Inorganic inference of chemical problems, and inorganic inference of chemical problems

I don't know if you want the answer, or you want to analyze, this question is really long.

Variant Training 3 What about your **, how do you do it without a picture???

Exercises: 1. (1) 4HCl (concentrated) + MNO2 = heating = MnCl2 + Cl2 + 2H2O

2) C is NH3

NH4 + H2O = Reversible = NH3·H2O + H+3) C + H2SO4 = Heating = CO2 + SO2 + H2O4) 3Cu + 8HNO3 = 3Cu(NO3)2 + 2NO + 4H2O

So the total number of electrons transferred is 3mol

2. There is no picture.

3、a:h2 b:n2 c:o2 d:na e:si f:s1)x:h2o2 y:h2o

Electronic type: 2) The one on the right is silicon.

3) N2(G) +3H2(G) = Reversible = 2NH3(G); H = -4)2SO2(G) +02(G) = Reversible = 2SO3(G) From: 0

Change: Flat: So the conversion rate of SO2 in equilibrium is 90%.

Wow, that's too much, right? Go home and open "Optimization**" (it should be the title of this book, right?) Look good!

a red powder A, add Hno3 to obtain brown precipitate B, and add K2CRO4 to the precipitated solution C to obtain a yellow precipitate D; If concentrated HCl is added to B, gas E will be released; Gas E is passed into solution C with an appropriate amount of NaOH to obtain B. Ask A, B, C, D, EA

pb3o4,b:

pbo2,c:

pb(no3)2

d:pbcro4,e:cl2

Common Organic Inference Question Breakthroughs, 1The transformation relationship between organic functional groups, especially olefins, alcohols, halogenated hydrocarbons, aldehydes, carboxylic acids, and esters, is bidirectional.

2.A common category of substances that react with sodium metal, sodium hydroxide, sodium carbonate, sodium bicarbonate.

Sodium metal, sodium hydroxide, sodium carbonate, sodium bicarbonate.

means reacting, - means not reacting)

Alcohols. Phenol.

Carboxylic acid. Esters.

3. The calculation and application of unsaturation can quickly determine the number of unsaturated bonds in the molecule

I is an unstable green oxidant composed of AC, I is [H2O2], A is [H] C is [O], I decomposes, K is [H2O], and A is [O2].

D can react violently with cold K to form A and L, L has a yellow flame color, L contains sodium, D is [Na], D is [Na], A is [H2], and L is [NaOH].

F is passivated in a cold concentrated solution of m, and the passivation can occur with Al and Fe, but only Al and F are [Al] in a short period

In addition, Al is passivated in concentrated nitric acid or concentrated sulfuric acid, and the atomic number of g is certain"13, so g is [s].

Then H can only be [Cl] and then back to the above, "H can be absorbed by its aqueous solution to form 18-electron compounds J and C".

It can be known that H2O2 + Cl2 == 2HCl + O2

So h is [Cl2] and J is [HCL].

A compound n composed of b and c does not support combustion, but e can be burned in n to form b and o

CO2 does not support combustion, B is [C] N is [CO2] E is burned in CO2 to produce C

Apparently e is [mg] o is [mgo].

In summary: a [h] b [c] c[o] d [na] e [mg].

f 【al】 g【s】 h【cl】 i【h2o2】 j【hcl】

k【h2o】 l【naoh】 m【h2so4】 n【co2】 o【mgo】

a：hb：c

c：od：na

e：mgf：al

g：sh：cl

i：h2o2

j：hclk：h2o

l：naoh

m：h2so4

n：co2o：mgo

Write the chemical formula yourself, children have to study hard, this kind of question has to come up with it by themselves to make sense, many things can be asked for help from Du Niang, but the learning is more profound by yourself.

a, hydrogen, b, c, c, c, d, d, e, aluminum, g, sulfur, h, chlorine, l, sodium hydroxide, i, hydrogen peroxide, n, carbon dioxide, m, sulfuric acid, j, hydrochloric acid.

1:3 and A is negative, then A is -3 valence, and C and D are +1 nh3 li3n

A is n2 and 1:2 B is negative and B is -2 valence.

li2o h2o

B is O2 and C is H2 then D is Li

N2 +3H2== Catalyst heating ==2NH3

o2 +2li ==2li2o

If C is li, then D is h2

N2 +6Li == Combustion ==2Li3N

O2 +2H2== Combustion ==2H2O

a bicl3

b biocl

c bi(oh)3

d nabio3

e bi2s3

F Bi elemental.

There are two more days, I hope it can be useful to you, and it is not easy to get into graduate school.

Look, I hope it helps.

First of all, the flame color is red, and the metal is the CA element.

Secondly, we can be sure that E is NH3, because making litmus blue is obviously alkaline, and the only obvious alkaline gas is ammonia.

This also shows that calcium reacts with oxygen to form CaO, and reacts with nitrogen to form calcium nitride Ca3N2, which is similar to the reaction with magnesium and nitrogen.

The carbon dioxide dissolved in D has a white precipitate F, indicating that D is calcium hydroxide Ca(OH)2, and F is calcium carbonate CaCO3.

b,c can't be determined, only know that it is cao and ca3n2 mentioned earlier. Because there is no narrative about them.

That's the process, hope it helps!

Remember to give it an adoption! Hee-hee.

kio3, and cai2

Only these two can acid react to produce iodine element, and there is a white precipitate.

Add a small amount of dilute H2SO4, the solution turns yellow-brown, [the color of iodine element], and there is a white precipitate.

5cai2+2kio3+6h2so4=5caso4↓+k2so4+6h2o+6i2

Hope it helps you o(o

Calcium iodide and potassium iodate. Iodine ions and iodate groups are classified into iodine elements under acidic conditions and disproportionated under alkaline conditions. Sulfuric acid and calcium ions form calcium sulfate microsolutes.

There are bacl2 and ki

Because the colorless solution cannot have Ca2+ ions, both H2SO4 and KIO3 are oxidizing and cannot reduce I. So there are only two left.