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Thepotemich [5.8K]
3 years ago
15

Determine the net change in NADHNADH , FADH2FADH2 , and GTPGTP from the passage of two acetyl-SCoA molecules through the citric

acid cycle. Degradation of one molecule of glucose yields two acetyl-SCoA molecules. How many molecules of each of the cofactors are produced when these two acetyl-SCoA molecules pass through the citric acid cycle?
Chemistry
1 answer:
CaHeK987 [17]3 years ago
8 0

Answer:The citric acid cycle is the final common pathway for the aerobic oxidation of fuel molecules. Moreover, as we will see shortly (Section 17.3) and repeatedly elsewhere in our study of biochemistry, the cycle is an important source of building blocks for a host of important biomolecules. As befits its role as the metabolic hub of the cell, entry into the cycle and the rate of the cycle itself are controlled at several stages.

Explanation:

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fomenos
I think I it’s gonna be C.
8 0
3 years ago
If 16.29 grams of Na2SO4 is mixed with 3.697 grams of C and allowed to react according to the balanced equation: Na2SO4(aq) + 4
abruzzese [7]

<u>Answer:</u> The limiting reagent in the given chemical reaction is carbon metal.

<u>Explanation:</u>

Excess reagent is defined as the reagent which is present in large amount in a chemical reaction.

Limiting reagent is defined as the reagent which is present in small amount in a chemical reaction. Formation of product depends on the limiting reagent.

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

  • <u>For sodium sulfate:</u>

Given mass of sodium sulfate = 16.29 g

Molar mass of sodium sulfate = 142 g/mol

Putting values in equation 1, we get:

\text{Moles of sodium sulfate}=\frac{16.29g}{142g/mol}=0.115mol

  • <u>For carbon:</u>

Given mass of carbon = 3.697 g

Molar mass of carbon = 12 g/mol

Putting values in equation 1, we get:

\text{Moles of carbon}=\frac{3.697g}{12g/mol}=0.31mol

For the given chemical reaction:

Na_2SO_4(aq.)+4C(s)\rightarrow Na_2S(s)+4CO(g)

By Stoichiometry of the reaction:

4 moles of carbon reacts with 1 mole of sodium sulfate

So, 0.31 moles of carbon will react with = \frac{1}{4}\times 0.31=0.0775mol of sodium sulfate

As, given amount of sodium sulfate is more than the required amount. So, it is considered as an excess reagent.

Thus, carbon metal is considered as a limiting reagent because it limits the formation of product.

Hence, the limiting reagent in the given chemical reaction is carbon metal.

4 0
3 years ago
The rate constant for the reaction 3A equals 4b is 6.00×10 how long will it take the concentration of a to drop from 0.75 to 0.2
Lelu [443]

This question is incomplete, the complete question is;

The rate constant for the reaction 3A equals 4B is 6.00 × 10⁻³ L.mol⁻¹min⁻¹.

how long will it take the concentration of A to drop from 0.75 to 0.25M ?

from the unit of the rate constant we know it is a second reaction order

OPTIONS

a) 2.2×10^−3 min

b) 5.5×10^−3 min

c) 180 min

d) 440 min

e) 5.0×10^2 min

Answer:

it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

Explanation:

Given that;

Rate constant K =  6.00 × 10⁻³ L.mol⁻¹min⁻¹

3A → 4B

given that it is a second reaction order;

k = 1/t [ 1/A - 1/A₀]

kt = [ 1/A - 1/A₀]

t = [ 1/A - 1/A₀] / k

K is the rate constant(6.00 × 10⁻³)

A₀ is initial concentration( 0.75 )

A is final concentration(0.25)

t is time required = ?

so we substitute our values into the equation

t = [ (1/0.25) - (1/0.75)] / (6.00 × 10⁻³)

t = 2.6666 / (6.00 × 10⁻³)

t = 444.34 ≈ 440 min     {significant figures}

Therefore it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

6 0
3 years ago
What is the equation of sodium phosphide
BlackZzzverrR [31]
Na3P is the formula if that helps
6 0
3 years ago
Help?......please???
Flura [38]

Answer:im in middle school in know this but w x y = d

Explanation:so your answer is d :)

5 0
2 years ago
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