If 4 g of O_2 and 2 g of H_2 are filled in | vedclass

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(A) $1$. Calculate the number of moles for each gas: $n(O_2) = \frac{4 \ g}{32 \ g/mol} = 0.125 \ mol$ $n(H_2) = \frac{2 \ g}{2 \ g/mol} = 1.0 \ mol$

Vedclass · Accepted Answer

$25.215$

Vedclass · Answer

(A) $1$. Calculate the number of moles for each gas: $n(O_2) = \frac{4 \ g}{32 \ g/mol} = 0.125 \ mol$ $n(H_2) = \frac{2 \ g}{2 \ g/mol} = 1.0 \ mol$ $2$. Total moles $(n_{total})$ = $0.125 + 1.0 = 1.125 \ mol$ $3$. Use the ideal gas equation $PV = nRT$: $P = \frac{n_{total} 	imes R 	imes T}{V}$ $P = \frac{1.125 \ mol 	imes 0.0821 \ L \cdot atm \cdot K^{-1} \cdot mol^{-1} 	imes 273 \ K}{1 \ L}$ $P = 1.125 	imes 22.4133 = 25.215 \ atm$

If $4 \ g$ of $O_2$ and $2 \ g$ of $H_2$ are filled in a $1 \ L$ bulb at $0 \ ^oC$,the total pressure of the mixture will be...... (in $atm$)

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