A 9.40-L container holds a mixture of two gases at 45 degree C. The partial pres
ID: 842687 • Letter: A
Question
A 9.40-L container holds a mixture of two gases at 45 degree C. The partial pressures of gas A and gas B, respectively, are 0.300 atm and 0.601 atm. If 0.240 mol of a third gas is added with no change in volume or temperature, what will the total pressure become? Under identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simultaneously. After a certain amount of time, it was found that 5.39 mL of O2 had passed through the membrane, but only 3.62 mL of of the unknown gas had passed through. What is the molar mass of the unknown gas? Find the rms speed of the molecules of a sample of N2 (diatomic nitrogen) gas at a temperature of 35.7 degree C. When aluminum is placed in concentrated hydrochloric acid, hydrogen gas is produced. A sample of N2 gas has a volume of 11.0 L at a pressure of 1.50 atm and a temperature of 23 degree C. What volume, in liters, will the gas occupy at 3.50 atm and 271 degree C? Assume ideal behavior.Explanation / Answer
1) total pressure= p of A + p of B + p of C
GIVEN pressure of A =0.300 atmp
pressure of B=0.601atmp
pressure of C= nRT/V = 0.240*0.0821*318/9.40 = 0.666 atmp
total pressure= p of A + p of B + p of C= 0.300+ 0.601+0.666= 1.567atmp
total pressure= 1.567atmp
2)
(V1/ V2 )= (M2/M1 )1/2
(5.39/3.62)= (M2/32)1/2
M2=70.94 gr/mole
3)
R.MS= (3RT/M)^1/2
=(3*8.314*10^7*308.7/28)^1/2
=52439.06cm/sec
4)
according to equation
54 grams of aluminium gives-----------------------------------------3*22.4 lit of H2 GAS
7.20 grams of aluminium gives-----------------------------------------------------?
=3*22.4*7.20/54
=8.96 lit
5)
P1V1/T1 = P2V2/T2
1.50*11/298 = 3.50*V2/544
V2=8.605lit
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