± Core Chemistry Skill: Calculating Partial Pressure Resources « previous| 20 of
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± Core Chemistry Skill: Calculating Partial Pressure Resources « previous| 20 of 23 | next Core Chemistry Skill: Calculating Partial ressure Part A A mixture of He, Ar, and Xe has a total pressure ot 2.90 atm. The partial pressure of He is 0.350 atm, and the partial pressure of Ar is 0.350 atm. What is the partial pressure of Xe? Learning Goal: To use partial pressure in gas law calculations. Express your answer to three significant figures and include the appropriate units. Hints In a mixture of gases, the total pressure of the gas mixture is equal to the sum of the partial pressures of the individual gases. For example, if you have a mixture of helium at 2 atm and argon at 4 atm, then Figure 1 of 1 PeValue Units Submit My Answers Give Up Part B A volume of 18.0 L contains a mixture of 0.250 mole N2,0.250 mole O2, and an unknown quantity of He. The temperature of the mixture is o C, and the total pressure is 1.00 atm How many grams of helium are present in the gas mixture? Express your answer to three significant figures and include the appropriate units. Hints mass of He-Value UnitsExplanation / Answer
A)
total pressure = sum of individual partial pressures
2.90 atm = p(He) + p(Ar) + p(Xe)
2.90 = 0.350 + 0.350 + p(Xe)
p(Xe) = 2.20 atm
Answer: 2.20 atm
B)
1st find the total mol of gas:
we have:
P = 1.0 atm
V = 18.0 L
T = 0.0 oC
= (0.0+273) K
= 273 K
find number of moles using:
P * V = n*R*T
1 atm * 18 L = n * 0.08206 atm.L/mol.K * 273 K
n = 0.8035 mol
now use:
total mol = mol of N2 + mol of O2 + mol of He
0.8035 mol = 0.250 mol + 0.250 mol + mol of He
mol of He = 0.3035 mol
Molar mass of He = 4.003 g/mol
we have below equation to be used:
mass of He,
m = number of mol * molar mass
= 0.3035 mol * 4.003 g/mol
= 1.215 g
Answer: 1.21 g
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