Addition of 12 M (concentrated) HCl to saturated NaCl results in precipitation of NaCl. Yet when 1 M HCl is added to the same NaCl solution, no precipitate forms. Explain.

When we add 12M HCl, reaction quotient Qsp becomes greater than Ksp of NaCl leading to precipitation

Explanation:

When sodium chloride is dissolved in water, the equilibrium dissolution process known as Ksp is set up as typified by the equation;

NaCl(s)> Na^+(aq) + Cl^-(aq)

According to Le Chatelier's principle, this system will remain in equilibrium after the addition of the 1M HCl. This explains why there will not be any change in the system after adding this solution.

However, addition of 12 M HCl will significantly distort the equilibrium position since there is now more Cl^- ion in the system. The system now seeks to relieve this stress by recrystallizing more sodium chloride leading to the formation of a precipitate. Succinctly put, when we add the 12M HCl, the reaction quotient Qsp becomes greater than Ksp of NaCl leading to precipitation of NaCl crystals.

(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