Desperate for a chemistry problem to solve! Chemistry problem analysis, everyone help! Homework is u

W: Sodium, X: Aluminum, Y: Sulfur, Z: Chlorine.

1)al(oh)3+naoh=naalo2+2h2o2)na2s

3)so?+cl?+2h?

o=h?so?+2HCLHClLop + SO2 + H2O == HCL+ H2SO4 Detailed Answer:

1) From the first piece of information, our first reaction wx is acid and base, but the title says that wx is a metal element, so our first reaction is wrong, and then think that the most ** oxide corresponding hydrate of the two metal elements can react like this, then there is only amphoteric oxide alumina and another strong base, and the atomic number of x is larger than w, so we can know that w is sodium, and x is aluminum.

2) Since we have deduced that WX is sodium and aluminum, and the atomic number of YZ is larger than them, then the remaining scrap metal elements in this cycle are, silicon, phosphorus, sulfur, chlorine, etc., since WY can form W2Y, then we clearly know that Y is chlorine and W2Y is Na2S

3) If we already know that y is sulfur, then z can only be chlorine.

Very detailed Oh I hope it helps you, I hope you too! ~

A, false, did not say strong alkali or weak alkali, if it is a strong alkali, the alkali is not too much B, wrong, also if it is a weak alkali, then the alkali is excessive.

c, false, if it is a weak base, then the cation will be hydrolyzed.

D, yes, after the reaction, there are only M+ and A- and H+ and Oh- in the solution, at this time M+ +H+ = A- +OH-, because the solution is neutral, so H+ = OH-, so the concentration of A ions in the reaction solution is equal to the concentration of M ions.

The alkali excess solution is not neutral.

If it is weakly alkaline, it is not neutral.

The resulting salts are hydrolyzed.

The fourth one doesn't need to be explained.

From the conservation of charge C(M+)+C(H+)=C(A+)+C(OH-), it can be seen that when the concentration of a ion is equal to the concentration of M ion, the concentration of hydrogen ion is equal to the concentration of hydroxide ion, and the solution is neutral. As for ABC, the strength of the base is unknown, and there are different situations.

a If it is a strong base, it is the same amount.

b If it is a weak alkali, it should be less alkaline.

c It depends on which case it is AB.

d am equal, h+ and oh- can be equal, can be neutral I don't know if this explanation is correct, you can see for yourself.

No, the obvious answer is * 8 = mol.

s2- to s6+, loss of 8 electrons; N5+ to N4+, 1 electron is obtained.

There is a concentrated nitric acid in the reaction.

b According to the conservation of electrons s2- to s6+, 8 electrons are lost; N5+ to N4+, 1 electron is obtained.

So s:n=8:1, s=8n, i.e., the restored.

Answer: No answer.

It should be Moore.

The nitric acid that participates in the reaction is.

16 o2-

Note that the title says that Zheng Pituan is a particle grip, not an atom.

The number of outermost electrons of the ion is 8, so it is stable to call orange, so y=8, x=16

According to the title, the number of electrons in the first and second shells is 8, which is impossible, so the question is wrong.

There are three layers, which means that the first two layers are 2 and 8

The key to Namor is the last potato steak layer, x-y=8

2+8+y=x

So 2+8+y=y+8

Therefore, this number disturbs Chang and Li Sen is wrong.

It is the sixteenth element AR "Argon". There is a problem with your words, the nucleus of the Kaishan only has protons and neutrons, and outside the nucleus are electrons. You should say that the quiet withering is:

The number of electrons outside the nucleus is x or the number of nucleons is x. The idea is: the regularity of the arrangement of atoms, the first layer is 2, the second layer is 8.

Multiply the relative molecular mass of CO and subtract this number to get the mass of O that is more than CO in CO2, so dividing the number obtained at this time by the relative atomic mass of the O atom is the amount of matter of CO2!

The second question is, after the reaction between CO2 and NaOH, there is only CO!

Calculate all the CO first, and then find the part with more CO than CO2! Actually, I'll give you another mathematical method.

Let the quantity of the CO substance be x the amount of the CO2 substance is y

Establish the equation: 28x+44y=mass relation).

x+y=quantity relation of matter).

Solve the inequality again.

The answers you find, take your time.

Equation: 2NaOH + CO2 = Na2CO3 + H20NaOH + CO2 = NaHCO3

Because the amount of NaOH is sufficient, the carbon dioxide must be completely reacted, and the remaining gas is carbon monoxide, so Equation 1 is ignored.

Because carbon monoxide does not react, let carbon monoxide be xmol and carbon dioxide be ymol vertical; x+y=

28x+44y=

The final remaining gas must be xmol (the number of molecules contained) with a mass of 28x

Let there be x molco2,y molco in the mixture, and the system of equations is obtained: x+y=; 44x+28y=solution.

x=,y=After the mixed gas slowly passes through a sufficient amount of sodium hydroxide solution, the remaining gas is Co, CO2+2NaOH=Na2CO3+H2O, and the mass is m(Co)=; The number of molecules is n(co)=pcs).

The remaining gas is carbon monoxide, carbon dioxide and sodium hydroxide react and are absorbed.

Let the amount of carbon dioxide be x and the amount of carbon monoxide be y, x+y=

44x+28y=

The solution is x= y=

Mass: m=n*m=

Number of molecules: n=n*na=

Answer this question. You can't do it yourself.

Let me tell you about the knowledge.

Question 1: Equilibrium constant.

Question 2: Carbonate hydrolysis reaction (because bicarbonate hydrolysis reaction is the second step of carbonate hydrolysis reaction, it is OK to master carbonate hydrolysis).

There is no knowledge point in the third question: look at the question answer: (Obviously, it is all lead carbonate Basic lead carbonate and lead carbonate Lead carbonate and lead oxide are all lead oxide There are other Here are only four examples. Can you answer it yourself, and you don't need any knowledge points? ）

Question 4: Hint: Carbonate is heated and turns into an oxygen (Title: It decomposes into lead oxide no matter what is heated) (the mass is reduced by a mass of carbon dioxide 44, you can see the change in its mass by proportion).

k= c(so42－)/c(co32－)

Na2CO3 The large amount of C(CO23) in Na2CO3 solution is a mixture of 2PBC3·PB(OH)2 PBC3 and 2PBC3·PB(OH)2.

Experimental Procedure. Expected experimental phenomena and conclusions.

The sample is heated sufficiently to decompose, and the resulting gas is passed through a drying tube containing anhydrous copper sulfate.

If anhydrous copper sulphate does not change blue, then hypothesis one is true;

If anhydrous copper sulphate turns blue, then the assumption is not true.

Agree If all PBCo3 is used, the solid mass of MG is G after complete decomposition