High School Chemistry Please talk about item B, high school chemistry, how to find item B of this qu

The equation is 3S+6OH- *****SO3 2-+2S2-+3H2O

If you look at the equation, 1mols is oxidized, and 4mol of s is generated, and 4mol of electrons is lost, so 2mol of s2- is generated

3s+6oh- *****so3 2-+2s2-+3h2o

Because 2 S are oxidized and 4 electrons are transferred to S2-, if 1 mols, 2 molS2-

There is already a sulfur element on the left side of the equation, such as putting sulfur ions or sulfite ions on the left side of the equation and the other on the right side of the equation cannot conform to the redox rule, so the reaction is a disproportionation reaction, and the sulfide ions and sulfite ions are on the right side of the equation.

Water has been formed on the right side of the equation, so there should be hydroxide on the left side of the equation.

Trim, the equation should be 3s + 6oh- = 2s2- +so32- +3h2o

3 sulfurs, one is oxidized to +4 valence is sulfite ions, and two are reduced to -2 valence, resulting in 2 sulfur ions (principle of conservation of electrons gained and lost).

The equation is 3S+6OH =2S2 +SO32 +3H2O, see that two of the first three S of the equation are reduced and one is oxidized, so when one mole is oxidized, two moles of sulfide ions are generated, b is correct.

b is wrong, and the test point of this question is that the ratio of the measurement number of the cracks is equal to the rate ratio.

<>Hey, Kaiming, I'm sorry to stare at the eggplant, and I don't seem to be right.

The brightness attraction decreases, that is, the current decreases and the conductive particles in the solution decrease.

If it is a, the first reaction is to produce sodium sulfate and water, and the cation has always been only sodium ion, and its number is the same, but the amount of water increases, so the reading will decrease. As for whether it will become larger later, it should be related to the concentration of the dropping solution, and if the concentration is very thin, it will never become larger.

b It is impossible to have the process of first becoming smaller and then larger, either decreasing all the time, or increasing all the time, because the number of several ions in the solution will not change, if the concentration of the dropping solution is larger than the original solution, it will increase, and vice versa.

This is determined by the conductivity of the solution.

The acid produced in a is a strong electrolyte, but the volume of the solution continues to increase, so the reading of the sensitive spring scale first decreases, and then gradually increases after the reaction.

B does not react, so it is still a strong electrolyte solution, and the conductivity does not change much.

C and A are similar.

d consumes a weak electrolyte to form a strong electrolyte (ammonium acetate), so the conductivity increases, and finally the conductivity weakens, so the reading of the sensitive spring scale first becomes larger, and then gradually decreases.

b.Keeping the temperature and container pressure unchanged, filling 1molSO3(g) with extreme assumption analysis is actually equivalent to adding 2molSO2 and 1 molO2, while the temperature and pressure remain unchanged, which is equivalent to doubling the size of the mentions, so the concentration does not change, so the equilibrium does not change.

b.Keep the temperature and container pressure constant, and fill 1molSO3 (g) because the container pressure remains unchanged, then after filling SO3, it is actually an equivalent equilibrium!

2molSO3 and 2molSO2 and 1 mol O2 are equivalent!

And 1molSO3 and 2molSO3 are actually only different in volume under isobaric conditions, and the final balance obtained is also equivalent!

The filling of 1molSO3 (g) is analyzed by the extreme assumption method, which is actually equivalent to adding 1molSO2 and mol O2, while the temperature and pressure remain unchanged, which is equivalent to doubling the mentioned expansion, so the concentration does not change, so the equilibrium remains unchanged.

a On the basis of **, the different boiling points of gasoline and kerosene are used to separate b Alkanes with 4 or less carbon numbers are gaseous at room temperature, so the process is a cracking process, and the temperature requirements are more demanding.

c is the purpose of cracking.

d Reforming of alkanes.

Cracking --- breaks chain hydrocarbons into smaller gaseous hydrocarbons, generally to obtain ethylene.

So b belongs to cleavage.

A. OH- is formed by ionization and hydrolysis of water molecules, acetic acid and hypochlorous acid, respectively. However, acetic acid is more acidic than hypochlorous acid, so the degree of hydrolysis of hypochlorous acid is relatively large, and the concentration of hypochlorous acid and ions will be lower. Concentration C from large to small (Na+>CH3COO->CLO->OH->H+).

B, according to the principle of strong acid to weak acid, carbonic acid is more acidic than hypochlorous acid (the first-order ionization constant of carbonic acid is greater than hypochlorous acid), and a small amount of CO2 is introduced into the solution, which is equivalent to the formation of carbonate H2CO3, and carbonic acid can acidify CIO- and form HCL.

Hello, KA2 is the second step of the reaction HCO3- dissociates an ionization equilibrium constant of H+, the lower this constant value, the weaker its ionization ability, that is, the stronger the non-separation ability of CO3 ions and H+ to form HCO3-, compared with the ionization equilibrium constant in option B, it can be seen that CO3 ions cannot be generated.

Hypochlorous acid is weaker acidic than carbonic acid.

ka= hypochlorous acid, ka= carbonic acid. It can be shown that the acidity of carbonate is stronger than that of hypochlorous acid, and weak acid cannot make strong acid.

The KA of hypochlorous acid is larger than the secondary ionization constant of carbonic acid, so the acidification of hypochlorous acid by carbonation can only stay in the stage of bicarbonate, and bicarbonate cannot continue to acidify hypochlorite.

Sodium bicarbonate should be generated.

Binding proton capability.

co3 2->clo->hco3-

naclo+co2+h2o=nahco3+hclo

Sodium atoms are metallic bonds between them, and the force is high. Sulfur is an intermolecular force that is less important. So sodium has a higher boiling point than sulfur.

However, the melting point is different from the boiling point, and the melting point of sodium is smaller than that of sulfur, because the interatomic force of metal crystals is greater than that of molecular crystals, and the melting only breaks about 5% of the metal bonds.