A container contains 4 g of H_2,4 g of He,a | vedclass

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(D) According to Dalton's Law of Partial Pressures,the partial pressure of a gas is directly proportional to its number of moles in a mixture at const

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$20$

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(D) According to Dalton's Law of Partial Pressures,the partial pressure of a gas is directly proportional to its number of moles in a mixture at constant temperature and volume $(P_i = n_i 	imes \frac{RT}{V})$.For the partial pressure of $Ne$ to be equal to the partial pressure of $He$,their number of moles must be equal $(n_{Ne} = n_{He})$.First,calculate the moles of $He$ $(n_{He})$:$n_{He} = \frac{	ext{mass}}{	ext{molar mass}} = \frac{4 \ g}{4 \ g/mol} = 1 \ mol$.Since $n_{Ne} = n_{He}$,the required moles of $Ne$ is $1 \ mol$.Now,calculate the mass of $Ne$:$	ext{Mass of } Ne = n_{Ne} 	imes 	ext{molar mass of } Ne = 1 \ mol 	imes 20 \ g/mol = 20 \ g$.

$A$ container contains $4 \ g$ of $H_2$,$4 \ g$ of $He$,and a certain amount of $Ne$ at a certain temperature. What is the mass of $Ne$ required so that the partial pressure exerted by $Ne$ is equal to the partial pressure of $He$ (in $g$)?

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