Please talk about these two chemistry questions

The first experimental design purpose deviates from nature. Sulfur dioxide gas is dissolved in water to form sulfurous acid, sulfurous acid has strong reducing property, bromine water has strong oxidation, the two react, bromine water fades, showing sulfur dioxide reducing property rather than bleaching. If you want to verify the bleaching properties of sulfur dioxide, you must pass sulfur dioxide gas into the magenta solution, and the fading of the magenta solution indicates that the sulfur dioxide gas is bleaching.

The second experiment also ignores the nature of the products produced in the reaction. According to the different reaction temperatures, the reaction between ethanol and concentrated sulfuric acid may not only produce ethylene gas, but also redox reaction may occur to generate sulfur dioxide gas, or ether vapor and other complex products, which have strong oxidation and can be reduced and faded by sulfur dioxide gas, ether vapor, ethylene gas and other substances, so the phenomenon of potassium permanganate solution cannot indicate that the product is ethylene gas. The method of testing ethylene gas is to inject a carbon tetrachloride solution of bromine, if the solution fades, it is ethylene gas, and sulfur dioxide gas can fade bromine water, but not the carbon tetrachloride solution of bromine.

The number of the beginning of the heavy head, because x is a negative divalent short-period element, x may be oxygen and sulfur, if x is oxygen, then y is sodium (both are neon electron shell structure), then z atomic number 36-8-11 = 17, is chlorine.

If x is sulfur, then y needs to be potassium, which is not a short period, thus.

xoxy, sodium z, chlorine.

The radius of sodium on the far left is obviously greater than that of chlorine, a false.

The leftmost sodium metallic of the same period is the strongest, b pair.

Oxygen is at the top of the same main group, c pair.

Perchloric acid is the most acidic in the same cycle, d pair.

SO2, but bromine water is more oxidizing than SO2 and will oxidize SO2.

2H2O+BR2+SO==2HBR+H2SO4 and thus the cause of the discoloration of bromine water is not the bleaching of SO2.

Ethanol volatilizes when coheated, and the volatile vapor also reacts with the acidic potassium permanganate

5c2h5oh + 4kmno4 + 6h2so4 ==== 5ch3cooh + 4mnso4 + 11h2o + 2k2so4

As a result, the color will fade regardless of whether there is an ethylene solution or not.

The short cycle is the first three cycles.

The first question is not clear, the second question is like this, Cl2 will react with Br to form hydrochloric acid and HBR, maybe HBro4 can't, in short, the second is because H2SO4 will produce SO2 and can also fade KMno4.

Select A, due to the excess of AgNO3, the AGI colloids adsorb AG+ and have a positive charge, so the precipitation ability of the electrolyte with the largest number of anion charges is the strongest, and the phosphate has the most three negative charges, so A is selected

Because of the excess of silver nitrate, silver iodide colloids adsorb silver ions ag and have a positive charge, and the electrolyte sodium phosphate contains the largest negative charge of PO4, so the ability to accumulate and settle silver iodide colloids is the largest. So choose A

1.The molecular formula is the number of c, h, and o atoms in the number. Note that there is carbon at the endpoints and inflection points in the bond type, and the number of bonds plus hydrogen is equal to four.

2. Hydrolysis under acidic conditions is two ester bonds. After hydrolysis, it is three molecules, and the same on both sides is acetic acid 3The fair side of the molecule is the two benzene rings and the connected carbon atom 4

The carbon-oxygen double bond on the ester bond cannot be added, so here only the benzene ring can be added. So it's 5mol. The two benzene rings are combined together, and the common parts can no longer be added.

5.It reacts with sodium hydroxide as a hydrolysis reaction. Sodium acetate and phenol are formed by hydrolysis under alkaline conditions, and phenol reacts with sodium hydroxide. So it's 4mol. The hydrolysis of the halogen atoms on the phenol ester and the benzene ring consumes 2 sodium hydroxide.

Quite simply, the benzene molecule is a planar structure, and the naphthalene molecule and the carbon on the methyl group have a total of 11 coplanarities;

The ester groups on both sides of K4 have a symmetrical structure, and there are two kinds of hydrolysates.

Only the double bond on the benzene ring can be added, that is, it can be added with 5mol of hydrogen;

After the ester group is hydrolyzed, the hydroxyl group on the benzene ring can also react with NaOH as phenol, so 2+2 gets 4.

The first one is not to be said, the question of counting; Compounds contain ester groups, which are hydrolyzed by acid to form acetic acid and compounds containing phenolic hydroxyl groups; The benzene ring and the atoms directly connected to it are coplanar, count; With hydrogen addition, do not consider the double bond in the ester group; In the reaction with sodium hydroxide, in addition to considering the hydrolysis of ester groups, why should the phenolic hydroxyl group in the product be considered to react with sodium hydroxide! Hope for good reviews, I'm tired to death!

1.The molecular formula is to look up the number of atoms;

2.The hydrolysate is phenol and two acetic acids;

3.two benzene rings of coplanar carbon atoms and 11 methyl carbons on them;

4.There is a 5mol double bond, 5molH2 is required;

Hydrolysis under the action, the product has two phenolic hydroxyl groups and two carboxyl groups, and 4 NaOH are required.