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Savatey [412]
3 years ago
7

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 = ?

so

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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Answer:

Nucleus

Explanation:

Nucleus is your blueprint for the cell. It has all the directions packaged in tiny DNA molecules. The nucleus of a cell is an organelle that stores the cell's hereditary material, or DNA, and it coordinates the cell's activities, which include growth, intermediary metabolism, protein synthesis, and reproduction.

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Question 11 (10 points)
andrey2020 [161]

Answer:

The correct answer is option a.

Explanation:

When aluminum hydroxide reacts with of nitrous acid it gives of aluminum nitrite and of water.

Al(OH)_3+3HNO_2\rightarrow Al(NO_2)_3+3H_2O

According to above reaction ,when 1 mole of aluminum hydroxide reacts with 3 moles of nitrous acid it gives 1 mole of aluminum nitrite and 3 moles of water.

Hence, the correct answer is option a.

4 0
3 years ago
Predict the bond polarity of the following bonds:
Nataliya [291]
CH is nonpolar
NH is polar
CCl is polar
SiO is polar
SCl is polar
CO is polar
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Remember that polarity results from unequal electron sharing
8 0
3 years ago
How is the way a mixture is combined different from how a compound is combined?
artcher [175]

Mixture is composed of molecules of different types. A compound can only be separated through chemical means. While a compound is a pure substance that contains 2 or more elements chemically combined together while a mixture are formed when two substances are added together without chemical bonds being formed.

7 0
3 years ago
Balance the following redox reaction if it occurs in acidic solution. What are the coefficients in front of h2c2o4 and h2o in th
shusha [124]

Answer:

See explanation below

Explanation:

The overall reaction, is:

MnO4⁻(aq) + H₂C₂O₄(aq) ---------> Mn²⁺(aq) + CO₂(g)

Balancing this redox reaction means that one compound is reducting while the other is oxidizing. So, we need to separate both compounds into 2 semi equations and balance both of them, per separate and then, we can join them.

As we want to balance in acid medium, means that we need to add water and H⁺ in both reactions. Doing that we have the following:

MnO₄⁻   ---------------> Mn²⁺

In this reaction, we can clearly see that it's not balanced. To balance this semi equation, let's see the elements. In the reactans we have Mn and O, but in the products we only have Mn, the atom of oxygen where could it be? As we are doing acid medium, if in the reactants we have oxygen, this oxygen can be as products in the form of water, so we add water there.

MnO₄⁻   ---------------> Mn²⁺ + H₂O

Now, the water has hydrogen atoms, and if we are in acid medium, the hydrogen can only come from the acid medium, and in this case H⁺ so:

H⁺ + MnO₄⁻ ----------> Mn²⁺ + H₂O

Now, it's time to balance the charges. First Mn²⁺ is the lowest oxidation state of the manganese, this means that in the reactants Mn is passing from a higher state to a lower state, therefore, this compouns is reducting. How many electrons? well, in this case, we know that oxygen usually have the oxidation state -2, so the manganese would be:

(-2 * 4) + x = -1

-8 + x = -1 -------> x = +7

Therefore, manganese passes from 7+ to 2+, it's gaining 5 electrons so:

H⁺ + MnO₄⁻ + 5e⁻ ----------> Mn²⁺ + H₂O

Finally, we just balance the masses and charges:

8H⁺ + MnO₄⁻ + 5e⁻----------> Mn²⁺ + 4H₂O

Now, we just do the same thing with the other semi equation which is oxydizing. The explanation of that, is similar to this, so I'm gonna do it directly:

C₂O₄²⁻ -----------> CO₂

In this case, we can easily see that carbon is losing 2 electrons, so, let's put the 2 electrons on the product to balance the charges, and then, the masses:

C₂O₄²⁻ -----------> CO₂ + 2e⁻

C₂O₄²⁻ -----------> 2CO₂ + 2e⁻

Let's join both equations and do the sum of them:

8H⁺ + MnO₄⁻ + 5e⁻----------> Mn²⁺ + 4H₂O

C₂O₄²⁻ -----------> 2CO₂ + 2e⁻

As we can see, we do not have the same electrons on both equations, we need to equal those values so:

2 * (8H⁺ + MnO₄⁻ + 5e⁻----------> Mn²⁺ + 4H₂O)

5 * (C₂O₄²⁻ -----------> 2CO₂ + 2e⁻)

16H⁺ + 2MnO₄⁻ + 10e⁻----------> 2Mn²⁺ + 8H₂O

5C₂O₄²⁻ -----------> 10CO₂ + 10e⁻

Now, let's sum both equations:

16H⁺ + 2MnO₄⁻ + 10e⁻----------> 2Mn²⁺ + 8H₂O

5C₂O₄²⁻ -----------> 10CO₂ + 10e⁻

___________________________________

16H⁺ + 2MnO₄⁻ + 5C₂O₄²⁻ -----------> 2Mn²⁺ + 10CO₂ + 8H₂O

This would be the balanced reaction, however, let's put it as it was originally with the H2 in the C2O4 and balance it:

<h2>2MnO₄⁻ + 5H₂C₂O₄ + 6H⁺ -----------> 2Mn²⁺ + 10CO₂ + 8H₂O</h2>
7 0
3 years ago
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