High One Chem Jd P 101 2 6

1 Answer C because HCO3- can react with acids and bases.

2 dActually, mg al and water have to react, but both prevent the reaction because they form substances that adhere to their metal surfaces, so they don't need to be distinguished by anyone to exclude AB

They react violently with hydrochloric acid, so they don't use c

d al can reflect mg is not good so we can use d to distinguish it

The first question is to know the substances that can react with both acids and alkalis, choose c, hehe.

The second question is possible in boiling water, to see that kind of metal dissolves that is mg, b is not reacting, c has gas, although it is possible to see the rate, but the best method to be used in the question, so choose d pull.

Also, the mg in boiling water seems to be able to react, while the oxide film seems to be unable to prevent the reaction between mg and boiling water.

2。Put boiling water into both can react, put in cold water after the reaction is relatively slow, put into hydrochloric acid can generate hydrogen, and magnesium is not with caustic soda reaction but aluminum is amphoteric compounds can react with caustic soda to produce hydrogen, the properties of magnesium and aluminum are similar, the difference is that magnesium is an alkaline compound, aluminum is an amphoteric compound, aluminum can react with acid or alkali.

You can write these equations and look at the properties of matter.

SiO2+3C=High temperature SiC +2Co

Is it that only 3 units of C have a valency of -4 to get SiC as an oxidant.

The C of the other two increases the valence to obtain a CO of +2 as a reducing agent, so the ratio of the amount of oxidant to the reducing agent substance is 1 : 2

SiO2+3C=High temperature SiC +2Co

Oxidant: The valency of the reaction increases.

Reducing agent: The valency is reduced in the reaction.

Apparently C 0 sic -4

c 0 →co +2

It is a disproportionation reaction (one rise and one drop).

The CO coefficient is 2, 2 parts of reducing agent, and 1 part of oxidant (both reducing agent and oxidant are C) is 1 to 2

Learning progress If you are satisfied, please remember to adopt it

In 3C, the carbon in 2CO is increased from 0 valence to +2 valence, which is the reducing agent, and the carbon in SiC is changed from 0 valence to -4 valence, which is the oxidant, so the ratio of the amount of oxidant to the reducing agent substance is 1:2

Remember: in a compounding reaction, it is the reducing agent that increases the valency and the oxidant that decreases.

cu2+ ag+ fe3+

The oxidative ag+ of the threeFe3+>Cu2+, so the first reaction that occurs after adding Fe is 2AG+ +Fe =Fe2+ +2AG

If there is enough of it, then Fe + 2Fe3+=3Fe2+, and finally Fe + Cu2+= Fe2+ +Cu(1) Obviously, if Fe is left, according to the above reaction, there must be no Cu2+ and Fe3+ Cu2+ in the solution

2) When there is no fe in the beaker At this time, the possible scenario is 1 Only the first reaction has occurred.

2 It is also possible that the first reaction partially occurred.

3 Only the first and second reactions occurred.

4 The first one completely occurs and the second part reacts.

5 Or the first two reactions occur completely, and the third reaction occurs partially, so the following are all possible scenarios.

1 ag+ fe2+ fe3+ cu2+

2 fe2+ fe3+ cu2+

3 fe2+ cu2+

4 fe2+

So there must be fe2+

3) According to reaction (2), if there is Cu2+ in the solution, it may contain Fe2+ AG+ Fe3+ Cu2+

Personally, I think there is something wrong with your question

The original question should be that when there is no Cu2+ in the solution, must it not be contained?

The answer is Fe3+ AG+

Iron ions react with iron to form ferrous ions; Iron ions will react with copper to form ferrous ions and copper ions, so iron elements react with iron ions first, so there are ferrous ions, if there are copper ions, it means that the iron elements have not replaced all the copper ions, and the iron ions and silver ions have been reacted first.

It is judged by the order of oxidation of the metal.

Can ionize OH-: Therefore, Al3+ can react with the formation of slag-resistant Al(OH)3Al3+ +==Al(OH)3 + 3NH4 + SO2 reduction is stronger than Changshen, Cl2 oxidation is stronger than Liang Zhen, and the reaction between the two: SO2 + Cl2 + 2H2O ==H2SO4 + 2HCl

Answer 1 A oh- k+ cl-

B Fe3+ AG+ NO3-

2 plus Fe ** metal silver.

3 React ag+ Fe3+ Cl-OH- in wastewater to produce a precipitate (AGCL Fe(OH)3

After treatment, it mainly contains KNO3

The wastewater of factory A is alkaline, indicating that there is OH-

What can coexist with OH is K+ Cl- NO3- Ag+ and Fe3+ will precipitate with Oh-, and Ag and Cl- do not coexist.

So cl only with k+.

Because the solution must contain anions and cations to be conserved in charge , NO3- is put into B.

So methyl oh- k+ cl-

B Fe3+ AG+ NO3-

2. Activated carbon can adsorb those large impurities, and cannot react with metal ions, and cannot achieve the purpose of **metal.

**Metal has to be added to it to be able to replace other metal things that don't react and it's useless to add them.

The feso4 here can't be added, only the iron powder can ** silver ag.

3 Ag+ Fe3+ Cl- OH- reaction produces a precipitate (AGCL Fe(OH)3

So, after mixing in proportion, K+ and NO3 do not react to anyone.

Whether it's a solution or not, it's equivalent to a solution of kno3, which is an aqueous solution of fertilizer.

Because KN3 is fertilizer.

Because there are two kinds of solutions and each of them should have three ions, since A has OH-, then AG+ and Fe3+, which cannot coexist with OH-, can only be in B. Cl—cannot coexist with AG+, so Cl—is in A, and the solution cannot have only anions and no cations, so there is K+ in A, and NO3—in B

The latter ones are as others say.

Whether the chemical properties are similar or not is determined by the number of electrons in the outermost shell of that particle.

The sodium atom is reducible, and the atomic structure of the sodium ion is stable, so it is rare to obtain electrons, reflecting weak oxidation. Since it is the most **, there is no reduction.

The sodium ion is one electron less than the sodium atom, and its proton number does not change.

The sodium atom is unstable, while the outermost electron number of the sodium ion is 8, so it is stable.

A because it is a different substance, the chemical properties are different, wrong.

b sodium atom is reducible, sodium ion chemical valence can be reduced, electrons are obtained, oxidation, correct.

C false. D atom is not unstable how to say.

Whether the chemical properties of the main group elements are similar or not depends on the number of electrons in the outermost shell.

The oxidizing properties of sodium ions. Quite weak, the stronger the atomic reducibility, the weaker the oxidation of the corresponding ions.

How can the number of protons change when it's not all na?

D's apparent 8 electrons are stable.

The number of electrons in the outermost shell determines the chemical properties.

According to the phenomenon, the following reactions occur when precipitation occurs first and then partial dissolution.

AlCl3+3NaOH==Al(OH)3+3NaClAL(OH)3+NaOH===Naalo2+2H2O, so m>3nThe precipitate is not completely dissolved, it is m<4n

AlCl3+3NaOH ==Al(OH)3+3NaCln 3N N -- according to the excess of NaOH, all precipitate first.

Al(OH)3+NaOH==Naalo2+2H2OM-3N M-3N M-3N--- according to the amount of remaining NaOH, only Al(OH)3 of M-3N is dissolved

Conclusion: The amount of precipitated material = generated-dissolved = n - (m-3n) = 4n-m

According to the title, sodium hydroxide should be excessive. The excess sodium hydroxide then reacts with aluminum hydroxide to form sodium aluminate.

From the reaction equation, the answer to a can be obtained.

alcl3+3naoh=3nacl+al(oh)3↓n 3n n

al(oh)3+naoh=naalo2+2h20x m-3n

x=m-3n

Therefore there is m-3n of al(oh)3 to be reacted.

So the remaining al(oh)3=n-(m-3n)=4n-m

Since there is n mol al+, there is n mol al(oh)3 generated, and at the same time 3 mol oh- i.e. the remaining m-3n mol oh- is consumed

The remaining continues to react with Al(OH)3 and consumes M-3N molAL(OH)3

That is, there is 4n-m mol al(oh)3, so choose a

Mr. precipitate, and then dissolve, it proves that the first step of the reaction is the reaction of aluminum ions, and there are hydroxide ions left, so that the second step of the reaction can be carried out. Therefore, the hydroxide ion reacted in the first step is 3n moles, the hydroxide ion participating in the second step reaction is (m-3n) mol, and the aluminum hydroxide reacted is also (m-3n) mol, so the remaining precipitate is [n-(m-3n)] mol. (n is the amount of precipitation generated in the first part).

So choose A

This question is about the reaction process.

After Na2CO3 is converted to NaHCO3, HCl reacts with NaHCO3 to form CO2 gas.

na2co3 + hcl === nahco3 + nacl nahco3 + hcl === nacl + co2↑ +h2o

The amount of HCl is 6*10 -5 mol and the amount of Na2CO3 is 1*10 -4 mol

This means that Na2CO3 has not been completely converted to NaHCO3, so no CO2 gas is released.

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