I want all the chemistry equations in high school, and I will be paid a lot of money!!! 100

You go to the library, it's very comprehensive, really, you can take a look if you don't believe it, after all, a person is incomplete.

o(oThank you.)

The landlord sent an email address, and I sent it to you.

Solution: V liters of sodium sulfate solution contain a gram of sulfate ions, then v liters of sodium sulfate solution contain.

The amount of the substance of the sulfate ion = a

96 (96 is SO42-.)

molar mass) then v liters.

Sodium sulfate solution.

The amount of the substance of the sodium ion = (a 96) *2

Take this solution v 2 liters, then.

The amount of the substance of the sodium ion removed = a 96

Dilute with water to 2 v liters, then.

The amount of sodium ions in the diluted solution of the substance concentration =

a÷96)÷2v

Options. Hydrochloric acid reacts with sodium carbonate and sodium bicarbonate, so it cannot be selected.

Option b. 2NaHCO3 is heated to generate Na2CO3+CO2+H2O while sodium carbonate does not react. Therefore, choose B

c option. NAOH reacts with NAHCO3. However, the addition of NaOH introduces new impurities. Hence the error.

d option. Na2CO3 + CO2 + H2O to produce 2NaHCO3 which is the removal of Na2CO3. Therefore, it cannot be chosen.

Broad definition of combustion: combustion refers to any violent reaction of luminescence and heat, which does not necessarily require oxygen to participate, such as the reaction of sodium metal (Na) and chlorine (Cl2) to form sodium chloride (NaCl), which does not involve oxygen, but is a violent chemical reaction that emits heat and heat, and also belongs to the category of combustion. It doesn't have to be a chemical reaction, such as the combustion of nuclear fuel.

It contains Na+, which will interfere with the flame color reaction, and sulfuric acid in C does not volatile.

And d hydrochloric acid is volatile.

Heating NaHCO3 pyrolysis.

Combustion must have oxygen involved, because mg can be burned in CO2, H2 can be burned in Cl2, etc.

3 d.Dilute hydrochloric acid removes the platinum wire that has been oxidized, will not interfere with the flame color reaction, Na+ will cause interference, and sulfuric acid is easy to oxidize platinum.

1 Option B removes NaHCO3 and generates NA2CO3 at the same time

2 Choose d to give a counterexample, e.g. mg burns in n2.

3 Select DA and B contain Na+, which will interfere with the flame color reaction, C sulfuric acid is not volatile, high boiling point, and its salt is not easy to volatilize at high temperature, and it is impossible to wash the substances adhered to the platinum wire during washing. The hydrochloric acid is volatile, and its salt is also volatile at high temperature, and when burning, hydrochloric acid and its salts can be volatilized together to achieve the purpose of washing.

1a2d3d

After sodium bicarbonate is heated, it is decomposed into sodium carbonate, carbon dioxide, and water does not produce new impurities, and magnesium and carbon dioxide are also combustion reactions, but there is no oxygen to participate.

On the one hand, acid is used to remove metal oxides from the surface of the platinum wire.

On the other hand, hydrochloric acid is a volatile acid, and its salts are also easily volatilized and removed when burned.

In short, the impurities on the platinum wire can be well removed, which is convenient for completing the experiment.

no, that is, from the original no3-, to the current no, the total number of electrons = will be obtained.

In aluminum-mafic alloys, they were all elemental, but now they are all ions, and they should lose electrons in total, that is, they have a positive charge. So when it comes time to react with OH-, it should react with OH-.

But the solution may also contain H+, which is conserved by charge: n (positive charge) = n (negative charge), and n (positive charge) = n (h+) n (negative charge) = n (residual no3-) = n (original total hno3) - n(no) = - mol = mol

n(h+)= mol

So in the end, n(naoh) = n (negative charge) + n (h+) =

So V(NaOH) = 3 mol L = L = 150 ml

This question is calculated according to the conservation of electrons.

The electron loss of the metal is equal to the electron gain of the NO3- ion, and the metal ion in each valence state is combined with an oh- to form a precipitate! ）

no3- no (3 valence).

Hence no3- 3e- 3oh-

n(no3-)=

n(oh-)=

The amount of nitric acid can be obtained from n=c*v, so the amount of n element substances is Because the amount of nitric oxide substances can be found as, so the amount of n substances in nitric oxide is So the amount of total substances of n in the generated aluminum-mafic iron salts is Because the ratio of salt to alkali precipitate is anion and cation, so the amount of NaOH is equal to the amount of n in salt = Because sodium hydroxide is 3mol l, v=150ml is obtained from v=n c

According to the conservation of nitrogen, the nitrogen in nitric acid is eventually converted to NO and sodium nitrate.

Let the volume be v, then.

Get 150ml

There is no way to do this, and teaching you the method is not right away, mainly by writing more questions by yourself, and after writing more questions, you will form your own thinking pattern, and you will summarize the method.

(1) The amount of hydrochloric acid is:

The amount of the substance of iron is:

So there is an excess of acid.

Hydrogen is generated at 2min, that is, according to the reaction equation, the molar of ferrous chloride is calculated as:

So the reaction rate c= v=c t=. min

2) 4min is the iron powder is completely dissolved, should be calculated by the amount of iron substance, when consuming iron, consume hydrochloric acid.

So c= v=2 4=. min

n(hcl)=m/m= n(fe)= n(h2)=2hcl + fe ==fecl2 + h2v(fecl2 )=

After 4 min, Fe is completely reactive, and the amount of Fe is used to calculate the amount of the substance that reacts hydrochloric acid, that is, the amount of change.

v（hcl ）=

In fact, the total amount of hydrochloric acid is of little use in this question.

It seems that someone has already sent it. Don't explain it.

Suppose that 1 volume of water absorbs x volumes of ammonia.

The water is set to 1L, that is, 1L of water absorbs xL of ammonia.

Then, the mass of ammonia after the completion of absorption m 1000

Its volume v m 915(l)(

Its molar concentration c=(x =

Joining the above equations, we can solve x =

Therefore, 1 volume of water should absorb about the volume of ammonia.

Ask the teacher if you want it, and it will usually be sorted out.

Just memorize the chemical equations and ionic equations in the book.

Count you ruthless, how many equations are there in high school, how can you finish writing .........