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Chemistry Help. This is VERY hard for me? A spacecraft is filled with 0.500 atm of N2 and 0.500 atm of O2. Suppose a micrometeor strikes this spacecraft and puts a very small hole in its side. Under these circumstances, A) N2 is lost from the craft 6.9% faster than O2 is lost. B) N2 is lost from the craft 14% faster than O2 is lost. C)N2 and O2 are lost from the craft at the same rate. D) O2 is lost from the craft 6.9% faster than N2 is lost. E) O2 is lost from the craft 14% faster than N2 is lost.

By: Guest
Date: Unknown--
Response
2

Answer 1

Not so hard if you remember Grahams Law of diffusion through a small hole. For these gases the relative loss would be sqrt(32/28) or about 1.07. So the answer is A.

Answer 2

The rate of effusion (or the speed that the molecules will leave the spacecraft) is directly related to the molecule's molar mass, so immediately you can rule out C, D, and E. Now to do the math.

g/mol 02= 32

g/mol N2=28

Graham's law states that the effusion rate of a molecule is equivalent to the inverse square root of it's mass. So...

The squrare root of 32/28= 1.069

The answer is A.

Answer 3

C) The use of Graham's law is not relevant here. The gasses do not "diffuse" or "effuse" they are forced out into the vacuum of space at a terrific speed - the outside pressure being nearly zero atmospheres.

Since the partial pressures of the gases are equal ( .5 atm each) their relative amounts (moles) are equal so it is a 50-50 mixture of N2 and O2.

The key to this question is the fact that the pressure on the other side of the opening (no matter how small) is near zero.

[d] By: Guest
Date: Unknown---
Response
What is 1 + 100



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