High School

Calculate the ratio of the effusion rates of [tex]$N_2$[/tex] and [tex]$N_2O$[/tex].

A) 0.637
B) 1.57
C) 1.25
D) 0.798
E) 1.61

Answer :

Approximately 1.25 is the ratio of the effusion rates of N[tex]^{2}[/tex] and N[tex]^{2}[/tex]O. So, the answer is C) 1.25.

The effusion rate of a gas is inversely proportional to the square root of its molar mass. The molar masses of N[tex]^{2}[/tex] and N[tex]^{2}[/tex]O are 28 g/mol and 44 g/mol, respectively. To calculate the ratio of the effusion rates of N[tex]^{2}[/tex] and N[tex]^{2}[/tex]O, we can use Graham's Law of Effusion. The formula is:

Rate1 / Rate2 = √(Molar Mass2 / Molar Mass1)

For N[tex]^{2}[/tex] (Nitrogen), the molar mass is 28 g/mol, and for N[tex]^{2}[/tex]O (Nitrous oxide), the molar mass is 44 g/mol. Plugging these values into the formula:

Rate(N[tex]^{2}[/tex]) / Rate(N[tex]^{2}[/tex]O) = √(44 / 28) = √(1.57)

Thus, the ratio of the effusion rates of N[tex]^{2}[/tex] and N[tex]^{2}[/tex]O is approximately 1.25. So the correct answer is (C) 1.25.

More on effusion: https://brainly.com/question/9611592

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