Answer:
[tex]2.0489\times 10^{-6} cm[/tex] 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 = [tex]19.32g/cm^3[/tex]
[tex]D=\frac{m}{V}=19.32g/cm^3=\frac{1.78 g}{V}[/tex]
[tex]V = 0.09213 cm^3[/tex]
Area of the hammered gold sheet,A = [tex]48.4 ft^2=44,965.052 cm^2[/tex]
Thickness of the hammered gold = h
([tex]1 ft^2=929.03 cm^2[/tex])
Volume = Area × thickness
[tex]V= A\times h[/tex]
[tex]0.09213 cm^3=44,965.052 cm^2\times h[/tex]
[tex]h=2.0489\times 10^{-6} cm[/tex]
[tex]2.0489\times 10^{-6} cm[/tex] is the average thickness of the sheet.
