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2 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]2 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,

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

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1. Copper (__) is an element on the periodic table.

user100 [1]

Answer: Cu

Explanation: It is Cu because the origin of the word Copper comes from the latin word "Cuprum".

3 0

2 years ago

Caffeine, a molecule found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 10.0 g of caffeine pr

denis-greek [22]

Answer:

194 g/mol.

Explanation:

Hello,

In this case, one first must compute the mass of each element as shown below:

$C=18.13gCO_2*\frac{12gC}{44gCO_2} =4.945gC\\H=4.639gH_2O*\frac{2.016gH}{18.0152gH_2O}=0.519gH\\N=2.885gN_2\\O=10.0g-4.945g-0.519g-2.885g=1.651gO$

Next, the corresponding moles:

$C=4.945gC*\frac{1molC}{12gC}=0.412mol\\H=0.519gH*\frac{1molH}{1gH}=0.519mol\\N=2.885gN*\frac{1molN}{14gN}=0.206molN\\O=1.648gO*\frac{1molO}{16gO} =0.103molO$

Then, each element's subscripts is found to be:

$C=\frac{0.412}{0.103}=4\\H=\frac{0.519}{0.103}=5\\N=\frac{0.206}{0.103} =2\\O=\frac{0.103}{0.103}=1$

Therefore, the empirical formula is:

$C_4H_5N_2O$

Nonetheless, it has a molar mass of 97bg/mol, thereby, by multiplying such formula by 2 one gets:

$C_8H_10N_4O_2$

Which has a molar mass of 194 g/mol being correctly contained in the given interval.

Best regards.

8 0

2 years ago

Ethyl alcohol is produced by the fermentation of glucose, C6H12O6. C6H12O6 (s) → 2 C2H5OH (l) + 2 CO2 (g) ΔH° = – 69.1 kJ Given

Norma-Jean [14]

Answer:

-1273.3

Explanation:

Enthalpy of formation of a compound is the amount of heat absorbed or evolved when one mole of the compound is formed from other compounds.

enthalpy of formation Of CO2 = 2 X -393.5 = -787

enthalpy of formation Of C2H5OH = 2 X -277.7 = -555.4

enthalpy of formation Of C6H12O6 = 69.1 (reverse sign) + (-787 + -555.4) = - 1273.3 Joules

6 0

2 years ago

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple _______. Then, plot ln(K

Aloiza [94]

Answer:

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us use the thermodynamic definition of the Gibbs free energy and its relationship with Ksp as follows:

$\Delta G=-RTln(Ksp)\\\\\Delta G=\Delta H-T\Delta S$

Thus, by combining them, we obtain:

$-RTln(Ksp)=\Delta H-T\Delta S\\\\ln(Ksp)=-\frac{\Delta H}{RT} +\frac{T\Delta S}{RT} \\\\ln(Ksp)=-\frac{\Delta H}{RT} +\frac{\Delta S}{R}$

Which is related to the general line equation:

$y=mx+b$

Whereas:

$y=ln(Ksp)\\\\m=-\frac{\Delta H}{R} \\\\x=\frac{1}{T} \\\\b=\frac{\Delta S}{R}$

It means that we answer to the blanks as follows:

Regards!

8 0

2 years ago

32.7 grams of water vapor takes up how many liters at standard temperature and pressure (273 K and 100 kPa)?

kotegsom [21]

Under standard temperature and pressure conditions, it is known that 1 mole of a gas occupies 22.4 liters.

From the periodic table:
molar mass of oxygen = 16 gm
molar mass of hydrogen = 1 gm
Thus, the molar mass of water vapor = 2(1) + 16 = 18 gm

18 gm of water occupies 22.4 liters, therefore:
volume occupied by 32.7 gm = (32.7 x 22.4) / 18 = 40.6933 liters

5 0

2 years ago