for each pair of gases, select the one that most likely has the highest rate of effusion. use the periodic…

Explanation:

Step1: Recall 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 ($r\propto\frac{1}{\sqrt{M}}$), where $r$ is the rate of effusion and $M$ is the molar mass. So, the gas with the lower molar mass will have a higher rate of effusion.

Step2: Calculate molar masses and compare for oxygen and hydrogen

The molar mass of $O_2$: $M_{O_2}=2\times16.00\ g/mol = 32.00\ g/mol$. The molar mass of $H_2$: $M_{H_2}=2\times1.01\ g/mol=2.02\ g/mol$. Since $M_{H_2}<M_{O_2}$, hydrogen ($H_2$) has a higher rate of effusion.

Step3: Calculate molar masses and compare for methane and carbon - tetrachloride

The molar mass of $CH_4$: $M_{CH_4}=12.01 + 4\times1.01=16.05\ g/mol$. The molar mass of $CCl_4$: $M_{CCl_4}=12.01+4\times35.45 = 153.81\ g/mol$. Since $M_{CH_4}<M_{CCl_4}$, methane ($CH_4$) has a higher rate of effusion.

Step4: Calculate molar masses and compare for nitrogen and ammonia

The molar mass of $N_2$: $M_{N_2}=2\times14.01\ g/mol = 28.02\ g/mol$. The molar mass of $NH_3$: $M_{NH_3}=14.01+3\times1.01 = 17.04\ g/mol$. Since $M_{NH_3}<M_{N_2}$, ammonia ($NH_3$) has a higher rate of effusion.

Step5: Calculate molar masses and compare for fluorine and chlorine

The molar mass of $F_2$: $M_{F_2}=2\times19.00\ g/mol = 38.00\ g/mol$. The molar mass of $Cl_2$: $M_{Cl_2}=2\times35.45\ g/mol = 70.90\ g/mol$. Since $M_{F_2}<M_{Cl_2}$, fluorine ($F_2$) has a higher rate of effusion.

Answer:

Oxygen ($O_2$) or hydrogen ($H_2$): Hydrogen ($H_2$) Methane ($CH_4$) or carbon tetrachloride ($CCl_4$): Methane ($CH_4$) Nitrogen ($N_2$) or ammonia ($NH_3$): Ammonia ($NH_3$) Fluorine ($F_2$) or chlorine ($Cl_2$): Fluorine ($F_2$)