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Chem 10 – Exam 1, version A
Useful information: R = 0.08206 liter atm/ mol K = 8.3145 J/mol K
1 atm = 101325 Pa = 760 torr
1.
At 100 ºC, a 1.20 L flask contains 42 mg of N
2
, 80 mg of O
2
, 5 x 10
3
mol Ar and NO at a
concentration of 2.50 x10
18
molecules/cm
3
. What is the total pressure in the flask?
42 mg N
2
= 1.5 mmol
80 mg O
2
= 2.5 mmol
Ar is 5 mmol
2.5 x 10
18
/cm
3
= 2.5 x 10
21
/L, 2.5 x 10
21
/L x 1.2 L = 3 x 10
21
molecules = 5 mmol
Total = n = 14 mmol = 1.4 x 10
3
mol P= nRT/V = 0.357 atm
2. In the above question, what is the mole fraction of each O
2
?
2.5 mmol O
2
/ 14 mmol total = .178
3.
A bicycle tire is filled to 130 psi at 22 ºC. It’s a cold day, and while riding, the temperature
goes to 5 ºC. The tire also leaks and loses 7% of the air in it.
What is the new pressure?
Drop due to temperature a factor of (5 + 273) / (22 + 273) = 0.9085 ( about a 9% drop)
Drop due to leak is 7%, or the pressure remaining is 0.93 of the original.
Final P = 130 psi x 0.9085 x .93 = 109.8
4.
A 2.0 L stainless steel vessel is charged with 2.00 atm of hydrogen gas and 2.50 atm of
nitrogen gas at 22 ºC. A spark ignites the mixture and it reacts, producing ammonia, NH
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This note was uploaded on 07/02/2010 for the course CHEM 10 taught by Professor G during the Spring '10 term at UCM.
 Spring '10
 G
 Chemistry, Mole

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