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Leviafan [203]
3 years ago
10

The portion of the circulation governed by the right side of the heart is the ______ circulation.

Biology
1 answer:
AlekseyPX3 years ago
4 0
<span>The correct answer is B. Pulmonary. The answers C. Cerebral and D. Extremity, are baseless regarding this question; A. Systemic could be a good alternative, as the blood flows from the systemic to the pulmonary circulation in the right side of the heart; but the systemic circulation is in fact mostly handled in the left side of the heart, while the right side handles the pulmonary one. </span>
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7) Calculate ∆G^0' for the following reaction: glucose + 6O2 → 6 CO2 + 6 H2O. Is this reaction exergonic or endergonic?
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The reaction is endergonic.

<h2>Procedure - Determination of the Gibbs reaction type</h2>

<h3>Thermochemical model</h3>

First, we need to determine the Gibbs free energy function (\Delta G_{o}), in kilojoules, of the entire reaction by this formula:

\Delta G_{o} = \Sigma\limits_{j=1}^{m} n_{j}\cdot g_{o, j} - \Sigma\limits_{i=1}^{p}n_{i}\cdot g_{o,i} (1)

Where:

  • n_{i} - Number of kilomoles of the i-th reactive, in kilomoles.
  • n_{j} - Number of kilomoles of the j-th product, in kilomoles.
  • g_{o,i} - Specific Gibbs free energy function of the i-th reactive, in kilojoules per kilomole.
  • g_{o,j} - Specific Gibbs free energy function of the j-th reactive, in kilojoules per kilomole.

<h3>Data from thermochemical tables</h3>

From thermochemical tables we have the following information:

Reactives

g_{glu} = -2871596\,\frac{kJ}{kmol}, g_{O_{2}} = 0\,\frac{kJ}{kmol}

n_{glu} = 1\,kmol, n_{O_{2}} = 6\,kmol

Products

g_{CO_{2}} = -137150\,\frac{kJ}{kmol}, g_{H_{2}O} = -237180\,\frac{kJ}{kmol}

n_{CO_{2}} = 6\,kmol, n_{H_{2}O} = 6\,kmol

<h3>Determination of the Gibbs free energy function</h3>

Now we proceed to determine the Gibbs free energy function and later determine the nature of the <em>thermochemical</em> reaction:

\Delta G_{o} = (6\,kmol)\cdot \left(-137150\,\frac{kJ}{kmol} \right) + (6\,kmol)\cdot \left(-237180\,\frac{kJ}{kmol} \right)-(1\,kmol)\cdot \left(-2871596\,\frac{kJ}{kmol} \right) - (6\,kmol)\cdot \left(0\,\frac{kJ}{kmol} \right)

\Delta G_{o} = 625616\,kJ

The Gibbs free energy function of the <em>entire</em> reaction is 625616 kilojoules.

<h3>Analysis of the result</h3>

The Gibbs free energy function has a <em>positive</em> sign, which means that this reaction is endergonic.

To learn more on Gibbs free energy, we kindly invite to check this verified question: brainly.com/question/9552459

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