Class 12th
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New answer posted
a year agoContributor-Level 9
Normal to plane is n= (-4i+5j+7k).
Plane: -4 (x-3)+5 (y-1)+7 (z-1)=0 ⇒ -4x+5y+7z=0.
Passes through (α, -3,5) ⇒ -4α-15+35=0 ⇒ α=5.
New answer posted
a year agoNew answer posted
a year agoContributor-Level 9
A? A = I
⇒ a²+b²+c²=1 and ab+bc+ca=0
Now, (a+b+c)²=1 ⇒ a+b+c=±1
So, a³+b³+c³-3abc = (a+b+c) (a²+b²+c²-ab-bc-ca) = (±1) (1-0)=±1
⇒ 3abc = 2±1 = 3,1
⇒ abc = 1, 1/3
New answer posted
a year agoContributor-Level 10
This is electrophilic substitution reaction which is determine by electronic effect of 
New answer posted
a year 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
a year agoContributor-Level 10
E2 elimination. Most acidic proton is removed. Fluorine is more electronegative.
New answer posted
a year 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
a year agoContributor-Level 10
Seliwanoff's test distinguishes aldoses from ketoses. Sucrose hydrolyzes to glucose (aldose) and fructose (ketose). Fructose gives a red color.
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