In a gaseous mixture of a $2.0 \cdot L$ $2.0L$ volume, at $755 \cdot mm Hg$ $755"mm Hg"$ , and a temperature of $346 \cdot K$ $346K$ , $P_{dinitrogen} = 355 \cdot m m \cdot H g$ $P_"dinitrogen"=355mm*Hg$ . What are the molar quantities of helium and dinitrogen?

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In a gaseous mixture of a $2.0 \cdot L$ $2.0L$ volume, at $755 \cdot mm Hg$ $755"mm Hg"$ , and a temperature of $346 \cdot K$ $346K$ , $P_{dinitrogen} = 355 \cdot m m \cdot H g$ $P_"dinitrogen"=355mm*Hg$ . What are the molar quantities of helium and dinitrogen?

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Answer 1 · 2017-05-08T12:38:09

$Dalton's Law of Partial Pressures tells us that.........$ $"Dalton's Law of Partial Pressures tells us that........."$ , I get approx. $150 \cdot m g$ $150*mg$ of helium............

Explanation:

$In a gaseous mixture, the pressure exerted by a component gas$ $"In a gaseous mixture, the pressure exerted by a component gas"$
$is the same as the pressure if it ALONE occupied the container.$ $"is the same as the pressure if it ALONE occupied the container."$

$The total pressure is the sum of the individual partial pressures.$ $"The total pressure is the sum of the individual partial pressures."$

And this $P_{mixture} = P_{He} + P_{N_{2}}$ $P_"mixture"=P_"He"+P_("N"_2)$

And so $P_{He} = (755 - 355) \cdot m m \cdot H g = 400 \cdot m m \cdot H g$ $P_"He"=(755-355)*mm*Hg=400*mm*Hg$ .

Given that we now $P_{He}$ $P_"He"$ we can calculate its equivalent mass by means of the Ideal Gas Equation.......

$n = \frac{P V}{R T} = \frac{\frac{400 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t m^{- 1}} \times 2.0 \cdot L}{0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 346 \cdot K} = 0.0371 \cdot m o l$ $n=(PV)/(RT)=((400*mm*Hg)/(760*mm*Hg*atm^-1)xx2.0*L)/(0.0821*(L*atm)/(K*mol)xx346*K)=0.0371*mol$ .

Which represents a mass of $0.0371 \cdot m o l \times 4.0 \cdot g \cdot m o l^{- 1}$ $0.0371*molxx4.0*g*mol^-1$ with respect to helium gas.

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In a gaseous mixture of a $2.0 \cdot L$ $2.0L$ volume, at $755 \cdot mm Hg$ $755"mm Hg"$ , and a temperature of $346 \cdot K$ $346K$ , $P_{dinitrogen} = 355 \cdot m m \cdot H g$ $P_"dinitrogen"=355mm*Hg$ . What are the molar quantities of helium and dinitrogen?

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Explanation:

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In a gaseous mixture of a 2.0⋅L2.0*L volume, at 755⋅mm Hg755*"mm Hg", and a temperature of 346⋅K346*K, Pdinitrogen=355⋅mm⋅HgP_"dinitrogen"=355*mm*Hg. What are the molar quantities of helium and dinitrogen?

1 Answer

Explanation:

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In a gaseous mixture of a $2.0 \cdot L$ $2.0L$ volume, at $755 \cdot mm Hg$ $755"mm Hg"$ , and a temperature of $346 \cdot K$ $346K$ , $P_{dinitrogen} = 355 \cdot m m \cdot H g$ $P_"dinitrogen"=355mm*Hg$ . What are the molar quantities of helium and dinitrogen?