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Sveta_85 [38]
2 years ago
7

what mechanistic intermediate is used to explain the preference for addition of the br atom , of hbr, to the internal carbon of

a terminal alkyne
Chemistry
1 answer:
ra1l [238]2 years ago
7 0

The mechanistic intermediate that is used to explain the preference for addition of the Br atom , of Hbr, to the internal carbon of a terminal alkyne is free-radical conditions.

The addition of HBr to an alkyne under free-radical conditions is governed by the addition of a Br atom to the alkyne which helps in giving the most stable alkyl radical possible. The addition of HBr in the absence of radicals is governed by the addition of a proton to the double bond which helps in giving the most stable carbocation possible.

A molecule wherein a carbon atom has a fine fee and three bonds is known as a carbocation. We can basically say that they are carbon cations. Later, it forms into carbonium ion. Any even-electron cation that possesses a big effective charge at the carbon atom is known as an carbocation.

Learn more about mechanistic intermediate from the link given below.

brainly.com/question/7158628

#SPJ4

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Which pack has the greatest change in enthalpy, warm packs (such as hand warmers) or cold packs (such as an ice pack)?
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4 years ago
Which variables does equilibrium depend on?
Eddi Din [679]

Answer:

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7 0
3 years ago
Calculate the mass of each product formed when 84.3 g of silver sulfide reacts with excess hydrochloric acid: ag2s(s) + hcl(aq)
oee [108]

Answer:

              48.75 g of AgCl

               11.60 g of H₂S

Solution:

The Balance Chemical Equation is as follow,

                                     Ag₂S  +  HCl     →     AgCl  +  H₂S

<u>Calculate amount of AgCl produced</u><u>,</u>

According to equation,

                247.8 g (1 mol) of Ag₂S produces  =  143.32 g (1 mol) of AgCl

So,

                         84.3 g of Ag₂S will produce  =  X g of AgCl

Solving for X,

                      X  =  (84.3 g × 143.32 g) ÷ 247.8 g

                      X  =  48.75 g of AgCl

<u>Calculate amount of H</u><u>₂</u><u>S produced</u><u>,</u>

According to equation,

                247.8 g (1 mol) of Ag₂S produces  =  34.1 g (1 mol) of H₂S

So,

                         84.3 g of Ag₂S will produce  =  X g of H₂S

Solving for X,

                      X  =  (84.3 g × 34.1 g) ÷ 247.8 g

                      X  =  11.60 g of H₂S

6 0
3 years ago
An original sample of K-40 has a mass of 25.00 grams. After 3.9 m 109 years, 3.125 grams of the original sample remains unchange
Svetllana [295]

An original sample of K-40 has a mass of 25.00 grams. After 3.9 m 109 years, 3.125 grams of the original sample remains unchanged. What is the half-life of K-40?

First step is to determine the remaining decimal amount.

3.125 grams /25.00 grams = 0.125

Second step is to determine the number of half lives.

(1/2)^n = 0.125

N log (1/2) = log 0.125

N = 3 years

7 0
4 years ago
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