BCC Metal Molybdenum Temperature (C) 1202 Metal Lead Equilibrium Number of Vacan
ID: 534417 • Letter: B
Question
BCC Metal Molybdenum Temperature (C) 1202 Metal Lead Equilibrium Number of Vacancies (m 3) 8.32E+23 Temperature for Metal A 101 Metal B Tin 1) If the atomic radius of a metal is the value shown above and it has the face-centered cubic aystal structure, calculate the volume of its unit cell in nm 3? Your Answer -0.0482 Incorrect. What is the atomic packing factor for the BCC aystal stnucture? Your Answer 068 Correct! Exact Answer 0.68+/- Find the theoretical density for the FCC Element shown above in glam 3 Your Answer -894 Correct! Exact Answerw 8.8955 1.00E-01 4) Calculate the atomic radius, in nm, of the BCC Metal above utilizing the density and the atomic weight provided by examination booklet Your Answer Calculate the fraction of atom sites that are vacant for copper (Cu at the temperature provided above. Assume an energy for vacancy formation of 0.90 evlatom Incorrect. Your Answer 9.58E-4 Your Answer Incorrect. 133 TE44 Calculate the energy (in evlatom) for vacancy fommation for the Metal A and the equilibrium number of vacancies at the temperature provided above Incorrect Your Answer 28E25 8)Calculate the number of atoms per cubic meter in Metal B (units atoms/m 3) Your Answer -368E28 Correct! Exact Answers 3.6880E+28 1.00E+26 What is the composition in atom percent of an alloy that contains a 36 g Metal A and b) 47 gMetal B?Composition for Metal A Incorrect. Your Answer 56 62 What is the composition, in atom percent of an alloy that contains a 36 g Metal A and b 47gMetal B?Composition for Metal B Your Answer 43 37 Incorrect. Your Answer -2628 Correct! Exact Answer- 2.628680193 ti- 1.00E-01 2) What is the composition of Metal B in atom percent, ifthe alloy consists of 4.5 wt% Metal A and 95.5 wt% of Metal B? Your Answer -97 372 Correct Exact Answer 97.37131981 1.00E-01Explanation / Answer
Lead has an FCC crystal structure and an atomic radius of 8.32 m-3 e+23. The FCC unit cell volume may be computed from Equation
=16 * Radius ^ 3 * sqrt [ 2 ]
Radius = 8.32 m-3 e+23
=0.000832 nm3
= 16 *(0.000832)3 * sqrt [ 2 ]
=0.00000000921 * 1.41
= 1.30 nm3
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