Answer:
Option B. A
Explanation:
From the question given above, the following data were obtained:
C(s) + 2H₂ (g) —> CH₄ (g). ΔH = –74.9 kJ
From the reaction above, we can see that the enthalpy change (ΔH) is negative (i.e –74.9 KJ) which implies that the heat content of the reactants is greater than the heat content of the products. Thus, the reaction is exothermic reaction.
For an exothermic reaction, the energy profile diagram is drawn in such a way that the heat content of reactants is higher than the heat content of products because the enthalpy change
(ΔH) is always negative.
Thus, diagram A (i.e option B) gives the correct answer to the question.
Bread is the answer you are looking for please mark as brainliest! :)
<h2>Answer:</h2>
An increase in the population of prey means more food for predators, which will<u> increase</u> the predators' population. As a result, many more prey will be hunted, causing the prey population to<u> decrease.</u> In response, the predator population will <u>decrease</u>, resulting in an increase in the prey population.
<h3>Explanation:</h3>
The feedback loop is control of the system containing the prey and predator.
In this loop, the population of predator is directly proportional to prey but the population of prey is inversely proportional to a predators population.
As the population of prey increases, the population size of predator becomes large due to the availability of food. This inversely decreases the population of prey, as the predator do more hunt.
So the decrease in prey population results in a decrease in the predator population.
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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Minerals form by erosion of rocks.The erosion process includes water and other rocks rubbing against other rocks.
