There is a packet of powder, which may be one or several of the components of NaCl, Na2CO3, FeCl3, N

White powder dissolved in water, colorless transparent solution. Instructions do not have fecl3 because it has color.

Add a sufficient amount of BaCl2 solution to generate a white precipitate. Description: May contain Na2CO3 or Na2SO4. BACO3 and BASO4 are generated

Then it definitely contains Na2SO4 and definitely does not contain Na2CO3

After filtration, hydrochloric acid is added to the resulting precipitate, and the precipitate is insoluble. It shows that there is no BaCO3 and only dissolves in HCl.

AgNO3 solution was added to the filtrate to produce a white precipitate, and then a sufficient amount of dilute nitric acid was added to make the precipitate insoluble. It cannot be stated that NaCl is contained because step 1 adds BaCl2 and provides Cl-.

If NaCl is not certain, and to test the presence of a substance that cannot be determined, the reagent that should be added to Procedure 1 of the above experiment is BA(NO3)2. This will not provide cl-

1.Colorless solution, exclude ferric chloride, generally iron ions, copper ions, etc. are colored in the solution, add barium chloride, there is a white precipitate, it may be barium carbonate or barium sulfate, and sodium chloride can not be excluded at the same time.

2.The precipitation with hydrochloric acid does not dissolve, so it must not be barium carbonate, and barium carbonate meets hydrochloric acid to produce carbon dioxide gas, so it must contain Na2SO4

3.Silver nitrate is added to the filtrate, and the resulting white precipitate is insoluble in nitric acid, which proves that the silver chloride precipitate is generated (the insoluble precipitate in nitric acid is silver chloride and barium sulfate, and a sufficient amount of barium chloride is added in step 1, so there are no sulfate ions in the filtrate), the drug must contain NaCl, so the powder must contain Na2SO4 and NaCl

Typing is tiring and I hope to give points.

There are Na2SO4 and NaCl, what is not sure is Na2CO3, you can add acetic acid in the step to see if there are any bubbles, if there is, it contains Na2CO3, and vice versa, there is none. It definitely does not contain FeCl2, it is colored.

No, the first step is to add clarified lime water, because it is impossible to judge whether it is carbonated again, there is no green iron and green iron, because the addition of silver acid does not dissolve.

From 1, it is known that it contains one of sodium carbonate and sodium sulfate, and it certainly does not contain FeCl3 (the solution is yellow) combined2, and there is definitely no sodium carbonate. Definitely contains Na2SO4A large amount of Cl ions are introduced in the first and second steps, so it is no longer possible to determine whether NaCl is present in the third part of the white precipitate

1. The white powder is dissolved in water, first add barium nitrate, 2 add dilute nitric acid. In this way, if there is a white precipitate in 3, it contains sodium chloride, otherwise it does not contain it.

1) From the experiment, it can be seen that there must be no FeCl3 in the original powder, which can indicate that there is no MgCl2 in the original powder

Experiments have proved that there is no Na2CO3 in the original powder

2) According to the above experiments, it can be concluded that the substances that may be present in the original powder are NaCl and Na2SO4.

3) From the experiment, it can be seen that there must be NaCl in the original powder

From the experiment, it can be seen that there must be no Na2SO4 in the original powder

Try to write out the chemical formulas of the solutes in solutions A and B used in the two experiments: A Agno3 and B BaCl2

(1) Because it is completely soluble in water and the solution is colorless. The precipitate disappeared after the addition of hydrochloric acid) (2) Na2CO3

3)nacl

4) Take 2g of powder again. Add water to dissolve. Drop in excess dilute nitric acid until no more bubbles are produced.

Add a few drops of agno3 dropwise at this pointIf there is a precipitation generated. It means that the possible substance NaCl does exist.

If there is no obvious phenomenon. then there is no NACL

1.Colorless, so copper sulfate excluded (blue), solution, so calcium carbonate excluded (insoluble)2Barium ions plus carbonate ions or sulfate ions precipitate white precipitates, and copper sulfate is excluded, so sodium carbonate and sodium sulfate are at least one.

3.Barium sulfate is insoluble in water and insoluble in acid precipitate, because the hydrochloric acid precipitate disappears, so there is no so, copper sulfate calcium carbonate sodium sulfate must not be, sodium carbonate must have, sodium chloride maybe, test sodium chloride with hydrochloric acid acidification of silver nitrate solution, there is a white precipitate indicating that there is a part of this, the ion equation can be written by yourself, do not forget to mark the precipitation and gas symbols.

(1)caco3

na2so4

cuso42)na2co3

3) NaCl (4) take a small amount of test solution in the test tube, add nitrate silver solution dropwise to see if there is precipitation, and sodium chloride is produced. Chloride ions and silver ions react to form a precipitate (can't be beaten, mobile phone).

The flame color reacts, and the purple type is seen through the blue blue cobalt glass.

The white powder is dissolved in water to obtain a colorless solution. Note: Without fecl3, mountain code and k2so4, bacl2 can only have less than one.

NaOH solution was added to the resulting colorless solution and heated, only white precipitate A was observed, indicating that there was no NH4NO3 because no gas was produced, and at the same time, because there was only one precipitate, BaCl2 would not have a precipitate if there was no apparent precipitation, so there was BaCl2 (NaOH and NaHCO3 reacted with carbonate and BaCl2) and at the same time showed that there was no K2SO4

Conclusion: Only KCL cannot be determined.

The solution is colorless, indicating that there is no fecl3, adding sodium hydroxide only white precipitation means that there is no ammonium nitrate, and the white precipitation should be carbon Zen carrying barium acid, so there must be sodium bicarbonate and barium chloride in the end of the original powder slap socks, so there must be no potassium sulfate because there is barium. Because there are chloride ions in barium chloride, the presence of potassium chloride is uncertain, and the high school stage can only use the flame color reaction to determine whether there is potassium, and the purple flame seen through the cobalt glass indicates that there is potassium chloride.

1) From the experiment, it can be seen that there must be no FeCl3 in the original powder, which can indicate that there is no MgCl2 in the original powderExperiments have proved that there is no Na2CO3 in the original powder

2) According to the above experiments, it can be concluded that the substances that may be present in the original powder are NaCl and Na2SO4. If you need to further determine whether the original powder is pure or a mixture, the experiments you are ready to proceed with are:

Add Ba(NO3)2, if there is a white precipitate generated, there is Na2SO4, filter, take the filtrate and add AgNo3, if there is a white precipitate generated, there is NaCl

If there is no Na2SO4, add AgNO3, and if there is a white precipitate generated, then there is NaCl

First of all, student A put the label of CuSO4 on the reagent bottle of the light blue solution, and then took out a little of this light blue solution in a small test tube and then added another solution, and found that there was no phenomenon, so he immediately put the label of NaCl on the reagent bottle of this solution; The remaining two vials of A were experimented with the flame color reaction, and A attached the label of koh to the bottle that observed the flame reaction purple through the blue cobalt glass, and the last bottle of nails was labeled with the last remaining piece of Naoh.

(1)①fecl3

Because the ferric chloride solution is light yellow, take a small amount of powder and add water to dissolve it to obtain a colorless transparent solution, so the solution does not contain ferric chloride.

MgCl2 because magnesium chloride reacts with sodium hydroxide solution to form magnesium hydroxide white precipitate, take a little solution and add sodium hydroxide solution a little, no obvious phenomenon can be seen, so the solution does not contain magnesium chloride.

na2co3

Because sodium carbonate solution can react with dilute hydrochloric acid to form gas, it does not contain sodium carbonate due to the obvious phenomenon of hydrochloric acid reaction.

2) Sodium chloride, sodium sulfate.

Take a small sample and put it into a dry and clean test tube, then add a small amount of water to the test tube to prepare the solution, and then add a sufficient amount of barium nitrate solution to the solution, if there is a white precipitate generated, it means that the sample contains sodium sulfate; If there is no white precipitate, it means that the sample does not contain sodium sulfate, and it must contain sodium chloride;

If there is a white precipitate, filter after full reaction, add silver nitrate to the filtrate, if there is a white precipitate, the sample contains sodium chloride, and the sample is a mixture of sodium chloride and sodium sulfate; If no white precipitate is generated, the sample contains no sodium chloride and only sodium sulfate;

b) cuso4

Because the copper sulfate solution is light blue, the rest of the solution is colorless;

NaClBecause NaCl does not react with copper sulfate solution, sodium hydroxide and potassium hydroxide solution can react with copper sulfate solution to produce blue precipitate;

KOH because the flame color reaction experiment was observed through the blue cobalt glass, and the purple flame color reaction was observed through the blue cobalt glass, indicating that the solution contained K+, so this solution was KOH solution;

naoh

(1) From the experiment, it can be seen that there must be no MgCl2 in the original powder; (Because only magnesium carbonate is insoluble in water).

Experiments can show that there is no FeCl3 in the original powder; (Because only Fe3+ reacts with OH- to form a precipitate).

Experiments have proved that there is no Na2CO3 in the original powder(Because only CO3- reacts with H+ to form CO2 gas).

2) According to the above experiments, it can be concluded that the substances that may be present in the original powder are NaCl and Na2SO4.

To determine whether the original powder is a pure substance or a mixture, proceed with the following experiments:

take a little of the obtained solution, add solution A dropwise, and a white precipitate is generated; (The most typical precipitations in high school chemistry are silver chloride and barium sulfate.) Therefore, here we can infer that the added solution A is AGNO3, which generates a silver chloride precipitate. Why not barium sulfate?

Continue to add solution A dropwise until there is no white precipitate generated, let it stand, take a little supernatant and add solution B dropwise, and there is no obvious phenomenon. (Continue to add solution A dropwise until no white precipitate is formed, indicating that AG- has completed the reaction of Cl- in the solution.) At this time, we can add BaCl2 dropwise to check whether there is a white precipitate of BaSo4.

There is no obvious phenomenon, indicating that there is no Na2SO4 in the original powder. The inference here is in line with the meaning of (3) below, and if agno3 and na2so4 are interchangeable, it is not in line with (3).

3) From the experiment, it can be seen that there must be NaCl in the original powderFrom the experiment, it can be seen that there must be no Na2SO4 in the original powder

Try to write out the chemical formulas of the solutes in solutions A and B used in the two experiments: A Agno3 and B BaCl2

There must be : sodium carbonate and barium chloride.

There must be no : copper sulphate and sodium sulphate.

Maybe: None of the possibly have.

If the upper layer is colorless, then there is no CuSO4 precipitate and all dissolved, then the precipitate is BaCO3 and there is no Na2SO4

Is it wrong to add dilute nitric acid in the (3rd) condition of your silver old land, it is to add dilute sulfuric acid.

1) The medium precipitate is BaCO3

2) BaCO3 dissolves to produce CO2 bubbles.

3) BaSO4 precipitation.

4) AGCL precipitation.

The white powder is composed of a certain front, with Na2CO3 and BaCl2, and must contain no K2SO4

There may be BA(NO3)2

na2co3 +bacl2 ==baco3 ↓ 2naclnacl +agno3 ==agcl↓ +nano3

It's dilute nitric acid in the title, yes.

(1) Because it is completely soluble in water and the solution is colorless. After adding hydrochloric acid, the precipitate disappeared) (2) Na2CO3 (3) NaCl (4) and re-take 2g of powder. Add water to dissolve.

Drop in excess dilute nitric acid until no more bubbles are produced. Add a few drops of agno3 dropwise at this pointIf there is a precipitation generated.

It means that the possible substance NaCl does exist. If there is no obvious phenomenon. then there is no NACL

Colorless indicates that there is no CuSO4 because Cu ions dissolve in water and become blue; There is a white precipitate with the addition of BACL2, and the white precipitate is BaCO3 or Baso4, but the precipitate with HCl disappears, indicating that the precipitate is not a BaCO3 ion, because the carbonate ion and hydrochloric acid will react to form carbon dioxide and water, so there are no carbonate ions in the solution, that is, there is no sodium carbonate and calcium carbonate; There must be NaSO4, and NaCl can be present or not. So the answer is: (1) CuSO4, CaCO3, BaCO3 (2) BaSO4(3) NaCl