Question

) 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?

Answer 1

Step-by-step 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.

Answer 2

(a). 5.4 × 10^23 atoms.

(b). 0.8928 moles Ni.

Step-by-step explanation:

The first thing to do is to convert surface area from the in^2 to cm^2 and also the thickness unit to cm.

Therefore, 200 in^2 = 1290.3 cm^2 and 0.002 in = 0.00508 cm.

The next thing to do is to calculate the volume by using the formula below;

Volume= A × d. Where A = area and d = thickness.

Then, volume = 1290.3 × 0.00508= 6.55 cm^3.

Although not given but the Density if Nickel = 8 g/cm^3.

We know that the formula for Calculating density = mass / volume. So, we have ; mass = density × volume.

Mass= 8 × 6.55 = 52.4 g.

(a). The number of atoms of Nickel= moles × Avogrado's number.

Numbers of moles= mass / molar mass.

==> 52.4 / 58.6934 = 0.8928 moles.

Then, the number of atoms of Nickel = 0.8928 × 6.02 × 10^23.

= 5.4 × 10^23 atoms.

(b). The number of moles of Nickel has been Calculated in (a) above to be = 0.8928 moles Ni.