Summary of chemical equations. Substances are required to be classified according to Equation 10

Metal ions with positive valence, gas liquid ions with negative valence, landlords set it themselves.

Summarizing by oneself is good for oneself, and it is not good for learning chemistry by memorizing others' summaries.

Metal 1It can react with acids to form salts and hydrogen.

2.It is possible to replace the metal in the metal salt solution that is listed after it in the metal activity order table.

3.It can react with oxygen (or with non-metals).

4.Some reactive metals react with water.

Non-metallic 1Reacts with metals.

2.Reacts with non-metals.

Metal oxide 1React with water to form the corresponding alkali (except for insoluble alkali)2Reacts with acids.

3.Reacts with reducing agent.

Non-metallic oxides 1Reacts with water to form the corresponding acids (except insoluble acids)2Reacts with alkali to form salt and water.

General properties of acids:1Reaction with indicator 2Reaction with metals 3Reaction with metal oxides4Reaction with alkali 5Reacts with salt.

General properties of bases:1Reaction with indicator 2Reaction with non-metallic oxides3Reaction with acids4Reacts with salt.

Generality of salt:1Reacts with metals 2Reaction with acids3Reaction with alkali 4Reacts with salt.

Acid: Discolors the indicator (purple litmus solution).

Reaction with base: 3HCl+Al(OH)3==AlCl3+3H2O Reaction with metal oxide: Fe3O3+6HCl==2FeCl3+3H2O4

Reaction with salt: CaCO3+2HCl==CaCl2+H2O+CO2 5Reacts with metals.

Fe+2HCl==FeCl2+H2 Base: 1Discolor the indicator (purple litmus solution).

2.Reaction with acid: Al(OH)3+3HCl==AlCl3+3H2O is very much.

naoh+co2=na2co3+h2o

But this reaction is not a metathesis reaction.

In fact, the principle of this reaction consists of two counter-defibrillations.

The second reaction, the reaction between the base and the acid, is the metathesis reaction.

2. Correct. CuSO4·5H2O= =CuSO4+5H2O is the decomposition reaction of Yudong beam.

In addition to the pyrolysis of crystal water, there are.

Cu(NH3)4】SO4===CuSO4+4NH3 copper ammonium sulfate)Cu(HSO4)2===CuSO4+H2SO4Cu4(OH)6SO4 (basic copper sulfate)==3CuO+3H2O+CuSO4

3. It is heating, H2 + Cuo = heating = Cu + H2O reduction of copper oxide is high temperature.

Hello! The first one is not a meta-decomposition reaction, which does not meet the definition of "talking about fiber exchange Sun Dou" and the second one. It is a decomposition reaction.

The third is just heating.

Hydrogen is highly reducible.

It only represents a personal opinion, don't spray if you don't like it, thank you.

a)nacl,kcl,nabr,ki

Solution: First of all, the flame color reaction is used, and the above four substances are dipped in platinum wire and then burned on the flame of the alcohol lamp. NaCl or NABR is a yellow flame; KCL or Ki

Through the blue cobalt glass is a lilac flame; Then prepare the solution into a test tube, and add the agno3 solution dropwise. If there is a white precipitate, it is evidence of NaCl or KCL; If there is a yellowish precipitate, it is proven.

nabr；If there is a dark yellow precipitate, it is a sign of ki

Taken together, you can discern it.

b) MgSO4, CaSO4, MgCl2, CaCl2 First of all, CaSO4 is slightly soluble in water and can be identified as long as its substance is placed in water. Then, the remaining 3 substances were configured into 2 groups of solutions in 6 test tubes, and the BaNO3 solution was added dropwise in the first group, and MGSO4 could be identified if there was a white precipitate. In the second group, the identified excess MgSO4 solution was added to the remaining 2 test tubes, and if a white precipitate was produced, it was CaCl2, and if there was no phenomenon, it was MgCl2

The group can be identified by flame color reaction: dip it with copper and burn it on an alcohol lamp, yellow is sodium chloride or sodium bromide, and purple is potassium chloride; In the drop of silver nitrate solution, there is a white precipitate for chloride.

1.Add AgNO3 to generate white, white, Agbr light yellow NABR and AGI yellow precipitated NAI, respectively.

Then remove a small amount of the solution that generates the white precipitate, add HClo4 dropwise, and the white precipitate is KCL, and the rest is NaCl;

2.Ba(NO3)2 was added respectively, and the white precipitate was sulfate, and the hydrochloride had no obvious change, and then a few drops of NaOH were added to each of the two groups to produce the white precipitate of MG salt.

First of all, the flame color reaction, divide them into two groups, one is NaCl, NaBR (directly ** yellow), the other group is KCL, ,Ki (purple through blue cobalt glass), and then group dropwise silver nitrate solution, chloride ions are white precipitates, bromine ions are light yellow precipitates, and iodine ions are yellow precipitates!!

First, a small amount of sodium hydroxide solution was added, calcium ions and magnesium ions were separated, and a group of MGSO4 and MGCL2 (magnesium hydroxide white precipitate) was divided.

CaSO4, CaCl2 group, continue to add barium chloride solution dropwise to separate chloride ions and sulfate ions (barium sulfate white precipitate)!

1. C+2CuO==2Cu+CO2 (gas symbol).

2. (laboratory-made Alobe Cl2) 4HCl (concentrated imitation of Dan) + MNO2==MnCl2 + Cl2 (gas symbol) + 2H2O (heating under disturbance conditions).

H2O2==2H2O+O2 (gas symbol) is MnO2 as a catalyst (occurs on O).

a) NaCl, KCL, Nabr, Ki Chlorine plus CCL4 is brownish-red is Nabr purple-red, Ki

The other two are reacted with flame color, the yellow one is NaCl, the purple one is KCl, B) MgSO4, CaSO4, MgCl2, CaCl2 plus BaCl2, the white precipitate is MgSO4, CaSO4 plus a small amount of alkali, and the one that has the precipitate immediately is MgSO4, and the other two are also distinguished in this way.

There are too many mistakes upstairs.

A) The group first took a small amount and dissolved it in water in the test tube, and added nitric acid acidified AGNO3, NaCl, and KCl, all of which produced white precipitates, Nabr produced pale yellow precipitates, and Ki produced yellow precipitates.

Then the two solutions that produce white precipitate are used for flame color reaction, and the one that passes through the blue cobalt glass is purple is KCL

b) The group took a small amount in the test tube, added water, and the insoluble was CaSO4; The rest were added with BaCl2, and the white precipitate was MgSO4,; The remaining two add Na2SO4, and the white precipitate is CaCl2, and the non-symptomatic is MgCl2

It's all metathesis reactions, and at the junior high school level, I won't write about it.