) In order to plate a steel part having a surface area of 200 in.2 with a 0.002 in.-thick layer of nickel: (a) how many atoms of nickel are required? (b) How many moles of nickel are required?

Explanation:

Ar = 58.7 g/mol

= 0.0587 kg/mol

Na = 6.02 × 10^23 atoms/mol

Density = 8908 kg/m³

Volume = area × thickness

= 200 × 0.002

= 0.4 in^3

To m3,

0.4 in^3 × (0.0254 m)^3/(1 in)^3

= 5.53 × 10^-6 m^3

Number of atoms = (5.53 × 10^-6 × 8908 × 6.02 × 10^23)/0.0587

= 5.052 × 10^23 atoms.

B.

Moles = mass/molar mass

= (5.53 × 10^-6 × 8908)/0.0587

= 0.839 moles.

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