Answer:
In the third step of the citric acid cycle, the oxidation of isocitrate takes place and one molecule of carbon dioxide is released.
Explanation:
In the first step of citric acid cycle, acetylCoA combines with a four-carbon molecule, oxaloacetate, forming a six-carbon molecule, citrate.
In the second step, the citrate in the presence of enzyme anicotase is converted into isocitrate.
<u>In the third step, the oxidation of isocitrate takes place and one molecule of carbon dioxide is released leaving behind one five-carbon molecule called as α-ketoglutarate. During this step, NAD⁺ is reduced to form NADH. </u>
<u>This is first round of the citric acid cycle that could possibly release a carbon atom originating from this acetyl CoA.</u>
On series of reaction, another carbon dioxide molecule also being relased and oxaloacetate is regenerated again.
The molarity of the solution will be 0.72 m.
The majority of reactions take place in solutions, making it crucial to comprehend how the substance's concentration is expressed in a solution when it is present. The number of chemicals in a solution can be stated in a variety of ways, including.
The symbol for it is M, and it serves as one of the most often used concentration units. Its definition states how many moles of solute there are in a liter of solution.
Given data:

Molarity can be determined by the formula:

where, M is molarity and V is volume.
Put the value of given data in above equation.
57.3 × 0.497 m = M × 39.5 L
M = 0.72 m
Therefore, the molarity of the solution will be 0.72 m
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Answer:
3–ethyl–4–methylhexane.
Explanation:
To name the above compound, do the following:
1. Determine the functional group of the compound.
2. Locate the longest continuous carbon chain. This gives the parent name of the compound.
3. Identify the substituent group attached to the compound.
4. Give the substituent the lowest possible count.
5. Combine the above to name the compound.
Now, we shall name the compound given in the question above as follow:
1. The compound contains only single bond. Therefore, the compound belong to the alkane family.
2. The longest continuous carbon chain is 6 i.e hexane.
3. The substituent group attached are:
i. Methyl, CH3.
ii. Ethyl, CH2CH3.
4. we shall name the substituents alphabetically i.e ethly will come before methyl. Therefore,
Ethyl is located at carbon 3.
Methy is located at carbon 4.
5. Therefore, the name of the compound is:
3–ethyl–4–methylhexane.