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(C) According to the ideal gas law,$PV = nRT$. Since the volume $V$,temperature $T$,and gas constant $R$ are constant,we have $P \propto n$.Initial mo

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$20$

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(C) According to the ideal gas law,$PV = nRT$. Since the volume $V$,temperature $T$,and gas constant $R$ are constant,we have $P \propto n$.Initial moles of nitrogen $(n_{N_2})$: $n_{N_2} = \frac{	ext{mass}}{	ext{molar mass}} = \frac{35 \, kg}{28 \, g/mol} = \frac{35000 \, g}{28 \, g/mol} = 1250 \, mol$.Using the proportionality $P_1/P_2 = n_1/n_2$:$6/9 = 1250 / n_{total}$$n_{total} = 1250 	imes (9/6) = 1250 	imes 1.5 = 1875 \, mol$.The total moles $n_{total} = n_{N_2} + n_{O_2}$,so $n_{O_2} = 1875 - 1250 = 625 \, mol$.Mass of oxygen supplied = $n_{O_2} 	imes 	ext{molar mass of } O_2 = 625 \, mol 	imes 32 \, g/mol = 20000 \, g = 20 \, kg$.

$A$ rigid tank contains $35 \, kg$ of nitrogen at $6 \, atm$. $A$ sufficient quantity of oxygen is supplied to increase the pressure to $9 \, atm$,while the temperature remains constant. The amount of oxygen supplied to the tank is .... $kg$.

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