Answer:
GOOD MORNING DIDI be happy
The second thesis statement is perfect. It supports the claim and presents main idea.
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Answer: option C) II < III < I
i.e [OH−] < [H3O+] < I
Explanation:
First, obtain the pH value of I and II, then compare both with III.
For I
Recall that pH = -log (H+)
So pH3O = -log (H3O+)
= - log (1x10−5)
= 4
For II
pOH = - log(OH-)
= - log(1x10−10)
= 9
For III
pH = 6
Since, pH range from 1 to 14, with values below 7 to be acidic, 7 to be neutral, above 7 to be alkaline: then, 9 < 6 < 4
Thus, the following solutions from least acidic to most acidic is II < III < I
The protons of methylene group between the two carbonyl groups in ethylacetoacetate are acidic in nature. When compounds containing such acidic protons are treated with bases the loose proton and form enolates.
In this particular example when ethylacetoacetate is reacted with methyl magnesium bromide, the methyl group abstracts the acidic proton and converts into
methane gas. The enolate when hydrolyzed is again converted into ethylacetoacetate as shown below,