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maksim [4K]
3 years ago
9

When oxygen is depleted, the citric acid cycle stops. What could we add to the system to restore citric acid cycle activity (oth

er than oxygen)?
Chemistry
1 answer:
Hoochie [10]3 years ago
8 0

Answer:

NAD+, FAD.

Explanation:

The citric acid cycle is popularly known as the Kreb's cycle. The cycle involve the oxidation of acetyl-CoA to produce energy. The Kreb's cycle is a chemical process that produces produces two carbon dioxide molecules,NADH,FADH2 and one ATP.

When oxygen is depleted, the citric acid cycle stops, apart from oxygen NAD+ and FAD could be added to the system to restore citric acid cycle activity. NAD+ acts as an electron acceptor.

Citric acid cycle/Kreb's cycle is an aerobic process that occurs in the mitochondria and produces thirty-six(36) ATPs.

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What is the molar concentration of the acid if 35.18 mL of hydrochloric acid was required to neutralize 0.745 g of ALUMINUM hydr
Makovka662 [10]

The mass number of aluminium hydroxide is 78 thus, the number of moles in 0.745 g is:

no. of moles= mass/ RFM

= 0.745/78

=0.00955moles

Therefore the 0.00955 moles should be in the 35.18 ml

therefore 1000ml of the solution will have:

(0.00955ml×1000ml)/35.18

=0.2715moles

The  solution will be 0.27M hydrochloric acid  

3 0
3 years ago
If 5 grams of hydrogen reacted with 40 games of oxygen to form water, how much of water would be formed?
GREYUIT [131]
2H₂ + O₂ = 2H₂O

n(H₂)=m(H₂)/M(H₂)
n(H₂)=5g/2.0g/mol=2.5 mol

n(O₂)=m(O₂)/M(O₂)
n(O₂)=40g/32.0g/mol=1.25 mol

H₂ : O₂ = 2 : 1

2.5 : 1.25 = 2 : 1

n(H₂O)=n(H₂)=2n(O₂)=2.5 mol

m(H₂O)=n(H₂O)M(H₂O)

m(H₂O)=2.5mol*18.0g/mol=45.0 g
3 0
2 years ago
Check all statements below that are true.
docker41 [41]

Answer:

The answers are A,B,C.

Explanation: Just got it right on Edge 2020

3 0
3 years ago
Write an equilibrium expression for each chemical equation involving one or more solid or liquid reactants or products.
Alex_Xolod [135]

Answer:

a.

Keq=\frac{[HCO_3^-][OH^-]}{[CO_3^{2-}]}

b.

Keq=[O_2]^3

c.

Keq=\frac{[H_3O^+][F^-]}{[HF]}

d.

Keq=\frac{[NH_4^+][OH^-]}{[NH_3]}

Explanation:

Hello there!

In this case, for the attached reactions, it turns out possible for us to write the equilibrium expressions by knowing any liquid or solid would be not-included in the equilibrium expression as shown below, with the general form products/reactants:

a.

Keq=\frac{[HCO_3^-][OH^-]}{[CO_3^{2-}]}

b.

Keq=[O_2]^3

c.

Keq=\frac{[H_3O^+][F^-]}{[HF]}

d.

Keq=\frac{[NH_4^+][OH^-]}{[NH_3]}

Regards!

5 0
2 years ago
HURRRYYYY
Sedaia [141]

Answer:

1, 2, 1

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

4 0
3 years ago
Read 2 more answers
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