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:
M₂ = 1.9 M
Explanation:
Given data;
Volume of sodium chloride = 200 mL
Molarity of sodium chloride = 4.98 M
Volume of water = 532 mL
Final Molarity = ?
Solution:
M₁V₁ = M₂V₂
M₂ = M₁V₁ /V₂
M₂ = 4.98 M × 200 mL / 532 mL
M₂ = 996 mL. M /532 mL
M₂ = 1.9 M
