A 10 , L cylinder contains 0.4 , g of helium, | vedclass

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Question

(A) First,calculate the number of moles for each gas:$n_{He} = \frac{0.4 \, g}{4 \, g/mol} = 0.1 \, mol$$n_{O_2} = \frac{1.6 \, g}{32 \, g/mol} = 0.05

Vedclass · Accepted Answer

$0.492 \, atm, 0.246 \, atm$

Vedclass · Answer

(A) First,calculate the number of moles for each gas:$n_{He} = \frac{0.4 \, g}{4 \, g/mol} = 0.1 \, mol$$n_{O_2} = \frac{1.6 \, g}{32 \, g/mol} = 0.05 \, mol$$n_{N_2} = \frac{1.4 \, g}{28 \, g/mol} = 0.05 \, mol$Total moles $n_{total} = 0.1 + 0.05 + 0.05 = 0.2 \, mol$Using the ideal gas equation $PV = nRT$ at $T = 27 + 273 = 300 \, K$:$P_{total} = \frac{n_{total}RT}{V} = \frac{0.2 	imes 0.082 	imes 300}{10} = 0.492 \, atm$Partial pressure of helium $P_{He} = \chi_{He} 	imes P_{total}$$P_{He} = \left( \frac{0.1}{0.2} ight) 	imes 0.492 = 0.246 \, atm$

$A$ $10 \, L$ cylinder contains $0.4 \, g$ of helium,$1.6 \, g$ of oxygen,and $1.4 \, g$ of nitrogen at $27 \, ^\circ C$. Calculate the total pressure of the mixture and the partial pressure of helium gas. Assume ideal gas behavior. $(R = 0.082 \, L \, atm \, K^{-1} \, mol^{-1})$

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