A mixture of 28 g of nitrogen,4 g of hydrog | vedclass

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(C) First,calculate the number of moles for each gas:$n_{N_2} = \frac{28 \ g}{28 \ g/mol} = 1 \ mol$$n_{H_2} = \frac{4 \ g}{2 \ g/mol} = 2 \ mol$$n_{H

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$2 \ g/L$

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(C) First,calculate the number of moles for each gas:$n_{N_2} = \frac{28 \ g}{28 \ g/mol} = 1 \ mol$$n_{H_2} = \frac{4 \ g}{2 \ g/mol} = 2 \ mol$$n_{He} = \frac{8 \ g}{4 \ g/mol} = 2 \ mol$Total moles $n = 1 + 2 + 2 = 5 \ mol$.Total mass $M_{total} = 28 + 4 + 8 = 40 \ g = 0.04 \ kg$.Using the ideal gas equation $PV = nRT$,find the volume $V$:$V = \frac{nRT}{P} = \frac{5 	imes 8.3 	imes 400}{8.3 	imes 10^5} = 2 	imes 10^{-2} \ m^3$.Density $ho = \frac{M_{total}}{V} = \frac{0.04 \ kg}{2 	imes 10^{-2} \ m^3} = 2 \ kg/m^3$.Since $1 \ kg/m^3 = 1 \ g/L$,the density is $2 \ g/L$.

$A$ mixture of $28 \ g$ of nitrogen,$4 \ g$ of hydrogen,and $8 \ g$ of helium is kept in a vessel at a temperature of $400 \ K$ and a pressure of $8.3 \times 10^5 \ Pa$. What is the density of the mixture?

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