Class 12th

1° amine: Forms a precipitate that is soluble in NaOH.

2° amine: Forms a precipitate that is insoluble in NaOH.

3° amine: No reaction.

A function f (x) is continuous at x=1, so lim (x→1? ) f (x) = lim (x→1? ) f (x) = f (1).
Assuming a piecewise function like f (x) = { -x, x<1; ax+b, x1 } (structure inferred from derivative).
Continuity at x=1: f (1) = 1. a (1)+b = 1 => a+b=1.

The function is differentiable at x=1. The derivative of f (x) at x=1 from the left is -1. The derivative from the right is a.
So, a = -1. (The image has 2a = -1, which would imply a function like -x and ax²+b). Let's assume f' (x) = 2a for x>1.
2a = -1 => a = -1/2.

From a+b=1, b = 1 - a = 1 - (-1/2) = 3/2.
So, a = -1/2 and b = 3/2.

Please consider the following Image

Inversion of Sucrose:
C? H? O? (Sucrose) + H? O - (Invertase)-> C? H? O? (Glucose) + C? H? O? (Fructose)

Fermentation of Glucose:
C? H? O? (Glucose) - (Zymase)-> 2C? H? OH (Ethanol) + 2CO?

KMnO? oxidizes alkanes that contain a tertiary hydrogen (3°H) to form tertiary alcohols.

n-alkanes, which do not contain tertiary hydrogens, are not oxidized by KMnO? under these conditions.

Evaluate the integral:
∫ (2x-1)cos (√ (4x²-4x+6) / √ (4x²-4x+6) dx
∫ (2x-1)cos (√ (2x-1)²+5) / √ (2x-1)²+5) dx

Let (2x-1)² + 5 = t².
Differentiating both sides:
2 (2x-1)*2 dx = 2t dt
2 (2x-1) dx = t dt
(2x-1) dx = (t/2) dt

Substitute into the integral:
∫ cos (t)/t * (t/2) dt
= 1/2 ∫ cos (t) dt
= 1/2 sin (t) + C
= 1/2 sin (√ (2x-1)²+5) + C
= 1/2 sin (√ (4x²-4x+6) + C

Co (NH? )? ] [Cr (CN)? ]: Co-ordination isomerism

[Co (NH? )? (NO? )? ]: Linkage isomerism

[Cr (H? O)? ]Cl? : Solvate isomerism

Cis- [CrCl? (Ox)? ]³? : Optical isomerism

Please consider the following Image

CN? and NO? are examples of ambident nucleophiles.

The change in entropy for the following processes is negative (ΔS = -ve), indicating an increase in order:

Water (l) → Ice (s) at 0°C

H? O (l) → Ice (s) at -10°C

N? (g) + 3H? (g) → 2NH? (g)

Adsorption