A container consists of a mixture of 28 g N | vedclass

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(B) $1$. Calculate the number of moles for each gas: $n(N_2) = \frac{28 \ g}{28 \ g/mol} = 1 \ mol$ $n(He) = \frac{8 \ g}{4 \ g/mol} = 2 \ mol$ $n(Ne)

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$4$

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(B) $1$. Calculate the number of moles for each gas: $n(N_2) = \frac{28 \ g}{28 \ g/mol} = 1 \ mol$ $n(He) = \frac{8 \ g}{4 \ g/mol} = 2 \ mol$ $n(Ne) = \frac{40 \ g}{20 \ g/mol} = 2 \ mol$ $2$. Calculate the total number of moles: $n_{total} = 1 + 2 + 2 = 5 \ mol$ $3$. Calculate the mole fraction of $N_2$: $x(N_2) = \frac{n(N_2)}{n_{total}} = \frac{1}{5} = 0.2$ $4$. Calculate the partial pressure of $N_2$ using Dalton's Law: $P(N_2) = x(N_2) 	imes P_{total} = 0.2 	imes 20 \ bar = 4 \ bar$

$A$ container consists of a mixture of $28 \ g \ N_2$,$8 \ g \ He$,and $40 \ g \ Ne$ at $25^{\circ} C$. If the total pressure exerted by the gaseous mixture is $20 \ bar$,what is the partial pressure exerted by $N_2$ (in $bar$)?

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