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Troyanec [42]
3 years ago
13

Alveolar air (a mixture of nitrogen, oxygen, and carbon dioxide) has a total pressure of 0.998 atm. If the partial pressure of o

xygen gas is 0.198 atm and the partial pressure of nitrogen gas is 0.770 atm, what is the partial pressure of carbon dioxide in millimeters of mercury?
Chemistry
1 answer:
inn [45]3 years ago
6 0

Answer:

The partial pressure of carbon dioxide is 22.8 mmHg

Explanation:

Dalton's Law is a gas law that relates the partial pressures of the gases in a mixture. This law says that the pressure of a gas mixture is equal to the sum of the partial pressures of all the gases present.

In this case:

Ptotal=Pnitrogen + Poxygen + Pcarbondioxide

You know that:

  • Ptotal= 0.998 atm
  • Pnitrogen= 0.770 atm
  • Poxygen= 0.198 atm
  • Pcarbondioxide= ?

Replacing:

0.998 atm=0.770 atm + 0.198 atm + Pcarbondioxide

Solving:

Pcarbondioxide= 0.998 atm - 0.770 atm - 0.198 atm

Pcarbondioxide= 0.03 atm

Now you apply the following rule of three: if 1 atm equals 760 mmHg, 0.03 atm how many mmHg equals?

Pcarbondioxide=\frac{0.03 atm*760 mmHg}{1 atm}

Pcarbondioxide= 22.8 mmHg

<u><em>The partial pressure of carbon dioxide is 22.8 mmHg</em></u>

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Answer:

\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3

Explanation:

Hess's Law of Constant Heat Summation states that if a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation is equal to the sum of the enthalpy changes of the other chemical equations. Thus, the reaction that involves the conversion of reactant A to B, for example, has the same enthalpy change even if you convert A to C, before converting it to B. Regardless of how many steps it takes for the reactant to be converted to the product, the enthalpy change of the overall reaction is constant.

With Hess's Law in mind, let's see how A can be converted to 2C +E.

\bf{\text{A} \rightarrow 2\text{B}}                  (Δ\text{H}_1)  -----(1)

Since we have 2B, multiply the whole of II. by 2:

\bf{2\text{B} \rightarrow 2\text{C} +2\text{D}}       (2Δ\text{H}_2) -----(2)

This step converts all the B intermediates to 2C +2D. This means that the overall reaction at this stage is \text{A} \rightarrow 2\text{C} +2\text{D}.

Reversing III. gives us a negative enthalpy change as such:

\bf{2\text{D} \rightarrow \text{E}}                  (-Δ\text{H}_3) -----(3)

This step converts all the D intermediates formed from step (2) to E. This results in the overall equation of \text{A} \rightarrow 2\text{C} +\text{E}, which is also the equation of interest.

Adding all three together:

\text{A} \rightarrow 2\text{C}+\text{E}            (\bf{\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3 })

Thus, the first option is the correct answer.

Supplementary:

To learn more about Hess's Law, do check out: brainly.com/question/26491956

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