An analytical procedure requires a solution of chloride ions ,How many grams of CaCl2 must be disolved to make2.15L of 0.0520 M C

Answer 1

I assume you meant 0.0520 M Cl^-.

You want mols Cl^- = M x L = approx 0.12 but that's just an estimate.

Then mols CaCl2 = 0.118/2 since 1molecule of CaCl2 supplies 2 Cl^- ions.

g CaCl2 = mols CaCl2 x molar mass CaCl2.

Answer 2

you have exposed electrodes of light bulb in a solution of H2SO4 such that the light bulb is on you add dilute solution and bulb goes dim which of the following could be the solution

Ba(OH)2
NaNO3
K2SO4
Cu(NO3)2
none of the above

Answer 3

Thank you very much

Answer 4

You didn't express your post very well but I think you have H2SO4 and are adding one of the four to it. If the light dims (and in fact it will go essentially out if you add to the exact equivalence point) Ba(OH)2 this will form BaSO4 and that is insoluble. So the answer is a.
Ba(OH)2 + H2SO4 ==> BaSO4 + 2H2O