Calculating partial pressure. Help Please! I\'m stuck on a and b A mixture of He

Question

Calculating partial pressure. Help Please! I'm stuck on a and b A mixture of He, Ar. and Xe has a total pressure of 2.60 atm . The partial pressure of He is 0.400 atm , and the partial pressure of Ar is 0.200 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 0.250 mole N_2 0.250 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 helium are present in the gas mixture?

Explanation / Answer

A- total P = pHe+ pAr + pXe

2.60atm= 0.4atm+0.2atm + pXe

partial pressure of Xe= total pressure - (partial pressure of He+partial pressure of Ar)

pXe= 2.6atm - 0.6atm = 2 atm

B-PV = n RT,

P=1atm , V = 18L , R= 0.082L-atm/mol-K , T= 273+0=273

n = PV/ RT= 1x18 /0.082 x 273 = 18/22.386 = 0.8041mol=total mole

0.8041= mole of N2 + mole of O2 + mole He = 0.250+ 0.250+ mole of He

mole of He = 0.8041 - 0.5 = 0.3041 mole

gram of He = mwof He x mole of He = 4x 0.3041 = 1.2164g

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