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3 years ago

If a mixture contains 75% of one compound and 25% of its enantiomer, what is the e.e. of the mixture

Chemistry

1 answer:

VMariaS [17]3 years ago

7 0

Answer: The enantiomeric excess ee is 50%

Explanation:

Given that;

compound A = 75%, B = 25%

ee = ?

ee = [( A - B) / ( A + B)] × 100

= [( 75 - 25) / ( 75 + 25)] × 100

= [50 / 100] × 100

= 0.5 × 100

= 50%

Therefore enantiomeric excess ee is 50%

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Explanation:

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$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

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Read 2 more answers

What is the volume of 40.0 grams of argon gas at STP ?

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Answer:

24.9 L Ar

General Formulas and Concepts:

Atomic Structure

Reading a Periodic Table
Moles
STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

Aqueous Solutions

States of Matter

Stoichiometry

Using Dimensional Analysis

Explanation:

Step 1: Define

[Given] 40.0 g Ar

[Solve] L Ar

Step 2: Identify Conversions

[PT] Molar Mass of Ar - 39.95 g/mol

[STP] 22.4 L = 1 mol

Step 3: Convert

[DA] Set up: $\displaystyle 40.0 \ g \ Ar(\frac{1 \ mol \ Ar}{39.95 \ g \ Ar})(\frac{22.4 \ L \ Ar}{1 \ mol \ Ar})$
[DA] Divide/Multiply [Cancel out units]: $\displaystyle 24.9235 \ L \ Ar$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

24.9235 L Ar ≈ 24.9 L Ar

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