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devlian [24]
3 years ago
6

Pls don't answer if you done know. Which of the following best describes what would happen if mitosis did not take place?

Chemistry
1 answer:
mojhsa [17]3 years ago
8 0

Answer: B

Explanation:

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In the reaction catalyzed by the pyruvate dehydrogenase complex, the two carbons constituting the acetyl group are
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Transferred to the lipoamide by an earlier intermediate in the process.

The pyruvate dehydrogenase complex (PDC) is a mitochondrial multienzyme complex composed of three different enzymes

<h3>What reaction is catalyzed by enzyme 2 of the pyruvate dehydrogenase complex ?</h3>

the pyruvate dehydrogenase complex is the bridge between glycolysis and the citric acid cycle

  • Five coenzymes are used in the pyruvate dehydrogenase complex reactions: thiamine pyrophosphate or TPP, flavin adenine dinucleotide or FAD, coenzyme A or CoA, nicotinamide adenine dinucleotide or NAD, and lipoic acid.
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True or false , the viscosity of water is much greater than the viscosity ofhoney.
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False... Honey is greater in velosity then water
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4 years ago
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When 136g of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezin
loris [4]

The given question is incomplete. The complete question is:

When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.

Answer: The vant hoff factor for sodium chloride in X is 1.9

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=8.2^0C = Depression in freezing point

K_f = freezing point constant

i = vant hoff factor = 1 ( for non electrolyte)

m= molality =\frac{136g\times 1000}{950g\times 75.07g/mol}=1.9

8.2^0C=1\times K_f\times 1.9

K_f=4.32^0C/m

Now Depression in freezing point for sodium chloride is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=20.0^0C = Depression in freezing point

K_f = freezing point constant  

m= molality = \frac{136g\times 1000}{950g\times 58.5g/mol}=2.45

20.0^0C=i\times 4.32^0C\times 2.45

i=1.9

Thus vant hoff factor for sodium chloride in X is 1.9

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