A super difficult chemistry problem is advanced by the master! There are traps everywhere in the que

First of all, you can judge that B is AG+, Mr. into silver hydroxide, and then generate silver ammonium complex ions to dissolve, do not mistake it for Al3+, Al3+ is insoluble when it reacts with a weak base, F is Fe2+, and then the cation of ACE is a strong alkali metal ion, the anion is 0H- and the two weak acid groups of carbonate and acetate are right, metallic BA>K>CA, barium acetate, barium carbonate is insoluble, so it is barium hydroxide, acetic acid is stronger than carbonic acid, it should be potassium carbonate Yes, and then calcium acetate. D added barium nitrate does not react, indicating that it does not contain carbonate sulfate, and then it can be judged that Alcl3 is correct when excluding cations before, and then the remaining AGSO4 is insoluble, so it is silver nitrate, and finally ferrous sulfate. So you're doing the right thing.

Make a table, there are 36 possibilities.

1.The exclusion of insoluble or slightly soluble is shown in the table Microsoluble (1).

2.The solution prepared separately, Ca(OH)2 solubility is small, and the solution that cannot be prepared can be excluded, see the insoluble and slightly soluble (2) in the table

3.Solutions A, C, E, are alkaline and alkaline A e CTherefore, it is concluded that the anions in ACE are hydroxide ions, acetate ions, and carbonate ions.

4.Ammonia was added dropwise to the F solution to generate a white flocculent precipitate, which quickly turned gray-green and finally reddish-brown. The cation in the solution is Fe2+

Looking at the carbonate row, the cation in E can only be K+, and E is K2CO3The other 5 compounds do not contain k, see (4) in the table

5.Looking at the OH- line, the cation in A can only be Ba2+ and A is Ba(OH)2Other compounds do not contain Ba2+, see (5) in the table

6.Add dilute ammonia dropwise to solution B, precipitate first, continue to add dilute ammonia dropwise, and the precipitate disappears. The cation in b is ag+

Looking at the AG+ column, the anions in b can only be NO3- (AC- in C) and B is AGNO3Other compounds do not contain NO3-, see (6) in the table

7.Barium nitrate solution was added dropwise to solution d, and there was no obvious phenomenon. The anions in d are not sulfate ions, carbonate ions (nor oh-), but can only be chloride ions, acetate ions, and nitrate ions; The latter two are in C and B, respectively, and the anion in D can only be Cl-

So f can only be sulfate, and f is feso4At this time, there are several that can be excluded, see (7) in the table

8.There are two more possibilities left:

a.C is Caac2 and D is AlCl3, which are alkaline and acidic, respectively, which is in line with the title.

b.c is alac3 and d is cacl2, both of which are neutral and do not fit the topic.

a: ba(oh)2

b: agno3

c: caac2

d: alcl3

e: k2co3

f: feso4

1 alkaline AEC, so A, OH; e，co3 2－；C,AC2,B,Ag, considering anion, must be AgNO33,D,Cl

4, f is Fe2

It's going to turn off the lights, so continue tomorrow morning...

Agree with your answer

From 1, it can be seen that the anions in a, c, and e are hydroxide, carbonate, and acetate respectively;

From 2, it can be seen that the cation in b is a silver ion, and the only one in the silver ion that can coexist with the silver ion in the solution is the nitrate ion, so it can be determined that b is silver nitrate;

From 3, it can be seen that there is no sulfate ion in D, so the anion in D can only be chloride ion;

From 4, it can be seen that there are ferrous ions in f; The anion is sulfate ion by the exclusion method, so f is ferrous sulfate.

The following is determined by the method of exclusion from the remaining ions:

In the remaining cations, the energy of aluminum ions and chloride ions coexist, so d is aluminum chloride;

The ionized can only coexist with OH-, and a is BA(OH)2;

Calcium ions cannot coexist with carbonate, so C is calcium acetate and E is potassium carbonate.

2，(c2h4o)n

24+4+16)*n=88, the solution is n=2

So the molecular formula is C4H8O2

Propyl formate, isopropyl formate, ethyl acetate, methyl propionate, a 4-kind 3, potassium permanganate solution, the double bond of propylene will fade potassium permanganate, and cyclopropane will not work 4, (1) C3H4O2

2） ch2=ch-cooh

3) A is a monocarboxylic acid, so there is -COOH in A, and A can react with hydrogen, bromine, and hydrogen bromide, so there are carbon-carbon unsaturated bonds in the molecule. The simplest compounds containing carbon-carbon unsaturated bonds are acrylic acid and propynylic acid, but only acrylic acid has a carbon mass fraction of 50%. So a is acrylic

1.（1）.Hydroxyl.

2）.c3h6o ch2=ch-ch2oh2.c4h8o2 (write it yourself)3Bromine water fades to propylene.

4.(1) C3H4O2 (2) CH2=CH-COOH (3) First of all, it is a monobasic acid, and there are two oxygen and an addition reaction can occur It should contain a double bond so it is CXH2X-2O2 Then the carbon is 50%, and X=3 is obtained

If there is only one phosphazene BAI complex, it can be inferred that x=3 or 4If the first thing is to determine whether the general formula dao (pncl2)x is correct, 464 is not possible, because the amount of p at this time exceeds the amount before the reaction, if it is 3, it means that there are other products after the reaction, otherwise it cannot be explained.

But you're missing an important point. That is, there is more than one type of phosphazene compound in the product, which should be noticed when you know x=10 3. That is, there are two kinds of phosphazene compounds generated, one is x=3 and the other is x=4.

As for the relationship between them, it can be found very simply by the cross method, and we can know that generation (pncl2) 3 is and generation (pncl2) 4 is. Just the perfect explanation: the answer is "and" instead of "or". I hope my answer can help you.

N(Fes)=,N(HCl)=, according to the equation Fes+2H+=====Fe2++H2S, the excess of hydrochloric acid can be judged, so the calculation should be based on the amount of Fes substance.

1) Co-generation of H2S is.

The amount of substance dissolved in solution is:

mol/l×

l=mol so the amount of matter collected to H2S gas is.

mol = volume of H2S gas collected (standard condition) is:

l/mol×

mol = l (2 points).

2) Fe2+ is generated, and its concentration is:

The amount of H+ species before the H+ reaction is consumed is:

Then: The amount concentration of the substance of H+ in the solution is:

So ph=1

1.Two hydrogen peroxide molecules react to form 2 water molecules and 1 oxygen molecule.

2.Hydrogen peroxide reacts to produce water and oxygen.

3.The reactant is hydrogen peroxide, and the products are water and oxygen.

4.Their ratio of particles to particles is 2:2:1.

5.For every 68 parts of hydrogen peroxide, the reaction produces 36 parts of water and 32 parts of oxygen.

That's probably all there is to ......The teacher only talked about so many ......