NaOH + CH3COOH -> CH3COONa + H20
Answer:
100 j
Explanation:
bascially u j subtract 50 from 150 because the system lost 50 j but theres 150 in its surroundings
Let us say that R is the major enantiomer, while
S is the minor enantiomer, therefore the formula for enantiomeric excess (ee)
is:
ee = (R – S) * 100%
Let us further say that the fraction of R is x (R
= x), and therefore fraction of S is 1 – x (S = 1 – x), therefore:
75 = (x – (1 – x)) * 100
75 = 100 x – 100 + 100 x
200 x = 175
x = 0.875
Summary of answers:
R = major enantiomer = 0.875 or 87.5%
<span>S = minor enantiomer = (1 – 0.875) = 0.125 or
12.5%</span>
