded in a ferrite or pearlite matrix."
|
||
},
|
||
{
|
||
"question": "Compare gray and malleable cast irons with respect to mechanical characteristics",
|
||
"answer": "Gray iron--Relatively weak and brittle in tension; good capacity for damping vibrations. Malleable iron--Moderate strength and ductility."
|
||
}
|
||
]
|
||
处理第 784/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What are the distinctive features, limitations, and applications of titanium alloys?",
|
||
"answer": "Distinctive features: relatively low density, high melting temperatures, and high strengths are possible. Limitation: because of chemical reactivity with other materials at elevated temperatures, these alloys are expensive to refine. Applications: aircraft structures, space vehicles, and in chemical and petroleum industries."
|
||
},
|
||
{
|
||
"question": "What are the distinctive features, limitations, and applications of refractory metals?",
|
||
"answer": "Distinctive features: extremely high melting temperatures; large elastic moduli, hardnesses, and strengths. Limitation: some experience rapid oxidation at elevated temperatures. Applications: extrusion dies, structural parts in space vehicles, incandescent light filaments, x-ray tubes, and welding electrodes."
|
||
},
|
||
{
|
||
"question": "What are the distinctive features, limitations, and applications of superalloys?",
|
||
"answer": "Distinctive features: able to withstand high temperatures and oxidizing atmospheres for long time periods. Applications: aircraft turbines, nuclear reactors, and petrochemical equipment."
|
||
},
|
||
{
|
||
"question": "What are the distinctive features, limitations, and applications of noble metals?",
|
||
"answer": "Distinctive features: highly resistant to oxidation, especially at elevated temperatures; soft and ductile. Limitation: expensive. Applications: jewelry, dental restoration materials, coins, catalysts, and thermocouples."
|
||
}
|
||
]
|
||
```
|
||
处理第 785/821 条数据...
|
||
It's a single issue.
|
||
处理第 786/821 条数据...
|
||
[
|
||
{
|
||
"question": "What are the characteristics of sand casting technique?",
|
||
"answer": "For sand casting, sand is the mold material, a two-piece mold is used, ordinarily the surface finish is not an important consideration, the sand may be reused (but the mold may not), casting rates are low, and large pieces are usually cast."
|
||
},
|
||
{
|
||
"question": "What are the characteristics of die casting technique?",
|
||
"answer": "For die casting, a permanent mold is used, casting rates are high, the molten metal is forced into the mold under pressure, a two-piece mold is used, and small pieces are normally cast."
|
||
},
|
||
{
|
||
"question": "What are the characteristics of investment casting technique?",
|
||
"answer": "For investment casting, a single-piece mold is used, which is not reusable; it results in high dimensional accuracy, good reproduction of detail, and a fine surface finish; and casting rates are low."
|
||
},
|
||
{
|
||
"question": "What are the characteristics of lost foam casting technique?",
|
||
"answer": "For lost foam casting, the pattern is polystyrene foam, whereas the mold material is sand. Complex geometries and tight tolerances are possible. Casting rates are higher than for investment, and there are few environmental wastes."
|
||
},
|
||
{
|
||
"question": "What are the characteristics of continuous casting technique?",
|
||
"answer": "For continuous casting, at the conclusion of the extraction process, the molten metal is cast into a continuous strand having either a rectangular or circular cross-section; these shapes are desirable for subsequent secondary metal-forming operations. The chemical composition and mechanical properties are relatively uniform throughout the cross-section."
|
||
}
|
||
]
|
||
处理第 787/821 条数据...
|
||
It's a single issue.
|
||
处理第 788/821 条数据...
|
||
It's a single issue.
|
||
处理第 789/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is the process of compression molding used to form plastic materials?",
|
||
"answer": "For compression molding, both heat and pressure are applied after the polymer and necessary additives are situated between the mold members."
|
||
},
|
||
{
|
||
"question": "What is the process of transfer molding used to form plastic materials?",
|
||
"answer": "For transfer molding, the solid materials (normally thermosetting in nature) are first melted in the transfer chamber prior to being forced into the die."
|
||
},
|
||
{
|
||
"question": "What is the process of injection molding used to form plastic materials?",
|
||
"answer": "For injection molding (normally used for thermoplastic materials), the raw materials are impelled by a ram through a heating chamber, and finally into the die cavity."
|
||
}
|
||
]
|
||
```
|
||
处理第 790/821 条数据...
|
||
[
|
||
{
|
||
"question": "Why must fiber materials that are melt-spun be thermoplastic?",
|
||
"answer": "Fiber materials that are melt spun must be thermoplastic because they must be capable of forming a viscous liquid when heated, which is not possible for thermosets."
|
||
},
|
||
{
|
||
"question": "Why must fiber materials that are drawn be thermoplastic?",
|
||
"answer": "Fiber materials that are drawn must be thermoplastic because during drawing, mechanical elongation must be possible; inasmuch as thermosetting materials are, in general, hard and relatively brittle, they are not easily elongated."
|
||
}
|
||
]
|
||
处理第 791/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is the maximum thermal conductivity for a cermet that contains 90 vol% titanium carbide (TiC) particles in a nickel matrix, given thermal conductivities of 27 W/m·K for TiC and 67 W/m·K for Ni?",
|
||
"answer": "The maximum thermal conductivity k_max is calculated as follows: k_max = k_m V_m + k_p V_p = k_Ni V_Ni + k_TiC V_TiC = (67 W/m·K)(0.10) + (27 W/m·K)(0.90) = 31.0 W/m·K"
|
||
},
|
||
{
|
||
"question": "What is the minimum thermal conductivity for a cermet that contains 90 vol% titanium carbide (TiC) particles in a nickel matrix, given thermal conductivities of 27 W/m·K for TiC and 67 W/m·K for Ni?",
|
||
"answer": "The minimum thermal conductivity k_min is calculated as follows: k_min = (k_Ni k_TiC) / (V_Ni k_TiC + V_TiC k_Ni) = (67 W/m·K)(27 W/m·K) / [(0.10)(27 W/m·K) + (0.90)(67 W/m·K)] = 28.7 W/m·K"
|
||
}
|
||
]
|
||
```
|
||
处理第 792/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Estimate the upper limit for the elastic modulus of the composite consisting of tungsten particles within a copper matrix with volume fractions of 0.70 for tungsten and 0.30 for copper, given the modulus of elasticity for copper is 110 GPa and for tungsten is 407 GPa.",
|
||
"answer": "The upper limit for the elastic modulus of the composite is calculated as follows: E_c(u) = E_Cu V_Cu + E_W V_W = (110 GPa)(0.30) + (407 GPa)(0.70) = 318 GPa."
|
||
},
|
||
{
|
||
"question": "Estimate the upper limit for the specific gravity of the composite consisting of tungsten particles within a copper matrix with volume fractions of 0.70 for tungsten and 0.30 for copper, given the specific gravity for copper is 8.9 and for tungsten is 19.3.",
|
||
"answer": "The upper limit for the specific gravity of the composite is calculated as follows: ρ_c = ρ_Cu V_Cu + ρ_W V_W = (8.9)(0.30) + (19.3)(0.70) = 16.18."
|
||
},
|
||
{
|
||
"question": "Calculate the specific stiffness of the composite using the upper limits for the elastic modulus (318 GPa) and specific gravity (16.18) of the composite.",
|
||
"answer": "The specific stiffness is calculated as follows: Specific Stiffness = E_c(u) / ρ_c = 318 GPa / 16.18 = 19.65 GPa."
|
||
},
|
||
{
|
||
"question": "Calculate the specific stiffness of the composite using the specific stiffness-volume fraction product for both copper and tungsten, given the modulus of elasticity for copper is 110 GPa and for tungsten is 407 GPa, and the specific gravity for copper is 8.9 and for tungsten is 19.3, with volume fractions of 0.30 for copper and 0.70 for tungsten.",
|
||
"answer": "The specific stiffness is calculated as follows: Specific Stiffness = (E_Cu / ρ_Cu) V_Cu + (E_W / ρ_W) V_W = (110 GPa / 8.9)(0.30) + (407 GPa / 19.3)(0.70) = 18.47 GPa."
|
||
}
|
||
]
|
||
```
|
||
处理第 793/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For a continuous and oriented fiber-reinforced composite with moduli of elasticity in the longitudinal and transverse directions of 33.1 GPa and 3.66 GPa respectively, and a fiber volume fraction of 0.30, determine the modulus of elasticity of the fiber phase.",
|
||
"answer": "Using the longitudinal modulus equation: E_cl = E_m(1 - V_f) + E_f V_f, and the transverse modulus equation: E_ct = (E_m E_f) / [(1 - V_f) E_f + V_f E_m], we solve these simultaneously for E_f. The solution yields E_f = 104 GPa (15 x 10^6 psi)."
|
||
},
|
||
{
|
||
"question": "For a continuous and oriented fiber-reinforced composite with moduli of elasticity in the longitudinal and transverse directions of 33.1 GPa and 3.66 GPa respectively, and a fiber volume fraction of 0.30, determine the modulus of elasticity of the matrix phase.",
|
||
"answer": "Using the longitudinal modulus equation: E_cl = E_m(1 - V_f) + E_f V_f, and the transverse modulus equation: E_ct = (E_m E_f) / [(1 - V_f) E_f + V_f E_m], we solve these simultaneously for E_m. The solution yields E_m = 2.6 GPa (3.77 x 10^5 psi)."
|
||
}
|
||
]
|
||
```
|
||
处理第 794/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "In an aligned and continuous carbon fiber-reinforced nylon 6,6 composite, the fibers are to carry 97% of a load applied in the longitudinal direction. Using the data provided, determine the volume fraction of fibers required. The modulus of elasticity for carbon fiber is 260 GPa (37x10^6 psi) and for nylon 6,6 is 2.8 GPa (4.0x10^5 psi).",
|
||
"answer": "To compute the volume fraction of fibers (V_f) required for the fibers to carry 97% of the load, we use the equation: F_f/F_m = (E_f V_f)/(E_m (1-V_f)). Given F_f/F_m = 0.97/(1-0.97) = 32.3, E_f = 260 GPa, and E_m = 2.8 GPa, we substitute these values into the equation: 32.3 = (260 GPa V_f)/(2.8 GPa (1-V_f)). Solving for V_f yields V_f = 0.258."
|
||
},
|
||
{
|
||
"question": "What will be the tensile strength of this composite? Assume that the matrix stress at fiber failure is 50 MPa (7250 psi). The tensile strength of carbon fiber is 4000 MPa (580,000 psi) and of nylon 6,6 is 76 MPa (11,000 psi). The volume fraction of fibers (V_f) is 0.258.",
|
||
"answer": "The tensile strength of the composite (σ_cl*) is calculated using the equation: σ_cl* = σ_m* (1-V_f) + σ_f* V_f. Given σ_m* = 50 MPa, V_f = 0.258, and σ_f* = 4000 MPa, we substitute these values into the equation: σ_cl* = (50 MPa)(1-0.258) + (4000 MPa)(0.258) = 1070 MPa (155,000 psi)."
|
||
}
|
||
]
|
||
```
|
||
处理第 795/821 条数据...
|
||
[
|
||
{
|
||
"question": "Cite several reasons why fiberglass-reinforced composites are used extensively.",
|
||
"answer": "Reasons why fiberglass-reinforced composites are utilized extensively are: (1) glass fibers are very inexpensive to produce; (2) these composites have relatively high specific strengths; and (3) they are chemically inert in a wide variety of environments."
|
||
},
|
||
{
|
||
"question": "Cite several limitations of fiberglass-reinforced composites.",
|
||
"answer": "Several limitations of these composites are: (1) care must be exercised in handling the fibers inasmuch as they are susceptible to surface damage; (2) they are lacking in stiffness in comparison to other fibrous composites; and (3) they are limited as to maximum temperature use."
|
||
}
|
||
]
|
||
处理第 796/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "What is a hybrid composite?",
|
||
"answer": "A hybrid composite is a composite that is reinforced with two or more different fiber materials in a single matrix."
|
||
},
|
||
{
|
||
"question": "List two important advantages of hybrid composites over normal fiber composites.",
|
||
"answer": "Two advantages of hybrid composites are: (1) better overall property combinations, and (2) failure is not as catastrophic as with single-fiber composites."
|
||
}
|
||
]
|
||
```
|
||
处理第 797/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "(a) Write an expression for the modulus of elasticity for a hybrid composite in which all fibers of both types are oriented in the same direction.",
|
||
"answer": "For a hybrid composite having all fibers aligned in the same direction, the longitudinal modulus of elasticity may be determined using a modified form of Equation as follows: E_cl = E_m V_m + E_f1 V_f1 + E_f2 V_f2 in which the subscripts f1 and f2 refer to the two types of fibers."
|
||
},
|
||
{
|
||
"question": "(b) Using this expression, compute the longitudinal modulus of elasticity of a hybrid composite consisting of aramid and glass fibers in volume fractions of 0.25 and 0.35, respectively, within a polyester resin matrix [E_m=4.0 GPa (6x10^5 psi)]. The elastic moduli of aramid and glass fibers are, respectively, 131 GPa (19x10^6 psi) and 72.5 GPa (10.5x10^6 psi).",
|
||
"answer": "Now we are asked to compute the longitudinal elastic modulus for a glass- and aramid-fiber hybrid composite. Thus, from the previous expression E_cl = (4 GPa)(1.0 - 0.25 - 0.35) + (131 GPa)(0.25) + (72.5 GPa)(0.35) = 59.7 GPa (8.67x10^6 psi)."
|
||
}
|
||
]
|
||
```
|
||
处理第 798/821 条数据...
|
||
[
|
||
{
|
||
"question": "Briefly describe laminar composites.",
|
||
"answer": "Laminar composites are a series of sheets or panels, each of which has a preferred high-strength direction. These sheets are stacked and then cemented together such that the orientation of the high-strength direction varies from layer to layer."
|
||
},
|
||
{
|
||
"question": "What is the prime reason for fabricating laminar composites?",
|
||
"answer": "These composites are constructed in order to have a relatively high strength in virtually all directions within the plane of the laminate."
|
||
}
|
||
]
|
||
处理第 799/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Is a voltage generated between the two cell halves of an Fe/Fe2+ concentration cell where one half has an Fe2+ concentration of 0.5 M and the other has 2 x 10^-2 M?",
|
||
"answer": "Yes, a voltage is generated between the two cell halves."
|
||
},
|
||
{
|
||
"question": "What is the magnitude of the voltage generated in the Fe/Fe2+ concentration cell with Fe2+ concentrations of 0.5 M and 2 x 10^-2 M?",
|
||
"answer": "The magnitude of the voltage generated is 0.0414 V."
|
||
},
|
||
{
|
||
"question": "Which electrode will be oxidized in the Fe/Fe2+ concentration cell with Fe2+ concentrations of 0.5 M and 2 x 10^-2 M?",
|
||
"answer": "The electrode in the cell half with the lower Fe2+ concentration (2 x 10^-2 M) will be oxidized."
|
||
}
|
||
]
|
||
```
|
||
处理第 800/821 条数据...
|
||
[
|
||
{
|
||
"question": "For the aluminum and cast iron alloys that are coupled in seawater, predict the possibility of corrosion; if corrosion is probable, note which metal/alloy will corrode.",
|
||
"answer": "For the aluminum-cast iron couple, corrosion is possible, and aluminum will corrode."
|
||
},
|
||
{
|
||
"question": "For the Inconel and nickel alloys that are coupled in seawater, predict the possibility of corrosion; if corrosion is probable, note which metal/alloy will corrode.",
|
||
"answer": "For the Inconel-nickel couple, corrosion is unlikely inasmuch as both alloys appear within the same set of brackets (in both active and passive states)."
|
||
},
|
||
{
|
||
"question": "For the cadmium and zinc alloys that are coupled in seawater, predict the possibility of corrosion; if corrosion is probable, note which metal/alloy will corrode.",
|
||
"answer": "For the cadmium-zinc couple, corrosion is possible, and zinc will corrode."
|
||
},
|
||
{
|
||
"question": "For the brass and titanium alloys that are coupled in seawater, predict the possibility of corrosion; if corrosion is probable, note which metal/alloy will corrode.",
|
||
"answer": "For the brass-titanium pair, corrosion is possible, and brass will corrode."
|
||
},
|
||
{
|
||
"question": "For the low-carbon steel and copper alloys that are coupled in seawater, predict the possibility of corrosion; if corrosion is probable, note which metal/alloy will corrode.",
|
||
"answer": "For the low-carbon steel-copper couple, corrosion is possible, and the low-carbon steel will corrode."
|
||
}
|
||
]
|
||
处理第 801/821 条数据...
|
||
It's a single issue.
|
||
处理第 802/821 条数据...
|
||
[
|
||
{
|
||
"question": "Assuming that activation polarization controls both oxidation and reduction reactions, determine the rate of corrosion of metal M (in mol/cm2·s) given the following data: For Metal M: V(M/M2+) = -0.90 V, i0 = 10-12 A/cm2, β = +0.10; For Hydrogen: V(H+/H2) = 0 V, i0 = 10-10 A/cm2, β = -0.15.",
|
||
"answer": "The rate of oxidation for metal M is calculated by first establishing the potential expressions for both oxidation and reduction reactions. For hydrogen reduction: VH = V(H+/H2) + βH log(i/i0H). For M oxidation: VM = V(M/M2+) + βM log(i/i0M). Setting VH = VM and solving for log iC gives: log iC = [1/(βM - βH)][V(H+/H2) - V(M/M2+) - βH log i0H + βM log i0M]. Substituting the given values: log iC = [1/(0.10 - (-0.15))][0 - (-0.90) - (-0.15){log(10-10)} + (0.10){log(10-12)}] = -7.20. Thus, iC = 10-7.20 = 6.31 × 10-8 A/cm2 = 6.31 × 10-8 C/s·cm2. The oxidation rate is then: r = iC / (nF) = (6.31 × 10-8 C/s·cm2) / (2 × 96,500 C/mol) = 3.27 × 10-13 mol/cm2·s."
|
||
},
|
||
{
|
||
"question": "Compute the corrosion potential for the reaction involving metal M given the following data: For Metal M: V(M/M2+) = -0.90 V, i0 = 10-12 A/cm2, β = +0.10; For Hydrogen: V(H+/H2) = 0 V, i0 = 10-10 A/cm2, β = -0.15, and the corrosion current density iC = 6.31 × 10-8 A/cm2.",
|
||
"answer": "The corrosion potential Vc is calculated using the expression for hydrogen reduction: Vc = V(H+/H2) + βH log(iC/i0H). Substituting the given values: Vc = 0 + (-0.15 V) log[(6.31 × 10-8 A/cm2) / (10-10 A/cm2)] = -0.420 V."
|
||
}
|
||
]
|
||
处理第 803/821 条数据...
|
||
It's a single issue.
|
||
处理第 804/821 条数据...
|
||
It's a single issue.
|
||
处理第 805/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For copper, the heat capacity at constant volume \(C_{V}\) at \(20 \mathrm{~K}\) is \(0.38 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\) and the Debye temperature is \(340 \mathrm{~K}\). Estimate the specific heat at \(40 \mathrm{~K}\).",
|
||
"answer": "For copper, \(C_{V}\) at \(20 \mathrm{~K}\) may be approximated by Equation, since this temperature is significantly below the Debye temperature \((340 \mathrm{~K})\). The value of \(C_{V}\) at \(20 \mathrm{~K}\) is given, and thus, we may compute the constant \(A\) by rearranging Equation as follows: \[ A=\frac{C_{V}}{T^{3}}=\frac{0.38 \mathrm{~J} / \mathrm{mol}-\mathrm{K}}{(20 \mathrm{~K})^{3}}=4.75 \times 10^{-5} \mathrm{~J} / \mathrm{mol}-\mathrm{K}^{4} \] Therefore, at \(40 \mathrm{~K}\) the value of \(C_{V}\) is equal to \[ C_{V} =A T^{3} =(4.75 \times 10^{-5} \mathrm{~J} / \mathrm{mol}-\mathrm{K}^{4})(40 \mathrm{~K})^{3} =3.04 \mathrm{~J} / \mathrm{mol}-\mathrm{K} \] and the specific heat is computed using the following expression: \[ c_{V}=\left(\frac{C_{V}}{A^{\prime}}\right)\left(\frac{1000 \mathrm{~g}}{1 \mathrm{~kg}}\right) \] In this equation, the parameter \(A^{\prime}\) represents the atomic weight of the material. Therefore, the specific heat for copper at \(40 \mathrm{~K}\) is equal to \[ c_{V}=\left(\frac{3.04 \mathrm{~J} / \mathrm{mol}-\mathrm{K}}{63.55 \mathrm{~g} / \mathrm{mol}}\right)\left(\frac{1000 \mathrm{~g}}{1 \mathrm{~kg}}\right) =47.8 \mathrm{~J} / \mathrm{kg}-\mathrm{K} \]"
|
||
},
|
||
{
|
||
"question": "For copper, the heat capacity at constant volume \(C_{V}\) at \(20 \mathrm{~K}\) is \(0.38 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\) and the Debye temperature is \(340 \mathrm{~K}\). Estimate the specific heat at \(400 \mathrm{~K}\).",
|
||
"answer": "Because \(400 \mathrm{~K}\) is above the Debye temperature, a good approximation for \(C_{V}\) is as follows: \[ C_{V}=3 R = (3)(8.31 \mathrm{~J} / \mathrm{mol}-\mathrm{K})=24.9 \mathrm{~J} / \mathrm{mol}-\mathrm{K} \] We convert this \(C_{V}\) to specific heat using Equation as follows: \[ c_{V} =\left(\frac{C_{V}}{A^{\prime}}\right)\left(\frac{1000 \mathrm{~g}}{1 \mathrm{~kg}}\right) =\left(\frac{24.9 \mathrm{~J} / \mathrm{mol}-\mathrm{K}}{63.55 \mathrm{~g} / \mathrm{mol}}\right)\left(\frac{1000 \mathrm{~g}}{1 \mathrm{~kg}}\right) =392 \mathrm{~J} / \mathrm{kg}-\mathrm{K} \]"
|
||
}
|
||
]
|
||
```
|
||
API调用错误: Invalid \escape: line 3 column 71 (char 78)
|
||
处理第 806/821 条数据...
|
||
It's a single issue.
|
||
处理第 807/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "For the pair of materials pure silver and sterling silver (92.5 wt% Ag-7.5 wt% Cu), which has the larger thermal conductivity? Justify your choice.",
|
||
"answer": "Pure silver will have a larger conductivity than sterling silver because the impurity atoms in the latter will lead to a greater degree of free electron scattering."
|
||
},
|
||
{
|
||
"question": "For the pair of materials fused silica and polycrystalline silica, which has the larger thermal conductivity? Justify your choice.",
|
||
"answer": "Polycrystalline silica will have a larger conductivity than fused silica because fused silica is noncrystalline and lattice vibrations are more effectively scattered in noncrystalline materials."
|
||
},
|
||
{
|
||
"question": "For the pair of materials linear and syndiotactic poly(vinyl chloride) (DP =1000) and linear and syndiotactic polystyrene (DP =1000), which has the larger thermal conductivity? Justify your choice.",
|
||
"answer": "The poly(vinyl chloride) will have the larger conductivity than the polystyrene because the former will have the higher degree of crystallinity. Both polymers are syndiotactic and have the same degree of polymerization. However, with regard to side-group bulkiness, the PVC is more likely to crystallize. Since heat transfer is by molecular chain vibrations, and the coordination of these vibrations increases with percent crystallinity, the higher the crystallinity, the greater the thermal conductivity."
|
||
},
|
||
{
|
||
"question": "For the pair of materials atactic polypropylene (Mw=10^6 g/mol) and isotactic polypropylene (Mw=10^5 g/mol), which has the larger thermal conductivity? Justify your choice.",
|
||
"answer": "The isotactic polypropylene will have a larger thermal conductivity than the atactic polypropylene because isotactic polymers have higher degrees of crystallinity. The influence of crystallinity on conductivity is explained in part (c)."
|
||
}
|
||
]
|
||
```
|
||
处理第 808/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Briefly explain why thermal stresses may be introduced into a structure by rapid heating or cooling.",
|
||
"answer": "Thermal stresses may be introduced into a structure by rapid heating or cooling because temperature gradients are established across the cross section due to more rapid temperature changes at the surface than within the interior; thus, the surface will expand or contract at a different rate than the interior and since this surface expansion or contraction will be restrained by the interior, stresses will be introduced."
|
||
},
|
||
{
|
||
"question": "For cooling, what is the nature of the surface stresses?",
|
||
"answer": "For cooling, the surface stresses will be tensile in nature since the interior contracts to a lesser degree than the cooler surface."
|
||
},
|
||
{
|
||
"question": "For heating, what is the nature of the surface stresses?",
|
||
"answer": "For heating, the surface stresses will be compressive in nature since the interior expands to a lesser degree than the hotter surface."
|
||
}
|
||
]
|
||
```
|
||
处理第 809/821 条数据...
|
||
[
|
||
{
|
||
"question": "The magnetic flux density within a bar of some material is 0.630 tesla at an H field of 5 x 10^5 A/m. Compute the magnetic permeability for this material.",
|
||
"answer": "The magnetic permeability of this material may be determined as follows: mu = B / H = 0.630 tesla / 5 x 10^5 A/m = 1.260 x 10^-6 H/m"
|
||
},
|
||
{
|
||
"question": "The magnetic flux density within a bar of some material is 0.630 tesla at an H field of 5 x 10^5 A/m, and its magnetic permeability is 1.260 x 10^-6 H/m. Compute the magnetic susceptibility for this material.",
|
||
"answer": "The magnetic susceptibility is calculated as follows: chi_m = mu_r - 1 = mu / mu_0 - 1 = (1.260 x 10^-6 H/m) / (1.257 x 10^-6 H/m) - 1 = 2.387 x 10^-3"
|
||
},
|
||
{
|
||
"question": "The magnetic susceptibility of a material is 2.387 x 10^-3. What type(s) of magnetism would you suggest is (are) being displayed by this material? Why?",
|
||
"answer": "This material would display both diamagnetic and paramagnetic behavior. All materials are diamagnetic, and since chi_m is positive and on the order of 10^-3, there would also be a paramagnetic contribution."
|
||
}
|
||
]
|
||
处理第 810/821 条数据...
|
||
It's a single issue.
|
||
处理第 811/821 条数据...
|
||
It's a single issue.
|
||
处理第 812/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Calculate the number of Bohr magnetons per unit cell (n_B) given the saturation magnetization (1.35 x 10^5 A/m), the cubic unit cell edge length (1.2529 nm), and the Bohr magneton value (9.27 x 10^-24 A·m^2/BM).",
|
||
"answer": "n_B = (1.35 x 10^5 A/m) * (1.2529 x 10^-9 m)^3 / (9.27 x 10^-24 A·m^2/BM) = 28.64 Bohr magnetons/unit cell."
|
||
},
|
||
{
|
||
"question": "Determine the number of Bohr magnetons per formula unit given there are 8 formula units per unit cell and the previously calculated n_B (28.64 Bohr magnetons/unit cell).",
|
||
"answer": "Bohr magnetons per formula unit = 28.64 Bohr magnetons/unit cell / 8 formula units/unit cell = 3.58 Bohr magnetons/formula unit."
|
||
},
|
||
{
|
||
"question": "Calculate the net magnetic moment contribution per formula unit from the Fe^3+ ions, given there are 2 Fe^3+ ions on a sites and 3 Fe^3+ ions on d sites, each with 5 Bohr magnetons, and their magnetic moments are aligned antiparallel.",
|
||
"answer": "Net magnetic moment from Fe^3+ ions = (2 * 5 Bohr magnetons) - (3 * 5 Bohr magnetons) = -5 Bohr magnetons/formula unit."
|
||
},
|
||
{
|
||
"question": "Compute the number of Bohr magnetons associated with each Sm^3+ ion, given the net magnetic moment per formula unit (3.58 Bohr magnetons), the contribution from Fe^3+ ions (-5 Bohr magnetons), and there are 3 Sm^3+ ions per formula unit.",
|
||
"answer": "Number of Bohr magnetons/Sm^3+ = (3.58 Bohr magnetons + 5 Bohr magnetons) / 3 = 2.86 Bohr magnetons/Sm^3+ ion."
|
||
}
|
||
]
|
||
```
|
||
处理第 813/821 条数据...
|
||
It's a single issue.
|
||
处理第 814/821 条数据...
|
||
[
|
||
{
|
||
"question": "Cite the differences between hard and soft magnetic materials in terms of hysteresis behavior.",
|
||
"answer": "Relative to hysteresis behavior, a hard magnetic material has a high remanence, a high coercivity, a high saturation flux density, high hysteresis energy losses, and a low initial permeability; a soft magnetic material, on the other hand, has a high initial permeability, a low coercivity, and low hysteresis energy losses."
|
||
},
|
||
{
|
||
"question": "Cite the differences between hard and soft magnetic materials in terms of typical applications.",
|
||
"answer": "Soft magnetic materials are used in devices that experience alternating and when energy losses must be low-such as transformer cores, generators, motors (electromagnet), and switching circuits. Hard magnetic materials are used in applications that require permanent magnets-such as motors (permanent magnet motors), audio and video recorders, hearing aids, and computer peripherals."
|
||
}
|
||
]
|
||
处理第 815/821 条数据...
|
||
It's a single issue.
|
||
处理第 816/821 条数据...
|
||
It's a single issue.
|
||
处理第 817/821 条数据...
|
||
```json
|
||
[
|
||
{
|
||
"question": "Visible light having a wavelength of 5 × 10^-7 m appears green. Compute the frequency of a photon of this light.",
|
||
"answer": "v = c / λ = (3 × 10^8 m/s) / (5 × 10^-7 m) = 6 × 10^14 s^-1"
|
||
},
|
||
{
|
||
"question": "Visible light having a wavelength of 5 × 10^-7 m appears green. Compute the energy of a photon of this light.",
|
||
"answer": "E = h c / λ = (6.63 × 10^-34 J·s × 3 × 10^8 m/s) / (5 × 10^-7 m) = 3.98 × 10^-19 J (2.48 eV)"
|
||
}
|
||
]
|
||
```
|
||
处理第 818/821 条数据...
|
||
It's a single issue.
|
||
处理第 819/821 条数据...
|
||
It's a single issue.
|
||
处理第 820/821 条数据...
|
||
It's a single issue.
|
||
处理第 821/821 条数据...
|
||
It's a single issue.
|
||
处理完成，结果已保存到 /home/ubuntu/50T/fsy/layer2/QA/code/821_single_select_includes_process.json
|