Chemistry questions are asked for help to write out the thought process.

1 a+fecl3→b↓+c

2 c + d e (white) + f

3 F + agno3 g (white) + hno3

In reaction 3, G is AGCL, and the only AGCl-containing compound insoluble in HNO3 in secondary school is AGCL. According to the equation, F is HCL.

In Reaction 1, since B and G are two substances and B is not necessarily white, the case that B is also AGCL is ruled out. Only Fe3+ can be formed, so B is Fe(OH)3, A must contain OH-, and C must contain Cl-.

In reaction 2, since C contains Cl- and F is HCl, only the cation in C and the anion in D form a precipitate.

Finally, it was guessed that the precipitated E may be CaCO3 or BaSO4.

If E is CaCO3, CaCO3 and HCl do not coexist, excluded.

So E is BaSo4, then C is BaCl2, A is Ba(OH)2, and D is H2SO4.

Conclusion: a BA(OH)2

b fe(oh)3

c bacl2

d h2so4

e baso4

f hclg agcl

From the third one, it can be seen that G contains AG and is white, then G may be AGCL and F is HCl

There must be Cl in B and C in the first one, and if B has it, then it is AGCL, and because G is silver chloride, C contains chloride ions.

The second C contains chloride ions and F is hydrochloric acid, so D and E contain hydrogen ions.

Because F is an acid, there are only two precipitates that are insoluble in acids, barium sulfate and silver chloride, so E is barium sulfate.

c is barium chloride.

then a is barium hydroxide.

b is ferrous hydroxide.

Did you make a mistake in fecl3, is it 2?

G is AGCl, F is HCl, E is BaSO4, C is BACL, D is H2SO4, B is Fe(OH)3, and A is BA(OH)2

Thought Process: The reaction with silver nitrate in the third equation produces a white precipitate insoluble in nitric acid. It can be concluded that G is silver chloride and therefore F is HCl, and E can be judged by the second equation as a white substance insoluble in HCl.

At this time, we can know that there are many answers to satisfaction, and then look at equation one, we can know that there is OH- that precipitates with ferric iron ions, and we can know that B is iron hydroxide, and C must contain chloride ions, and in considering E is a white substance that is insoluble in HCL, it can be inferred that E can be barium sulfate, then C is barium chloride, and since B is iron hydroxide, A is silver hydroxide. Check it to see if there are any obvious errors.

(1) The larger the mass fraction, the greater the density The larger the mass fraction, the smaller the density.

3) I generally can't summarize the rules.

(1) The higher the concentration of sulfuric acid, the greater the density; The higher the concentration of ammonia, the lower the density.

2) a=20, >, 3) and other mass mixing, the concentration of the mixed solution is the middle value of the two; When mixing in equal volumes, if the concentration increases with density, it is greater than the median value; Conversely, it is less than the median value. (Note: The median value can also be called the average).

(10 24)n :(10 64)*n:(10 56)*n=56:21:24, where n is the number of molecules in one mole, it's useless, I forgot how much.

24/n）*10：（64/n）*n ：（56/n）*10=3：8：7

Do it with conservation of electrons of gain and loss.

It is not necessary to calculate the concentration of protoHno3.

The amount of 672ml O2 substance ==, the number of electrons obtained by O2 = = Then the sulfuric acid is dropped in the solution obtained by the reaction of iron and nitric acid, and the gas produced is NO, NO, the volume of NO==, the number of electrons obtained by NO==

So the reaction gets a total of electrons = =

Fe is eventually converted to Fe3+, so the amount of reaction Fe = = Mass of Fe = 56* = Option A

Conservation of electrons.

Iron is oxidized by oxygen and nitrate ions reduced to 224mlNO, and the amount of iron is x, then:

The solution is: x=, so the mass is:

That's an interesting question, huh.

1molCO2 is calculated, so the ratio of C,H atoms in organic matter is 2:3

So the mass of the organic matter consumed is the mass of 1molc and the mass of 12+ and the chemical formula is c2xh3x

According to the inscription, 24x+3x=54

x=2, so c4h6

Crystals precipitate after evaporation of potassium nitrate solution, indicating that the solution has reached saturation. Then calculate the solubility by calculating the solvent and solute at this time.

Solute mass after evaporation = 150g 20% 11g = 19g before evaporation 150g potassium nitrate with water 150 (1 20%) = 120g solvent mass after evaporation = 120g 50g = 70g

So the solubility is 19g 70g 100g = keep 2 decimal places)

150g of 20% potassium nitrate solution with a mass of 30 grams of potassium nitrate and 120 grams of water.

After evaporating 50 grams of water, 11 grams of potassium nitrate crystals are precipitated, and the solution is a saturated solution.

At this point, the mass of potassium nitrate contained in the solution is 19 grams. 70 grams of water.

Therefore, the solubility of potassium nitrate at this temperature is: 19*100 70=g.

As long as you know what is the amount of kno3 h2o at saturation, you can find it.

Water == 150 x (1 - 20%) 50 == 70 g

kno3 == 150 x 20 % 11 == 19 g

x / 100 == 19 / 70x == 27g / 100 g h20

A mol of sodium sulfate in a solution that happens to completely precipitate barium ions, then there is a mol of BA2+ in the solution of 1 2

If the solution of b mol of silver nitrate happens to precipitate the chloride ions completely, then the solution of 1 2 has b mol of cl-

The positive and negative charges in the solution are equal, and if there is k+ x mol in the solution of 1 2, then the sum of the positive charges is x +2a ==b

x==b-2a

Then there is 2x mol K+ in a 500ml solution, which is b-2a mol

The concentration is (B-2A) -4A mol l

Answer: A mol sodium sulfate proves that there are amol barium ions, and 2amol chloride ions and barium ions are salted.

Bmol of silver nitrate proves the presence of bmol chloride ions.

b-2a is potassium chloride, and the number of chloride ions is the number of potassium ions.

So the concentration is (b-2a) mol l