C) can give up electrons more readily than others.
Answer:
See figure 1
Explanation:
We have to remember that in the isomer structures we have to <u>change the structure</u> but we have to maintain the <u>same formula</u>, in this case
.
In the formula, we have 1 nitrogen atom. Therefore we will have as a main functional group the <u>amine group</u>.
In the amines, we have different types of amines. Depending on the number of carbons bonded to the "N" atom. In the <em>primary amines</em>, we have only 1 C-H. In the <em>secondary amines</em>, we have two C-N bonds and in the <em>tertiary amines</em>, we have three C-N bonds.
With this in mind, we can have:
-) <u>Primary amines:</u>
1) n-butyl amine
2) sec-butyl amine including 2 optical isomers
3) isobutyl amine
4) tert-butyl amine
-) <u>Secondary amines:</u>
5) N-methyl n-propyl amine
6) N-methyl isopropyl amine
7) N, N-diethyl amine
-) <u>Tertiary amines:</u>
8) N-ethyl N, N-dimethyl amine
See figure 1
I hope it helps!
Answer:
66.4 mL
Explanation:
A 75.7% (v/v) value, means that f<u>or every 100 mL of rubbing alcohol, there are 75.7 mL of isopropanol.</u>
With the above information in mind, we can s<u>olve the problem by multiplying 87.7 mL by 75.7 %</u>:
87.7 mL * 75.7 / 100 = 66.4 mL
So there are 66.4 mL of isopropanol in 88.7 mL of rubbing alcohol.
HI and I⁻ is the acid-conjugate base pair.
<h3>What is acid and conjugate base pair?</h3>
According to Bronsted-Lowry definition of acid and base, those pair which are differ in the single no. of proton or H⁺ ion is known as acid and conjugate base pair or base and conjugate acid pair.
Given chemical reaction is:
HI + H₂O → H₃O⁺ + I⁻
Here, HI = acid,
H₂O = base,
H₃O⁺ = conjugate acid and
I⁻ = conjugate base
Hence, option (d) is correct i.e. HI & I⁻ is the acid-conjugate base pair.
To know more about the acid-conjugate base pair, visit the below link:
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