Ask the master to do a high school chemistry problem, ask for a high school chemistry problem

1) The number of outermost electrons of the --- z atom is 3 times that of the subouter electron.

The number of electrons in the subouter shell can only be asked 2 or 8, 8 is obviously not right, so the number of electrons in the secondary outer shell 2 and the outermost shell is 6, so z is o;

Then find the breakthrough information--- y and z are adjacent elements with the same period; There are 3 covalent bonds in the hydride molecule of Y.

Only N and F are adjacent to O, and the atomic numbers of the four elements --- X, Y, Z, and W increase in turn--- so Y is n;

Knowing y and z, continue with the number of protons of the w atom equal to the sum of the outermost electrons of the y and z atoms...

W can only be na ......

Finally, x and w are the same as the main group, and the atomic numbers of x, y, z, and w increase sequentially.

x can only be h;

2) Look at the topic of the two elements; It can be dissolved in water to produce gas (such information can also be understood as: it can react with water to produce gas. That's what the title means); The aqueous solution is alkaline.

In relation to what you have learned, it can only be Na2O2, 2Na2O2 +2H2O=4NaOH+O2

3) The ionic compounds formed by x, y, and z are Hno3; The hydration of the highest oxide of W is a dilute solution of the oxide of W The oxides of W are Na2O and Na2O2, and their hydrates (products dissolved in water) are all NaOH;

Reaction formula: HNO3+NaOH=nano3+H2O

1) x is h, y is n, z is o, and w is na

The number of outermost electrons of the z atom is 3 times the number of electrons in the subouter shell of 2 electrons in the subouter shell and 6 in the outermost shell. So the proton is 8, element O

There are 3 covalent bonds in the hydride molecule of Y. y and z are adjacent elements of the same period. The outermost 5 electrons of y are of the same period as o and are n

The number of protons of the W atom is equal to the sum of the number of outermost electrons of the Y and Z atoms. o The outermost 6 electrons n the outermost 5 electrons, so w is element 11 na

The four elements x, y, z, and w are the elements of three consecutive short periods in the periodic table, yz is the second period, w is the third period, and the atomic number of x is small, so it is the first period. And x and w are the same as the main family, so they are h

2）na2o2 2na2o2 +2h2o=4naoh+o2

d Resolution:

According to the NaHCO3 solid is fully heated, the thermal decomposition of the base spring to generate carbon dioxide and water. Carbon dioxide and water react with sodium peroxide to form sodium carbonate and oxygen. When x, there is an excess of carbon dioxide in the gas, and the highest final content of the cavity has (, and the solid contains (x+

NH3 + HClL = H2O + NH2Cl (reversible reaction), HCl is an effective substance, and HCl is unstable, easy to decompose at the beginning of adding NH3 can make the reaction move to the right to generate a relatively stable NH2Cl, and when the HCl in the water is consumed, the balance begins to shift to the left, and continues to generate HCl, in order to achieve the effect of continuous sterilization and sterilization.

In fact, balance topics can be mastered by doing more and thinking carefully by yourself.

Because NH2Cl is more stable than HCl, NH2Cl+2H2O = NH3·H2O+HClL (reversible), NH2Cl is continuously hydrolyzed to form HCl, which can prolong the sterilization time.

HClO is an active substance that shifts to the left as the HClO decomposes and balances, allowing HClop to persist in water for a long time

Awesome ......The upstairs are so strong.

1) If the original oxygen is 1mol, the mass is 32g, and the residual oxygen is reacted to generate ozone, so the total gas in the system is.

2) Since 3 L of oxygen reacts to produce 2 L of ozone, the volume of reduction is 1 L. It is known that the volume of the reaction is reduced by (, then the ozone produced by the reaction is 3L.

3) The amount of copper is: 20g 64g mol=;

The material force of the gas mixture is: because ozone can only react with copper, from which it is judged that there is an excess of copper! The gas reacts completely.

From the conservation of mass, we can know the nature of the oxygen atom in the gas. Let oxygen be xmol and ozone be ymol, then there is:

x+y=；2x+3y= solution x=; y=。

Note that the third question must determine whether the gas is completely reactive, otherwise the answer is not rigorous.

1. It may be assumed that the original oxygen has 10a mol, the oxygen after the reaction has 7a mol, and the ozone has 2a mol, and the total mass of the original oxygen is 10a * 32 = 320a g, so the average molar mass of the mixed gas is now 320a (7a + 2a) = 320 9 = g mol

2. The gas is reduced, and there is a chemical equation that if 2L O3 is not generated, the volume of the gas will be reduced by 1L, so the volume of the generated O3 is x 2 = , x = 3L

3. The amount of substance of the mixed gas is = mol, and before and after the reaction, the mass of the solid increases = g

The average molar mass of the original mixture is = 40 g mol, and the volume fraction of O3 should be y.

48y+32(1-y) = 40, y=50%.

3）n=

m (gas mixture) =

m (gas mixture) =

If the volume ratio of ozone to oxygen is (40-32) (48-40)=1 1, the volume fraction of ozone in the original gas mixture is 50%.

2）o2---2/3o3

3 of which ozone 3 liters.

1) m (average) = n (mass and ) n (sum of the amount of matter) = 100 (70 32 + 30 48) =

It should not be said that adding saturated potassium bromide solution to the silver chloride suspension should be dropping a few drops of potassium bromide solution, observing the phenomenon, the white precipitate is transformed into a yellow precipitate, so as to get the conclusion that the solubility of silver bromide is smaller.

Because even if the solubility of silver chloride precipitation is small, there will still be ionization, and there will also be precipitation dissolution and conversion equilibrium. Since it is equilibrium, it will follow the principle of Le Chatre.

Precipitation transformation equilibrium.

AGCL + BR == AGBR + CL When the concentration of bromine ions is large, the equilibrium shifts back to the right, even if the solubility of silver bromide is less than that of silver chloride, it will still form a yellowish precipitate, and the solubility cannot be concluded.

This also has an application in the removal of impurities.

(1) A mixture of benzene and liquid bromine.

2) C6H6+BR2 (arrow, FE written above) C6H5BR+HBR reaction solution is slightly boiled, and reddish-brown gas fills the A container.

3) Removal of bromine dissolved in bromobenzene.

BR2+2NAOH ====NABr+NABro+H2O (or 3BR2+6NAOH==== 5NABr+NABro3+3H2O).

Ferrous ions, bromine ions.

Distillation (4) removes bromine vapor from hydrogen bromide gas.

5) Replace the reaction with the liquid into the pipette and stabilize at a tiny height.

Litmus solution. The solution turns red.

The ratio of the relative atomic masses of the atoms of elements A and B is 7:8, while the mass ratio of elements A and B in a compound is 7:4, indicating that A:B = 2:1

Option A is yes, choose A

ecln + n agno3 === e(no3)n + nagcl

1 n* can be found n=2 ecl2

Therefore, the valency of E is +2 at its boiling point and the melting point is low, which shows that it is a molecular crystal, and E is connected by a covalent bond with Cl, so it can be seen that E cannot be the second main group (if it is the second main group, it will form an ionic crystal). Depending on its melting point the value of the boiling point of Ecln may be Scl2

From the molecular formula, it can be known that the E element should be the second main group or the sixth main group. And because the melting and boiling point of chloride is low, the possibility of ionic compounds is excluded, so E cannot be the second main group element. Combining the previous two points, e is the sixth main group element (via).