O2(g) effuses at a rate that is ______ times that of xe(g) under the same conditions.
Asked by maham237 @ in Chemistry viewed by 411 People
Answered by maham237 @
Answer:
2.02 times
Explanation:
According to Graham's law, the rate of effusion of two gases can be expresed as:
Where M are the atomic/molecular masses of the gases.
In this problem, the subscript 1 refers to O₂ gas (32 g/mol), and the subscript 2 refers to Xe gas (131.293 g/mol).
Putting the data in the equation gives us:
So O₂ effuses at a rate that is 2.02 times faster than the rate of Xe.
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