A chemistry question for senior one, a chemistry question for senior one

Generally speaking, an outer electron number of 8 is a stable structure, and atoms have a tendency to make their outermost electrons become 8-electron stable structure. The X element with an outer electron number of 3 can gain 5 electrons or lose 3 electrons (the subouter shell is generally 8 electrons) can make itself a stable structure with the outermost shell of 8 electrons, because it is much more difficult to get 5 electrons than to lose 3 electrons, so it often loses the outermost 3 electrons and becomes a stable structure with the outermost 8 electrons, showing a valency of +3 valence.

In the same way, if the outermost electron number of the Y element is 6, it is common to get 2 electrons and become a stable structure with the outermost electron of 8 electrons, showing that the valency is -2 valence.

Questions added. The chemical formula of gaseous hydride of this element is.

Here x is +4 valence, but it cannot be said whether 4 electrons are gained or not lost. In the case of ionic compounds, the valency is expressed as the number of electrons gained and lost, and this XH4 is a covalent compound, and the valency is expressed as the number of shared electron pairs composed.

For some atoms (if the outermost electron number is 4, hydrogen outermost electron number is 1) is not easy to get a certain number of electrons, nor easy to lose a certain number of electrons, then it has to form a shared electron pair with the outermost electrons of other atoms, and the common electron pair formed is shared by two atoms, so as to meet the stable structure that both atoms can reach 8 electrons (only 2 electrons for hydrogen), and the electron pair is biased between the two nuclei, and the biased side shows negative valence, The side that deviates shows positive valence.

XH4 is a covalent compound composed of an X atom and 4 H atoms, the outermost 4 electrons of X and the outermost 4 electrons of 4 H atoms form a stable structure of 4 shared electron pairs (a total of 8 electrons) to meet the outermost 8 electrons of X, and at the same time, the outermost 1 electron of 1 H atom and one of the 4 electrons in the outermost shell of X atom form an electron pair, which is shared by X and H and satisfies the 2-electron stable structure of H atom. Since the nucleus of x is larger than that of h, the electron pairs are biased towards the x atom, so x shows -4 valence and h shows +1 valence.

In general, an element can remain in a steady state after it completely loses its outermost electrons or the number of outermost electrons reaches 8.

Element x has 3 electrons in the outermost shell, which is +3 valence after complete loss, and y has 6 electrons in the outermost shell, and 8 electrons after 2 electrons are obtained, which can remain stable at -2 valence. So the answer is b.

Its practical special case method can be considered as al2o3

In the supplementary problem, the outermost shell of x is 4 electrons, so you can get 4 electrons to become -4 valence, or you can lose 4 electrons to become +4 valence, depending on the redox ability of other elements in the compound.

Similarly, you can use c (carbon) to specialize x.

The outermost electron is less than 4 and is volatile; More than 4, it is easy to get electrons.

Therefore, the x element is +3 valence, and the outermost shell of the y element has six electrons, so it is easy to get 2 electrons to form a relatively stable structure, and the original y element if (there are 2+6=8 electrons), and then get 2, there are 10 electrons, 8-10=-2

So it's -2 valence.

And because the sum of the valencies of the elements in the compound is 0, so: (+3*2)+(2*3)=0

So choose B

The outermost electron of x is 3, the volatile electron x is +3 valence y is 6, and the easily obtained electron y is -2 valence. A x2y x2 is not the same as y with valency and cannot form a compound while b can.

The number of outermost electrons in x is 3, which loses 3 electrons and is +3 valence, and the outermost electron number of y is 6, which gives 2 electrons and is -2 valence.

The order of reactions is.

2Na2O2 + 2H2O = 4NaOH + O22AL + 2NaOH +2H2O = 2Naalo2 + 3H2 The resulting gas is So the total mass of H2 and O2 is, after which hydrochloric acid is added

NaOH + HCl = NaCl + H2O (The reaction of NaOH on the top side may be excessive.)

naalo2 + hcl + h2o = nacl + al(oh)3

Al(OH)3 + 3HCl = ALCl3 + 3H2O just makes the resulting precipitate completely dissolved, remember that in order for the precipitate to dissolve, you have to react NaOH to get it, and the final solution is a mixture of AlCl3 and NaCl, and according to the conservation of atoms, set the unknowns to Almol and Na2O2 to ymol

Gas mass column.

32*y/2+2*3*x/2=

Cl- atom is conserved, so (chlorine atoms are all from hydrochloric acid * mol) 3x+2y=

Simultaneous solving. x=，y=

x： y= ： =1 ：2

The ratio of the amount of Al to the amount of Na2O2 is 1:2

ps To do this problem, you have to analyze the process, and the goal is to analyze the gas and the final solution, and the problem is actually insignificant, and only two equations are needed: 32*y 2+2*3*x 2=

3x+2y=

Let the mass of aluminum be x and the mass of sodium peroxide be y

From the title, it can be seen that the solutes in the final solution are only aluminum chloride and sodium chloride, Al-3HCl, Na2O2-2HCl, 2Al-3H2, 2Na2O2-O2

x 3x y 2y x y

3x+2y=

2*x= y=

First, the gas is produced because of the reaction between sodium peroxide and water, the gas is oxygen, and there is enough water, which means that the reaction of sodium peroxide is complete, so the mass of the gas is used to calculate the quality of sodium peroxide.

Second, in addition to sodium peroxide and water to produce oxygen, sodium hydroxide is also produced, and sodium hydroxide reacts with aluminum to form sodium metaaluminate, forming a five-color solution, indicating that aluminum has also been consumed.

Thirdly, the aqueous solution of sodium metaaluminate and hydrochloric acid first produces aluminum hydroxide precipitation, and then continues to add hydrochloric acid to generate aluminum chloride, which is soluble in water. The amount of hydrochloric acid can be used to calculate the quality of aluminum.

If you set the precipitate to 100g, you can find that the sample o taken away by CO is 16g. The original sample was 58.

Fe is 58-16 = 42g.

Then the molecular weight ratio of Fe to O is 42 56:16:16=

That is, the molecular weight ratio of Fe to O is 3:4

See here. We need to pay attention to the characteristics of this question.

Item A is not right, why.

We can see that Fe3O4=FeO*Fe2O3, that is, the atoms contained in a molecule of Fe3O4 are the same as those of FeO and Fe2O3 in a molecule.

This represents the product that may be mixed with FeO and Fe2O3 in the raw material.

Here I will talk about molecular weight, or at the level of high school, I should say moles. Thank you for your understanding.

Item C is also incorrect, the theory is the same as A, which can be satisfied when FeO and Fe2O3 are mixed.

The D term is not true, because we calculate the mass ratio, if there is FeO, then the Fe content will increase and the oxygen content will decrease.

Because the Fe content in FEO is lower.

But this premise is fe we count according to 56. This also depends on the rigor of the question. The molecular weight of FE is not an integer. But that's probably not the point of consideration here.

So it can only be term B, which may be Fe3O4, or it may be a mixture of FeO2O3 of equal molecular weight, or it may be a mixture of Fe3O4 and FeO2O3 of equal molecular weight, that is, it may be a mixture of FeO, Fe3O4, Fe2O3.

To sum up, select item B.

I hope the landlord will adopt it. It's not easy to type these 500 words. Ha ha.

b may be ferric tetroxide.

Assuming that the original compound is Feox, its molar mass (relative molecular mass) is (56+16x)g mol, and the resulting CaCO3 is (100x) g. The following equation can be obtained, (56+16x)*50 29=100x, and x=4 3 is solved, so it may be Fe3O4 or a 1:1 mixture of FeO and Fe2O3.

Hopefully it will be adopted.

In this case, I think we should use the extreme value method!! First of all, I assume ferric tetroxide and do the math to see if it is the same as it!! (Given a relatively simple data, it's easy to calculate!!)

See if it's the same (I don't think it should be a so it should be different), if not, let's say it's iron oxide,,Calculate it again,,See if that value is in the middle of the two values you calculate,,If so, choose D!! (I didn't count the specifics),,You can also use that elimination method,,A is right, C must be right,,So don't choose both,(Chemistry questions are multiple choice questions) In this case, you can use the above method again!! You can figure out whether it's B or D!!

Hope it helps!!