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4 years ago

What fraction of the carbon dioxide exhaled by animals is generated by the reactions of the citric acid cycle

Chemistry

1 answer:

puteri [66]4 years ago

4 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

1 molecule of carbon dioxide is eliminated from 1 molecule of pyruvate.

Also,

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

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

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)

Thus,

Fraction = 2/6 = 1/3

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According to the octet rule, an atom with two electron shells is most stable when it contains eight

Lorico [155]

According to the octet rule, an atom with two electron shells is most stable when it contains eight electrons. The octet rule is always based on valence electrons and their tendency to form bonds.

4 0

4 years ago

The chemical formula of ordinary sugar is C12H22O11. Calculate the mass of 7.35 mol of sugar.

ch4aika [34]

First, you need to find the mass of 1 mol of sugar. Mass, or molar mass, can simply be found by adding the masses of the individual elements. These are given to you on the periodic table.

$C_{12}H_{22}O_{11}$

12 x 12.011 grams (molar mass of Carbon) = 144.132 g

22 x 1.008 grams (molar mass of Hydrogen) = 22.176 g

11 x 15.999 grams (molar mass of Oxygen) = 175.989 g

Add all of the pieces together.

144.132 g + 22.176 g + 175.989 g = 342.297 grams

So, if one mole has 342.297 grams, then 7.35 of that amount will be your answer.

342.297 g/mol x 7.35 mol = 2,515.88 grams

7 0

3 years ago

Vitellium (Vi) has the following composition:Vi-188: 187.9122 amu; 10.861%Vi-191: 190.9047 amu; 12.428%Vi-193: 192.8938 amu; 76.

yKpoI14uk [10]

10.861% / 100 = 0.10861

12.428 % / 100 = 0.12428

(0.10861 * 187.9122) + (0.12428 * 190.9407) + (0.76711 x 192.8938)

= 192.1100 amu .

hope this helps!

7 0

3 years ago

Calculate the standard molar enthalpy of formation, in kJ/mol, of NO(g) from the following data:

Paraphin [41]

The above question is incomplete, here is the complete question:

Calculate the standard molar enthalpy of formation of NO(g) from the following data at 298 K:

$N_2(g) + 2O_2 \rightarrow 2NO_2(g), \Delta H^o = 66.4 kJ$

$2NO(g) + O_2\rightarrow 2NO_2(g),\Delta H^o= -114.1 kJ$

Answer:

The standard molar enthalpy of formation of NO is 90.25 kJ/mol.

Explanation:

$N_2(g) + 2O_2 \rightarrow 2NO_2(g), \Delta H^o_{1} = 66.4 kJ$

$2NO(g) + O_2\rightarrow 2NO_2(g),\Delta H^o_{2} = -114.1 kJ$

To calculate the standard molar enthalpy of formation

$N_2+O_2\rightarrow 2NO(g),\Delta H^o_{3} = ?$ ...[3]

Using Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

[1] - [2] = [3]

$N_2+O_2\rightarrow 2NO(g),\Delta H^o_{3} = ?$

$\Delta H^o_{3} =\Delta H^o_{1} - \Delta H^o_{2}$

$\Delta H^o_{3}=66.4 kJ - [ -114.1 kJ] = 180.5 kJ$

According to reaction [3], 1 mole of nitrogen gas and 1 mole of oxygen gas gives 2 mole of nitrogen monoxide, So, the standard molar enthalpy of formation of 1 mole of NO gas :

= $\frac{\Delta H^o_{3}}{2 mol}$

$=\frac{180.5 kJ}{2 mol}=90.25 kJ/mol$

3 0

3 years ago

Please help

Sonbull [250]

Answer:

A is the answer.

Explanation:

4 0

3 years ago

Read 2 more answers