Recent content by aati2sh

Because mv^2 has the unit same as that of Energy.

I tried this: Ksp of Ag2CO3 = [2xSolub.]2x[Solub.] => Solubility = 1.270334x10-4 = [Ag+] Now, [Cl-]=0.003/35.5 M = 8.45 x 10-5 Ksp of AgCl = 1.270334x10-4 x 8.45 x 10-5 = 1.073 X 10-8. Why was concentration and volume of Na2CO3 was given if this is correct?

I don't know how to start.

Yes. I calculated the concentration by dividing gm/litre by 35.5 to give molarity. But, I don't know where to go beyond.

Homework Statement A sample of AgCl was treated with 5ml of 2M Na2CO3 solution to produce Ag2CO3. The remaining solution contained 0.003 g of Cl- per litre. Calculate the solubility product of AgCl.(Ksp of Ag2CO3 = 8.2 ×10-12) Homework Equations ksp of AgCl = [Ag+]×[Cl-] ksp of Ag2CO3 =...