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Niobium forms a substitutional solid solution with
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Answer #1

1. Nb = 1.55 x 10^22 atoms/cm^3

atomic mass Nb = 92.906 g/mol

Using densities,

the weight percent Nb to be added to vanadium would be be (C),

C = 100/[1+ (6.023 x 10^23 x 6.10/92.906 x 1.55 x 10^22) - (6.10/8.57)]

    = 35.23%

2. For gold

atomic mass Au = 196.97 g/mol

we get,

number of gold atoms per cm^3 (N)

N = 6.023 x 10^23 x 10/[(10 x 196.97/19.32) + (196.97 x 90/10.49)]

    = 3.36 x 10^21 atoms/cm^3

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Answer #2

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