Chemistry inference questions, high score rewards, 10 high prizes for solving chemistry problems

A is S (sulfur).

B is H2S (hydrogen sulfide).

C is SO2 (sulfur dioxide).

D is SO3 (sulfur trioxide).

E is H2SO4 (sulfuric acid).

I'm dizzy, I just did this today.

The answer is what the above said. That's right.

Admire, by: Artichoke 0 0 is really amazing.

A is sulfur and B is hydrogen sulfide.

c is sulfur dioxide.

d is sulfur trioxide.

e is sulfuric acid. Inference Steps:

As can be seen in the figure, C is the pivot of the entire chemical reaction, so we should start from it. First of all, from a to C and C to D these two processes, it can be seen that C must be a compound, and it contains at least one intermediate valence element, and then look at the process of E to C, copper can only be used as a reducing agent, in the order of activity, that is, silver, platinum, gold These three can be reduced by copper, but these three substances obviously do not conform to the equation, so C is a non-metallic oxide, in middle school chemistry, there are a few non-metallic compounds such as nitride and sulfide that can be reduced by copper, if it is nitrogen dioxideIf it is nitric oxide, then A is not a solid and contradicts the title, so C is sulfur dioxide. After that, it is very simple to deduce other substances.

A is sulfur, B is hydrogen sulfide, c is sulfur dioxide, and d is sulfur trioxide.

e is concentrated sulfuric acid.

Probably like this, I haven't studied for many years and forgot.

1. 10 GnaOH in 100g of NaOH solution with mass fraction of 10, 90g

H20,1) Add NaOH solid X to the solution set to 10

g,(10+x) (10+x)+90=20%The solution is:x=

Evaporated water YG10 100-Y

Solution: y=50g

The structure of a is simply as follows: (h3c)2c=c(ch3)-ch2ch2-cho, b is: h3c-(c=o)-ch2ch2-cooh, where b is a carbonyl c=o in parentheses, and it is not easy to enter here, so it is written like that.

The first reaction is the addition of the double bond in a with bromine, plus the -choaldehyde group is oxidized by bromine water to -cooh;

The second reaction is the reaction of the -choaldehyde group in a with 2,4-dinitrophenylhydrazine.

The third reaction is that the double bond in A is oxidized, and the carbonyl group C=O is formed on both Cs of the double bond, and the -choaldehyde group in A is also oxidized to carboxyl-Cooh, thus forming a molecule of acetone and compound B[H3C-(C=O)-CH2CH2-COOH].

The fourth reaction is actually carried out in two steps, see the introduction below:

1.The chloroform reaction is one of the haloform reactions.

2.Haloform reaction refers to the reaction of organic compounds with hypohalides to produce haloform. Methyl ketones and acetaldehyde react with chlorine, bromine and iodine under alkaline conditions to form chloroform, bromoform and iodoform, respectively.

The haloform reaction is carried out in two steps.

1) Complete halogenation (alkali catalysis of aldehydes and ketones) of -methyl groups with the following reaction formula RCOCH3+3NAoX ---RCOCX3+3NAOH

2) Carbon chain alkaline cleavage of trihaloaldehydes (ketones), the reaction formula is as follows.

rcocx3+naoh---chx3+rcoona

When x=cl, it is a chloroform reaction;

Therefore, after the above two steps of reaction, B is cleaved under alkaline conditions to form one molecule of chloroform and one molecule of disodium succinate salt.

a：(ch3)2c=chch2ch2c=och3b:hoocch2ch2cch3

OA+BR2=(CH3)2C(BR)CH(BR)CH2CH2C=OCH3A+2,4-Dinitrophenylhydrazide = (CH3)2C=CH2CH2C=N-NH-benzene ring (its ortho and para positions are each attached to a nitro group).

CH3 A+KMno4=(CH3)2C=O+BB+Naclo=CHCl3+NaoOCch2CH2Coona can react with 2,4-dinitrophenylhydrazine, indicating that A contains a carbonyl group, B has acid properties, haloform reaction indicates that B contains methyl ketone, and carboxyl group, write the -c=O of the carboxyl group of B and the oxygen of -C=O of acetone, remove the middle O and reduce it to a double bond. You will ask why not write the -c=o in methyl ketone against the -c=o oxygen in acetone, if then there should be two o's in a. Complete.

The oxyylene flame produced during d combustion is often used to weld or cut metal, and it is said to be "oxyylene flame", of course, D is acetylene.

In this way, it is also known that B is cac2.

The calcium carbide to acetylene will also generate Ca(OH)2, and the amount of such substances A and B can be fully mixed with a small amount of water to react completely, and the gas C generated by A and Ca(OH)2 can react with O2 to form two gases, and it is speculated that the former and C are NH3. Then we know that f is no, h is h2o, and i is CO2.

It is further known that g is NO2 and M is HNO3.

A must be NH4HCO3, E is CaCO3, J is CAO, and K is Ca(OH)2.

Write the chemical equation for the reaction of A and B.

The introduction of I gas into a NaCl solution dissolved in saturated C gas is a key step in Hou's alkali method, and the chemical equation of this reaction is written.

Write out the ion equation for the reaction of the lower Huixu:

A: Add M HCO3-+H+=CO2+H2O to solution A

B: Add excess sodium hydroxide solution NH4+ +HCO3- +2OH- =NH3·H2O + CO32- +H2O to solution A

C: It is known that ferric oxide can be dissolved in M solution and produce gas F, write the ion reaction 3Fe3O4 + 28H+ +NO3- =9Fe3+ +NO+14H2O

Need an answer, or a train of thought?

Some of the obvious ones in the title:

1) The oxyylene flame produced during d combustion is often used to weld or cut metals, and it can be concluded that d is C2H2 (acetylene).

2) The preparation of acetylene is obtained by the reaction of Cac2 with water, so B is Cac2

3) Another product of CAC2 to prepare acetylene is Ca(OH)2, and only NH4+ can produce gas with CA(OH)2, and this gas must be NH3, so C is NH3

4) C+O2---F F+O2---G G+H2O---F+M should go without saying.

5) E contains Ca The source species of Ca-containing substances are decomposed by heat to form I, and I is the combustion product of acetylene.

6) The later self should be able to deduce.

Don't give too specific answers, try to deduce them yourself.

1 A is carbon monoxide, CO

b is red phosphorus p

C is carbon and Cd is oxygen.

2 1 Phosphorus ignites in oxygen to produce phosphorus pentoxide2 carbon and carbon dioxide, which produces carbon monoxide at high temperatures.

3 Carbon and oxygen ignite in the absence of oxygen to form carbon monoxide.

Thank you!!! And a is definitely not carbon dioxide.

A is CO2, B is red phosphorus (P), C is carbon (C), D is oxygen (O2) ignition P+5O2*****=2P2O5 2, C+CO2====2CO (conditional high temperature) ignition C+O2====2CO (under the condition of insufficient oxygen).

A is CO2, B is red phosphorus (P), C is carbon (C), D is oxygen (O2), then the gas produced by 2 is carbon monoxide (CO P+5O2=2P2O5...).2. C+CO2=2CO (high temperature).

C+O2=2CO (under the condition of insufficient oxygen).

a CO2 b p (red phosphorus) c c (carbon) d o2

2 Generate CO

I'll write the equation myself.