Class 12th

In a C? fullerene molecule, all 60 carbon atoms are identical. Each carbon atom is bonded to three other carbon atoms via sigma bonds. The structure is composed of 20 six-membered rings and 12 five-membered rings.

In diacetamide (CH? CO)? NH), the lone pair of electrons on the nitrogen atom is delocalized through resonance with both adjacent carbonyl groups. This extensive resonance greatly decreases the electron density on the nitrogen atom.

Only 1° amine react with benzene sulphonyl chloride to give a compound which is soluble in alkali

Metal sulphide sols are negatively charged while metal oxide sols are positively charged sols. So, CdS is negative while TiO? is positively charged sol.

Colloidal particles are small enough to pass through an ordinary filter but can be stopped by an ultrafilter paper due to their specific particle size range.

P (x) = f (x³) + xg (x³) is divisible by x²+x+1. The roots of x²+x+1=0 are the complex cube roots of unity, ω and ω².
P (ω) = f (ω³) + ωg (ω³) = f (1) + ωg (1) = 0 — (I)
P (ω²) = f (ω²)³) + ω²g (ω²)³) = f (1) + ω²g (1) = 0 — (II)
Subtracting (II) from (I): (ω - ω²)g (1) = 0. Since ω ≠ ω², we must have g (1) = 0.
Substituting g (1)=0 into (I) gives f (1) = 0.
We need to find P (1) = f (1³) + 1*g (1³) = f (1) + g (1).
P (1) = 0 + 0 = 0.

Please consider the following Image

In the ammonolysis reaction, HCl is produced as a byproduct. To neutralize this acidic impurity, the mixture is treated with NaOH.

P' (x) = a (x-1) (x+1) = a (x²-1).
P (x) = ∫ P' (x) dx = a (x³/3 - x) + b.
Given P (-3) = 0 ⇒ a (-9+3) + b = 0 ⇒ b = 6a.
Given ∫ P (x)dx = 18. Assuming the integration is over a symmetric interval like [-c, c] and using the fact that a (x³/3-x) is an odd function, ∫ (a (x³/3 - x)dx = 0. Then ∫ b dx = 18. If the interval is [-1, 1], this would be b (1 - (-1) = 2b = 18, so b=9.
With b=9, we find a = b/6 = 9/6 = 3/2.
So, P (x) = 3/2 (x³/3 - x) + 9 = x³/2 - 3x/2 + 9.
The sum of all coefficients is 1/2 - 3/2 + 9 = -1 + 9 = 8.