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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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Cloud [144]

2NaNO3(aq) + BaSO4 = Ba(NO3)2(aq) + Na2SO4(aq) (s)

Procedures involved:

The cations or anions may transfer positions in this twofold replacement/displacement reaction, which results in AB + CD AD + CB. In such a reaction, water, an insoluble gas, or an insoluble solid must be one of the byproducts (precipitate). The reaction in question has the following molecular equation:

2NaNO3(aq) + BaSO4 = Ba(NO3)2(aq) + Na2SO4(aq) (s)

Double displacement:

When two atoms or groups of atoms swap positions, a double displacement reaction occurs, creating new compounds. Typically, aqueous solutions are where it happens.

Na2SO4 + BaCl2 BaSO4 + 2NaCl is an example of a double displacement reaction.

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2 years ago
2. HBIO3 name this acid.​
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Answer:

Hydroxy(oxo)bismuthine oxide

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3 years ago
What volume of 3.00 MM HClHCl in liters is needed to react completely (with nothing left over) with 0.750 LL of 0.500 MM Na2CO3N
AlladinOne [14]

<u>Answer:</u> The volume of HCl needed is 0.250 L

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

<u>For sodium carbonate:</u>

Molarity of sodium carbonate solution = 0.500 M

Volume of solution = 0.750 L

Putting values in above equation, we get:

0.500M=\frac{\text{Moles of sodium carbonate}}{0.750}\\\\\text{Moles of sodium carbonate}=(0.500mol/L\times 0.750L)=0.375mol

The chemical equation for the reaction of sodium carbonate and HCl follows:

Na_2CO_3+2HCl\rightarrow 2NaCl+H_2CO_3

By Stoichiometry of the reaction:

1 mole of sodium carbonate reacts with 2 moles of HCl

So, 0.375 moles of sodium carbonate will react with = \frac{2}{1}\times 0.375=0.750mol of HCl

Now, calculating the volume of HCl by using equation 1:

Moles of HCl = 0.750 moles

Molarity of HCl = 3.00 M

Putting values in equation 1, we get:

3.00M=\frac{0.750mol}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.750mol}{3.00mol/L}=0.250L

Hence, the volume of HCl needed is 0.250 L

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3 years ago
For the net reaction: 2AB + 2C → A2 + 2BC, the following slow first steps have been proposed. AB → A + B 2C + AB → AC + BC 2AB →
puteri [66]

This is an incomplete question, here is a complete question.

For the net reaction: 2AB + 2C → A2 + 2BC, the following slow first steps have been proposed.

(1) AB → A + B

(2) 2C + AB → AC + BC

(3) 2AB → A₂ + 2B

(4) C + AB → BC + A

What rate law is predicted by each of these steps?

Answer : The rate law expression for the following reactions are:

(1) \text{Rate}=k[AB]

(2) \text{Rate}=k[C]^2[AB]

(3) \text{Rate}=k[AB]^2

(4) \text{Rate}=k[C][AB]

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

The general reaction is:

A+B\rightarrow C+D

The general rate law expression for the reaction is:

\text{Rate}=k[A]^a[B]^b

where,

a = order with respect to A

b = order with respect to B

R = rate  law

k = rate constant

[A] and [B] = concentration of A and B reactant

Now we have to determine the rate law for the given reaction.

(1) The balanced equations will be:

AB\rightarrow A+B

In this reaction, AB is the reactant.

The rate law expression for the reaction is:

\text{Rate}=k[AB]

(2) The balanced equations will be:

2C+AB\rightarrow AC+BC

In this reaction, C and AB are the reactants.

The rate law expression for the reaction is:

\text{Rate}=k[C]^2[AB]

(3) The balanced equations will be:

2AB\rightarrow A_2+2B

In this reaction, AB is the reactant.

The rate law expression for the reaction is:

\text{Rate}=k[AB]^2

(4) The balanced equations will be:

C+AB\rightarrow BC+A

In this reaction, C and AB are the reactants.

The rate law expression for the reaction is:

\text{Rate}=k[C][AB]

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