Chemistry is superior. High school chemistry master

D is Na, because the yellow solid Na2O2E is generated by burning in the air is H, and only the nuclide of H is neutron-free.

B and A can form Ba4, C has C2 elemental substance, A has A2 type elemental substance, A, B, and C are adjacent, so A is O, B is S, C is Cl, and 8+16+17=41

Answer: 2kmNO4=δ=K2mNO4+MNO2+O2 2(Cl-)+2H2O=Energized=Cl2 +H2 +2(OH-) First drop into the solution into dilute HCl, there is no obvious phenomenon, and then drop into the BACL2 solution, there is a white precipitate generated, which proves the existence of SO42-.

973190230 there is a low-level error, if acidification with nitric acid will oxidize SO32+ to SO42- resulting in a misjudgment, so the test of SO42- can only be acidified with hydrochloric acid.

First of all, d can be introduced as the element of na, and a nuclide of e without neutrons, then e is the element h. According to "in an adjacent position. And the atomic number of the three elements is between 41 "" and the atomic number of d, and "Ba4(2-)" can be deduced A as an O element, B as an S element, and further C as a Cl element.

So A2 is oxygen, and the equation: 2H2O=2H2+O2 Conditional electrification to chlorine gas: 2Cl- +2H2O==Cl2+H2+2OH- (gas symbol is added by yourself).

Detection of barium nitrate acidified with nitric acid for sulfate, and the presence of white precipitate can prove the presence of sulfate.

The conditions under which the metathesis reaction takes place: the reaction of precipitation, gas, weak electrolyte (water, weak acid, etc.) is generated, Na2FeO4 + 2 Koh = K2FeO4 + 2 NaOH

If there is no gas, weak electrolyte is produced, then there must be precipitation.

NaOH is not a precipitate, it is only possible that K2FeO4 is a precipitate.

Solubility of the precipitate Solubility of the soluble substance.

So the solubility of K2FeO4 is the solubility of Na2FeO4.

Generalization: Metathesis reactions are always carried out in the direction of decreasing solubility or decreasing ion concentration in solution.

Potassium ferrate can be prepared by adding a certain amount of KOH to the sodium iron rate solution at low temperature, and the reason why it can be prepared is because the solubility of potassium iron rate is smaller than that of sodium iron rate.

One. Take it all.

Two. : 30 * 10000 m2 * 5g m2 = 3 * 10 6g three. 20 drops of water are about 1ml, and if the density of water is 1g, 20 drops of water per milliliter are about 1g, and if each drop contains x water molecules, there are:

20*x*pcs=1g=10 -3kg, which yields: x=

1. Water freezing (physical change); Iron rust (chemical change) two: 3 * 10 to the 5th power * 5g = 6 grams of grams.

Three: first convert the mass of the water molecule to the minus 23 power of 3*10, and calculate that the mass of 20 drops of water is 1g, then the mass of one drop of water is.

Finally, the minus 23rd power of the minus 21st power.

In the HCl solution with pH = 5, in the NaOH solution with C(H+) = 10 -5 mol LPh=13, C(OH-) = mol L If the volume of the mixed HCl solution and the NaOH solution are both V L, then there is a surplus of OH- after mixing, and the remaining N(OH-) 10 -5 * V = H+ in HCl can be ignored).

c(oh-) = after mixing

Dilute another 5-fold postc'(oh-) =

The resulting solution pH = 12

pH = 5 H+ in HCl solution.

pH = 13 for NaOH solution OH- is.

After the reaction, oh- remains.

After dilution, oh-=

Then the corresponding h+ is 5*10e-13

Take -log to have a logarithmic table that should be given).

The pH of the resulting solution is.

Mix in equal volumes to obtain a hydroxide ion concentration in the solution [oh-]=10 -1 - 10 -5)2 approximately.

Dilute 5-fold again, and the pH is approximately 12.

Exact solution: ph=14+lg[ (10 -1 - 10 -5) 2 ].

It seems that you should belong to the kind of students with better grades, and I would like to give you some advice, I hope it will help you.

1.The difficult questions should be done appropriately, but don't do it excessively, you can analyze the chemistry questions of the high school entrance examination in various places, the difficulty is only medium, and the high school examination questions in Changde, Hunan this year are super simple.

2.It is necessary to pay attention to the basic knowledge in the books, there are many reference books now, and there may be many things that are different from what the books say, so we must follow the books at this time.

3.Regarding the choice of reference books, it is recommended that you ask your teacher, which is not easy to say, don't buy too many reference books, that is to say, I don't think the sea tactics are suitable. The key is to master the question type and grasp the basic knowledge.

I recommend doing more basic questions, because most of the high school entrance examinations are basic questions, and there are one or two questions at most about 10 points.

I deducted 2 points for chemistry in the high school entrance examination. However, I never did too many questions and only completed the tasks assigned by the teacher. Chemistry in junior high school is actually not difficult at all, and there is no need to be nervous at all.