## Chemistry and Chemical Reactivity (9th Edition)

a) $65\ m^3$, $65\times10^3\ L$ b) $65\ kg$, $140\ lb$ *The answer may be different depending on the value used for density of air. Be sure to use the density you are given to calculate.
a) The volume of the room is given by $V=L\times w\times h,\ L=18\ ft,\ w=15\ ft,\ h=8\ ft\ 6\ in=8.5\ ft$: $V=2295\ ft^3$ In cubic meters: $2295\ ft^3 \times \left(\frac{0.3048\ m}{1\ ft}\right)^3=65\ m^3$ In liters: $65\ m^3 \times \frac{1000\ L}{1\ m^3}=65\times10^3\ L$ b) Density is given $\rho = 1\ g/L=1\times10^{-3}\ kg/L$, and, since we have the volume, the mass is: $m=65\ kg$ In pounds: $65\ kg \times \frac{1\ lb}{0.4539\ kg} = 140\ lb$