A neon-dioxygen mixture contains 64 g O_2 an | vedclass

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(A) First,calculate the number of moles for each gas: $n_{O_2} = \frac{64 \ g}{32 \ g/mol} = 2 \ mol$ $n_{Ne} = \frac{160 \ g}{20 \ g/mol} = 8 \ mol$

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$5$

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(A) First,calculate the number of moles for each gas: $n_{O_2} = \frac{64 \ g}{32 \ g/mol} = 2 \ mol$ $n_{Ne} = \frac{160 \ g}{20 \ g/mol} = 8 \ mol$ Next,calculate the mole fraction of dioxygen $(x_{O_2})$: $x_{O_2} = \frac{n_{O_2}}{n_{O_2} + n_{Ne}} = \frac{2}{2 + 8} = \frac{2}{10} = 0.2$ Finally,calculate the partial pressure of dioxygen $(P_{O_2})$ using Dalton's Law: $P_{O_2} = x_{O_2} 	imes P_{Total} = 0.2 	imes 25 \ bar = 5 \ bar$

$A$ neon-dioxygen mixture contains $64 \ g$ $O_2$ and $160 \ g$ $Ne$. If the total pressure is $25 \ bar$,calculate the partial pressure of dioxygen. (in $bar$)

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