O2(g effuses at a rate that is times that of xe(g under the same conditions.

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.