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(B) Using the ideal gas equation $PV = nRT$,where $n$ is the total number of moles.Given: $P = 100 \; kPa = 10^{5} \; Pa$,$V = 2000 \; cm^{3} = 2 	im

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$3/1$

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(B) Using the ideal gas equation $PV = nRT$,where $n$ is the total number of moles.Given: $P = 100 \; kPa = 10^{5} \; Pa$,$V = 2000 \; cm^{3} = 2 	imes 10^{-3} \; m^{3}$,$T = 300 \; K$,$R = 8.314 \; J/(mol \cdot K) \approx 25/3 \; J/(mol \cdot K)$.Total moles $n = \frac{PV}{RT} = \frac{10^{5} 	imes 2 	imes 10^{-3}}{(25/3) 	imes 300} = \frac{200}{2500} = 0.08 \; mol$.Let $n_{1}$ be the moles of $H_{2}$ and $n_{2}$ be the moles of $O_{2}$.$n_{1} + n_{2} = 0.08$ (Equation $1$).The total mass is $m = n_{1}M_{1} + n_{2}M_{2} = 0.76 \; g$.$2n_{1} + 32n_{2} = 0.76$ (Equation $2$).From Equation $1$,$n_{1} = 0.08 - n_{2}$.Substitute into Equation $2$: $2(0.08 - n_{2}) + 32n_{2} = 0.76$.$0.16 - 2n_{2} + 32n_{2} = 0.76 \implies 30n_{2} = 0.60 \implies n_{2} = 0.02 \; mol$.Then $n_{1} = 0.08 - 0.02 = 0.06 \; mol$.The ratio $n_{1}/n_{2} = 0.06/0.02 = 3/1$.

$A$ mixture of hydrogen and oxygen has a volume of $2000 \; cm^{3}$,a temperature of $300 \; K$,a pressure of $100 \; kPa$,and a mass of $0.76 \; g$. The ratio of the number of moles of hydrogen to the number of moles of oxygen in the mixture is:

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