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son4ous [18]
3 years ago
9

If total pressure of multiple gases is 512 mmHg, and the pressure of oxygen gas is 332 mmHg and the pressure of carbon monoxide

is 113 mmHg, what is the remaining pressure using Dalton’s Gas Law?
Chemistry
1 answer:
deff fn [24]3 years ago
5 0

Answer: 67 mmHg

Explanation:

According to Dalton's Gas Law, the total pressure of a mixture of gases is the sum of the pressure of each individual gas.

i.e Ptotal = P1 + P2 + P3 + .......

In this case,

Ptotal = 512 mmHg

P(oxygen) = 332 mmHg

P(carbon mono-oxide) = 113 mmHg

Remaining pressure (P3) = ?

To get P3, apply Dalton's Gas Law formula

Ptotal = P(oxygen) + P(carbon mono-oxide) + P3

512 mmHg = 332 mmHg + 113 mmHg + P3

512 mmHg = 445 mmHg + P3

P3 = 512 mmHg - 445 mmHg

P3 = 67 mmHg

Thus, the remaining pressure is 67 mmHg

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Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride
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The question is incomplete, here is the complete question:

Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride would be produced by this reaction if 1.1 mL of methane were consumed? Be sure your answer has the correct number of significant digits.

<u>Answer:</u> The volume of hydrogen chloride produced in the reaction will be 4.4 mL

<u>Explanation:</u>

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Volume of methane gas = 1.1 mL

The chemical equation for the reaction of methane gas and chlorine gas follows:

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Moles of methane gas = 1 mole

Moles of hydrogen chloride gas = 4 moles

The relationship of number of moles and volume at constant temperature and pressure was given by Avogadro's law. This law states that volume is directly proportional to number of moles at constant temperature and pressure.

The equation used to calculate number of moles is given by:

\frac{V_1}{n_1}=\frac{V_2}{n_2}

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V_1\text{ and }n_1 are the volume and number of moles of methane gas

V_2\text{ and }n_2 are the volume and number of moles of hydrogen chloride

We are given:

V_1=1.1mL\\n_1=1mol\\V_2=?L\\n_2=4mol

Putting values in above equation, we get:

\frac{1.1}{1}=\frac{V_2}{4}\\\\V_2=\frac{1.1\times 4}{1}=4.4mL

Hence, the volume of hydrogen chloride produced in the reaction will be 4.4 mL

5 0
3 years ago
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