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DanielleElmas [232]
3 years ago
12

You are given a solution containing a pair of enantiomers (A and B). Careful measurements show that the solution contains 98% A

and 2% B. What is the ee of this solution
Chemistry
1 answer:
Andreyy893 years ago
8 0

Answer:

ee = 96%

Explanation:

Enantiomeric excess, ee, is a way to express a mixture that is not enantiomerically pure. It is defined as 100 times the ratio between the  differences of amounts of enantiomers and the total amunt. that is:

ee = |A-B|/ A+B * 100

ee = |98%-2%| / 98+2 * 100

<h3>ee = 96%</h3>
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Now we know that

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Let Q1 be the specific heat of the lighter ball.

c1 be the specific heat of the lighter ball.

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Let Q2 be the specific heat of the heavier ball.

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