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katrin [286]
3 years ago
9

You collect a sample of gases from an indoor pool area. The sample contains air and water vapor. The total pressure is 100.18 ki

lopascals, and the partial pressure of the water vapor is 3.36 kilopascals. What is the partial pressure of the air in the sample?
A. 29.8 kPa
B. 51.77 kPa
C. 96.82 kPa
D. 103.54 kPa
E. 337 kPa
Chemistry
2 answers:
nikitadnepr [17]3 years ago
7 0

Here we have to choose the partial pressure exerted by the air in the given sample among the given options.

The partial pressure of the air in the sample is C. 96.82 kPa.

As per the Dalton's law of partial pressure the total pressure of the gas in the pool (P) is equivalent to the summation of the partial pressure of water vapor and air pressure.

Mathematically we may write:

P_{Total}= P_{water vapor} +  P_{Air}

Now the P_{Total} = 100.18 kilopascalsP_{water vapor} = 3.36 kilopascals.On plugging the values:100.18 kilopascals = 3.36 kilopascals + P_{Air}

Or, P_{Air}[/tex] = (100.18 - 3.36) kilopascals

Or, P_{Air}[/tex] = 96.82 Kilopascals.

Thus the partial pressure of the air in the sample is C. 96.82 kPa.

gogolik [260]3 years ago
4 0

Thus the partial pressure of the air in the sample is C. 96.82 kPa.

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The disposal method used for high-level nuclear waste Concentrate and contain

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High intensity nuclear waste is an underlying issue for the world where the generated nuclear waste is one side very hazardous and on other side would help us  in many viable processes but the negative sides of a consequence  just outnumber the positive sides of the situation.

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Ratling [72]

Hello!

1.00 L of a gas at STP is compressed to 473 mL. What is the new pressure of gas?

  • <u><em>We have the following data:</em></u>

Vo (initial volume) = 1.00 L  

V (final volume) = 473 mL → 0.473 L  

Po (initial pressure) = 1 atm (pressure exerted by the atmosphere - in STP)  

P (final pressure) = ? (in atm)

  • <u><em>We have an isothermal transformation, that is, its temperature remains constant, if the volume of the gas in the container decreases, so its pressure increases. Applying the data to the equation Boyle-Mariotte, we have:</em></u>

P_0*V_0 = P*V

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P = \dfrac{1}{0.473}

\boxed{\boxed{P \approx 2.11\:atm}}\:\:\:\:\:\:\bf\green{\checkmark}

<u><em>Answer:  </em></u>

<u><em>The new pressure of the gas is 2.11 atm  </em></u>

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\bf\blue{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}

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