Which of the following is an example of a precipitate?

Chemistry

1 answer:

Bogdan [553]3 years ago

3 0

I think should be d. Please give me brainlest let me know if its correct or not okay thanks bye

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The Process of forming a new eukaryotic cell is called

gulaghasi [49]

Answer:

Mitosis I think

Explanation:

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3 years ago

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

diamong [38]

Answer:

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

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

Explanation:

For A:

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

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

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

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.

For B:

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:

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