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Stels [109]
3 years ago
6

A sample of an unknown gas takes 371 s to diffuse through a porous plug at a given temperature.

Chemistry
1 answer:
blsea [12.9K]3 years ago
5 0

Answer:

125.84 g/mol is the molar mass of the unknown gas.

Explanation:

Let the volume of the gases effusing out be V.

Effusion rate of the unknown gas = R=\frac{V}{371 s}

Effusion rate of the nitrogen gas = r=\frac{V}{175 s}

Molar mass of unknown gas = m

Mass of nitrogen gas = 28 g/mol

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

\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

\frac{R}{r}=\sqrt{\frac{28 g/mol}{M}}

\frac{\frac{V}{371 s}}{\frac{V}{175s}}=\sqrt{\frac{28 g/mol}{M}}

\frac{175 s}{371 s}=\sqrt{\frac{28 g/mol}{M}}

M=\frac{28 g/mol\times 371\times 371}{175\times 175}=125.84 g/mol

125.84 g/mol is the molar mass of the unknown gas.

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Answer:

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First, we have to write the chemical equation for the reaction between ammonia gas (NH₃) and oxygen gas (O₂) to give nitrogen gas (N₂) and water (H₂O), as follows:

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In the balanced equation, we can see that 1 mol of N₂ is produced from 2 moles of NH₃. We convert the moles of NH₃ to grams by using its molecular weight (MW):

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To calculate how many moles of N₂ are produced from 4.0 of NH₃, we multiply the mass by the conversion factor:

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We introduce the data in the equation:

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P = 2.6 atm

R = 0.082 L.atm/K.mol (is the gas constant)

n= 0.1176 moles

⇒ V = nRT/P = (0.1176 mol x 0.082 L.atm/K.mol x 305 K)/(2.6 atm)

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Answer:

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b) A saturated solution is a solution that has the maximum amount of salt dissolved, so, the concentration dissolved is solubility. As we can notice from the reactions, the concentration of M⁺² is the same for both salts.

c) The equilibrium will be not modified because the salts have the same solubility. So, let's suppose that the volume of each one is 1 L, so the number of moles of the cation in each one is 4x10⁻⁴ mol. The total number of moles is 8x10⁻⁴ mol, and the concentration is:

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