Answer:
A. 3-chloro-1-methylcyclobutane.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to infer that the name of this compound is A. 3-chloro-1-methylcyclobutane because of the fact that the parent chain is a cyclobutane which starts by the methyl radical as it has the priority over the chlorine radical which is actually named first at the third carbon (clockwise).
Therefore the name is given in A, accordingly to the IUPAC rules of nomenclature.
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Answer:
I think this answer should be 2.33 g H2O
The phosphorylation of fructose 6-phosphate to fructose-1,6-bisphosphate is the committed step in glycolysis because. it is the rate-limiting step
<h3>What is
phosphorylation?</h3>
The first step in the metabolism of carbohydrates is frequently their phosphorylation. Because the phosphate group stops the molecules from migrating back across the transporter, phosphorylation enables cells to store carbohydrates. Glucose phosphorylation is a crucial step in the metabolism of sugar. In the first phase of glycolysis, D-glucose is converted to D-glucose-6-phosphate using the chemical equation D-glucose + ATP D-glucose-6-phosphate + ADP G° = 16.7 kJ/mol (° signifies measurement under standard conditions).
The rate-limiting stage in the liver's metabolism of glucose is the initial rate of phosphorylation of glucose (ATP-D-glucose 6-phosphotransferase) and non-specific hexokinase. Hepatic cells are freely permeable to glucose (ATP-D-hexose 6-phosphotransferase).
encouraging certain glucose transporters to translocate to the cell membrane.
To learn more about phosphorylation from the given link:
brainly.com/question/2138188
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