Class 11th

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8 months ago

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V
Vishal Baghel

Contributor-Level 10

(c) In homolytic cleavage, one of the electrons of the shared pair in a covalent bond goes with each of the bonded atoms. Thus, in homolytic cleavage, the movement of a single electron takes place instead of an electron pair.

A heterolytic cleavage yields carbocations or carbanions, while a homolytic cleavage gives free radicals as reactive intermediate.

New answer posted

8 months ago

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V
Vishal Baghel

Contributor-Level 10

(b) It is a nucleophilic substitution reaction. KOH (aq) provides OH- ion for the nucleophile attack.

New answer posted

8 months ago

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V
Vishal Baghel

Contributor-Level 10

(d) Is the correct answer.

New answer posted

8 months ago

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V
Vishal Baghel

Contributor-Level 10

(b) Is the most stable since it is a tertiary carbocation.

New answer posted

8 months ago

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V
Vishal Baghel

Contributor-Level 10

(b) Iron (III) hexacyanidoferrate (II) (or ferriferrocyanide) Fe4 [Fe (CN)6]is the correct answer.

New question posted

8 months ago

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New answer posted

8 months ago

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V
Vishal Baghel

Contributor-Level 10

Mass of the compound = 0.468 g
Mass of barium sulphate= 0.668 g

% of sulphur = 

                     

New answer posted

8 months ago

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V
Vishal Baghel

Contributor-Level 10

Mass of the compound = 0.3780 g
Mass of silver chloride = 0.5740 g

% of chlorine =

New answer posted

8 months ago

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V
Vishal Baghel

Contributor-Level 10

Step I: Calculation of volume of unused acid i.e. V2 =?

V1 = Volume of NAOH solution required = 60 cm3 

N1 = Normality of NaOH solution = ½ N

N2 = Normality of H2SO4 = 1N

Applying N1V1 = N2V2

½ N x 60 cm3 = 1N x V2

Or V2 = 30 cm3

Step II: calculation of volume of acid used

Volume of acid added = 50 cm3

Volume of unused acid = 30 cm3

Volume of acid used = 50 – 30 = 20 cm3

Step III: Calculation of % of nitrogen

Mass of compound = 0.50 g

Volume of acid used = 20 cm3

Normality of acid used = 1 N

% of nitrogen = (1.4 x 20 x 1) / 0.50 = 56%

New answer posted

8 months ago

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V
Vishal Baghel

Contributor-Level 10

Step 1: Calculation of mass of CO2 produced

Mass of compound = 0.20 g

% of carbon = 69%

i.e. 12/44 x  = Mass of Carbondioxide formed / Mass of Compound = 69/100

Therefore, mass of CO2formed = (69 x 44 x 0.20) / (12 x 100) = 0.506 g

Step 2: Calculation of mass of H2O produced

Mass of compound = 0.20 g

% of hydrogen = 4.8%

i.e. 2/18 x Mass of Water Formed/Mass of Compound = 4.8/100

Therefore, mass of H2O formed = 4.8*18*0.20/2*100 = 0.0864 g

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