## Chemistry and Chemical Reactivity (9th Edition)

$2\times10^{-9}\ m$
Volume of oil in $m^3$: $1\ t.s\times\frac{5\ cm^3}{1\ t.s.}\times\left(\frac{1\ m}{100\ cm}\right)^3=5\times10^{-6}\ m^3$ Area of water in m²: $0.5\ acre\times \frac{1\times10^4\ m^2}{2.47\ acres}=2\times10^3\ m^2$ The volume being given by: $V=A.t$ $t=5\times10^{-6}\div2\times10^3 = 2\times10^{-9}\ m$ Given that this is in the range of nm, this is propably the molecular size.