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.
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