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ira [324]
3 years ago
15

Oxidative phosphorylation could not occur without glycolysis and the citric acid cycle, because _____.

Chemistry
1 answer:
katrin [286]3 years ago
8 0

Answer:

these two processes supply the electrons that are needed for the electron transport chain

Explanation:

Oxidative phosphorylation is the process in which electrons transfer from electron donors to electron acceptors (usually oxygen). These reactions are called redox reactions, and they provide energy used to form ATP.

Electron donors (NADH and FADH2) used in oxidative phosphorylation are produces in some of the catabolic biochemical processes, such as glycolysis, the citric acid cycle, and beta oxidation. The NADH and FADH2 are energy-rich molecules because each of them contains a pair of electrons thus having a high transfer potential. Because of that, oxidative phosphorylation could not happen without first obtaining electron donors in glycolysis and citric acid cycle.

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What is a particle diagram to illustrate the composition of air
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A diagram of the composition of air would be the air's chemical formula and it's structure. 
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4 years ago
As water is cooled from 4º C to 0°C, its density
frosja888 [35]

Answer:

decreases

Explanation:

3 0
4 years ago
Free-energy change, ΔG∘, is related to cell potential, E∘, by the equationΔG∘=−nFE∘where n is the number of moles of electrons t
mart [117]

Answer:

a)\Delta G=372490 J

b)\Delta G=-568614 J

Explanation:

a) The reaction:

Mg(s) +Fe^{2+}(aq) \longrightarrow Mg^{2+}(aq) + Fe(s)

The free-energy expression:

\Delta G=-n*F*E

E=E_{red}-E_{ox]

The element wich is reduced is the Fe and the one that oxidates is the Mg:

-0.44V=E_{red}-(-2.37V)=1.93V

The electrons transfered (n) in this reaction are 2, so:

\Delta G=-2mol*96500 C/mol * 1.93 V

\Delta G=-372490 J

b) If you have values of enthalpy and enthropy you can calculate the free-energy by:

\Delta G=\Delta H - T* \Delta S

with T in Kelvin

\Delta G=-675 kJ*\frac{1000J}{kJ} - 298K*-357 J/K

\Delta G=-568614 J

7 0
3 years ago
How many joules of energy is required to bring a five-pound (1 pound = 454 g) bag
sweet-ann [11.9K]
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5 0
3 years ago
Calculate each of the following quantities.<br> (a) mass in kilograms of 3.7 x 1020 molecules of NO2
HACTEHA [7]

Answer:

The answer to your question is:  0.028 kg of NO2

Explanation:

Data

3.7 x 10²⁰ molecules of NO2 in kg

MW of NO2 = 14 + (16 x 2) = 14 + 32 = 46 kg

                   1 mol of NO2 ---------------------  6.023 x 10 ²³ molecules

                   x                     --------------------- 3.7 x 10²⁰ molecules

                   x = 3.7 x 10²⁰ x 1 / 6.023 x 10 ²³

                   x = 0.00061 mol

         

                 1 mol of NO2 ---------------------  46 kg of NO2

                 0.00061 mol     ------------------    x

                 x = 0.00061 x 46/1

                x = 0.028 kg of NO2

7 0
4 years ago
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