Chemistry questions ask for a master to come to the rescue!!

If you choose B, it is easy to choose D by mistake

You should understand according to the conditions given in the question.

Rather than pre-existing knowledge.

If you choose A, it can be seen from the 3 conditions that it will not be. It's not a big problem.

If C is chosen, the magnesium and water react under heating conditions, so even if the reaction is with the H ion of the acid instead of water, as for whether it reacts with the H ion of the acid depends on the explanation of D.

If you choose D, you have to know how the H ions in bromine water come from, which are ionized by HBR, and although HBR is a strong acid, the solubility of BR2 is small, so there are very few HBRs generated, so there are very few H ions. Obviously not right.

Therefore, choose B. Here to correct the mistake of the "12th class discipline inspection committee", may I ask if BR2 has a high solubility in water, if it is not big, where can there be so many h ions, take a good look at the book. In addition, you are as stupid as others as you are, who doesn't know that HBR is a strong acid.

Class 12 Discipline Inspection Committee, you should do more exercises.

d correct. 1) Emphasis on cold water while others without emphasis on temperature, suggesting that it may react with hot water.

2) produces gas, most likely hydrogen, and 3) does not react with liquid bromine (or pure bromine).

4) The color fades quickly, indicating that the reaction is related to the concentration of bromine.

The essence of the reaction is the reaction of magnesium with the acid in bromine water, and the formation of hydrobromic acid and hypobromic acid by bromine and water with bromine, etc.

A false reaction is hydrogen ions and not bromine.

B False: Water is not usually used as a catalyst.

c False: The hydrogen ions in the reactants are mainly from acids; But strictly speaking, it can be right, because the hydrogen ions in the system are constantly exchanged, and in the final analysis, they come from water.

Am you mistaken?! Of course, choose D!!

A and C have been explained to the point that they are beyond reproach, and I will not dwell on them.

First of all, to correct one mistake of the Kairs: HBR is a complete strong acid! (BR is the same family as CL, HCL is a weak acid?) ）

Then, from the "gas discharged" in (2), it can be deduced that it is hydrogen, so it can be deduced as a result of the reaction between mg and H+.

And by the equation:

BR2 + H2O == (reversible here) HBR + HBRO 1

mg + 2h+ ==mg2+ +h2 『2』

It can be seen that when Mg consumes H+, the equilibrium in 1 shifts to the right, so the concentration of Br2 decreases, so the phenomenon of problems occurs!

In B, if H2O is a catalyst, it should remain unchanged before and after the reaction. And the fact is that the hydrogen produced can only come from water, so b is incorrect.

I'm also a 07 candidate, so I'll give me a lot of advice

BBR2 reacts with H2O to change the balance of water.

Is your answer to this question written incorrectly, it should be a (2a+2b).

You don't have to think about this reaction at all, because the question asks about the ratio of atoms of sulfur to oxygen, not the ratio of molecules. The reaction may change the number of molecules, but the atoms remain constant, so only the ratio of atoms introduced at the beginning needs to be considered.

AMOL SO2 and BMOL O2 contain sulfur atom A mol and oxygen atom 2A+2B mol, so the answer is A (2A+2B).

1.Let the relative molecular mass of saturated chain hydrocarbon A be m, then the average relative molecular mass of the gas mixture formed by complete cracking under certain conditions is m 2, and because the ratio of relative density is the ratio of relative molecular weight, therefore.

m/2:2=

The solution is m=58

From the general formula of saturated chain hydrocarbons (cnh2n+2), it can be seen that 12n+2n+2=58 is solved to obtain n=4, so the molecular formula of saturated chain hydrocarbon a is c4h10

2.Let b be a CMH (2M+2) and M an integer.

Because C is an olefin, C contains at least 2 carbon atoms, so 0cmH(2M+2)+(3M+1) 2O2=MCO2+(M+1)H2O Because the amount of the substance that requires oxygen is an integer multiple of it, that is, (3M+1) 2 is an integer, so M=1, B is CH4, which can be known from the conservation of atoms, C is C3H6, so C is propylene. bar.

The average molar mass of the gas mixture is 29, and only methane is below 29 in the alkanes, so alkane b is methane, and then the molar mass of c is 42, so c is propylene, and a is ethylene.

Question 2 can be found without answering.

(1) When the conversion of Fe2 Fe3 is carried out in the human body, Fe2 is used as a reducing agent, and Fe3 is used as an oxidant.

2) The sentence "taking vitamin C can reduce Fe3 to Fe2 in food" points out that vitamin C acts as a reducing agent in this reaction and has reducing properties.

3) A certain cereal on the market contains trace particles of reduced iron powder, which is converted into ferrous salt under the action of human gastric acid (the main component is hydrochloric acid). The ionic equation for this reaction is Fe+2H+=Fe2++H2 (gas).

22 (7 points) (1) From the comparison of the following sets of reactions, determine which particle has the strongest oxidation and which particle has the strongest reducibility.

After the iron nail is immersed in CuSO4 solution, a red substance will be attached to the surface; When the copper wire is immersed in the AGNO3 solution, a silvery-white substance will be attached to the surface. Among Cu, Fe and Ag, Fe has the strongest reductibility. Among Cu2, Fe2 and Ag, Ag+ has the strongest oxidizing activity.

Iron nails are rusted into a tan substance (FeCl3) in chlorine gas, while a pale green solution (FeCl2) is formed in hydrochloric acid. Among the chlorine molecules (Cl2), chloride ions, and hydrogen ions, Cl2 H+ has oxidizing properties, and Cl2 has the strongest oxidizing activity.

2) In order to prevent and treat iodine deficiency diseases, a small amount of potassium iodate (kio3) is usually added to table salt. Potassium iodate and potassium iodide can undergo the following reactions in solution: kiO3 5ki 3H2SO4===3I2 3K2SO4 3H2O

In this reaction, the oxidizing agent is KiO3 and the oxidation product is I2

When 3 mol of I2 is generated, 5 mol of reducing agent is oxidized.

Friend, there are no pictures in 23 questions, it's not easy to do.

Reducing agent Oxidizing agent Reducing agent Oxidizing Fe + H+ = Fe2+ + H2Fe Ag2+ Cl2 Cl2KLO3 I2 5

fe h2so4 co2 cao nahco3

The answer is reducing agent, oxidant, reducing agent, reducibility, fe + 2h + (hydrogen ions) = fe2 + (ferrous ions) + h2 (hydrogen, followed by a gas symbol) The symbol is too difficult to play, children, these contents are very basic, you will learn hard.

caco3+2hcl=cacl2+h2o+co2↑10073

x60*10%

x=60*10%*100/73=

The quality of the calcium carbonate reflected in the participation is.

Put manganese dioxide under the separating funnel, put hydrogen oxide on it, open the water stop, hydrogen oxide flows down, and turn off the water stop clip at the end of production.

Carbon monoxide is toxic, nitrogen oxides and sulfur dioxide are harmful substances that form acid rain, and soot is an inhalable particle that pollutes the air, so only carbon dioxide is not a harmful substance. The answer is d.

Cars burn gasoline, which is smelted from petroleum, and there are carbon, nitrogen, and sulfur in petroleum.

Under combustion conditions, all kinds of oxides are generated, ABCD will be produced, plus the restriction of harmful components, then there is only D,

Answer: d

A is toxic and B can form acid rain or photochemical smog.

c. Acid rain and dust are formed.

Step 1: White precipitate and colorless solution indicate the absence of CuSO4 and FeCl3.

Step 2 precipitate partial dissolution indicates that there are BaCl2, Na2CO3, Na2SO4, because BaSo4 and BaCO3 precipitate are generated, the former is insoluble in hydrochloric acid solution, and the latter is soluble in hydrochloric acid solution, so it is partially dissolved.

Step 3 does not explain anything because BaCl2 contains Cl-, so adding a silver nitrate solution to the colorless filtrate B to produce a white precipitate insoluble in dilute nitric acid does not indicate the presence of NaCl.

Therefore, it must contain BACL2, Na2CO3, and Na2SO4.

May contain KNO3 and NaCl.

2hcl+baco3=bacl2+h2o

bacl2,na2so4,na2co3

kno3,nacl

Barium carbonate reacts with hydrochloric acid to produce water, carbon dioxide, and barium chloride.

Analysis: Colorless filtrate: no FeCl3, CuSO4, white precipitate: barium carbonate, barium sulfate are all possible.

From step 2, it is known that barium carbonate and barium sulfate exist, then, sodium sulfate also exists, and from step 3, the white precipitation knows that there are cl ions and may have sodium chloride.

And copper carbonate in the solver **, this substance does not exist.

(1) The solubility of Ba(OH)2 is soluble, and a slightly soluble salt CaSO4 (chemical formula) is found from the above table

2) Whether the two solutions of Zn(NO3)2 and CuCl2 can react cannot be (fill in "can" or "can't"), on the grounds that there is no precipitation, gas or water formation or does not meet the conditions for metathesis reaction

3) There is a white solid mixture, which may be mixed from several mixtures of BaCl2, Kno3, CuSO4, FeCl3, Na2CO3, Na2SO4, NaCl, etc., in order to prove its composition, the following experiments are done:

Step 1: take a certain amount of the solid mixture, add a sufficient amount of water, stir fully, and obtain a white precipitate and a colorless solution, filter to obtain a white precipitate, and a colorless filtrate B

Step 2: Add excess hydrochloric acid solution to the white precipitate A, and the precipitate is partially dissolved.

Step 3: Add silver nitrate solution to colorless filtrate B to produce a white precipitate insoluble in dilute nitric acid;

then the solid mixture must contain BaCl2, Na2SO4, Na2CO3, and may contain NaCl and KNO3; There must be no CuSO4, FeCl3 (fill in the blanks with chemical formulas).

Test Points: Conditions and Essence of Metathesis Reactions; identification and inference of substances; Writing and meaning of chemical formulas; The connection between the phenomenon of reaction and the essence

Special topic: Inference of mixture composition; Information combined with textbook knowledge

Analysis: This question mainly examines the condition and essence of the metathesis reaction

1) To test students' understanding of the solubility of substances formed by ionic bonding in **

2) Examine the conditions under which the metathesis reaction occurs: there is gas or water or precipitation formation

3) Use the metathesis reaction to infer the composition of the mixture, and use the information white precipitate and colorless solution, and the precipitate is partially dissolved to solve the problem

Answer: Solution: (1) It can be found directly from **, there are 2 kinds of microsolubles, but calcium hydroxide is a base, so the answer is: soluble; caso4．

2) The recombination of ions in Zn(NO3)2 and CuCl2 solutions cannot obtain gas or water or precipitation, so there is no reaction, so the answer is: no; There is no precipitation, gas or water formation or does not meet the conditions for metathesis reactions

3) the solid mixture is added with a sufficient amount of water, stirred fully, and a white precipitate and a colorless solution are obtained, because the solution of CuSO4 and FeCl3 is blue and yellow respectively, there must be no CuSO4 and FeCl3, and the white precipitate can be known to be barium carbonate or barium sulfate from the substances in the mixture, and an excessive amount of hydrochloric acid solution is added to the white precipitate a, and the precipitate is partially dissolved and known as a mixture of the two, that is, the original solid mixture must have BACL2, Na2CO3, Na2SO4, Step 3 only shows that the precipitate is silver chloride, and the solution has chloride ions, but BaCl2 also has chloride ions when dissolved in water, so there may be NaCl, and Kno3 cannot be judged whether there is or not, so BaCl2, Na2SO4, and Na2CO3 are selected; nacl、kno3；cuso4、fecl3．

( Lalalala.)

There must be Na2CO3, Na2SO4, Baci2, and possibly NaCl Kno3

Filtration yields a white precipitate, colorless filtrate b The white precipitate is either BaSO4 or BaCO3 or both, and the colorless filtrate indicates the absence of FeCl3 and CuSO4

Excessive hydrochloric acid solution was added to the white precipitate A, and the partial dissolution of the precipitate proved that both had 2BaCO3+6HCI=2BAC2+3H2O+2CO2

The addition of silver nitrate solution to the colorless filtrate B to produce a white precipitate insoluble in dilute nitric acid proves that there are CI ions in the solution.