Answer:
A variable shape that adapts to fit its container.
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Answer:
The ratio of HC2H3O2(aq) in the flask after the addition of 5.0mL of NaOH(aq) to HC2H3O2(aq) in the flask after the addition of 1.0mL of NaOH(aq) is 15 : 19 .
Explanation:
HC2H3O2 is CH₃⁻ COOH, which is also known as Acetic acid.
IUPAC name of this compound is Ethanoic acid.
Acetic acid has a basicity of 1. so there is one acidic hydrogen is acetic acid.
Given that, equivalence point was reached when 20.0mL of NaOH is added.
let the normality of acetic acid is N₁ and that of NaOH is N₂.
volume of acetic acid is V₁ and that of NaOH is V₂.
Equivalence point occurs when, N₁ × V₁ = N₂ × V₂.
⇒ N₁ × V₁ = N₂ × 20.
after the addition of 5.0mL of NaOH(aq), remaining N₁ × V° = N₂ × (20 - 5).
= N₂ × 15.
after the addition of 1.0mL of NaOH(aq), remaining N₁ × Vˣ = N₂ × (20 - 1).
= N₂ × 19.
⇒ V° : Vˣ = 15 : 19 .
⇒
The product of this reaction is a halohydrin as shown here.
<h3>What are the products?</h3>
We have a reaction that first involves the formation an alkene as the bases are used on the first substrate. The alkene that is thus created is now able to react with the bromine in water.
The first step of the reaction is where the multiple bond is created and this multiple bond is what can now go on to participate in a chemical reaction in the next step of the process of reaction as shown in the image.
Bromine in water is also hat we call bromine water. This bromine water is able to add across a double bond and when that happen we will have a saturated compound. This could also be regarded as halohydrin reaction.
Recall that the first step of the reaction of the alkene with bromine is the formation of a dibromide via the brominium intermediate. This is now followed by reaction with water to form the halohydrin product.
Learn more about organic reaction:brainly.com/question/9585105
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