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3 years ago

What is the preasure in atmospheres of 20 mol of nitrogen gas in 36.2 L cylinder at 25 degrees C?

Chemistry

1 answer:

Savatey [412]3 years ago

7 0

Answer:

P = 13.5 atm

Explanation:

Given that

No. of moles, n = 20 moles

Volume of nitrogen gas = 36.2 L

Temperature = 25°C = 298 K

We need to find the pressure of the gas. Using the ideal gas equation

PV = nRT

Where

R is gas constant, $R=0.082057\ L-atm/K-mol$

So,

$P=\dfrac{nRT}{V}\\\\P=\dfrac{20\times 0.082057\times 298}{36.2 }\\\\P=13.5\ atm$

so, the pressure of the gas is equal to 13.5 atm.

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Answer:

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<u>Explanation:</u>

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By Stoichiometry of the reaction:

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3 years ago

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tigry1 [53]

Answer:

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Explanation:

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Answer:

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3 years ago

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ehidna [41]

Picture credit - Google images

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