Ab2 has a molar solubility of 3.72×10−4 m. what is the value of the solubility product constant for ab2? express your answer numerically.

Answer is: Ksp for AB₂ is 2.06·10⁻¹⁰.
Chemical reaction: AB₂(s) → Aˣ⁺(aq) + 2Bⁿ⁻(aq).

s(AB₂) = 3,72·10⁻⁴ M.
s(AB₂) = s(Aˣ⁺).

s(Bⁿ⁻) = 2s(AB₂).

Ksp(AB₂) = s(Aˣ⁺) · (s(Bⁿ⁻))².
Ksp(AB₂) = 3,72·10⁻⁴ · (7,44·10⁻⁴)².

Ksp(AB₂)= 2,06·10⁻¹⁰.

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied