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algol13
3 years ago
12

Nicotine is a toxic substance present in tobacco leaves. There are two lone pairs in the structure of nicotine. In general, loca

lized lone pairs are much more reactive than delocalized lone pairs. With this information in mind, do you expect both lone pairs in nicotine to be reactive?
A. Both lone pairs are delocalized and, therefore, both are expected to have the same reactivity.
B.Lone pair in pyrrolidine ring is localized and, therefore, is expected to be more reactive.
C. Both lone pairs are localized and, therefore, both are expected to be reactive.
D. Lone pair in pyridine ring is localized and, therefore, is expected to be more reactive.
Chemistry
1 answer:
Harlamova29_29 [7]3 years ago
8 0

Answer:

B.Lone pair in pyrrolidine ring is localized and, therefore, is expected to be more reactive.

Explanation:

There are two nitrogen atoms bearing lone pairs of electrons in the structure of nicotine as shown in the image attached.

One nitrogen atom is found in the pyrrolidine ring. The lone pair on this nitrogen atom is localized hence it is more reactive than the lone pair of electrons found on the nitrogen atom in the pyridine ring which is delocalized a shown in the image attached to this answer.

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8 0
3 years ago
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The concentration of hydrogen peroxide (H2O2) can be determined by titrating it with an acidified MnO4− solution. The following
Vaselesa [24]

Answer:

A) {MnO_4}^- + 3 H^+ + \frac{5}{2} H_2O_2 \rightarrow Mn^{2+} + 4 H_2O + \frac{5}{2} O_2

B) 7.5 molar

Explanation:

A) <u>Reduction</u>

{MnO_4}^- + 8 H^+ + 5e^- \rightarrow Mn^{2+} + 4 H_2O

<u>Oxidation</u>

H_2O_2 \rightarrow O_2 + 2 H^+ + 2e^-

Multiplying the oxidation reaction by 5/2 and adding it to the reduction equation:

{MnO_4}^- + 8 H^+ + 5e^- \rightarrow Mn^{2+} + 4 H_2O

+

\frac{5}{2} H_2O_2 \rightarrow \frac{5}{2} O_2 + 5 H^+ + 5e^-

-----------------------------------------------------------------------------------------------------

{MnO_4}^- + 3 H^+ + \frac{5}{2} H_2O_2 \rightarrow Mn^{2+} + 4 H_2O + \frac{5}{2} O_2

B) 10 ml = 0.01 L

20 ml = 0.02 L

mol of MnO4− = molarity*volume = 1.5*0.02 = 0.03

1 mol of MnO4− reacts with 5/2 mol of H2O2, then:

mol of H2O2 = 0.03*5/2 = 0.075

molarity = mol/volume = 0.075/0.01 = 7.5 molar

5 0
3 years ago
In a reaction, a + b + c→ d, it is found that the reaction is first order in terms of a and second order in terms of b and half
Vikentia [17]
Rate=[a]*([b]^2)*([c]^(1/2)]
rate=[2a]*([b]^2)*([2c]^(1/2)]= 2*(2^(1/2)[a]*([b]^2)*([c]
it increases  times 2*(2^(1/2)=2√2
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3 years ago
How many grams of perchloric acid, HClO4, are contained in 39.1 g of 74.9 wt% aqueous perchloric acid
Leno4ka [110]

Answer:

29.3 g HClO₄

Explanation:

We have 39.1 grams of 74.9 wt% aqueous perchloric acid solution, that is, there are 74.9 grams of perchloric acid in 100 grams of perchloric acid solution. The mass, in grams, of perchloric acid contained in 39.1 grams of perchloric acid solution is:

39.1 g Solution × (74.9 g HClO₄/100 g Solution) = 29.3 g HClO₄

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PLZ HELP
riadik2000 [5.3K]
Too many to know in the world.
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