Class 12th

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

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.