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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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Answer: 17.78g

Explanation:

Assume there is no heat exchange with the environment, then the amount of heat taken by the steel rod, Q(s), is equal to the amount of heat lost by the water, Q(w), but with opposite sign.

Q(s) = -Q(w)

Remember, Q = mc(ΔΦ)

Where Q = amount of heat

m = mass of steel

c = specific heat capacity of steel

ΔΦ = Initial temperature T1 - Final temperature T2

Q = mc(T1-T2)

Recall, Q(s) = -Q(w). Then,

m(s)*c(s)*(T1s - T2s) = - m(w)*c(w)*(T1w - T2w)

Substituting each values

Note: m(w) = volume of water*density = 75mL*1g/mL = 75g

m(s)*0.452*(21.5-2) = -75*4.18*(21.5-22)

m(s)*8.814 = 156.75

m(s) = 156.75/8.814

m(s) = 17.78g

Therefore, the mass of steel is 17.78g

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