Urgent need for chemistry reasoning questions for junior high school, chemistry reasoning questions

The above is not from the first semester of the third year of junior high school, I haven't even studied it!

Colorless gases are carbon dioxide and a filtered (separation of solids from liquids).

There must be sodium carbonate (which produces the substance that precipitates in lime water, "you may learn later that sulfur dioxide is fine, but now it's just carbon dioxide").

There must be sodium sulfate (barium chloride is added to the colorless solution to produce precipitation, not chloride, "hydrochloric acid has chloride ions before", the first thing that comes to mind is barium sulfate, only sodium sulfate has sulfate).

There is not necessarily sodium chloride (although the addition of silver nitrate solution produces a precipitate "silver chloride", but hydrochloric acid has been previously added to contain chloride ions).

There is not necessarily potassium nitrate (no reaction in the total process).

Sodium chloride can only be detected Chloride ions Add silver nitrate to a colorless solution with a precipitate (silver chloride) indicating the presence of sodium chloride.

Nitrate There will be no precipitation Test Potassium nitrate I'll think again.

Carbon dioxide standing must contain sodium sulfate, sodium carbonate, possibly sodium chloride, add silver nitrate solution, dilute nitric acid to the colorless solution, and if there is a precipitate then it contains sodium chloride.

CO2 filtration.

Sodium carbonate, sodium sulfate, sodium chloride potassium nitrate.

CO2, static filtration,

Sodium carbonate, sodium sulfate, sodium chloride, add silver nitrate solution to the colorless solution and add dilute nitric acid dropwise, if there is a precipitate, then it contains sodium chloride.

Analysis: A and B are oxides, D and E are elemental, C is a gas that can supply breathing, and the reaction of D to E has the formation of light green liquid; then c is oxygen, a and b are oxides, calcium oxide or carbon dioxide, water, respectively; D and E are elemental iron, hydrogen, respectively; Substitute the conclusion into the original question for verification, which is in line with the topic.

Answer: b is calcium oxide (or carbon dioxide);

The chemical equation for the reaction of A and B is CaO+H2O=Ca(OH)2 (or CO2+H2O=H2CO3);

e may be h2

Question 2: Substance D is calcium carbonate, which is assumed by the known hypothetical method: a, calcium hydroxide, b, sodium hydroxide, c, sodium carbonate, and e hydrochloric acid; Therefore, the chemical equation of the reaction between D and E is CaCO3+2HCl, CaCl2+H2O+CO2; The chemical equation for converting a to b is ca(oh)2 + na2CO3 = CaCO3 + 2NaOH

The chemical formula of C is O2

The chemical equation for the reaction of A and B is CaO+H2O=Ca(OH)2 (or CO2+H2O=H2CO3);

e may be h2

The chemical formula of d is CaCO3, and the chemical equation of the reaction between D and E is CaCO3+2HCl, CaCl2+H2O+CO2

The chemical equation for converting a to b is ca(oh)2 + na2CO3 = CaCO3 + 2NaOH

I'm not sure if the push was right.

A Co A H2 B Fe C C Since B and C are both metals, then B and C are displacement reactions, the latest thought of iron and copper, and A can go to B and C, then A is also a reducing agent, think of Co, the same reason A and D are also reducing agents, then A is H2, D is C.

AH2 B Al Propylene Fe Butyl C

A Co to C H2 reduces Fe2O3

B to A2AL + 2NaOH +2H2O =2Naalo2 + 3H2

B to C 2AL + Fe2O3 = 2Fe + Al2O3 D to Propylene Carbon reduces Fe2O3

A to C Co reduces Fe2O3

Ding to A2C+O2=2Co

A to B CO reduces Al2O3

I don't know if it's right. It doesn't seem to be the third year of chemistry anymore.

I'm a junior in high school, hehe. My train of thought is this, first of all, the common elements in high school are chlorine, oxygen, hydrogen, C, fluorine, Na, Al, silicon, sulfur, Fe, Cu. First of all, the breakthrough point of this question is ethylene propylene, metal for metal, we must think of al heat and the replacement of metal and salt, okay.

Secondly, a, Ding A can react to form C, then C must be Cu,In this way, it is clear that B is Fe, and Fe is exchanged for hydrogen, so A is hydrogen. Then if the non-metal is changed to Cu, there should be ammonia, c, co, h2

So the question is solved. Do you have to ask me, my QQ is 867259407Hehe.

A is H2, B is Fe, C is Cu, D is C, A is Co

Carbon dioxide.

Copper magnesium, sodium carbonate, sodium sulfate, copper oxide.

Sodium ion Ca(OH)2 + CO2 = CaCO3 + H2Ops The filter residue may be magnesium reducing copper ions to copper.

1.The gas produced b is: CO2 2Filter residue c is:

3.In mixture A, the substances that are definitely present are: Na2CO3, Na2SO4

4.In filtrate D, the metal cation that is definitely present is: Na+

5.Write the chemical equation for the chemical reaction that occurs during the experiment

co2+ca（oh)2==caco3↓+h2o___

These substances are soluble in hydrochloric acid, there is no filter residue, if you have any questions, you can ask me, unless the hydrochloric acid is not enough, otherwise Shenma is a floating cloud.

1.Carbon dioxide.

2.No. 4.Sodium ions.

One of the most important things I do when I do the inference questions is to analyze the questions first. For example, according to the reaction, we can know that there must be carbonate in A, so there must be Na2CO3 in A. According to the reaction, the filtrate is known to be Na2SO4

To do the inference problem with drawings, it is best to use a pen and draw a flowchart or MG on paper. Ion.

Ca(OH)2 == CaCO3 H2O must be clear about the displacement reaction, metathesis reaction conditions, and which salts are soluble. This is the premise and important condition for making acid and alkali salts.

If there is anything inappropriate, please criticize.

Gas carbon dioxide.

The filter residue is copper carbonate.

There are definitely sodium carbonate, copper oxide, sodium sulfate.

Sodium ion CO2 + Ca(OH)2 = CaCO3 + H2O

A must have dilute hydrochloric acid, (only dilute hydrochloric acid is acidic among the four substances) B must have soda ash and may have caustic soda (there can be no dilute hydrochloric acid and barium chloride with soda ash, if there is, there will be gas or precipitation, and it is impossible to coexist).

BaCl2 goes out of which soda ash.

Phenolphthalein does not change color.

(1) dilute hydrochloric acid (because if it does not contain dilute hydrochloric acid, the solution will not be acidic: only sodium chloride may be present in the solution - neutral, caustic soda, sodium carbonate or sodium bicarbonate are also alkaline; In summary, it can only be dilute hydrochloric acid).

2) Group B inferred that there must be soda ash (sodium carbonate), and there may be caustic soda (sodium hydroxide) Na2CO3 + 2HCl = 2NaCl + H2O + CO2 Because there is sodium carbonate and barium ions, precipitation will occur.

ba2+ +co32- =baco3

Add excess barium chloride, BaCl2 + Na2CO3 = BaCO3 + 2NaCl, remove soda ash (or carbonate ions, see requirements).

The phenomenon is that phenolphthalein does not change its red color, indicating that there is no caustic soda.

Caustic soda is NaOH and soda ash is Na2CO3

Group A inferred that there must be dilute hydrochloric acid.

Group B inferred that there must have been pure barium, possibly barium chloride, and excessive CaCl2 was added to NaOH to remove caustic soda.

Colorless (Note: There is a problem with this question, if there is HCL in the waste liquid, it is impossible to have soda ash and caustic soda).

pH<7 Acidic There must be acid HCl

If there are bubbles, it means that there is gas formation Only NaCO3 in the waste liquid will react with HCl to form CO2 gas. So there must be naco3

There may be BACL2

Add (Na)2SO4 to remove BaCl2

If it doesn't change color, there must be no NAOH

group A is dilute hydrochloric acid; Group B must have soda ash, possibly barium chloride and caustic soda.

Add excessive salts containing sulfate but cannot react with sodium hydroxide, such as sodium sulfate, potassium sulfate, etc.

Remove the barium chloride from it.

The phenomenon is that the solution is colorless, indicating that there must be no such substance.

pH 7, then group A inferred that there must be a solute——— dilute hydrochloric acid, group B added hydrochloric acid dropwise to the waste liquid and found that there were bubbles, then group B inferred that there must be ——— soda ash in the waste liquid.

There may be ——— caustic soda.

In order to prove the possible ingredients, students in Group B are asked to complete the following blanks.

Take a small amount of waste liquid and add excess ——— barium chloride to remove the ——— soda ash and then add phenolphthalein to the waste liquid, the phenomenon is ——— colorless, indicating that there must be no such substance.

From the experiment of group A, it can be obtained: the pH of the waste liquid is <7, so there must be dilute hydrochloric acid (HCl) in it.

From the experiment of group B, it can be obtained: soda ash reacts with dilute hydrochloric acid and produces bubbles, so there must be CaCO3, and because the metathesis reaction of barium chloride and sodium carbonate will form precipitate, it cannot coexist stably in large quantities, so there may be caustic soda (NaOH).

Sodium carbonate can be removed by first adding excess barium chloride.

The phenomenon is still colorless.

Solution, group A deduced that there must be HCL and NaCl, and group B inferred that there must be Na2CO3, NaCl, and may have NaOH in the waste liquidA small amount of waste liquid was added to excess barium chloride to remove Na2CO3The phenomenon is not discolored.

pH<7 indicates that the solution is acidic, and there must be HCL in group A, and sodium carbonate in group B with hydrochloric acid.

Add excess calcium chloride to make the sodium carbonate completely turn into a precipitate, and then add phenolphthalein if it is red, indicating that there is sodium hydroxide.

Group A inferred that there must be dilute hydrochloric acid, because only dilute hydrochloric acid pH 7 Group B inferred that there must be soda ash (sodium carbonate), Na2CO3 + 2HCl = 2NaCl + H2O + CO2

There may be caustic soda (sodium hydroxide).

Add excess barium chloride, BaCl2 + Na2CO3 = BaCO3 + 2NaCl, remove soda ash (or carbonate ions, see requirements).

The phenomenon is that phenolphthalein does not change its red color, indicating that there is no caustic soda.

Today's homework.

I don't know if it's right.

A must have HCL NACL

B must have Na2CO3 NaCl and may have NaOH plus BaCl2 to remove Na2CO3

The solution does not turn red.

Because NaCl is formed as long as there is a reaction during the experiment, NaCl must be present in both groups.

The rest is the same as everyone else would have inferred.

(1) HCl is acidic.

2) Soda ash Caustic soda hydrochloric acid reacts with soda ash.

3）h2so4;Barium chloride Barium sulfate is precipitated (4) colorless; No caustic soda phenolphthalein does not change color.

There must be sodium carbonate and there may be barium chloride and sodium hydroxide.

A: dilute hydrochloric acid B: There must be soda ash and there may be caustic soda.

1 BaCl2 soda ash.

2 Phenolphthalein does not change color.

pH 7 solution is definitely acidic and must not be alkaline. There must be hydrochloric acid.

B has air bubbles, indicating that there is sodium carbonate, and there may be sodium bicarbonate.

HCL (ask about solutes, not dilute hydrochloric acid); soda ash, caustic soda; calcium chloride, soda ash (or carbonate ions); There was no significant change in the solution.