Chemical problem, please explain the reason thanks

1 is D, mainly because the valency CR of this question changes from +3 valence to +6 valence, the valency of oxygen in SO4 and OH has not changed, only the valency of H2O2 oxygen has decreased, which reflects the oxidation.

2 is c, and the valence of oxygen rises and falls.

3 is A, which is a typical reaction of HCL-H2O2 from strong acid to weak acid, and the valence does not change.

Merely oxidation means that the valency of the elements of the substance is only reduced.

What can reflect both oxidation and reduction is that the valency of material elements is increased in part and decreased in part.

The weakly acidic is reflected as long as the aqueous solution of the salt formed with a strong alkali is alkaline, such as the salt NaA formed by the reaction of HA and sodium hydroxide, if the aqueous solution is alkaline, it means that A is hydrolyzed in the aqueous solution, which means that HA is a weak acid.

The reason is the above.,I don't write the answer if you know it (in fact, I'm not sure about my answer.,But I still understand the principle).

If there is anything else I don't know how to send a message to ask me.

The reaction that only reflects oxidation is d

It is c that can reflect both oxidation and reducibility

Weakly acidic is A

A reaction can be seen as a strong acid to weak acid H2O2, the oxidation and reduction of H2O2 is not reflected.

B reaction H2O2 is only reducible.

c. In the reaction, the valency of oxygen elements can both increase and decrease, which can reflect both oxidation and reduction.

The valency of oxygen in d only decreases.

Observe the change in valency

The valency of H2O2 is only reduced.

The valence of H2O2 increases and falls.

The valency of H2O2 remains unchanged.

Amount of %Na2S2O3 substance = mol

2 Na2S2O3 + I2 = Na2S4O6 + 2 Nai, indicating the formation of I2 with mol.

ca(clo)2 + 4 ki + 2 h2so4 = caso4 + k2so4 + 2 kcl + 2 i2 + 2 h2o

It is indicated that Ca(clo)2 with mol is involved in the reaction.

Mass = mol 143 g mol = g And because there is only g of original bleaching powder in 25 ml of Ca(ClO)2 solution, the mass fraction of Ca(ClO)2 in the bleaching powder is = %.

kclo3

Potassium chlorate KMNO4

Potassium permanganate. Multiply 180 by each percentage, that is, the mass of each element, and then divide it by its own atomic weight, which is the number of atoms.

h)180*

c)8/1=8

h)108/12=9

c) So C9H8O2

This method is relatively simple.

Potassium chlorate, potassium permanganate.

c：h：o=60/12：：

then the simplest form is.

Then the molecular formula should be C9H8O4

First of all, sodium percarbonate is a strong oxidizing agent.

Then, trichloroisocyanuric acid is also an extremely strong oxidizing agent, as well as a chlorinating agent.

Mixed together, the two are likely to produce the explosive gas nitrogen trichloride.

Trichloroisocyanuric acid should be stored in a cool, dry and well-ventilated warehouse to prevent moisture, waterproof, waterproof, fireproof, and isolate fire and heat sources.

It is forbidden to mix with flammable, explosive, spontaneous combustion and self-explosion and other substances.

It is absolutely forbidden to mix and store with liquid ammonia, ammonia, ammonium carbonate, ammonium sulfate, ammonium chloride urea and other inorganic salts and organic substances containing ammonia, ammonium and amines.

Otherwise, it is prone to ** or burning.

Do not come into contact with nonionic surfactants, otherwise it is flammable.

1. There must be ammonia, nitric oxide, and carbon dioxide; There must be no hydrogen or carbon dioxide.

Explanation: The original gas is a search gas, and there must be no nitrogen dioxide, because nitrogen dioxide is reddish-brown; The gas is reduced by the volume of concentrated sulfuric acid, indicating that it contains the alkaline gas ammonia, and there must be ammonia; Turning reddish-brown through sodium peroxide means that there must be nitric oxide, and nitric oxide is oxidized by sodium peroxide to reddish-brown nitrogen dioxide, so there must be nitric oxide; The gas passes into the test tube filled with water and the liquid level fills the test tube at the end, indicating that there is no hydrogen, because hydrogen will not react with the previous conditions, and is insoluble in water, and there is no gas left at the end, so there must be no hydrogen; The last phenomenon also shows that there must be carbon dioxide, which reacts with sodium peroxide to form oxygen, and oxygen dissolves in water together with the nitrogen dioxide generated, so there must be carbon dioxide;

2. Nitric oxide carbon dioxide 2:1

no+na2o2=no2+na2o 1no:1no22co2+2na2o2→o2+2na2co3 2co2:o24no2+o2+2h2o=4hno3 4no2：

O2 yields a volume ratio of 2:1 to two gases

Must ammonia, nitric oxide, carbon dioxide must not be hydrogen, carbon dioxide.

Description: To determine the raw gas of the gas must not be nitrogen dioxide, nitrogen dioxide is a reddish-brown color; Reducing gas (volume) concentrated sulfuric acid contains the alkaline gas ammonia, which must have ammonia; By sodium peroxide to reddish brown, nitric oxide, nitric oxide, nitrogen dioxide, sodium peroxide reddish-brown oxidation must be done in order to ensure nitric oxide; Gas is filled with water into a test tube, and the final liquid level is filled with the test tube, indicating that there is no hydrogen, because hydrogen does not react and under previous conditions, it is insoluble in water, and finally no gas and the other must not be hydrogen; The final phenomenon must also have carbon dioxide, carbon dioxide, and sodium peroxide reaction, oxygen, oxygen and dissolved in water to form nitrogen dioxide, so there must be carbon dioxide;

2. Nitric oxide carbon dioxide 2:1

NO+NaSO2 = NO2 + Na2O: 1NO22CO2 + 2NO2 NO2CO2:O24NO2 + O2+2H2O = 4NO3 4NO2:

O2 provides two gases with a volume ratio of 2:1

1 Nitric Oxide Nitrogen dioxide.

2 Nitric oxide hydrogen.

Because of the withering of re3, hard acid tends to combine with hard alkali, i.e., oxygen atoms, while carbon is soft acid.

The following are 7 coordination modes of co-unshot transition metals:

See question 12 for the method.