Calculate the partial pressure exerted by dio | vedclass

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(B) Step $1$: Calculate the number of moles of each gas.$n_{O_2} = \frac{32 \ g}{32 \ g/mol} = 1 \ mol$$n_{Ar} = \frac{80 \ g}{40 \ g/mol} = 2 \ mol$$

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$2$

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(B) Step $1$: Calculate the number of moles of each gas.$n_{O_2} = \frac{32 \ g}{32 \ g/mol} = 1 \ mol$$n_{Ar} = \frac{80 \ g}{40 \ g/mol} = 2 \ mol$$n_{H_2} = \frac{4 \ g}{2 \ g/mol} = 2 \ mol$Step $2$: Calculate the total number of moles.$n_{	ext{total}} = n_{O_2} + n_{Ar} + n_{H_2} = 1 + 2 + 2 = 5 \ mol$Step $3$: Calculate the mole fraction of dioxygen $(x_{O_2})$.$x_{O_2} = \frac{n_{O_2}}{n_{	ext{total}}} = \frac{1}{5} = 0.2$Step $4$: Calculate the partial pressure of dioxygen $(P_{O_2})$.$P_{O_2} = x_{O_2} 	imes P_{	ext{total}} = 0.2 	imes 10 \ bar = 2 \ bar$

Calculate the partial pressure exerted by dioxygen from a mixture of $32 \ g$ $O_2$,$80 \ g$ $Ar$ (molar mass $40 \ g/mol$) and $4 \ g$ dihydrogen $(P_{\text{total}} = 10 \ bar)$. (in $bar$)

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