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soldier1979 [14.2K]
3 years ago
12

If glucose is the sole energy source, what fraction of the carbon dioxide exhaled by animals is generated only by the reactions

involved in oxidation of pyruvate to acetyl coa?
Chemistry
2 answers:
White raven [17]3 years ago
7 0

We have to know what fraction of the carbon dioxide exhaled by animals is generated only by the reactions involved in oxidation of pyruvate to acetyl coa.

The fraction is: 1 CO₂ per puruvate molecule.

Pyruvate oxidation takes place in the mitochondrial matrix, before chemical reactions begins, pyruvate enters the mitochondrion crossing the inner membrane and arriving at the matrix.

Oxidation of pyruvate results acetyl coenzyme A after release of one molecule of CO₂.


Flauer [41]3 years ago
3 0

Answer:

1/3

Explanation:

Pyruvate is produced by the glycolysis in cytoplasm. The oxidation of pyruvate takes place in mitochondrial matrix.  

Pyruvate is converted to acetyl-CoA in the reaction given below:

Pyruvate + NAD⁺ + CoA-SH ⇒ acetyl-CoA + NADH + CO₂

<u> 1 molecule of carbon dioxide is eliminated from 1 molecule of pyruvate.</u>

<u>Also, </u>

<u>2 molecules of carbon dioxide is eliminated from 2 molecules of pyruvate (as glucose on glycolysis yields 2 molecules of pyruvate).</u>

Also, acetyl-CoA further goes into the citric acid cycle and produces 2 molecules of carbon dioxide.

<u>Thus pyruvate produces total 3 molecules of CO₂  and hence glucose produces 6 molecules of CO₂ (as glucose on glycolysis yields 2 molecules of pyruvate)</u>

Thus,

<u>Fraction = 2/6 = 1/3</u>

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An organic compound contains , , , and . Combustion of 0.1023 g of the compound in excess oxygen yielded 0.2587 g and 0.0861 g .
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The question displayed below shows the missing information which therefore completes the question.

An organic compound contains C, H, N and O. Combustion of 0.1023 g of the compound in excess oxygen yielded 0.2587 g of CO2 and 0.0861 g of H2O. A sample of 0.4831 g of the compound was analyzed for nitrogen by the Dumas method. The compound is first reacted by passage over hot: The product gas is then passed through a concentrated solution of to remove the. After passage through the solution, the gas contains and is saturated with water vapor. At STP, 38.9 mL of dry N2 was obtained. In a third experiment, the density of the compound as a gas was found to be 2.86 g/L at 127°C and 256 torr. What are the empirical and molecular formulas of the compound? (Enter the elements in the order: C, H, N, O.)

Answer:

the empirical formula = \mathbf {C_3H_6O_{12}N}

the molecular formula = \mathbf {C_3H_6O_{12}N}

Explanation:

From the given information:

\bigg ( 0.2587 \ g \ of CO_2 \bigg) \times \dfrac{1 \ mol \ of CO_2}{44 \ of \ CO_2} \times \dfrac{1 \ mol \ of \  C}{1 \ mol \ of CO_2}

= 0.00588 \ mol \ of \ C \times \dfrac{12.01 \ g \ of \ C}{1 \ mol \ of \ C }

= 0.0706g of C

\bigg ( 0.0861\ g \ of H_2O \bigg) \times \dfrac{1 \ mol \ of H_2O}{18.02 \ g  \ of \ H_2O} \times \dfrac{2 \ mol \ of \  H}{1 \ mol \ of H_2O}

=0.0096 \ mol \times \dfrac{1.008 \ g \ of \ H}{1 mol \ H}

0.0097g of H

Given that N2 at STP = 1 atm, 273 K and V = 0.0389 L

PV = nRT

n = PV/RT

n = \dfrac{1 \ atm \times 0.0389 \ of \ H_2}{0.0821 \ L.atm /mol.K \times 273 \ K }

n = 0.00173 mol of N2

The oxygen in the sample = The total grams in sample -  gram in H - gram in C

The oxygen in the sample = 0.1023 g - 0.0097 g - 0.706 g

The oxygen in the sample = 0.022 g of O

The number of  moles of O_2 = \dfrac{0.02}{16}

= 0.001375 mol of O

O \ in \ product = (0.00588 \ mol \ of \ C ) \times \dfrac{2 \ mol \ of \ O }{1 \ mol \ of \ C }+ \bigg ( 0.0096 \ mol \ of \ H ) \times \dfrac{1 \ mol \ of  \ O }{1 \ mol \ of \ H}

O in product = 0.02136 mol of O

∴

we are meant to divide the moles of each compound by the smallest number of  moles; we have:

C = \dfrac{0.00588}{0.00173} \simeq 3

H = 0.0096 = \dfrac{0.0096}{0.00173} \simeq 6

O = 0.0199= \dfrac{0.0199}{0.00173} \simeq 12

N = 0.00173= \dfrac{0.00173}{0.00173} \simeq 1

Thus; the empirical formula = \mathbf {C_3H_6O_{12}N}

To estimate the molecular formula;  we have:

MM = \dfrac{dRT}{P}

MM = \dfrac{2.80 \ g/ L \times 0.0821 \ L.atm /mol.K \times 400 \ K }{0.337 \ atm}

MM = 272.86 g/mol

Also; the molar mass of \mathbf {C_3H_6O_{12}N} = 248 g/mol

∴

= \dfrac{272.86 \ g/mol}{248 \ g/mol}

=1

Thus; we can conclude that empirical formula as well as the molecular formula are the same.

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Explain :

As there are 500 moles of hydrogen gas in a kilogram, this means that burning a kilogram of hydrogen gas releases 500 times as much energy or 121 million joules .

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