Chemistry simple questions for high school students will be required tonight

One question = power multiplied by 10 (30*.)

1.First of all, you have to explain what exactly is the substance? Liquid solution can calculate the number of ions, right? Where do solid ions come from?

Assuming that the concentration of this topic is 1mol ml, then the concentration of Na+ is: Alfogadro's constant.

2.Let the diluted concentration be x, then.

30*, so x=30*

The calculation process for the fourth question is similar to that of the second question.

Question 1: The amount of 1mol of substance is to the 23rd power, which is about 3mol*10 to the 23rd power....

Another molecule, Na2CO3, has two Na ions, so its number is also to the 23rd power.

1.What is NH3T+ in the title?

2.The reaction of matter obeys the law of conservation of energy and mass, the total number of atoms does not change, the total number of protons does not change, the total number of electrons does not change, the molecule does not necessarily, so it is the total number of molecules that may change.

3.First of all, the sum of the outermost electrons of xyz is 13, the outermost electrons of z are 3 times that of the inner electrons of x, then the inner shell of x can only be 2, and the x atom has only two layers of structure, then the outermost electrons of z are 6, and the outermost electrons of y are only 2 electrons that can be obtained from this, therefore, the outermost electrons of x have 13-2-6=5, so x is n, because the original quarrel number of the ordinal number increases, so y is mg, then z is s, so the choice is b, a is incorrect, NH3 is alkaline when dissolved in water, C is incorrect, it will be volatile, the car can also have a redox reaction with O2, it will deteriorate, D is obviously wrong, it is a strong acid.

4.From the composition of the compound, it can be seen that the valency of x is +3, then it may be n or p or al or b (boron closed bi), then y may be o or s or, so a may be b(5) and s(16), b may be al and s, d may be al and o, so c is not possible.

The relative density of a hydrocarbon to hydrogen is 36, then the molecular weight of the hydrocarbon is 72 and only c is compliant.

The same volume of gas indicates that the amount of matter is the same, assuming that they are all 1mol gas, and CO burns C from +2 to +4, so the number of transferred electrons is 2mol and C in CH4 is -4, which becomes +4 in CO2, so the number of transferred electrons is 8mol, so the ratio of the number of transferred electrons is 1 to 4

1.1:4 The same volume of gas indicates that the amount of matter is the same, assuming that they are all 1mol gas, CO burns C from +2 to +4, so the number of transferred electrons is 2mol and C in CH4 is -4, which becomes +4 in CO2, so the number of transferred electrons is 8mol, so the ratio of the number of transferred electrons is 1 to 4

2.c The relative density of a hydrocarbon to hydrogen is 36, that is, the relative molecular formula of a hydrocarbon is 72, so C is chosen

C5H12 + 11O2 = = 5CO2 + 6H2O g C5H12 i.e., CO2 produced, i.e., 22g

1. 1:4 (after writing the chemical equation, it can be compared quickly).

2. c (The relative density of a hydrocarbon to hydrogen is 36, then its relative molecular mass is 72.)Let the number of c atoms in the molecular formula of the hydrocarbon be x, then 72 solution yields x=5Therefore, the hydrocarbon has 5 c atoms and 12 h atoms in its molecular formula. ）

Complete reaction, i.e., complete precipitation of silver ions, in silver nitrate solution, c(ag+)=

The volume is, so n(ag+)= is determined by ag++cl-=agcl so n(cl-)=

The volume of the BaCl2 solution is so c(Cl-)=

bacl2 + 2agno3 = 2agcl ↓ ba(no3)2x

The solution x=c=the concentration of the substance is .

Use conservation to solve it.

When the reaction is complete, the silver ions and chloride ions are completely precipitated, and AGCL, there are as many silver ions as there are chloride ions. The amount of silver ions is 0 01mol, then the chloride ion is also 0.01 mol, so c 0 01 50 10 (3) 0 2mol l

The outermost number of electrons of an atom c is equal to four times that of d, so d is the first main group and c is the fourth main group; ac is the same main group, and the atomic radius c a, so, c is silicon and a is carbon; c The number of protons in the nucleus is equal to the sum of the number of ab capitalists, and b is oxygen; The atomic radius d c, d is sodium.

h2o2co+sio2=2co2+si

Draw a location map. a b same period on a horizontal line.

Radius a》b indicates that a is on the left side of b.

ac is on the same vertical line as the main family a c.

The number of outermost electrons of a c atom is four times that of d, and the number of electrons in the outermost shell of c is either 4 or 8 8, and the noble gas is not interesting, so c is carbon or silicon, and it is easy to continue the analysis.

a:cb:o

c:sid:na where h20 is the most stable liquid state, c02 is the most stable gas, if you can't push it out, you can take care of the periodic table, he is quite useful, because the time is limited, there may be something inappropriate, please understand.

1. A is carbon, B is oxygen, C is silicon, and D is sodium. 2. They are: CH4, H2O, NaH, silicon has no stable hydride CO2 + 2 Na2O2 = 2 Na2CO3 + O2

C is silicon, D is sodium, A is carbon, B is oxygen, CH4

2co+sio2=2co2+si

A is carbon, B is oxygen, C is silicon, D is sodium, CO2, H2O,

It is an oxidation reaction.

b is the galvanic cell principle.

c. Oxidation reaction occurs between red-hot iron and water.

d is the electrolytic cell principle.

Plastics are difficult to corrode.

b Corrosion is the fastest, because iron is more reactive than tin.

c Corrosion is slower, so zinc is more active than iron, and it corrodes zinc first.

Both of these substances are relatively stable and generally difficult to corrode.

When the volume of the gas decreases, the h-coefficient of the average molecular formula of these three hydrocarbons is less than 4

FeCl3, 3-valent iron ions oxidize elemental copper to 2-valent copper ions under acidic conditions, 2Fe3+ Cu===2Fe2+ Cu2+

Copper nitrate and potassium nitrate bar. NO3- and SO4

2+ ions can form nitric acid, which is more oxidizing than dilute sulfuric acid.

Answers 1, 2, 3, 4

After adding these, the copper can be dissolved.

1, 2, can have trivalent iron, copper and ferric iron reaction to form ferric iron and copper.

3, 4, dissolved in solution, equivalent to nitric acid, copper and nitric acid reaction.

All FeCl3 + Cu==CuCl2 + iron oxide is first dissolved in dilute sulfuric acid to form Fe2(SO4)3, which can undergo redox reaction with copper to generate FeSo4 and CuSo4. The above two reactions take advantage of the oxidizing properties of Fe3+ containing nitrate ions, which have strong oxidizing properties (nitric acid) in solution and can dissolve copper. Thank you!

Option 34 introduces NO3- to form nitric acid, and HNO3 can be drunk CU reaction.

cu(no3)2

Kno3 is ionized by the combination of NO3- and H+ ionized by sulfuric acid to form nitric acid.

1) Dissolve the powder in water to obtain a colorless solution and white powder.

The colorless solution means that there is no CuCl2, and if there is a precipitate, it contains Ba(NO3)2, and only its cation can generate a precipitate.

2) Dilute nitric acid is added to the filtered precipitate, and some of the precipitate is dissolved, and colorless gas is produced at the same time, indicating that the precipitate contains BaCO3 and BaSO4

3) The filtrate was taken for color detection reaction, which could prove that the filtrate contained Na+ and did not contain K+, which was inferred from the above phenomenon.

1) The mixture must contain Na2CO3, Na2SO4, BA(NO3)2

It must not contain CuCl2, K2CO3, and K2SO4 may contain NaCl

2) How to check if it exists.

Take the filtrate in (1) and add an appropriate amount of silver nitrate and dilute nitric acid, if there is a precipitate formed, the solution contains NaCl

Copper chloride is excluded from the colorless solution.

Does not contain potassium ions, excludes potassium carbonate and potassium sulfate.

The precipitate contains salts insoluble in nitric acid, there must be barium nitrate and sodium sulfate (barium sulfate is insoluble in nitric acid) precipitate contains carbonate, and there must be sodium carbonate (barium carbonate insoluble in water) sodium chloride is unknown.

Sodium chloride test:

The solution is acidified with nitric acid to add silver nitrate, which, if precipitated, contains sodium chloride.

2kmno4 + 16hcl ==2kcl + 2mncl2 + 5cl2↑ +8h2o

Potassium permanganate so chlorine ie.

ag(+)cl(-)agcl↓

There are chloride ions left.

It turns out that there are chloride ions in total.

Concentration of raw hydrochloric acid.