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5 moles of $\mathrm{SO}_{2}$ and 5 moles of $\mathrm{O}_{2}$ are allowed to react. At equilibrium, it was found that $60 \%$ of $\mathrm{SO}_{2}$ is used up. If the partial pressure of the equilibrium mixture is one atmosphere, the partial pressure of $\mathrm{O}_{2}$ is
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$0.41$ atm

$\therefore \quad \mathrm{pO}_{2}=\frac{3.5 \times 1}{8.5}=0.41 \mathrm{~atm}$
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