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Andreyy89
3 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]3 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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A sample of water vapor has a volume of 3.15 L, a pressure of 2.40 atm, and a temperature of 325 K. What is the new temperature,
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Answer:

The answer to your question is:   T2 = 235.44 °K

Explanation:

Data

V1 = 3.15 L                    V2 = 2.78 L

P1 = 2.40 atm               P2 = 1.97 atm

T1 = 325°K                    T2 = ?

Formula

\frac{P1V1}{T1} = \frac{P2V2}{T2}

Process

            T2 = (P2V2T1) / (P1V1)

            T2 = (1.97x 2.78x 325) / (2.40 x 3.15)

            T2 = 1779.895 / 7.56

            T2 = 235.44 °K

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3 years ago
24 POINTS!!!!!!!!!!!!!!!!!!!!!!!!!!!
erastovalidia [21]
I'd say b, but i'm not 100 percent sure.<span />
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