Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Let's Review: The Tools of Quantitative Chemistry - Study Questions - Page 43e: 54


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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