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A mixture of He, Ar and Xe has a total pressure of 3 00 atm. The partial pressure of He is 0 350 atm, and the partial pressure of Ar is 0.300 atm. What is the partial pressure of Xe? Express your answer to three significant figures and include the appropriate units. A volume of 18.0 L contains a mixture of 0.25 mole N_2 0.25 mole O_2 and an unknown quantity of He. The temperature of the mixture is 0 degree C and the total pressure is 1.00 atm./How many grams of
Expert Answer
A)
total pressure = sum of all partial pressures
3.00 atm = p(He) + p(Ar) + p(Xe)
3.00 atm = 0.350 + 0.300 + p(Xe)
p(Xe) = 2.35 atm
Answer: 2.35 atm
B)
1st find the total mol of all the gas
Given:
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 * 0.0821 atm.L/mol.K * 273 K
n = 0.8031 mol
total mol = n(N2) + n(O2) + n(He)
0.8031 = 0.250 + 0.250 + n(He)
n(He) = 0.303 mol
molar mass of He = 4 g/mol
so,
mass of He = n(He)*molar mass
= 0.303 mol * 4 g/mol
= 1.21 g
Answer: 1.21 g
