Chemistry helps explain it faster!

Hello classmates The polyester reaction is generated by the polymerization of multiple monomers, which is a polycondensation reaction with the formation of small molecules, such as (1) the formation of polylactic acid.

NCH3CH(OH)COOH *****====>-[OCh3)Co-]N- +NH2O (2): Polycondensation reaction of phenol and formaldehyde, N A-OH N HCO*****====> (conditional concentrated hydrochloric acid heating).

C6H5OH2]-N NH2O (3): esterification reaction *-COOH+CH3CH2OH*****===>*Co-OCH2CH3+H2O

where * denotes benzene ring (4): adipic acid and ethylene glycol: nch3cooh+nhoch2ch2oh*****===>n[coo-ch2-ch2-coo-ch2-ch2-o]n+2nh2o (note that here is the polyacid and polyol reaction Remember!)

The conditions for esterification are concentrated sulfuric acid, heating, reversible reaction.

All esterification reactions are the removal of hydroxyl groups from acids, the removal of H from alcohols, and the formation of esters and water. Hope!

The dehydration and condensation of hydroxyl and carboxyl groups into a polyester group is a polyester reaction.

For example, the polymerization of ethylene glycol and adipic acid.

Polymerization of phthalic acid and hexylene glycol.

Both hydroxyl and carboxyl groups are dehydrated and polymerized.

In the third question of the conjecture, because the conjecture is copper oxide as the catalyst, the chemical properties and quality of the following fills remain unchanged.

The first blank in the form is (the wooden strip with sparks is rekindled, so it is judged that there is oxygen production,) in a large number of bubbles are generated.

As a result of step 6, the solid mass remains in grams (because copper oxide is the catalyst, the chemical properties and mass are constant).

2) The purpose of the medium is to serve the purpose of comparative experiments (to illustrate the catalytic effect of copper oxide by comparing it with the phenomenon after adding copper oxide in 3).

3) Glass rod (plays a role in drainage during the filtration process).

4)2h2o2==cuo==2h2o+o2

In the expansion, to verify that copper oxide is still used, you can do another catalytic hydrogen peroxide decomposition experiment, and you can also use a reducing agent to reduce copper oxide to obtain copper and another substance, so as to prove that it is still copper oxide.

1. The chemical properties are unchanged before and after the reaction, so there is no need to explain this 2. There are bubbles released in step 3. (catalyst acceleration) 3, step 6 quality unchanged (the following ** has been suggested that it is a catalyst). 4. Step 7: Re-ignition of the wooden strip with Mars (proving that not only the quality has not changed, but also the nature has not changed).

2) Demonstrate that the speed of the reaction is accelerated in CUO step 2.

3) Filter paper. 4) 2 hydrogen peroxide 2 water + oxygen (the equation is not easy to play, just know it) (5) hydrogen reduction copper oxide experiment.

Answer: (1) There are bubbles; The quality of copper oxide remains unchanged;

2) verify that hydrogen peroxide decomposes in the absence of a catalyst;

3) glass rods;

4) Copper oxide, which has been used as a catalyst, is reduced with hydrogen to see if its properties change.

Analysis: According to the analysis of the topic, it can be seen that this question is to verify whether copper oxide can catalyze the decomposition of hydrogen peroxide, so there are three conjectures, and the third conjecture is that copper oxide is a catalyst, so the mass and chemical properties before and after the reaction remain unchanged, therefore, the following experiments are used.

Because the phenomenon of step 3 is "re-ignition of the wooden strip with sparks", there is oxygen generation, and of course the phenomenon is "bubble generation". In addition, the conclusion is that "conjecture 3 is true", so the mass and chemical properties of copper oxide before and after the reaction are unchanged, so the result of step 6 is "the mass of copper oxide remains unchanged".

Ask a well-wisher to help explain your thoughts, be clear, thank you!! After the reaction, it is all to produce water, and the chemistry of the high school has not been unfamiliar for a long time. My explanation for this kind of topic is: first of all, you burn completely.

1. There is CO above the coal seam, and the flame it burns is blue.

C+O2=Ignition=CO2 2C+O2=Ignition=2CoC+CO2=High Temperature=2Co2Co+O2=Ignition=2CO22,2Cuo+C=High Temperature=2Cu+CO2

x y z 50-39=11g

Column proportions, solved.

x=40g y=3g z= 32g

Of the remaining 39 grams of solids, 32 grams are copper and 7 grams are either carbon or copper oxide.

1:2co+o2=2co2 2h2+o2=2h2o 2 c+2o2=2co+co2

2: Generate copper and carbon monoxide, carbon monoxide further reacts with copper oxide to form carbon dioxide and copper, so the product is copper, carbon monoxide and carbon dioxide, but it depends on the amount of copper oxide and carbon, if there is more carbon, copper oxide has been completely converted into copper, then excess carbon monoxide will exist, if the generated carbon monoxide is just after the reaction with copper oxide, the generated gas is completely carbon dioxide, anyway.

I haven't studied chemistry for 20 years, and I don't remember the molecular weight, but it's still easy to solve according to the idea.

See for yourself. The cow man.

Perfect answer、,

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According to the equation, the 10GCACO3 produced has , then CO2 also, for, it is known that the weight gain is generated, then the rest is water, for, . Then you can only have options, a is c2h6o2 indicator 62, for, produce CO2 yes, produce h2o yes in line with the topic. Then verify B c D, it can be seen that B must not be compliant, because A is compliant, B has less oxygen, the mass fraction of C and H is large, and more CO2 and H2O are produced, comparing A and C, it is obvious that the mass fraction of C is very large, and it does not match, D can, methanol CH4O, glycerol C3H8O3, and ethylene glycol in A is C2H6O2, which is CH3O and C3H9O3 after simplification, and the methanol deformation is C3H12O3, and glycerol C3H8O3

Both will h can be averaged to 9, so holds.

The precipitate is calcium carbonate, then CO2 is found according to calcium carbonate. So CO2 is and then we can find the mass of the water according to the weight gain of lime water, then we can find that the ratio of c and h of the sample is x2=1:3 and then according to the conservation of mass, and the water has c and h is so o mass is so finally we get the sample.

B is acidic to A.

c Oxidation.

d exhibits oxidation without acidity This contrast Cu does not react with other acids.

Select A analysis: Item A, benzene and ethylene are both planar molecules, and all atoms in the molecule are on the same plane.

The abbreviated formulas of item b are ch and ch2 respectively, which is obviously false.

C, ethylene can react with bromine water, but benzene can not react with bromine water, it can only extract bromine in bromine water and fade it, it should be noted that benzene can react with liquid bromine, obviously c is also wrong.

In item d, ethylene can be oxidized by an acidic potassium permanganate solution, but benzene cannot react with an acidic potassium permanganate solution.

I choose AC.

Needless to say, it's in the textbook.

d。Benzene cannot be oxidized by potassium permanganate.

A is false because the h atom and the c atom are not in the same plane.

B error 2C6H6+15O2====12CO2+6H2O 78g mol

C2H4+4O2====2CO2+2H20 28g mol of benzene and ethylene of the same mass The oxygen consumption of ethylene is greater than that of benzene C, and the substitution reaction occurs in the same way.

Under normal conditions, benzene cannot be oxidized by strong oxidants, while ethylene can be acidic and oxidized by potassium permanganate solution.

Therefore, I hope this question C will help you in your studies

If you don't understand AC, you can remember that ethylene is a typical six-atom coplanar structure, while benzene is a typical 12-atom coplanar structure. For B, you can remember that the higher the proportion of the substance, the more oxygen is consumed at the same mass, and it is clear that ethylene has a higher proportion of hydrogen than benzene. For c, it is very important, ethylene is addition, benzene is substitution, in addition, it should be noted that the substitution of benzene needs to use ferric bromide as a catalyst, speaking of which I will add another sentence, if you encounter the substitution of benzene homologues, you must pay attention to the position of substitution, if it is on the benzene ring, the condition is halogenated iron, if it is on the branched chain, the condition is light, and the substitution position can be controlled by controlling the conditions.

For d, acidic potassium permanganate is used to test double and triple bonds, benzene cannot fade acid potassium permanganate, while ethylene can. In addition, benzene can discolor bromine water, but the reason is due to extraction, not reaction.

Single or multiple?

I think BCD

13，d

14, C element symbols: C carbon, Ca calcium, Ne atmosphere, Na sodium, AI aluminum, C|Chlorine, N nitrogen, Cu copper, Ag silver, Hg mercury, Zn zinc, Ba barium.

What a question! Is it okay to take some responsibility?