Chemistry

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New answer posted

10 months ago

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A
alok kumar singh

Contributor-Level 10

Acidity: b (carboxylic acid)>c (phenol)>d (enol)>a (alkyne).

New answer posted

10 months ago

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A
alok kumar singh

Contributor-Level 10

This is electrophilic substitution reaction which is determine by electronic effect of

 

 

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10 months ago

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A
alok kumar singh

Contributor-Level 10

From I&II, rate∝ [B]². From I&III, rate∝ [A]¹.
From IV: 7.2e-2 = k (X) (0.2)². From II: 2.4e-2=k (0.1) (0.2)². X=0.3.
From V: 2.88e-1=k (0.3) (Y)². k=2.4e-2/ (0.1*0.04)=6.
2.88e-1 = 6 (0.3)Y². Y²=0.16. Y=0.4.

New answer posted

10 months ago

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alok kumar singh

Contributor-Level 10

Cast iron is used to make wrought iron and steel.

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alok kumar singh

Contributor-Level 10

E2 elimination. Most acidic proton is removed. Fluorine is more electronegative.

 

New answer posted

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alok kumar singh

Contributor-Level 10

Ligand field strength: NH? > NCS? > F? Stronger ligand, higher Δ, lower λ_max.
So λ (NH? ) < (NCS? ) < (F? ). A= (F? ), B= (NCS? ), C= (NH? ). A-ii, B-i, C-iii.

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10 months ago

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alok kumar singh

Contributor-Level 10

Seliwanoff's test distinguishes aldoses from ketoses. Sucrose hydrolyzes to glucose (aldose) and fructose (ketose). Fructose gives a red color.

New answer posted

10 months ago

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alok kumar singh

Contributor-Level 10

On moving left to right in a period.

Acidic character of oxides is increase.

3rd period element oxides.

(i) Acidic character

(ii) Atomic No?

So  have minimum Atomic No

&Z have maxima Atomic No

So correct order is 

 

 

New answer posted

10 months ago

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alok kumar singh

Contributor-Level 10

gives iodoform test and slow Lucas test, so it's a methyl secondary alcohol. D gives fast Lucas test, so it's a tertiary alcohol. The Grignard products must be tertiary and secondary alcohols. So A and B must be an aldehyde and a ketone.

 

 

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10 months ago

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alok kumar singh

Contributor-Level 10

Li and Mg form nitrides and have bicarbonates that are unstable

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