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 s
mall hole in its side. Under these circumstances,
A. O2 is lost from the craft 6.9% faster than N2 is lost.
B. O2 is lost from the craft 14% faster than N2 is lost.
C. N2 is lost from the craft 6.9% faster than O2 is lost.
D. N2 is lost from the craft 14% faster than O2 is lost.
E. N2 and O2 are lost from the craft at the same rate.
2 answers:
Answer:
N2 is lost from the craft 6.9% faster than O2 is lost.
Explanation:
<u>Step 1:</u> Data given
0.500 atm of N2
0.500 atm of 02
Molar mass of 02 = 2*16 g/mole = 32 g/mole
Molar mass of N2 = 2* 14 g/mole = 28g/mole
<u>Step 2:</u> Calculate the rate of loss
r1N2 / r2O2 = √(M2(02)/M1(N2)) = √(32/28) = 1.069
1.069-1)*100= 6.9%
This means N2 is lost from the craft 6.9% faster than O2 is lost.
Answer:
C. N₂ is lost from the craft 6.9% faster than O₂ is lost.
Explanation:
The rate of effusion of a gas (r) is inversely proportional to the square root of its molar mass (M), as expressed by the Graham's law.
The rate of effusion of N₂ is 1.069 times the rate of effusion of O₂, that is, N₂ effuses 6.9% faster than O₂.
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