Chemistry homework help?

[匿名]

2017-07-06 8:44 pm

1. Calculate the density of PF5(g) at 50 degrees Celsius and a pressure of 3.0 atm

2. A mixture of gases (Ar, He, and Xe) has a total pressure of 1000 mmHf at a temperature of 50 degrees Celsius and a volume of 10.0 L. The partial pressures of Ar and He are each 400 mmHg.

A. Calculate the partial pressure of Xe in mmHg.
B. Calculate the mole fraction of Xe.

3. Consider a 5.0L sample of SF6 (g) at 25 degrees Celsius and a pressure of 1.00 atm. If the temperature is increased to 50 degrees Celsius and the pressure is increased to 2.5 atm, what is the new volume?

Thanks.

回答 (1)

wanszeto

2017-07-06 11:32 pm

✔ 最佳答案

1.
For the PF₅ :
Pressure, P = 3.0 atm
Density, d = m/V
Temperature, T = (273 + 50) K = 323 K
Gas constant, R = 0.08206 L atm / (mol K)
Molar mass, M = (31.0 + 19.0×5) g/mol = 126.0 g/mol

Gas law : PV = nRT and n = m/M
Then, PV = (m/M)RT
Then, PM = (m/V(RT
Then, PM = dRT
Then, d = PM/(RT)

Density of PF₅, d = 3.0 × 126.0 / (0.08206 × 323) g/L = 14.3 g/L

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2.
A.
Total pressure is the sum of all the partial pressures.
P(Ar) + P(He) + P(Xe) = P(Total)
(400 mmHg) + (400 mmHg) + P(Xe) = 1000 mmHg
Partial pressure of Xe, P(Xe) = (1000 - 400 - 400) mmHg = 200 mmHg

B.
P(Total) × X(Xe) = P(Xe)
Hence, X(Xe) = P(Xe) / P(Total)

Mole fraction of Xe, X(Xe) = 200 /1000 = 0.2

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3.
Initial : P₁ = 1.00 atm, V₁ = 5.0 L, T₁ = (273 + 25) K = 298 K
Final : P₂ = 2.5 atm, V₂ = ? L, T₂ = (273 + 50) K = 323K

P₁V₁/T₁ = P₂V₂/T₂
Then, V₂ = V₁ × (P₁/P₂) × (T₂/T₁)

Final volume, V₂ = (5.0 L) × (1.00/2.5) × (323/298) L = 2.17 L

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