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Naily [24]
3 years ago
5

Nuehhdhdhdhdje hwhwje

Chemistry
2 answers:
Komok [63]3 years ago
4 0

Answer:

What is the question?

Explanation:

Natasha2012 [34]3 years ago
3 0

Answer:

:p

Explanation:

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vovikov84 [41]
The answer is silicon.

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4 years ago
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Write a mechanism for the esterification of propanoic acid with 18O-labeled ethanol. Show clearly the fate of the 18O label. (b)
tatuchka [14]

Answer:

See explanation and images attached

Explanation:

a) In the mechanism for the acid catalysed esterification of propanoic acid using ethanol, we can see that the first step is the protonation of the acid followed by nucleophillic attack of the alcohol. Loss of water and consequent deprotonation regenerates the acid catalyst. We can see the fate of the 18O labelled ethanol in the mechanism shown.

b)  In the second mechanism, an unnamed ester is hydrolysed using an acid catalyst. The attack of the acid and subsequent nucleophillic attack of water labelled with 18O leads to the incorporation of this 18O into the product acid as shown in the mechanism attached to this answer.

5 0
3 years ago
In each reaction box, place the best reagent from the list below. draw the intermediate compound.
Fynjy0 [20]

Answer:

Reagent A: PBr₃

Reagent B: Mg in Et₂O.

Explanation:

Hello,

In this case, your facing a problem in which a carboxylic acid is produced starting by an alcohol. More specifically, cyclopentanol must react with phosphorous tribromide in order to yield bromocyclopentane which is more likely to produce a carboxylic acid, therefore, reagent A is PBr₃.

On the other hand, by means of the production of the specified product, bromocyclopentane must react with carbon dioxide and magnesium in diethyl ether in acidic media to promote the production of the cyclopentanoic acid via the grignard reaction (substitution of the bromine by the carboxyle group), therefore, reagent B is Mg in Et₂O.

Best regards.

8 0
3 years ago
Where would you find almost no wind at all?
suter [353]

Answer:

A. the prevailing westerlies

7 0
3 years ago
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If the caffeine concentration in a particular brand of soda is 1.87 mg/oz, drinking how many cans of soda would be lethal? Assum
Drupady [299]
2.77mg caffeine / 1oz12oz / 1canLethal dose: 10.0g caffeine = 10,000mg caffeine First, find how much caffeine is in one can of soda, then divide that amount by the lethal dose to find the number of cans. (2.77mg caffeine / 1oz) * (12oz / 1can) = 33.24mg caffeine / 1can. (10,000mg caffeine) * (1can / 33.24mg caffeine) = 300.84 cans. Since we can't buy parts of a can of soda, then we have to round up to 301 cans. Notice how all the values were set up as ratios and how the units cancelled.
4 0
4 years ago
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