A mixture of He He , N 2 N2 , and Ar Ar has a pressure of 24.1 24.1 atm at 28.0 28.0 °C. If the partial pressure of He He is 301 3 3013 torr and that of Ar Ar is 2737 2737 mm Hg, what is the partial pressure of N 2 N2
1 answer:
Answer:
16.5 atm
Explanation:
<em>A mixture of He, N₂, and Ar has a pressure of 24.1 atm at 28.0 °C. If the partial pressure of He is 3013 torr and that of Ar is 2737 mm Hg, what is the partial pressure of N₂?</em>
The total pressure of a gaseous mixture is equal to the sum of the partial pressures .
P = pHe + pN₂ + pAr
pN₂ = P - pHe - pAr [1]
We need to express pHe and pAr in atm.
From [1],
pN₂ = 24.1 atm - 3.96 atm - 3.60 atm = 16.5 atm
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Answer: Option B. 76.83L
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1 mole of a gas occupy 22.4L at stp. This implies that 1mole of Radon also occupy 22.4L at stp.
If 1 mole of Radon = 22.4L
Therefore, 3.43 moles of Radon = 3.43 x 22.4 = 76.83L