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e-lub [12.9K]
2 years ago
6

A sample of carbon dioxide gas occupies a volume of 250 mL at 25c what volume will it occupy at 95c

Chemistry
1 answer:
anzhelika [568]2 years ago
6 0

Considering the Charles's law, the sample of carbon dioxide gas will occupy 308.72 mL.

<h3>Charles's law</h3>

Charles's law establishes the relationship between the temperature and the volume of a gas when the pressure is constant. This law says that the volume is directly proportional to the temperature of the gas: for a given sum of gas at constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases.

Mathematically, Charles's law states that the ratio between volume and temperature will always have the same value:

\frac{V}{T} =k

Considering an initial state 1 and a final state 2, it is fulfilled:

\frac{V1}{T1} =\frac{V2}{T2}

<h3>Final volume in this case</h3>

In this case, you know:

  • V1= 250 mL
  • T1= 25 C= 298 K (being 0 C=273 K)
  • V2= ?
  • T2= 95 C= 368 K

Replacing in Charles's law:

\frac{250 mL}{298 K} =\frac{V2}{368 K}

Solving:

V2=368 K\frac{250 mL}{298 K}

<u><em>V2= 308.72 mL</em></u>

Finally, the sample of carbon dioxide gas will occupy 308.72 mL.

Learn more about Charles's law:

brainly.com/question/4147359

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n(Ba(OH)2) = 0.014

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How many grams of oxygen gas will be produced when 2.50 moles of potassium chlorate is decomposed?
yawa3891 [41]

Answer:

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Explanation:

Hello!

In this case, since the decomposition of potassium chlorate is:

2KClO_3\rightarrow 2KCl+3O_2

We can see a 2:3 mole ratio between potassium chlorate and oxygen (molar mass 32.0 g/mol), thus, via stoichiometry, we compute the mass of oxygen that are produced by the decomposition of 2.50 moles of this reactant:

m_{O_2}=2.50molKClO_3*\frac{3molO_2}{2molKClO_3} *\frac{32.0gO_2}{1molO_2}\\\\m_{O_2}=120gO_2

Best regards!

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3 years ago
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