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Andreyy89
2 years ago
11

Describe how you would prepare your assigned ester from a carboxylic acid and an alcohol. You do not need to include a detailed

procedure, but you should include any necessary reagents or catalyst (solvents are not needed).
Chemistry
1 answer:
vesna_86 [32]2 years ago
8 0

Answer:

The general preparation of esters( for example ethyl ethanoate) is through a process known as ESTERIFICATION.

Explanation:

The formation of an ester by the reaction between an alkanol and an acid is known as esterification. This reaction is extremely slow and reversible at room temperature, and is catalyzed by a high concentration of hydrogen ions.

In the preparation of one of the simpler esters known as ETHYL ETHANOATE the reactants include ethanol(an alcohol) and glacial ethanoic acid(a carboxylic acid) in the presence of concentrated tetraoxosulphate VI acid as a CATALYST. Note that, a catalyst is any substance that is able to increase the rate of a chemical reaction.

The mixture is warmed in a water bath( hot but not boiling) for about 25 minutes. The mixture is poured into a beaker partially filled with a sodium or calcium chloride to remove interacted ethanol. The ethyl ETHANOATE floats on the mixture as oily globules.

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Vitreous humor is located wear?​
Vikentia [17]

Eye

Explanation:

Vitreous humor is found in the human eye and other animals.

It is a gel that fills the space between the lens and retina of the eye. This matter helps to keep the shape of the eye in place by maintaining a constant pressure in the eye.

It is typically made up of water, gelatinous and transparent.

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3 0
2 years ago
How does a system at equilibrium respond to the addition of more reactant
grin007 [14]

Answer:

It favors the forward reaction.

Explanation:

According to Le Chatelier's Principle, when a system at equilibrium suffers a perturbation, the system will react in order to counteract the effect of such perturbation.

If more reactant is added, the system will try to decrease its concentration. It will do so by favoring the forward reaction, decreasing the concentration of the reactant  and increasing the concentration of the products, in order to re-establish the equilibrium.

7 0
3 years ago
Iron and Oxygen form rust. Compare the mass of iron and oxygen before the reaction with the mass of rust resulting from the reac
LekaFEV [45]
The mass will stay the same because of the conservation of mass
3 0
3 years ago
Read 2 more answers
Suppose that 0.1000 mole each of H2and I2are placed in a 1.000-L flask, stoppered, and the mixture is heated to 425oC. At equili
Katen [24]

<u>Answer:</u> The value of equilibrium constant for the given reaction is 56.61

<u>Explanation:</u>

We are given:

Initial moles of iodine gas = 0.100 moles

Initial moles of hydrogen gas = 0.100 moles

Volume of container = 1.00 L

Molarity of the solution is calculated by the equation:

\text{Molarity of solution}=\frac{\text{Number of moles}}{\text{Volume}}

\text{Molarity of iodine gas}=\frac{0.1mol}{1L}=0.1M

\text{Molarity of hydrogen gas}=\frac{0.1mol}{1L}=0.1M

Equilibrium concentration of iodine gas = 0.0210 M

The chemical equation for the reaction of iodine gas and hydrogen gas follows:

                         H_2+I_2\rightleftharpoons 2HI

<u>Initial:</u>                0.1    0.1

<u>At eqllm:</u>          0.1-x   0.1-x   2x

Evaluating the value of 'x'

\Rightarrow (0.1-x)=0.0210\\\\\Rightarrow x=0.079M

The expression of K_c for above equation follows:

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

[HI]_{eq}=2x=(2\times 0.079)=0.158M

[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M

[I_2]_{eq}=0.0210M

Putting values in above expression, we get:

K_c=\frac{(0.158)^2}{0.0210\times 0.0210}\\\\K_c=56.61

Hence, the value of equilibrium constant for the given reaction is 56.61

6 0
3 years ago
What is the density of a liquid if it's volume is 125 mL and it's mass is 50g?
masha68 [24]

Answer:

0.4g/ml

Explanation:

density= mass/volume

density=50g/125ml

density=0.4g/ml

6 0
2 years ago
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