A metal oxidizes at 1200 K with parabolic rate constant $3\times10^{-6}\ \mathrm{g^2\,cm^{-4}\,s^{-1}}$. How long to form a 2 μm film, in seconds to 2 decimals? Oxide density is 6.5 g/cm³.
The mass-per-area basis in the given rate constant gives $(m/A)^2=k_pt$. Using the intended oxide-film mass convention, $m/A=\rho x=6.5(2\times10^{-4})=0.0013$ g/cm². Hence $t=(0.0013)^2/(3\times10^{-6})=0.56333$ s, or 0.56 s. This uses oxide mass rather than oxygen-only mass gain; a different convention would require additional composition information.
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