# Let's Review: The Tools of Quantitative Chemistry - Study Questions - Page 43d: 52

$9,600\ m^2$

#### Work Step by Step

Gold's mass and density: $m=279\ kg=279\times10^3\ g,\ \rho=19.3\ g/cm^3$ Gold's volume: $V=279\times10^3\div19.3=1.44\times10^4\ cm^3$ Layer thickness of $0.0015 mm=1.5\times10^{-4}\ cm$, volume given by $V=A.t$ Area covered: $A=9.6\times10^7\ cm^2$ Area in m²: $9.6\times10^7\ cm^2\times\left(\frac{1\ m}{100\ cm}\right)^2=9600\ m^2$

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