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alexgriva [62]
3 years ago
10

How many equivalents of pyruvate are needed to generate 1 equivalent of glucose?

Chemistry
1 answer:
kodGreya [7K]3 years ago
4 0
Gluconeogenesis is the process by which the body produce glucose from non-carbohydrate substrates such as pyruvate. To produce glucose from pyruvate, two moleucules of pyruvate is needed. The reaction for the gluconeogenesis reaction is as follow:
2 pyruvate + 4 ATP + 2 GTP + 2 NADH = Glucose + 4 ADP + 2 GDP + 2 NAD + 6Pi. 

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Stored mechanical energy is energy stored and awaiting to be used and mechanical energy is the energy that was stored being used.

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3 years ago
How does air pressure affect the boiling point of a liquid?
Agata [3.3K]

Answer:

D. Air pressure lowers the temperature of the liquid molecules.

3 0
2 years ago
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Find the atomicity of water.​
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One molecule of water contains two atoms of hydrogen and one atom of oxygen, the atomicity of water is three.
4 0
2 years ago
Determain the number of moles in 2.24l of ch4 at stp
valkas [14]

Answer:

0.1 mole of CH₄

Explanation:

From the question given above, the following data were obtained:

Volume of CH₄ = 2.24 L

Number of mole of CH₄ =?

The number of mole of CH₄ can be obtained as follow:

Recall:

1 mole of a gas occupy 22.4 L at stp. This implies that 1 mole of CH₄ occupies 22.4 L at stp.

22.4 L = 1 mole of CH₄

Therefore,

2.24 L = 2.24 × 1 mole of CH₄ / 22.4

2.24 L = 0.1 mole of CH₄.

6 0
3 years ago
When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr
Amanda [17]

<u>Answer:</u> The value of K_c for the given reaction is 1.435

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Given mass of N_2O_4 = 9.2 g

Molar mass of N_2O_4 = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M

For the given chemical equation:

                 N_2O_4(g)\rightleftharpoons 2NO_2(g)

<u>Initial:</u>          0.20

<u>At eqllm:</u>     0.20-x        2x

We are given:

Equilibrium concentration of N_2O_4 = 0.057

Evaluating the value of 'x'

\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143

The expression of K_c for above equation follows:

K_c=\frac{[NO_2]^2}{[N_2O_4]}

[NO_2]_{eq}=2x=(2\times 0.143)=0.286M

[N_2O_4]_{eq}=0.057M

Putting values in above expression, we get:

K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435

Hence, the value of K_c for the given reaction is 1.435

6 0
3 years ago
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