(1) magnesium ion is precipitated by hydroxide ion, magnesium ion is precipitated by excessive sodium hydroxide, sulfate ion is precipitated by barium ion, and sulfate ion is precipitated by excessive barium chloride. As for removing magnesium ions or sulfate ions first, calcium ions and carbonate ions precipitate. In addition to calcium ions, sodium carbonate is added to transform into precipitates, but the added sodium carbonate should be placed after the added barium chloride. In this way, sodium carbonate will remove the residual barium chloride from the reaction and the ions will precipitate out. After filtration, hydrochloric acid will be added to remove the residual hydroxyl ions and carbonate ions in the reaction, and sodium chloride will be obtained by evaporation and crystallization, so the answer is: excessive barium chloride solution; Excess sodium carbonate solution; Filtering; Evaporation crystallization;

(2) Evaporation pollutes the environment due to the volatilization of hydrogen chloride, so the answer is: evaporation pollutes the environment due to the volatilization of hydrogen chloride;

(3) Because some trace elements in seawater still remain in sea crystals, which can promote the growth of marine life, the answers are as follows: some trace elements in seawater still remain in sea crystals, and crude sea salt may not contain these elements;

(4) ① The valence of I increased by 5 from 0→+5, and the valence of Cl decreased by 65,438+0 from 0 →-65,438+0. According to the conservation of gain and loss electrons and mass, I2+5cl2+6H2O ═ 2hio3+65,438+00HCl is obtained.

So the answer is: I2+5cl2+6H2O ═ 2hio3+10hcl;

(2) Adding starch into the solution and slowly introducing chlorine. The solution turns blue first, then colorless, and then orange. These reactions will occur continuously:

Cl2+2NaI═I2+2NaCl, the solution turns purple;

I2+5cl2+6H2O ═ 2hio3+10hcl faded to colorless;

Cl2+2NaBr═Br2+2NaCl, the color of the solution turns orange;

The reducibility of reducing agent is greater than that of reducing product, which shows that the order of reducibility is I-> I2 > Br-.

So the answer is: I-> I2 > br-;

③ Because the lower reddish-brown carbon tetrachloride is insoluble in water, it can be separated by liquid separation, so the answers are: separating funnel, beaker;

④ A. Bromine can react with Na2CO3 solution to generate sodium bromide, sodium hypobromide and carbon dioxide, and bromine is not completely transferred from CCl4 layer to water layer in the form of Br-, so A is wrong;

B, bromine can react with Na2SO3 solution to generate sodium sulfate and hydrogen bromide, and bromine is only transferred from CCl4 layer to water layer in the form of Br-, so b is correct;

C, bromine does not react with KMnO4(H2SO4), and bromine is not completely transferred from CCl4 layer to water layer in the form of Br-, so c is wrong;

D, bromine can react with Na2S solution to generate sodium bromide and elemental sulfur, and bromine is only transferred from CCl4 layer to water layer in the form of Br-, so D is correct;

E bromine can react with phenol solution to generate tribromophenol and hydrogen bromide. Tribromophenol is dissolved in CCl4, and bromine is not completely transferred from CCl4 layer to water layer in the form of Br-, so e is wrong;

So: BD.