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Flura [38]
3 years ago
15

Compared to atoms of metals atoms of nonmetals generally have

Chemistry
1 answer:
Arturiano [62]3 years ago
5 0
B.) Have higher electronegativities
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A) The average molecular speed in a sample of Ar gas at a certain temperature is 391 m/s. The average molecular speed in a sampl
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<u>Answer:</u>

<u>For A:</u> The average molecular speed of Ne gas is 553 m/s at the same temperature.

<u>For B:</u> The rate of effusion of SO_2 gas is 1.006\times 10^{-3}mol/hr

<u>Explanation:</u>

<u>For A:</u>

The average molecular speed of the gas is calculated by using the formula:

V_{gas}=\sqrt{\frac{8RT}{\pi M}}

     OR

V_{gas}\propto \sqrt{\frac{1}{M}}

where, M is the molar mass of gas

Forming an equation for the two gases:

\frac{V_{Ar}}{V_{Ne}}=\sqrt{\frac{M_{Ne}}{M_{Ar}}}          .....(1)

Given values:

V_{Ar}=391m/s\\M_{Ar}=40g/mol\\M_{Ne}=20g/mol

Plugging values in equation 1:

\frac{391m/s}{V_{Ne}}=\sqrt{\frac{20}{40}}\\\\V_{Ne}=391\times \sqrt{2}=553m/s

Hence, the average molecular speed of Ne gas is 553 m/s at the same temperature.

<u>For B:</u>

Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. The equation for this follows:

Rate\propto \frac{1}{\sqrt{M}}

Where, M is the molar mass of the gas

Forming an equation for the two gases:

\frac{Rate_{SO_2}}{Rate_{Xe}}=\sqrt{\frac{M_{Xe}}{M_{SO_2}}}          .....(2)

Given values:

Rate_{Xe}=7.03\times 10^{-4}mol/hr\\M_{Xe}=131g/mol\\M_{SO_2}=64g/mol

Plugging values in equation 2:

\frac{Rate_{SO_2}}{7.03\times 10^{-4}}=\sqrt{\frac{131}{64}}\\\\Rate_{SO_2}=7.03\times 10^{-4}\times \sqrt{\frac{131}{64}}\\\\Rate_{SO_2}=1.006\times 10^{-3}mol/hr

Hence, the rate of effusion of SO_2 gas is 1.006\times 10^{-3}mol/hr

8 0
2 years ago
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