If the enantiomeric excess of a mixture is 81 %, what are the percent compositions of the major and minor enantiomer?

major enantiomer = 90.5 %

minor enantiomer = 9.5 %

Explanation:

Assuming that the  major isomer is in excess;

From, ee/2 + 50

ee = enantiomeric excess = 81%

%  major =81/2 + 50 = 90.5 %

%  minor = 100% - 90.5 %

%  minor = 9.5 %

the answer is 1260

you spend 7 hours in school for 1 school day with 5 school days in 1 school week with 36 school weeks in a school year.

7*5*36= 1260 it's all about multiplying