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Alla [95]
1 year ago
5

Challenge Air is a mixture of gases. By percentage, it is roughly 78 percent nitrogen, 21 percent oxygen, and 1 percent argon. (

There are trace amounts of many other gases in air. If the atmospheric pressure is 760 mmHg, what are the partial pressures of nitrogen, oxygen, and argon in the atmosphere?
Chemistry
1 answer:
Softa [21]1 year ago
4 0

The partial pressures of gases in air at 760 mmHg are N₂ (592.8 mmHg), O₂ (159.6 mmHg) and Ar (7.6 mmHg).

<h3>What is partial pressure?</h3>

Partial pressure is the pressure of a gas in a mixture of gases.

Air is a mixture of gases with 78% N₂, 21% O₂, 1% Ar and other gases in trace amounts. The partial pressure of a gas is proportional to its percentage.

Let's calculate the partial pressures of these gases, considering that the atmospheric pressure is 760 mmHg.

pN₂ = 760 mmHg × 78% = 592.8 mmHg

pO₂ = 760 mmHg × 21% = 159.6 mmHg

pAr = 760 mmHg × 1% = 7.6 mmHg

The partial pressures of gases in air at 760 mmHg are N₂ (592.8 mmHg), O₂ (159.6 mmHg) and Ar (7.6 mmHg).

Learn more about partial pressure here: brainly.com/question/19813237

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Correct Question: what is the oxidizing agent in the reaction.

2MnO4–(aq) +10Cl–(aq) + 16H+(aq) --------> 5Cl2(g) + 2Mn2+(aq) +8H2O(l)

Answer: MnO4-is the oxidizing agent

Explanation:

In the reaction 2MnO4–(aq) +10Cl–(aq) + 16H+(aq) --------> 5Cl2(g) + 2Mn2+(aq) +8H2O(l)

Oxidizing agent oxidizes other molecules while the themselves get reduced.

oxidizing agents give away Oxygen to other compounds.

MnO4-is the oxidizing agent because

On the reactants side

Oxidation number of Mn in 2MnO4- is +7

Oxidation number of Cl- is -1

On the products side

Oxidation number of Mn is +2

While oxidation number of Cl is zero

Therefore the oxidizing agent is 2MnO4 because is oxidizes Chlorine from -1 to 0 while itself got reduced from oxidation state of +7 to +2

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A 10,0-L cylinder of gas is stored at room temperature (20.0°C) and a pressure of 1800 psi. If the gas is
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Considering the Charles' law, the gas would have a temperature of -109.2 C.

<h3>Charles' law</h3>

Finally, Charles' law establishes the relationship between the volume and temperature of a gas sample at constant pressure. This law says that the volume is directly proportional to the temperature of the gas. That is, if the temperature increases, the volume of the gas increases, while if the temperature of the gas decreases, the volume decreases.

Charles' law is expressed mathematically as:

\frac{V}{T} =k

If you want to study two different states, an initial state 1 and a final state 2, the following is true:

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

<h3>Temperature of the gas in this case</h3>

In this case, you know:

  • P1= 1800 psi
  • V1= 10 L
  • T1= 20 C= 293 K (being 0 C= 273 K)
  • P2= 1800 psi
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You can see that the pressure remains constant, so you can apply Charles's law.

Replacing in the Charles's law:

\frac{10 L}{293 K} =\frac{6 L}{T2}

Solving:

\frac{10 L}{293 K} T2=6 L

T2=\frac{6 L}{\frac{10 L}{293 K} }

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The gas would have a temperature of -109.2 C.

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