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Feliz [49]
3 years ago
7

Why is it important to neutralize the reaction mixture during the workup? 1. It ensures that the ester is uncharged and therefor

e insoluble in water so that the product can be filtered. 2. If the reaction were not neutralized, the final product would not be in the desired oxidation state. 3. Neutralization prevents the ester from undergoing the reverse reaction to regenerate the carboxylic acid starting material. 4. Sulfuric acid is strong enough to dissolve glass and needs to be neutralized before the glassware loses its structural integrity.
Chemistry
1 answer:
DENIUS [597]3 years ago
3 0

Answer: the correct option is 1 (It ensures that the ester is uncharged and therefore insoluble in water so that the product can be filtered.)

Explanation:

Esters are formed through a process known as esterification. This involves the heating of a carboxylic acid with an alcohol while the water formed is continuously removed. In order for the reaction to occur at a maximum rate, an acidic catalyst is used. The most common acid catalysts includes:

- hydrochloric acid and

-sulfuric acid.

These catalyst needs to be neutralized in order to isolate the product. This is so because the solution begins to precipitate when the reaction mixture is neutralized.

Thus it is important neutralize the moisture in order to filter out the ester from the mixture, which otherwise will be in a dissolved state and will not be recoverable.

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As less rain fall,does more plants grow?
Vladimir79 [104]
More rain makes for more plant growth, so the answer is no because without water(rain) and nutrients plants can't really grow.

I hope that this helps you !
6 0
3 years ago
Definition: The phase of matter where particles at extremely high temperatures become ionized. Example: The state of matter of t
GalinKa [24]

Answer:

Plasma

Explanation:

Plasma is a state of matter that in some cases is a synonym of ionized gas. It consists of groups of ions (particles with a net charge) with some of their outer electrons removed.

How many electrons have been removed depends on the type of plasma, since the most energetic plasmas are completely free of orbiting electrons. This is the case of the interiors of the sun and other stars.

7 0
3 years ago
2 HI(g) ⇄ H2(g) + I2(g) Kc = 0.0156 at 400ºC 0.550 moles of HI are placed in a 2.00 L container and the system is allowed to rea
Alex_Xolod [135]

<u>Answer:</u> The concentration of hydrogen gas at equilibrium is 0.0275 M

<u>Explanation:</u>

Molarity is calculated by using the equation:

\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}

Moles of HI = 0.550 moles

Volume of container = 2.00 L

\text{Initial concentration of HI}=\frac{0.550}{2}=0.275M

For the given chemical equation:

                          2HI(g)\rightleftharpoons H_2(g)+I_2(g)

<u>Initial:</u>                  0.275

<u>At eqllm:</u>           0.275-2x      x         x

The expression of K_c for above equation follows:

K_c=\frac{[H_2][I_2]}{[HI]^2}

We are given:

K_c=0.0156

Putting values in above expression, we get:

0.0156=\frac{x\times x}{(0.275-2x)^2}\\\\x=-0.0458,0.0275

Neglecting the negative value of 'x' because concentration cannot be negative

So, equilibrium concentration of hydrogen gas = x = 0.0275 M

Hence, the concentration of hydrogen gas at equilibrium is 0.0275 M

4 0
3 years ago
To calculate the enthalpy change for the reaction, 2CO (g) + O2 (g) Imported Asset 2 CO2 (g), you can use ΔHf0 values for each r
svetoff [14.1K]

Answer:

ΔH0reaction = [ΔHf0 CO2(g)] - [ΔHf0 CO(g) + ΔHf0 O2(g)]

Explanation:

Chemical equation:

CO + O₂   →  CO₂

Balanced chemical equation:

2CO + O₂   →  2CO₂

The standard enthalpy for the formation of CO = -110.5 kj/mol

The standard enthalpy for the formation of O₂  = 0  kj/mol

The standard enthalpy for the formation of CO₂  = -393.5 kj/mol

Now we will put the values in equation:

ΔH0reaction = [ΔHf0 CO2(g)] - [ΔHf0 CO(g) + ΔHf0 O2(g)]

ΔH0reaction = [-393.5 kj/mol] - [-110.5 kj/mol + 0]

ΔH0reaction = [-393.5 kj/mol] - [-110.5 kj/mol]

ΔH0reaction = -283 kj/mol

7 0
3 years ago
Meta-bromoaniline was treated with nano2 and hcl to yield a diazonium salt. Draw the product obtained when that diazonium salt i
Y_Kistochka [10]

Answer:

I believe is the correct answer

Explanation:

100bromoaniline+nano2+hcl

4 0
2 years ago
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