The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.
C6H12O6(aq) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)

Calculate the volume of dry CO2 produced at body temperature (37°C) and 0.960 atm when 24.5 g of glucose is consumed in this reaction.

Answers

Answer 1

Answer:

[tex]\large \boxed{\text{21.6 L}}[/tex]

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:        180.16

         C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g:      24.5

(a) Moles of C₆H₁₂O₆

[tex]\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}[/tex]

(b) Moles of CO₂

[tex]\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}[/tex]

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37  °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

[tex]\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}[/tex]


Related Questions

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Answer:

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The pollution can be controlled with the help of reducing the need for the manufactured good and transportation.

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A pollution can be given as the presence of the harmful substances in the environment. The harmful substances known as pollutants deteriorate the quality of the environment and effect the health of living organism as well.

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If a body was lying on their back when they died, and was moved to their front after 12 hours, where would one see the skin discoloration from pooling blood?

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Answer:

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Explanation:

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Answers

Answer:

MRCORRECT has answered the question

Explanation:

13 electrons

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Why do limiting reactants occur?

Identification of the limiting reactant makes it possible to calculate the theoretical yield of a reaction. The reason there is a limiting reactant is that elements and compounds react according to the mole ratio between them in a balanced chemical equation.

How do you get the limiting reactant?

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Answer:

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P represents the atmosphere and Q represents the ocean.

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Answer:

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Answer:

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Answers

Answer:

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Explanation:

Step 1:

Data obtained from the question. This includes:

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This is illustrated below:

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Answer:Releasing insulin to decrease blood sugar and releasing glucagon to increase blood sugar

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Answer:

Answer:Releasing insulin to decrease blood sugar and releasing glucagon to increase blood sugar

Explanation:

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D.10.0 mol is the correct

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Answer:

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Explanation:

Create the chemical compound and find each individual element's molar mass. Lastly, add them up.

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Answers

The balanced equation based on half-reactions is 2Cu + 8HNO3 + 6H+ -> 2Cu(NO3)2 + 2NO + 4H2O.

How to determine the balanced equation based on half-reactions

To complete the balanced equation for the given reaction Cu + HNO3 -> Cu(NO3)2 + NO + H2O using half-reactions, we need to break down the overall reaction into separate oxidation and reduction half-reactions.

1. Oxidation Half-Reaction:

Cu -> Cu2+ + 2e-

In this step, copper (Cu) is oxidized, losing two electrons to form copper(II) ions (Cu2+).

2. Reduction Half-Reaction:

HNO3 + 3H+ + 2e- -> NO + 2H2O

In this step, nitric acid (HNO3) is reduced, gaining two electrons to form nitric oxide (NO) and water (H2O).

Now, to balance the half-reactions, we need to make sure that the number of electrons gained in the reduction half-reaction is equal to the number of electrons lost in the oxidation half-reaction. In this case, we can achieve this by multiplying the oxidation half-reaction by two.

Balanced Half-Reactions:

Oxidation: 2Cu -> 2Cu2+ + 4e-

Reduction: HNO3 + 3H+ + 2e- -> NO + 2H2O

Finally, we can combine the balanced half-reactions to obtain the balanced equation for the overall reaction:

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Therefore, the balanced equation based on half-reactions is 2Cu + 8HNO3 + 6H+ -> 2Cu(NO3)2 + 2NO + 4H2O.

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Answer: mercury

Explanation: I looked up the answer to your question.

Answer:

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I did this question and I got it right.

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Answer:

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