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mylen [45]
3 years ago
7

g, Assuming the precipitate is totally insoluble in water, which aqueous ions will be present in the solution (collected in the

beaker) after the equivalence point (after the limiting reagent is used up)? Question 4 options: 1) Cl1- (aq) 2) Na1+ (aq) 3) OH1- (aq) 4) Cu2+ (aq)
Chemistry
1 answer:
Allushta [10]3 years ago
3 0

Answer:

Cl⁻, Na⁺, OH⁻

Explanation:

The titration is:

CuCl₂(aq) + 2 NaOH(aq) → Cu(OH)₂(s) + 2 NaCl(aq)

In solution, before the reaction, the ions are Cu²⁺ and Cl⁻. The addition of NaOH (Na⁺ + OH⁻) produce the precipitation of Cu²⁺ forming Cu(OH)₂(s). When you reach the equivalence point, there is no Cu²⁺ because precipitates completely. All OH⁻ ions reacts when are added but when Cu²⁺ is finished, excess OH⁻ ions still in solution helping to detect the equivalence point.

Thus, ions present after the equivalence point are:<em> Cl⁻, Na⁺</em> (Don't react, spectator ions), and <em>OH⁻</em>.

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Read 2 more answers
How many particles are present in0.24moles of carbon?
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1.45 x 10²³ particles

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Given parameters:

Number of moles of carbon  = 0.24moles

Unknown:

Number of particles = ?

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7 0
2 years ago
Write the structure of the product that would be formed from the S(
prisoha [69]

Answer:The product formed on reaction with hydroxide ion as nucleophile is 2R-hexane-2-ol.

The product formed on reaction with water would be a 50:50 mixture of

2S-hexane-2-ol. and 2R-hexane-2-ol.

Explanation:

2S-iodohexane on reactiong with hydroxide ion would undergo SN² substitution reaction that is substitution bimolecular. Hydroxide ion has a negative charge and hence it is a quite good  nucleophile .

The rate of a SN² reaction depends on both the substrate and nucleophile . Here the substrate is a secondary carbon center having Iodine as a leaving group.SN² reaction  takes place here as hydroxide ion is a good nucleophile and it can attack the secondary carbon center from the back side leading to the formation of 2R-hexane-2-ol.

In a SN² reaction since the the nucleophile attacks from the back-side so the product formation takes place with the inversion of configuration.

When the same substrate S-2-iodohexane undergoes a substitution reaction with water as a nucleophile then the reaction occurs through (SN¹) substitution nucleophilic unimolecular mechanism .

The rate of a SN¹ reaction depends only on the nature of substrate and is independent of the nature of nucleophile.

The SN¹ reaction is a 2 step reaction , in  the first step leaving group leaves leading to the formation of a carbocation and once the carbocation is formed then any weaker nucleophile or even solvent molecules can attack leading the formation of products.

In this case a secondary carbocation would be generated in the first step and then water will attack this carbocation to form the product in the second step.

The product formed on using water as a nucleophile would be a racemic mixture of R and S isomers of hexane -2-ol in 50:50 ratio. The two products formed would be 2R-hexane-2-ol and 2S-hexane-2-ol.

Kindly refer the attachment for reaction mechanism and structure of products.

8 0
3 years ago
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