Step 1

$\mathbf{volume}\mathbf{}\mathbf{=}\mathbf{}\mathbf{14}\mathbf{}\mathbf{ft}\mathbf{\times}\mathbf{22}\mathbf{}\mathbf{ft}\mathbf{}\mathbf{\times}\mathbf{}\mathbf{11}\mathbf{}\mathbf{ft}\phantom{\rule{0ex}{0ex}}\mathbf{volume}\mathbf{}\mathbf{=}\mathbf{}\mathbf{3388}\overline{)\mathbf{}{\mathbf{ft}}^{\mathbf{3}}}\mathbf{}\mathbf{\times}{\left(\frac{1\overline{)m}}{3.281\overline{)\mathrm{ft}}}\right)}^{\overline{)\mathbf{3}}}\mathbf{}\mathbf{\times}\mathbf{}\frac{{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{L}}{\mathbf{1}\mathbf{}\overline{){\mathbf{m}}^{\mathbf{3}}}}$

**volume = 95923 L**

Step 2

T = 24°C + 273.15 = 297.15 K

$\overline{){\mathbf{PV}}{\mathbf{\hspace{0.17em}}}{\mathbf{=}}{\mathbf{}}{\mathbf{nRT}}}\phantom{\rule{0ex}{0ex}}\mathbf{n}\mathbf{}\mathbf{=}\frac{\mathbf{PV}}{\mathbf{RT}}\phantom{\rule{0ex}{0ex}}\mathbf{n}\mathbf{}\mathbf{=}\mathbf{}\frac{(1\overline{)\mathrm{atm}})(95923\overline{)L})}{(0.0821{\displaystyle \frac{\overline{)\mathrm{atm}}\xb7\overline{)L}}{\mathrm{mol}\xb7\overline{)K}}})(297.15\overline{)K})}$

Nickel carbonyl, Ni(CO)_{4} is one of the most toxic substances known. The present maximum allowable concentration in laboratory air during an 8-hr workday is 1 ppb (parts per billion) by volume, which means that there is one mole of Ni(CO)_{4} for every 10^{9} moles of gas. Assume 24 ^{o}C and 1.00 atm pressure. What mass of Ni(CO)_{4} is allowable in a laboratory that is 14ft x 22ft x 11ft?

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