The pressure exerted by a mixture of 3.2 g o | vedclass

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(B) First,calculate the number of moles of each gas: $n_{CH_4} = \frac{3.2 \ g}{16 \ g/mol} = 0.2 \ mol$ $n_{CO_2} = \frac{4.4 \ g}{44 \ g/mol} = 0.1

Vedclass · Accepted Answer

$0.82$

Vedclass · Answer

(B) First,calculate the number of moles of each gas: $n_{CH_4} = \frac{3.2 \ g}{16 \ g/mol} = 0.2 \ mol$ $n_{CO_2} = \frac{4.4 \ g}{44 \ g/mol} = 0.1 \ mol$ Total moles $n_{total} = 0.2 + 0.1 = 0.3 \ mol$ Using the ideal gas equation $PV = nRT$: $P = \frac{n_{total}RT}{V}$ Given $T = 27^{\circ}C = 300 \ K$,$V = 9 \ dm^3 = 9 \ L$,and $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$: $P = \frac{0.3 	imes 0.0821 	imes 300}{9} \ atm$ $P = \frac{7.389}{9} \ atm = 0.821 \ atm$ Thus,the total pressure is approximately $0.82 \ atm$.

The pressure exerted by a mixture of $3.2 \ g$ of methane $(CH_4)$ and $4.4 \ g$ of $CO_2$ contained in a $9 \ dm^3$ flask at $27^{\circ}C$ is: (in $atm$)

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