Question

a 1.40 g sample of gold having a density of 19.32 g/cm^3 is hammered into a very thin sheet. if the area of the sheet is 35.0ft^2, what is the average thickness of the sheet expressed in centimeters?

Answer 1

Using back the formula,
\[\rho= \frac{m}{V}\]
\[V= \frac{m}{\rho}\]
\[V= \frac{1.40 g }{19.32 g/cm^3}\]
\[V= 0.072cm^3\]
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Changing 35.0ft^2,
\[35.0ft^2=\frac{929cm^2}{1ft^2}=32515cm^2\]
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We assume that the metal sheet is either cubed or cuboid,
\[V=Length \times Base \times Height\]
\[V=Aree \times Height\]
\[Height=\frac{Volume}{Area}\]
\[Height=\frac{0.072cm^3}{32515cm^2}\]
\[Height=2.21 \times 10^{-6}cm\]

Answer 2

thanks can you explain how you get from v=basexhightxlength to v=areaxheight?

Answer 3

got it thanks