Class 12th

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

Given equations:
3x + 4y = 9
y = mx + 1 (assuming this from the substitution)

Substitute y into the first equation:
3x + 4 (mx + 1) = 9
3x + 4mx + 4 = 9
x (3 + 4m) = 5
x = 5 / (3 + 4m)

For x to be an integer, (3 + 4m) must be a divisor of 5, i.e., ±1, ±5.

• 3 + 4m = 1 => 4m = -2 => m = -1/2
• 3 + 4m = -1 => 4m = -4 => m = -1 (Integer value)
• 3 + 4m = 5 => 4m = 2 => m = 1/2
• 3 + 4m = -5 => 4m = -8 => m = -2 (Integer value)

The two integral values for m are -1 and -2.

The reaction shown is:
Anisole diazonium chloride (A) + Ethanol → Anisole + Acetaldehyde (X) + HCl + N? (Y)

f (x) + g (x) = √x + √1-x. The domain requires x ≥ 0 and 1-x ≥ 0, so x ≤ 1. Domain is [0,1].

f (x) - g (x) = √x - √1-x. Domain is [0,1].

f (x)/g (x) = √x / √1-x. Requires x ≥ 0 and 1-x > 0, so x < 1. Domain is [0,1).

g (x)/f (x) = √1-x / √x. Requires 1-x ≥ 0 and x > 0. Domain is (0,1].

The common domain for all these functional forms to be considered is (0,1).

A line passes through (1,3). Its equation is y - 3 = m (x - 1) or y = mx + (3-m).
The angle θ between this line and the line y = 3√2x - 1 (with slope m? = 3√2) is given by tanθ = √2.

tanθ = | (m - m? )/ (1 + m*m? )|
√2 = | (m - 3√2) / (1 + m*3√2)|

This gives two cases:

Case 1 (+ve):
√2 = (m - 3√2) / (1 + 3√2m)
√2 (1 + 3√2m) = m - 3√2
√2 + 6m = m - 3√2
5m = -4√2
m = -4√2 / 5

Case 2 (-ve):
-√2 = (m - 3√2) / (1 + 3√2m)
-√2 (1 + 3√2m) = m - 3√2
-√2 - 6m = m - 3√2
7m = 2√2
m = 2√2 / 7