There are four colorless solutions A, B, C, D, and they are one of Al NO3 3 Ba OH 2 NH4 2SO4 NH4Cl

1)a：k2co3 d：bacl2

2) The white precipitate is gradually produced in the solution, and the white precipitate is insoluble Al3+ 3NH3·H2O===Al(OH)3 +3NH4+

3)/1.Neuter.

oh- +so42- +h+ =h2o+baso4↓

A, B, C, D, and E were K2CO3, AlCl3, AgNO3, BaCl2, and NaHSO4, respectively

You should know the rest.

Since the pH value of A B is equal, it is known that sodium acetate is alkaline, ammonium chloride is acidic, sodium bisulfate is acidic, and barium chloride is neutral, then A and B must be ammonium chloride and sodium bisulfate solution A C becomes turbid after mixing, then C must be BaCl2, A is NaHSO4, B is ammonium chloride, and D is sodium acetate.

So a is wrong c is a strong acid and strong alkali salt, which is not hydrolyzed, so b is wrong d solution b does not contain na ions so wrong.

Only the answer C is correct A is NaHSO4, which reacts with HCO3- to form CO2

A NaHSO4 B NH4Cl C BaCl2 D CH3Coona Because ammonium chloride and sodium bisulfate are both acidic, sulfate ions and barium ions produce barium sulfate precipitates Answer: choose C

If the pH of CaB solution is the same as that of AC and the solution becomes turbid after mixing, then A is NaHSO4 solution, B is NH4Cl solution, and C is BaCl2 solution.

D is CH3Coona solution.

A is NaHSO4, B is NH4Cl, and AB is acidic.

C is BACL2 and D is CH3Coona

A is wrong B is wrong C is true.

A is NH4CL

b is ch3coona

C is HCLD and Na2SO4

So A is correct.

b is false, hydrochloric acid inhibits the ionization of water.

c correct. D is correct, because oh- and h+ are both ionized by water, so oh=h is chosen b for the answer

This question. There are two answers.

I chose ABPH, the same only possible ammonium chloride and hydrochloric acid.

Because ammonium acetate is alkaline sodium sulfate is neutral.

Only ammonium chloride and hydrochloric acid are acidic.

And the same degree of ionization is only hydrochloric acid and sodium sulfate.

Because they are all completely ionized, a hydrochloric acid b ammonium chloride c sodium sulfate d sodium acetate.

So option A is wrong and option B is wrong.

N2 is oxidized to N2O5 and then converted to nitric acid.

（1）（nh4）2so4 naoh nh4no3bacl2 (2) +oh-nh3↑+h2o +ba2+====baso4↓

Analysis: A and B, B and C are heated together, which can produce a gas that turns the wet red litmus paper blue, so B is NaOH, A and C are one of (NH4)2SO4 and NH4NO3, A and D are mixed with white precipitate to generate, so A is (NH4)2SO4.

Since B can make AC release ammonia, B is sodium hydroxide. Whereas, a c is one of ammonium sulfate and ammonium nitrate. d must be barium chloride.

Since ammonium sulfate and barium chloride can react, C is ammonium nitrate and A is ammonium sulfate.

The equation solves itself.

1: A D = solution + gas, there is a gas production description is based on the four colorless solutions can know that it is acid that dissolves the salt to produce CO2, so a and d two solutions are Na2CO3 and NaHSO4

2: According to one, we can know that b and c are na2so4, bacl2, and b+c can also produce precipitation, which is in line with the meaning of the question and shows that the reasoning is correct.

According to five, it can be known that the precipitate X is dissolvable, so D is Na2CO3 and B is BaCl2 and the precipitate X is BaCO3.

Then we know that A is NaHSO4 and C is Na2SO4.

Thus: a, b, c, d all come out.

B and the other three all produce precipitation, so B is barium chloride BaCl2Ad reaction with gas, indicating that one is Na2CO3 and the other is NaHSO4, so C is Na2SO4

The precipitate X and the solution Y produce gas, indicating that the precipitate X is barium carbonate, and the solution Y contains hydrogen ions, so D is Na2CO3 and A is NaHSO4

The solution is a colorless solution, judging that it must not contain mNO4

A part of the solution was taken, and the NaOH solution was dropped drop by drop until it was excessive, and only the white precipitate was observed to increase first and then partially dissolve, and there was no other obvious phenomenon; The description must contain al3+

mg2+；According to the ion coexistence, it must not contain HCO3; Because magnesium ions and hydroxide ions form precipitates that are insoluble in alkali, the resulting aluminum hydroxide is soluble in alkali; The combination of ammonium ions and hydroxide ions may form a gaseous ammonia gas, so it must be NH4-free

Take another part of the solution and add Hno3

Acidified BA (NO3

Solution, no precipitation is produced, and it is judged that there must be no SO4

According to the electrical neutrality of the solution, the anion in the solution must be Cl- In summary: the solution must contain Al3+

Mg2+, Cl-, must be NH4 free

so4hco3、mno4

There may be K+, so choose A

B and C are mixed to produce a white precipitate, and when heated, a gas is produced that turns the paper blue.

B, C are the mixture of Ba(OH)2, (NH4)2SO4A and B of the Wulu group to produce a white precipitate with no obvious change after heating.

A is Na2SO4

b is ba(oh)2

c is NH4)2SO4

There is no significant change in the mixing of B and E, and when heated, a gas is produced that turns the wet red litmus paper blue.

E is NH4CI

D is NACI