A sample of solution has a mixture of 0.01 M Aluminum Chloride and 0.05 M Magnesium Chloride. What is the molarity of Silver that is required to precipitate all the chloride ions present in the solution?

Explanation:

AlCl₃ ⇒Al⁺³ + 3Cl⁻¹

.01M                 3 x .01 M

So .01 M AlCl₃ will give 3 x .01 M Cl⁻¹

MgCl₂ ⇒ Mg⁺² + 2Cl⁻¹

.05 M                   2 x .05M

.05M MgCl₂ will give  2 x .05 M Cl⁻¹

Total Cl⁻¹ formed = 3 x .01 + 2 x .05 M

= .03 + .1 M

= .13 M

AgCl ⇒ Ag⁺ + Cl⁻¹

So AgCl required = .13 M . to precipitate all chloride ions.

i do not know why. good luck

by: nicholas

ek = (m·v2) / 2

v - velocity of system after the collision

m = m1 + m2

ek = (m1 + m2) ·v2/2

ek = (200 + 250) ·2,32/2 = 450·5,29/2 = 1190,25 j