A high school chemistry problem

Since HCl is a strong acid, the starting concentration of HCl HCl with pH=2 is, while ammonia is a weak alkali, and the initial concentration of ammonia with pH=12 is much greater. If the ammonia is neutralized with hydrochloric acid to form NH4Cl solution, due to NH4+ hydrolysis, the solution is acidic, and the title says that the solution is neutral, then the ammonia should be slightly excessive. Therefore, since the initial concentration of ammonia is almost 100 times, even if the ammonia is slightly excessive, the volume of ammonia consumed is still much smaller than that of hydrochloric acid.

HCI with pH equal to 2 does not hydrolyze, so the hydrogen ion concentration is determined as. But ammonia is a weak base NH3'Aqueous solution of H2O, weak NH3'The ionization of H2O is NH3'H2O-reversible-NH4+ +OH-.OH-

So when we see a weak alkaline solution, pH = so-and-so, in fact, the solute in the solution is in excess compared to the solute of the completely ionized solution.

Because ammonia is also hydrolyzed, H ions are ionized in the water.

Ammonia is a weak base, and some of the ammonia water with pH = 12 is not ionized, and it will ionize some hydroxide ions when the pH changes.

According to the topic, the sodium salt of X and Y can be searched for this bond to react with carbonic acid to form acid, indicating that their acid is not as strong as carbonic acid, and the corresponding sodium salt of X produces a weaker alkaline bicarbonate, indicating that the alkalinity of the sodium salt of X is not as strong as the alkalinity of the sodium salt of Y, so we can know that the acidity is hx hy. (A False).

The ability to bind hydrogen ions is the size of alkalinity, and according to the title, you can get the nax nay error in bc.

d. There is a strong or weak acidity in the medium.

According to the meaning of the topic, there are the following relations:

4kmno4 ~ 5 khc2o4·h2c2o4·2h2o

According to the principle of neutralization reaction, there is the following relation:

3naoh ~ khc2o4·h2c2o4·2h2o

If there is 5mol of KhC2O4·H2C2O4·2H2O in the reaction, then 15mol NaOH should be involved in the reaction, so the volume of sodium hydroxide solution is: 15mol = 100L

So the volume of potassium permanganate solution is 100 3 l The concentration of the solution is: 4mol 100 3 l =

The problem is solved, this kind of problem mainly considers several conservations: conservation of materials (conservation of quantity of matter), conservation of charge, conservation of the number of electrons gained and lost during redox, and sometimes conservation of mass, and then use these conservation relations to establish equations

Answer: B alcl3

MgCl2 powder 95mg reacts with a sufficient amount of AgNO3, which should be generated: 287mg, 300mg more, indicating that the proportion of Cl in the impurities is large.

MgCl2: 71-95

So only alcl3

Let NH4AL(SO4)2*12H2O1 Mo, the first step of heating must be to lose water and become 1molNH4AL(SO4)2, that is, A, (according to the solid residue rate, the solid at A is 237, which is consistent with the mass of 1molNH4AL(SO4)2) and then NH4AL(SO4)2=1 2AL2(SO4)3+1 2(NH4)2SO4 is B, (according to the solid residue rate, the solid at B is 171, which is equal to 1 2molAl2(SO4)3) and finally 1 2Al2(SO4)3 becomes 1 2Al2O3, i.e., C, (the solid mass at C is 51, which is equal to 1 2molAl2O3).

The most important thing in this kind of problem is to know what substances may be produced by heating the substance, and then find the mass from the residual rate and compare it with the mass of the substance that may be produced.

12. Hydrate aluminum sulfate ammonia, water heating disappears first, then ammonia, aluminum reaction, ammonia volatilizes by heating, and aluminum sulfate is left. In this order and graphically, the answer is obvious......

Please see the small details of the attention to the question [3 a 5].

The extreme value method considers that the amount of magnesium contained in the G alloy is 3, then the amount of the metal aluminum substance is: 97% 27 = mol, 2Al + 2Naoh + 2H2O = 2Naalo2 + 3H2 The amount of the substance that requires sodium hydroxide n(Naoh) = n(Al) = mol, and the volume is: l = = l = 97 ml

Explanation: Since magnesium does not react with sodium hydroxide, only aluminum reacts with sodium hydroxide in aluminum-magnesium alloy, because the content of magnesium in magnesium-aluminum alloy is known to be 3%-5%, that is to say, the aluminum contained in it is up to 97%, the reaction equation for aluminum dissolved in sodium hydroxide is: 2al+2naoh+2h2o=2naalo2+3h2 (arrow), there is an equation to know that in order to ensure that the aluminum in the alloy can be completely dissolved, the required sodium hydroxide is; Multiply 97% by 27 and divide by 2 by 1000ml to get 97ml, I don't know if you are satisfied with my explanation, please adopt if you are satisfied, thank you.

Select a Parse:

A: Putting iron nails into copper sulfate solution will cause the reaction of Fe+CuSO4===FeSo4+Cu, and iron is only used as a reducing agent.

B: Concentrated nitric acid has strong oxidation, ferrous ions have reducibility, after the two meet, nitric acid will oxidize ferrous ions, and ferrous ions are only used as reducing agents. Ion equation: Fe2+ +NO3- +2H+===Fe3+ +NO2 +H2O.

The equation for c: is Cu+2FeCl3===CuCl2+2FeCl2, not a displacement reaction.

D: From Fe+CuSO4===FeSo4+Cu, we can see that the oxidant is CuSO4, and the oxidation product is FeSO4, and the oxidation of the oxidant is greater than that of the oxidation product, so the oxidation Cu2+ Fe2+, so the D option is wrong.

It should be 33 kinds. It's just a few minor mistakes upstairs, though. Although the structure of H2O2 is polylinear, H-O-O-D is the same substance as D-O-O-H (which can be flipped).

H2O(16), HDO(16), HTO(16), DTO(16), H2O(18), HDO(18), HDO(18), DTO(18), D2O(16), T2O(16), T2O(18), D2O(18), Water 12 kinds.

h-o-oh, h-o-od, h-oot, d-o-od, t-o-ot, d-oot, and then say that the consideration of o2 is o2 (16), o2 (18), o(16)-o(18), that is, 3 * 6 = 18 kinds of h2o2

There is a special case h-o(18)-o(16)-d that is not the same as d-o(16)-o(18)-h, so add 3 more.

Therefore, the number of compounds formed is 12 + 18 + 3 = 33.

In terms of H2O alone, there are 12 types and 18 types of H2O2, making a total of 30 types (it is worth noting that the structure of H2O2 is polygonal, and H-O-O-D and D-O-O-H are different substances).

Now h, d, t, instead of 1 h, 2 h, 3 h three isotopes.

H2O(16), HDO(16), HTO(16), DTO(16), H2O(18), HDO(18), HDO(18), D2O(16), T2O(16), T2O(18), D2O(18), H-O-OH, H-O-OD, H-Oot, D-O-OD, D-O-OH, D-Oot, Consider O2 (16), O2(18), O(16)- O(18), and 18 isomers of H2O2 can be obtained.

t-o-ot,t-o-od,t-ooh

Therefore, the number of compounds formed is 30