Yes it is correct for that answer
Answer:
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Answer:
1) Conversion of glucose to glucose 6-phosphate by hexokinase
2) Conversion of fructose 6-phosphate to fructose 1,6-biphosphate by phosphofructokinase
3) Conversion of phosphoenolpyruvate to pyruvate by pyruvate kinase
Explanation:
There are 10 steps in the glycolysis pathway, three of which are irreversible. The enzymes controlling these reactions have not only catalytic properties but the irreversibility of the reaction gives them regulatory properties as well. These reactions serve as control points in the pathway.
Answer : The process is not spontaneous.
Explanation :
As, we know that:
Change in entropy = Change in entropy of system + Change in entropy of surrounding
As we are given in question, the entropy of surroundings decrease by the same amount as the entropy of the system increases.
For the given reaction to be spontaneous, the total change in entropy should be positive.
Given :
Entropy change of system = +125J/K
Entropy change of surroundings = -125J/K
Total change in entropy = Entropy change of system + Entropy change of surroundings
Total change in entropy = 125 J/K + (-125 J/K)
Total change in entropy = 0
The process is at equilibrium because the entropy change is equal to zero. So, the process is not spontaneous.
Answer:
Yes, the chemist can determine which compound is in the sample.
Explanation:
In 1 mole of K₂O, the mass of K is 2 × 39.1 g = 78.2 g and the mass of K₂O is 94.2 g. The mass ratio of K to K₂O is 78.2 g / 94.2 g = 0.830.
In 1 mole of K₂O₂, the mass of K is 2 × 39.1 g = 78.2 g and the mass of K₂O₂ is 110.2 g. The mass ratio of K to K₂O₂ is 78.2 g / 110.2 g = 0.710.
If the chemist knows the mass of K and the mass of the sample, he or she must calculate the mass ratio of K to the sample.
- If the ratio is 0.830, the compound is pure K₂O.
- If the ratio is 0.710, the compound is pure K₂O₂.
- If the ratio is not 0.830 or 0.710, the sample is a mixture.