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What is the percent enantiomeric excess (ee) of a mixture that has 86% of one enantiomer and 14% of the other
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What is the percent enantiomeric excess (ee) of a mixture that has 86% of one enantiomer and 14% of the other

User Dangling Cruze
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1 Answer

12 votes

Answer:

86% - 14% = 72%

Step-by-step explanation:

User Nigel Benns
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