1, there is something wrong with this question. On the one hand, Pb(2+) will react with Cl(-) to form precipitated PbCl2, so it can't be a solution from the beginning. On the other hand, of course, it is only related to Ksp, and it can be completely precipitated only when Ksp is extremely small, because acid will be produced during the introduction of H2S, and sulfide with high Ksp will be dissolved with the increase of H(+) concentration. At this time, it is impossible to form MnS, CuS will be completely precipitated, but CdS will only be basically precipitated. It seems that D is a better answer.

2. Hypervalent chromate is a strong oxidant. As an oxyacid salt, it has strong oxidation in strong acid solution, so it is suitable to prepare it under alkaline conditions. However, CrO3 is the highest oxidation state, so it is difficult to understand because it does not have to use oxidant. If the raw material is Cr2O3, the answer is B.

3. The crystal field theory is adopted: Co(3+) is d(6) electronic configuration, and D orbital splitting is related to the strength of ligand. The stronger the ligand, the greater the splitting energy and the shorter the wavelength of absorbed light, so [Co(CN)6](3-) will absorb light with short wavelength.

4、A:cr2o 7(2-)+ 14H(+)+6e = 2Cr(3+)+7H2O

b:cro 4(2-)+H2O+3e = cro 2(-)+2OH(-)

5.a, Fe(OH)3: Because the latter two hydroxides are extremely strong oxidants under acidic conditions, chlorine gas is released when Cl(-) is oxidized.

6、2Co(2+)+Br2+6OH(-)= 2Co(OH)3+2Br(-)

The last problem is simply to calculate the equilibrium (precipitation equilibrium and coordination equilibrium), which should be solved after thinking for yourself.