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kodGreya [7K]
3 years ago
6

Which 2 resonance forms destablize the carbocation intermediate if bezonitrile undergoes chlronation at the ortho or para positi

ons
Chemistry
1 answer:
Anna11 [10]3 years ago
8 0

The question is incomplete, the complete question is shown in the image attached

Answer:

A and B

Explanation:

The electrophilic substitution of arenes yields a cation intermediate. The positive charge of the cation is delocalized over the entire ring.

The -CN group directs incoming electrophiles to the ortho/para position. The resonance structures for the chlorination of benzonitrile are shown in the question.

Recall that -CN is an electron withdrawing group. The resonance forms that destablize the carbocation intermediate are those in which the -CN group is directly attached to the carbon atom bearing the positive charge as in structures A and B.

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Answer the questions about the characteristics of the elements in group 1 (the alkali metals). What happens when the elements in
irinina [24]

Answer:

See explanation

Explanation:

The elements in group form univalent positive ions and element in group 17 form univalent negative ions. Hence, when a group 1 element reacts with a group 17 element, a compound of the sort MX is formed. Hence, when a group 1 element reacts with bromine, a salt is formed with the general formula MBr.

Elements of group 1 are highly electro positive metals. They react with water to form the metal hydroxide and release hydrogen gas. Hence, when group 1 elements react with water, hydrogen gas is released.

A group 1 element forms a univalent positive ion while a group 16 element forms a divalent negative ion. Hence, when a groups 1 element reacts with oxygen, the compound formed must have the general formula M2O.

The reactivity of group 1 metal increases down the group hence Cs is the most reactive group 1 element.

Lithium displays a slightly different chemistry from other group 1 elements because of its small size.

6 0
3 years ago
Metal or Non-metal?
goldenfox [79]

Answer:

Fluorine:non-metal

Bromine:non-metal

Hydrogen:non-metal

Beryllium:metal

Nitrogen:non-metal

5 0
3 years ago
Find the percent ionization of a 0.337 m hf solution. the ka for hf is 3.5 x 10-4. 1.1 % 1.2 x 10-2 % 3.2 % 3.5 x 10-2 % 4.7 %
Ugo [173]

To determine the percent ionization of the acid given, we make use of the acid equilibrium constant (Ka) given. It is the ration of the equilibrium concentrations of the dissociated ions and the acid. The dissociation reaction of the HF acid would be as follows:<span>

HF = H+ + F-

The acid equilibrum constant would be expressed as follows:

Ka = [H+][F-] / [HF] = 3.5 x 10-4

To determine the equilibrium concentrations we use the ICE table,
         HF             H+              F-
I      0.337           0                 0
C      -x              +x               +x
---------------------------------------------
E    0.337-x        x                   x 

3.5 x 10-4 = [H+][F-] / [HF] 
3.5 x 10-4 = [x][x] / [0.337-x] </span>

Solving for x,

x = 0.01069 = [H+] = [F-]

percent ionization = 0.01069 / 0.337 x 100  = 3.17%

8 0
3 years ago
Read 2 more answers
Which of these is correct?
Viktor [21]

Answer:

1.89 nol Cu(NO3)2

Explanation:

if you calculate it it will be 1.89

5 0
3 years ago
How many grams of NH3 can be produced from 2.30 mol of N2 and excess H2.
MrRa [10]
<h3>Answer:</h3>

78.34 g

<h3>Explanation:</h3>

From the question we are given;

Moles of Nitrogen gas as 2.3 moles

we are required to calculate the mass of NH₃ that may be reproduced.

<h3>Step 1: Writing the balanced equation for the reaction </h3>

The Balanced equation for the reaction is;

   N₂(g) + 3H₂(g) → 2NH₃(g)

<h3>Step 2: Calculating the number of moles of NH₃</h3>

From the equation 1 mole of nitrogen gas reacts to produce 2 moles of NH₃

Therefore, the mole ratio of N₂ to NH₃ is 1 : 2

Thus, Moles of NH₃ = Moles of N₂ × 2

                                  = 2.3 moles × 2

                                  = 4.6 moles

<h3>Step 3: Calculating the mass of ammonia produced </h3>

Mass = Moles × molar mass

Molar mass of ammonia gas = 17.031 g/mol

Therefore;

Mass = 4.6 moles × 17.031 g/mol

         = 78.3426 g

         = 78.34 g

Thus, the mass of NH₃ produced is 78.34 g

3 0
3 years ago
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