O2 effuses at a rate that is _____ times that of Xe under the same conditions.
rate 02/rate x2 = ?
To determine the ratio of the effusion rates of oxygen (O2) to xenon (Xe), we need to use Graham's Law of Effusion.
According to Graham's Law of Effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
The molar mass of oxygen (O2) is 32 g/mol, and the molar mass of xenon (Xe) is 131 g/mol.
Now, let's calculate the ratio of the effusion rates:
Rate O2 / Rate Xe = √(Molar mass Xe / Molar mass O2)
= √(131 g/mol / 32 g/mol)
≈ √4.094
≈ 2.02
Therefore, the rate of oxygen effusing is approximately 2.02 times that of xenon under the same conditions.
To determine the ratio of the effusion rates of O2 and Xe, we can use Graham's law of effusion. According to Graham's law:
rate1/rate2 = sqrt(M2/M1)
Where rate1 and rate2 represent the effusion rates of two different gases, and M1 and M2 represent their respective molar masses.
In this case, we want to compare the effusion rates of O2 (rate1) and Xe (rate2).
Let's assign O2 as gas 1, and Xe as gas 2.
Molar mass of O2 (M1) = 32 g/mol
Molar mass of Xe (M2) = 131.3 g/mol
Now we can substitute these values into the formula:
rate O2/rate Xe = sqrt(M2/M1)
rate O2/rate Xe = sqrt(131.3/32)
By evaluating the square root (using a calculator or math software), we find:
rate O2/rate Xe ≈ 2.58
Therefore, the effusion rate of O2 is approximately 2.58 times that of Xe under the same conditions.
the rates are inversely proportional to the square root of molecular masses
rateO2/rateXe=sqrt ( 131/32)
you probably ought to be more accurate than that.