Question

Gold is a very soft metal that can be hammered into extremely thin sheets known as gold leaf. If a 1.78 g piece of gold is hammered into a sheet whose area is 48.4 ft2, what is the average thickness of the sheet? (The density of gold is 19.32 g cm-3.) 9.917*10^-5 cm

Answer 1

Answer:

$2.0489\times 10^{-6} cm$ is the average thickness of the sheet.

Explanation:

Mass of gold ,m= 1.78 g

Volume of the gold = V

Density of the gold = D = $19.32g/cm^3$

$D=\frac{m}{V}=19.32g/cm^3=\frac{1.78 g}{V}$

$V = 0.09213 cm^3$

Area of the hammered gold sheet,A = $48.4 ft^2=44,965.052 cm^2$

Thickness of the hammered gold = h

($1 ft^2=929.03 cm^2$)

Volume = Area × thickness

$V= A\times h$

$0.09213 cm^3=44,965.052 cm^2\times h$

$h=2.0489\times 10^{-6} cm$

$2.0489\times 10^{-6} cm$ is the average thickness of the sheet.