CO(g) effuses at a rate that is ______ times that of Xe(g) under the same conditions.
2.16posted by Alex
Let's make up a number for the rate of Xe gas. Let's call it 10 mL/second.
(rate CO/rate Xe) = sqrt(M Xe/M CO where M stands for molar mass Xe or molar mass CO.
(10/x) = sqrt(131.29/28)
Solve for x for rate of CO then calculate how much faster/slower this is than Xe. You know CO should have a higher rate since the molar mass is lower.posted by DrBob222