Chemistry Olympiad questions, the master enters a high school chemistry competition question, the ma

Upstairs is right, but it's not right! The mistake is that Cu2O is not called copper dioxide, it is called cuprous oxide!! Sweat!

Yes, in the "Is the red all oxide copper?" Copper is also red", but this problem breaks the convention, according to the reaction equation, no copper will be generated!

Here's another idea for homeowners: copper oxide decomposes at high temperatures, producing oxygen and cuprous oxide.

I'll teach you.

Solution: First of all, let's be clear, the red solid in the question is cuprous oxide.

1.Cuo overdose: 4cuo+c===2cu2o+CO2, 4cuo===2cu2o+O2

In this case, no matter which reaction is related: 4cuo---2cu2o, you see the reaction as this simple relationship!

The quality of CO2 is calculated by all the brothers, so that the mass of C is found, and the composition of the mixture is known, and the quality of Cuo is known, and the quality of Cu2O is OK.

The answer to this situation is: grams.

2.C excess: 4cuo + c = = = 2 cu2o + co2, so it is simpler, first find the quality of co2, the quality of c is there, and then cuo, cu2o! The answer to this situation is also grams.

This contest question examines the discovery of relational quantities.

And the answer in both cases is the same, this is a coincidence.

Upstairs drops, if you think about it carefully, how can there be copper!?I also won the provincial 1st prize in the chemistry competition! Don't say I'm talking nonsense!

It is impossible to overdose.

2.The red ones may be Cu2O and Cu, because Cu2O may decompose at high temperatures.

3.The gas may have co, because c+c02=2c0 then the problem cannot be solved, so I have to give up 2, that is, cu2o will not decompose at high temperature.

Knowing the precipitation, we can know the CO2, and the difference can know the mass of Cu and Co, and because of the oxygen in the original Cu2O and the oxygen in CO2 plus Co to set the amount of copper substance, we can find the oxygen in Cu2O, and C, and then. Equations can be listed.

You've been in the Oseiban, so you can take care of the rest.

Can I do that?

Not that they count. This is a complicated question. It is necessary to analyze what exactly the final product is.

One way to consider the excess of C, is that the last remaining Cu and the gas is Co, Co2, and the result of this situation is that if the Cuo is considered to be excessive, the final gas generated is only Co2, which is what was said upstairs, and there are two situations to consider. Right?

Upstairs! \"The product of CuO being reduced by C is Cu2O (Red).\'You see, the question has already said, the product is cu20, there is no cu, it seems that you are really ashamed of the teacher

The product of cuo being reduced by c is cu, right?

0~1：fe + 4hno3 == fe(no3)3 + no + 2h2o

1mol ∴4mol ∴1mol

1~2：2fe(no3)3 + fe == 3fe(no3)2

2mol 1mol - original solution Fe(NO3) 3: 2mol - 1mol ==1mol

2~3：cu(no3)2 + fe == fe(no3)2 + cu

1mol 1mol

So choose d without cu(no)2 thing, enter the wrong thing.

It will be CU(NO3)2.

When iron is added to 0 1mol, iron reacts with nitric acid.

fe + 4hno3 == fe(no3)3 + no + 2h2o

1mol originally has 4mol of nitric acid, which produces Fe(NO3)3,1mol

When iron is added to 1 2mol, iron reacts with ferric nitrate.

2fe(no3)3 + fe == 3fe(no3)2

Iron reacts with 2molFe(NO3)3, of which 1mol is obtained by the previous reaction.

When iron is added to 2 3mol, the iron and copper nitrate undergo a displacement reaction.

cu(no3)2 + fe == fe(no3)2 + cu

So it turns out that there is Cu(NO3)2,1mol

So choose D

This is the 2007 Bengbu No. 2 Middle School Experimental Class Enrollment Paper, right? Hehe, this is the 2007 Bengbu No. 2 Middle School Experimental Class Enrollment Paper, right? Huh 1The solution that can make phenolphthalein reddened contains a large amount of HCO3- of OH-,A that reacts with OH-

hco3-oh-

co32-h2o;

Cu2+ in C reacts with OH-

cu2+2oh-

cu(oh)2;

Fe3+ reacts with OH- in D:

fe3+3oh-

Fe(OH)3, so choose B

2.Let the alloys be magnesium, then hydrogen mass =

Assuming that the alloys are all aluminum, then the mass of hydrogen = is generated, and the actual result should be somewhere in between.

Therefore, C. is chosenThis question is not qualified enough to be quantitatively calculated, and the answer can only be selected by considering the limit results.

fecl2h2,fe2o3

6hcl2fecl3

3h2o,fe3o4

8hclfecl2

2fecl3

4h2o,naoh

hclnacl

H2O, so choose B

From the relative density of the equation c is , then the mass of b is , which may be 4 F or 1 br.

Re-analysis d: It is assumed that d contains 1, 2, 3, 5, and 7 O atoms, respectively, and only Li2O and SiO2 match. But Libr is an ionic crystal, so A can only be Si, then C is SiF4.

1）a：si；b：f；c：sif4；D: SiO2(2)SiH4 + 2AgnO3 = 2AG + SiO2 + 2O + 2H2O (I feel that only this is more reasonable).

Of the 5 symmetry planes, except for 3 straight ones, only 2 are oblique to them, and only 2 parallel to the z-axis are due to the orientation of the C2 ions.

Bao2 is [O2]2-ion, and the O-O bond level is 1. KO2 is [O2]- ion, and the O-O bond order is , so A