Q3/ The atomic radius of FCC Aluminum is 0.143 nm, and the atomic mass is 26.98 g/mol, determine the number of vacancies needed for a FCC Aluminum crystal to have a density of 2.67 g/cm3.
Q3/ The atomic radius of FCC Aluminum is 0.143 nm, and the atomic mass is 26.98 g/mol, determine the number of vacancies needed for a FCC Aluminum crystal to have a density of 2.67 g/cm3.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Q3/ The atomic radius of FCC Aluminum is 0.143 nm, and the atomic
mass is 26.98 g/mol, determine the number of vacancies needed for a
FCC Aluminum crystal to have a density of 2.67 g/cm3.
the number of vacancies
O 9872 *10^17 vacancy/ cm3
O 8872 *10^17 vacancy/ cm3
8072 *10^17 vacancy/ cm3
oonn *10017voeenou/om2
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