## Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning

# Let's Review: The Tools of Quantitative Chemistry - Study Questions: 54

#### Answer

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.

#### Work Step by Step

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: $65\ kg \times \frac{1\ lb}{0.4539\ kg} = 140\ lb$

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