If you covered Canada with 1 mole's worth of golf balls, how thick would the layer of golf balls be? I know moles are a chemistry thing, but only really the quantity matters in this question.
Canada, without counting the water, has an area of $9,093,507 \text{ km}^2$. A golf ball (pretending it is square) has a width/height of $4.318 \times 10^{-5}$ km. If I covered Canada with 1 mole ($6.022 \times 10^{23}$) of golf balls, how thick would the pile of golf balls be?
 A: The answer will depend on how the golf balls are packed. A random packing of equally sized spheres has packing density $0.6400$, that is, it would occupy $64.00\%$ of the space that contains it (see MathWorld: Sphere Packing). A hexagonal close packing, on the other hand, is far more efficient and has packing density $0.7405$. If there are $n$ many balls, each of volume $v$, packed together with density $\rho$, they will have a total volume that satisfies ${\rho}=\frac{nv}{V}$. If those packed balls are amassed up to an approximately level height $h$ on a surface with area $A$, then they will approximately form a prism (with an irregular base) of volume $V=hA$. It follows from elementary algebra that $h=\frac{nv}{\rho A}$. In your case, a random packing (hence, density $\rho=0.6400$) of $n=6.0\times10^{23}$ approximately spherical balls of diameter $4.3\,\mathrm{cm}$ (hence, volume $v=41\,\mathrm{cm}^3$) on a surface of area $9\times10^{6}\,\mathrm{km}^2$ will have height $4\times10^{3}\,\mathrm{km}$.
