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2 years ago

The immediate electron acceptor for the majority of the oxidative reactions of the citric acid cycle is.

Chemistry

1 answer:

Anna [14]2 years ago

6 0

NAD serves as the bulk of the oxidative processes in the citric acid cycle's initial electron acceptor.

<h3>What are electron acceptors in citric acid cycle?</h3>

In the Krebs cycle, which transfers electrons via the electron transport chain with oxygen as the final acceptor, coenzymes like FAD and NAD+ are reduced.
In a single cycle, three NADH+ and one FADH2 are produced, and when the cycle enters the electron transport chain, 10 ATP is produced.
The final electron acceptor in the electron transport chain is oxygen. The proton gradient in the intermembrane gap is produced by NADH molecules donating electrons that are then transmitted through a number of different proteins.

<h3>What occurs throughout the citric acid cycle?</h3>

The cycle of citric acid: In the citric acid cycle, a six-carbon citrate molecule is created when an acetyl group from acetyl CoA is joined to a four-carbon oxaloacetate molecule.

Citrate is oxidized over a number of steps, generating two molecules of carbon dioxide for each acetyl group added to the cycle.

learn more about citric acid cycle here

<u>brainly.com/question/14900762</u>

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This is an example of a mixture that will separate______.

devlian [24]

B is the correct answer

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3 years ago

(b) The conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 S m–1. If the molar conduct

Zarrin [17]

Explanation:

The given data is as follows.

$\Lambda^{o}_{m}$ (NaCl) = $1.264 \times 10^{-2}$

$\Lambda^{o}_{m}$ (H-O=C-ONO) = $1.046 \times 10^{-2}$

$\Lambda^{o}_{m}$ (HCl) = $4.261 \times 10^{-2}$

Conductivity of monobasic acid is $5.07 \times 10^{-2} S m^{-1}$

Concentration = 0.01 $mol/dm^{3}$

Therefore, molar conductivity ( $\Lambda_{m}$ ) of monobasic acid is calculated as follows.

$\Lambda_{m} = \frac{conductivity}{concentration}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol/dm^{3}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol \times 10^{3}}$

= $5.07 \times 10^{-3} S m^{2} mol^{-1}$

Also, $\Lambda^{o}_{m}$ = $\Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}$

= $4.261 \times 10^{-2} + 1.046 \times 10^{-2} - 1.264 \times 10^{-2}$

= $4.043 \times 10^{-2} S m^{2} mol^{-1}$

Relation between degree of dissociation and molar conductivity is as follows.

$\alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{4.043 \times 10^{-2} S m^{2} mol^{-1}}$

= 0.1254

Whereas relation between acid dissociation constant and degree of dissociation is as follows.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

Putting the values into the above formula we get the following.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

= $\frac{0.01 \times (0.1254)^{2}}{1 - 0.1254}$

= $0.017973 \times 10^{-2}$

= $1.7973 \times 10^{-4}$

Hence, the acid dissociation constant is $1.7973 \times 10^{-4}$ .

Also, relation between $pK_{a}$ and $K_{a}$ is as follows.

$pK_{a} = -log K_{a}$

= $-log (1.7973 \times 10^{-4})$

= 3.7454

Therefore, value of $pK_{a}$ is 3.7454.

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3 years ago

26.0 g of a liquid that has a density of 1.44 g/mL needs to be measured out in a graduated cylinder. what volume should be used

Vaselesa [24]

As Density = Mass/Volume
Mass = 26.0g
Density = 1.44g/mL
Therefore Volume = Mass/Density
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3 years ago

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Which term describes the weather in an area over a long period of time?

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Answer:

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Explanation:

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3 years ago

The smallest unit of an element that can exist either alone or in combination with other such particles of the same or different

Mekhanik [1.2K]

Answer

im not quite sure but I think the answer is <em>D atom</em><em> </em>

Explaination

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