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const2013 [10]
3 years ago
11

Please match each cofactor with its function. 1) thiamine pyrophosphate A) CO2 activation/transfer 2) coenzyme A B) oxidation/re

duction 3) biotin C) activation of aldehydes 4) NAD D) acyl group transfer
Chemistry
1 answer:
yanalaym [24]3 years ago
8 0

Answer:

1) thiamine pyrophosphate -activation of aldehydes

2) coenzyme A -acyl group transfer

3) biotin -CO2 activation/transfer

4) NAD -oxidation/reduction

Explanation:

1. Thiamine pyrophosphate: This is a derivative of Vitamin B1 also known as thiamine. It contains a pyrimidine group linked to the thiazole ring. This connection is further linked to the pyrophosphate group. It functions as a coenzyme in all reactions involving alpha-keto acids. This produces activated aldehydes that could be subject to oxidation.

2. Coenzyme A: This cofactor is a thiol that reacts with carboxylic acids to form thioesters. In so doing, it carries the acyl group. In this condition, it can also be referred to as acyl CoA.

3. Biotin: Also known as Vitamin B7, biotin consists of an ureido ring merged with tetrahydrothiophene. The ureido ring contains the CO2 that can be transferred or activated. Five carboxylase enzymes use biotin as a cofactor in processes such as fat synthesis, glucose generation and the breakdown of sugar.

4. NAD:  Nicotinamide adenine dinucleotide consists of two dinucleotides connected to each other at their phosphate groups. NAD exists in two states which are the NAD+ and NADH states. These two states serve as oxidizing and reducing agents respectively. The oxidizing agent becomes reduced to NADH after accepting electrons from other compounds. NADH donates an electron and becomes oxidized to NAD+.

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Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of water is pro
sergeinik [125]

The question is incomplete, here is the complete question:

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water. If 12.5 g of water is produced from the reaction of 72.6 g of sulfuric acid and 77.0 g of sodium hydroxide, calculate the percent yield of water. Be sure your answer has the correct number of significant digits in it.

<u>Answer:</u> The percent yield of water in the reaction is 46.85 %.

<u>Explanation:</u>

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

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

  • <u>For NaOH:</u>

Given mass of NaOH = 77.0 g

Molar mass of NaOH = 40 g/mol

Putting values in equation 1, we get:

\text{Moles of NaOH}=\frac{77.0g}{40g/mol}=1.925mol

  • <u>For sulfuric acid:</u>

Given mass of sulfuric acid = 72.6 g

Molar mass of sulfuric acid = 98 g/mol

Putting values in equation 1, we get:

\text{Moles of sulfuric acid}=\frac{72.6g}{98g/mol}=0.741mol

The chemical equation for the reaction of NaOH and sulfuric acid follows:

H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O

By Stoichiometry of the reaction:

1 mole of sulfuric acid reacts with 2 moles of NaOH

So, 0.741 moles of sulfuric acid will react with = \frac{2}{1}\times 0.741=1.482mol of NaOH

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

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

By Stoichiometry of the reaction:

1 mole of sulfuric acid produces 2 moles of water

So, 0.741 moles of sulfuric acid will produce = \frac{2}{1}\times 0.741=1.482moles of water

Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 1.482 moles

Putting values in equation 1, we get:

1.482mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(1.482mol\times 18g/mol)=26.68g

To calculate the percentage yield of water, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of water = 12.5 g

Theoretical yield of water = 26.68 g

Putting values in above equation, we get:

\%\text{ yield of water}=\frac{12.5g}{26.68g}\times 100\\\\\% \text{yield of water}=46.85\%

Hence, the percent yield of water in the reaction is 46.85 %.

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Bond energy of one X-X bond + 2370 kJ mol⁻¹ = 2775 kJmol⁻¹

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Bond energy of one X-X bond = +405 kJmol⁻¹

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