The steps to solve this problem are to seek the hegemony of chemistry

Impure reagent Na2CO3

Possible impurities: Na2SO4, CaCO3, CuSO4, NaCl, CaCl2

Experiment Exclude CuSO4 (blue-green solution) and exclude substances containing Ca ions (CaCO3 itself is precipitate and CaCl2), because it will react with sodium carbonate to form CaCO3 precipitate! A false!

There must be BAC3 or BASO4 produced.

The white precipitate did not disappear completely (BaCO4), and bubbles were produced (BaCO3).

It can't be said that the original solution has CL-, thinking that the BACL2 solution is added, and chloride ions are introduced. c correct (adding barium nitrate will not introduce chloride ions, and the resulting white precipitate is agcl, so it must contain nacl).

In summary, the reagent contains sodium carbonate, sodium sulfate; Assuming it does not contain NaCl!

The added BaCl2 reacts completely, and the precipitate is a mixture of barium sulfate and barium carbonate with a mass of m1;

After adding dilute nitric acid, barium carbonate reacts completely, and only barium sulfate is precipitated, and the mass is m2; The mass of barium carbonate is (m1-m2).

The mass of the last step that actually produces AGCL is denoted as m3; Where Cl comes entirely from the added barium chloride, then the mass of the total chloride ions can be calculated according to the mass of barium carbonate and barium sulfate, and the mass of the theoretical production of AGCL is recorded as M4, if M4=M3, it does not contain NaCl. If M4 > M3 contain NaCl

Dissolved in water to obtain a colorless solution, it means that there is no calcium carbonate, copper sulfate, calcium chloride (calcium chloride can react with sodium carbonate to form calcium carbonate precipitation). The white precipitate of barium chloride indicates the presence of carbonate or the presence of carbonate and sulfate. The addition of dilute nitric acid precipitate partially dissolves indicates that there is barium sulfate in the precipitate, so there are carbonate and sulfate in the solution, so there must be sodium sulfate.

The fourth part proves that there are chloride ions, so the impurities have sodium chloride. After knowing the accurate mass of the precipitate, such as the relative molecular mass, calculate the mass of sodium carbonate and sodium sulfate, and compare it with the mass of the sample to know whether there is sodium chloride. Barium nitrate in item C can play a role in removing sodium sulfate and sodium carbonate, at this time, only sodium chloride is left in the solution, and silver nitrate is added to test it.

This question is the electrolysis of water, and sulfuric acid is just a conductive medium.

X is connected to the negative electrode, and its reaction is 2 H+ +2E=H2 Y is connected to the positive electrode, and its reaction is 4oh- -4E=2H2O + O2 and its electrolytic reaction is 4H+ +4Oh- =2H2O+H2 +O2

Excessive amounts of Hno3 have gases produced both by CO32-

If there is a white precipitate of excessive Ba(NO3)2, there will be SO4 2-, and then if there is a white precipitate of excessive AgNO3, there will be Cl-

Therefore, 132 if the replacement will produce agso4 unfavorable identification, please adopt.

213 is added to 2 to produce a white precipitate with cl-

Add 1, there is gas, there is CO3-

Add 3, with occasional precipitation, there is SO42-

co, cu (superscript 2+), fe2o write superscript) 12, c d

Oxygen is isolated.

Molecules are in constant motion.

Potassium fertilizer. I can't see it in the back...

You don't say which question it is!!

Question 11: Positive bivalent, Cu2+ (2+ is the upper corner mark), Fe2O3 (2 and 3 are the lower corner mark).

12 Questions: (1) C, (2) D, (3) Isolation of oxygen.