# Gas A effuses 0.68 times as fast as Gas B. If the molar mass of Gas B is 17 g, what is the mass of Gas A?

Graham's law of effusion, which links the relative rates of gas effusion to their molar masses, can be applied.

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To find the mass of Gas A, we can use Graham's law of effusion:

Rate of effusion A / Rate of effusion B = sqrt(Molar mass B / Molar mass A)

Given: Rate of effusion A / Rate of effusion B = 0.68 Molar mass B = 17 g

Let's solve for the molar mass of Gas A:

0.68 = sqrt(17 / Molar mass A)

Squaring both sides:

0.4624 = 17 / Molar mass A

Multiplying both sides by Molar mass A:

Molar mass A = 17 / 0.4624

Molar mass A ≈ 36.73 g/mol

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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