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choli [55]
3 years ago
10

C6H12O6 + 6O2 + 38ADP + 38Pi => 6CO2 + 6H2O + 38ATP is the chemical equation for

Chemistry
1 answer:
Jobisdone [24]3 years ago
3 0

Answer:

C6H12O6 + 6O2 + 38ADP + 38Pi => 6CO2 + 6H2O + 38ATP is the chemical equation for cellular respiration

Explanation:

Cellular respiration is the process by which cells breakdown glucose molecules to produce energy in the form of ATP molecules and release waste products such as carbon dioxide and water molecules. Cellular respiration involves a series of reaction pathways such as glycolysis, pyruvate oxidation, citric acid cycle and the oxidative phosphorylation pathway.

The first step of glycolysis breaks down a glucose molecule to release two pyruvate molecules.

In pyruvate oxidation, two molecules of pyruvate are oxidized to acetyl-CoA molecules.

In the citric acid cycle, the acetyl-CoA molecules are used to produce the electron carriers NADH and FADH2.

In the oxidative phosphorylation pathway, NADH and FADH2 donate their electrons to oxygen and ATP molecules are produced using the energy of electron transfer and proton-pumping.

The overall equation for cellular respiration is given as:

C6H12O6 + 6O2 + 38ADP + 38Pi => 6CO2 + 6H2O + 38ATP

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At 500K, the equilibrium constant for the reaction N2O4D2NO2 is 1.5 x 10^3. What is the equilibrium constant for the reaction: 6
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Answer:

K = 2.96x10⁻¹⁰

Explanation:

Based on the initial reaction:

N2O4 ⇄ 2NO2; K = 1.5x10³

Using Hess's law, we can multiply this reaction changing K:

3 times this reaction:

3N2O4 ⇄ 6NO2; K = (1.5x10³)³ =3.375x10⁹

The inverse reaction has a K of:

6NO2 ⇄ 3N2O4 K = 1/3.375x10⁹;

<h3>K = 2.96x10⁻¹⁰</h3>

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Solid, Liquid, and Gas

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Use the following half-reactions to write three spontaneous reactions, calculate E°cell for each reaction, and rank the oxidizin
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Answer:

See explaination

Explanation:

1)

we know that

half cell with higher reduction potential is cathode

so

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

anode :

Cr(s) ---> Cr+3 + 3e-

so

overall reaction is

3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3

now

Eo cell = Eo cathode - Eo anode

so

EO cell = 1.77 + 0.74

Eo cell = 2.51 V

now

in this case

oxidizing agents are N20 and Cr+3

reducing agents are Cr and N2

higher the reduction potential , stronger the oxidizing agent

lower the reduction potential , stronger the reducing agent

so

oxidzing agents

N20 > Cr+3

reducing agents

Cr > N2

2)

cathode :

Au+ + e- --> Au

anode :

Cr ---> Cr+3 + 3e-

overall reaction

3Au+ + Cr ---> 3Au + Cr+3

Eo cell = 1.69 + 0.74

Eo cell = 2.43

now

oxidizing agents :

Au+ > Cr+3

reducing agents :

Cr > Au

3)

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

andoe :

Au ---> Au+ + e-

overall

2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20

Eo cell = 1.77 - 1.69

Eo cell = 0.08

oxidizing agents

N20 > Au+

reducing agents

Au > N2

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