Chemistry
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New answer posted
10 months agoNew answer posted
10 months agoContributor-Level 10
This is electrophilic substitution reaction which is determine by electronic effect of 
New answer posted
10 months agoContributor-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 agoContributor-Level 10
E2 elimination. Most acidic proton is removed. Fluorine is more electronegative.
New answer posted
10 months agoContributor-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.
New answer posted
10 months agoContributor-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 agoContributor-Level 10
On moving left to right in a period.
Acidic character of oxides is increase.
3rd period element oxides.
New answer posted
10 months agoContributor-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.
New answer posted
10 months agoContributor-Level 10
Li and Mg form nitrides and have bicarbonates that are unstable
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