Calculate the radius of a palladium atom in nm, given that Pd has an FCC crystal structure, a density of 12.0 g/cm3 and an atomic weight of 106.4 g/mol. A. 138 B. 0.138 C. 1.38 D. 1.38 x 10-8
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Step 1
0 g/cm^3 Atomic weight (AW) = 106.4 g/mol Avogadro's number = 6.022 x 10^23 atoms/mol a = (Z * AW / (d * Avogadro's number))^(1/3) a = (4 * 106.4 / (12.0 * 6.022 x 10^23))^(1/3) a ≈ 3.89 x 10^-8 cm Show more…
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