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alexandr402 [8]
2 years ago
8

Aqueous sodium phosphate and aqueous iron (III) chloride react to produce aqueous sodium chloride and solid iron (III) phosphate

.
Chemistry
1 answer:
mojhsa [17]2 years ago
4 0

Answer:

this one is hard

Explanation:

but it's iron because the sodium so yea there u go.

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In acetyl CoA formation, the carbon-containing compound from glycolysis is oxidized to produce acetyl CoA. From the following co
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Answer:

- Net Input: NAD⁺, coenzyme A, pyruvate

- Net Output: NADH, acetyl CoA, CO₂

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Explanation:

Hello,

In this case, it is important to recall that acetyl-CoA is produced either by oxidative decarboxylation of pyruvate derived from glycolysis, which is carried out into the mitochondrial matrix, by cause of the oxidation of high-order fatty acids, or by oxidative degradation of very specific amino acids. Acetyl-CoA then enters in the citric acid cycle where it is oxidized in the light of energy production.

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Carbon-14 is a common isotope of the element. How many electrons
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Consider the reaction: N2(g) 2 O2(g)N2O4(g) Write the equilibrium constant for this reaction in terms of the equilibrium constan
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Answer : The equilibrium constant for this reaction is, K=\frac{(K_b)^2}{K_a}

Explanation :

The given main chemical reaction is:

N_2(g)+2O_2(g)\rightarrow N_2O_4(g);  K

The intermediate reactions are:

(1) N_2O_4(g)\rightarrow 2NO_2(g);  K_a

(2) \frac{1}{2}N_2(g)+O_2(g)\rightarrow NO_2(g);  K_b

We are reversing reaction 1 and multiplying reaction 2 by 2 and then adding both reaction, we get:

(1) 2NO_2(g)\rightarrow N_2O_4(g);  \frac{1}{K_a}

(2) N_2(g)+2O_2(g)\rightarrow 2NO_2(g);  (K_b)^2

Thus, the equilibrium constant for this reaction will be:

K=\frac{1}{K_a}\times (K_b)^2

K=\frac{(K_b)^2}{K_a}

Thus, the equilibrium constant for this reaction is, K=\frac{(K_b)^2}{K_a}

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