Key Concept Problem 8.22 Part A Assume that you have a mixture of He MW = 4 amu)
ID: 551056 • Letter: K
Question
Key Concept Problem 8.22 Part A Assume that you have a mixture of He MW = 4 amu) and Xe MW = 131 amu) at 300 K. The total pressure of the mixture is 600 mmHg. What are the partial pressures of each of the gases? (blue = He; green = Xe)? What is the partial pressure of He? Express your answer to three significant figures and include the appropriate units, HA he O o ? 0 0 Pte = | | Value | Units submit my answer one in O Incorrect; Try Again; 5 attempts remaining Oo Part B What is the partial pressure of Xe? Express your answer to three significant figures and include the appropriate units. CHWA he CD ? | Pxe = ( Value | Units Submit My Answers Give UpExplanation / Answer
A)
PHe = total pressure * mole fraction of He
= total pressure * NHe / total number of molecules
= 600 mmHg * 8 / 12
= 400 mmHg
Answer: 400 mmHg
B)
remaining pressure is due to Xe
PXe = total pressure - PHe
= 600 mmHg - 400 mmHg
= 200 mmHg
Answer: 200 mmHg
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