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Yuki888 [10]
4 years ago
7

Enzyme A and Enzyme B both catalyze the same reaction. If Enzyme A has a larger turnover number than Enzyme B, then which enzyme

, if either, has a higher catalytic efficiency?
Enzyme B is more efficient because it's turnover number is smaller.
Enzyme A is more efficient because it's turnover number is larger.
Not enough information is provided to know for certain.
Chemistry
1 answer:
lina2011 [118]4 years ago
3 0

Answer:

Enzyme A, if we assume the concentration of the enzymes (A and B) are the same

Explanation:

As Enzyme turnover number is defined as the number of chemical conversions of substrate per second that a single enzyme can perform in its catalytic site, for a given enzyme concentration, we can say that the Enzyme A has a higher catalytic efficiency. This is true only if [Enzyme A] = [Enzyme B]

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The flask contains 10.0mL of HCl and a few drops of phenolphthalein indicator. The buret contains 0.320 M NaOH. What is the init
yan [13]

This question is incomplete, the complete question is;

The flask contains 10.0mL of HCl and a few drops of phenolphthalein indicator.

The buret contains 0.320 M NaOH,

It requires 11.2 mL of the NaOH solution to reach the end point of the titration

What is the initial concentration of HCI?

Answer:  

initial concentration of HCl is 0.3584 M

Explanation:

Given that;

the reaction between HCl and NaOH is 1:1 molar

HCl(aq) + NaOH(aq) -------> NaCl(aq) + H2O(l)

we know that molarity is the number of moles of solute per liter of solution

so moles of NaOH consumed

= ( 0.320 mol / 1000ml) × 11.2ml

= 0.003584 mol

now

0.003584 moles of NaOH reacts with 0.003584 moles of HCl

moles of HCl present in the 10.0ml solution = 0.003584 mol

therefore the initial concentration of HCl

= ( 0.003584 mol / 10.0ml )× 1000ml

= 0.3584 M

the initial concentration of HCl is 0.3584 M

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