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Daniel [21]
3 years ago
10

When ethyl-2-methylpropanoate is treated with sodium ethoxide in ethanol, there is no observable reaction. Explain this observat

ion.

Chemistry
1 answer:
den301095 [7]3 years ago
8 0

Answer:

No net change in reaction occurs in this nucleophilic acyl subtitution reaction

Explanation:

Sodium ethoxide in ethanol gives nucleophilic acyl substitution reaction with ethyl-2-methylpropanoate.

Here ethoxide group replaces an ethoxide group from ester through addition-ellimination pathway.

So, ultimately, the product of this reaction is identical with reactant i.e. ethyl-2-methylpropanoate is reproduced.

Hence one might observe no change during reaction as product and reactant of this reaction are same.

Mechanistic pathway has been shown below.

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What is the mass of 975 mL of mercury? Its density is 13.5 g/mL
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Answer:

72.22 g

Explanation:

975 mL Mercury× 13.5 g/mL = 72.22 g

7 0
3 years ago
What process is being shown by water being given off from each bond site?
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The process that is being shown by water being given off from a bond site is DEHYDRATION SYNTHESIS.
Dehydration synthesis is the process of joining two molecules or compounds together as a result of removal of water.
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3 years ago
Which solute would be more effective at lowering the freezing point of water: MgCl2 and KNO3? Explain.
Phantasy [73]

Answer:

AlCl₃.

Explanation:

Adding solute to water causes depression of the boiling point.

The depression in freezing point (ΔTf) can be calculated using the relation:

ΔTf = i.Kf.m,

where, ΔTf is the depression in freezing point.

i is the van 't Hoff factor.

van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

Kf is the molal depression constant of water.

m is the molality of the solution (m = 1.0 m, for all solutions).

(1) NaCl:

i for NaCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

∴ ΔTb for (NaCl) = i.Kb.m = (2)(Kf)(1.0 m) = 2(Kf).

(2) MgCl₂:

i for MgCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

∴ ΔTb for (MgCl₂) = i.Kb.m = (3)(Kf)(1.0 m) = 3(Kf).

(3) NaCl:

i for KBr = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

∴ ΔTb for (KBr) = i.Kb.m = (2)(Kf)(1.0 m) = 2(Kf).

(4) AlCl₃:

i for AlCl₃ = no. of particles produced when the substance is dissolved/no. of original particle = 4/1 = 4.

∴ ΔTb for (CoCl₃) = i.Kb.m = (4)(Kf)(1.0 m) = 4(Kf).

So, the ionic compound will lower the freezing point the most is: AlCl₃

4 0
3 years ago
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Most lakes have rivers flowing out of them, carrying water to the ocean. However, some lakes, including Great Salt Lake and the
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Since the only way of water flow to these lakes or bodies of water is through evaporation, I would expect an increase in unknown substances in the composition of the lakes due to the amount of contamination that globalization produces and affects terribly the surroundings when these unknown substances travel through evaporation as the outlet of these bodies of water. Therefore I think continuous contamination is what to expect after many more years of inflow and evaporation.

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