There is a chemistry question in the third year of high school, and a chemistry question in the thir

Second question: The hydrolysis is weak, but CO32- and Cu2+, a weak acid group, a weak alkali metal ion, both will undergo double hydrolysis, which will be violent! The formation of Cu(OH)2 should be a double hydrolysis;

Q4: Cu(OH)2 H2O CuCO3 CO2X Y

x= y=i.e. n[cu(oh)]:n[cuco3]=1:1, so cu(oh) i.e. cu2(oh)2CO3 is the total mass of cuo and equipment, don't forget the glass tube

It is hydrolysis, which is not enough to generate hydroxide ions and hydrogen ions, and when the hydrogen ions reach a certain solubility, they will reflect with the carbonic acid and ions, and then promote the occurrence of hydrolysis;

Cu(OH)2 and patina are two different concepts, I forgot the molecular structure of patina You can check the book yourself, this device is heated in addition to generating Cuo and Cu(Co)3, which is also a substance that will precipitate,

Fe2O3 and NaOH to Fe(OH)3 precipitation.

Na2SiO3 hydrochloride was added to become H2SiO3 precipitate, and finally ammonia gas was added to make Al3+ become Al(OH)3 precipitate.

In the second step, there must be an excess of hydrochloric acid, and metaaluminate has become aluminum ions.

Hydrochloric acid silicic acid precipitation can separate silica. Excess sodium hydroxide and iron hydroxide precipitation can be added to separate ferric oxide. Finally, aluminum oxide can be separated by adding carbon dioxide aluminum hydroxide precipitation.

In this problem, the conservation of charge is used, not the conservation of electrons, Cr has no valence, and the cross method cannot be used, and the reaction of H+ and CRO42- with the excess of the anode, makes the ratio of the Cr element to the negative charge from 1:2 to 1:1, so that K+ moves to the cathode.

You can write the remainder as k6cr4o12, ===, so the ratio is 1:2, and the transfer electrons are counted as 1 according to h+; 1

1) The anode produces colorless gas, and the solution gradually changes from yellow to orange;

4OH -4E = O2 +2H2O or 2H2O-4E =O2 +4H+ (2) 2:1 ; mol

Chemical formula trim: 2SO2 + O2 + 2H2O = 2H2SO4 mass 64 * 2 18 * 2 98 * 2

50% sulfuric acid is generated, the mass of sulfuric acid is 196, and the water quality is 196; The total water consumption is 196+18*2, and the sulfur dioxide consumption is 64*2

Sulphur dioxide:water = 16:29

2so2+o2+2h2o=2h2so4

Quality 128 36 196

50% sulfuric acid is generated, the mass of sulfuric acid is 196, and the water quality is 196; The total water consumption was 232 and the sulfur dioxide consumption was 128

Sulphur dioxide:water = 16:29

First of all, H2S and SO2 cannot exist at the same time, and O2 and NO cannot exist at the same time.

If the gas is colorless, it is not Cl2 (yellow-green gas) or NO2 (reddish-brown gas);

The density is larger than that of air, indicating that there is at least one gas of H2S, SO2, CO2 and O2.

If the mixed gas cannot fade the wet magenta test solution, it does not contain SO2, and the gas color does not change after opening the cork, it does not contain NO;

Through the concentrated sulfuric acid, there is turbidity, and the gas volume remains unchanged, indicating that there is H2S and no NH3 (alkaline gas).

H2SO4 + H2S (gas) = SO2 (gas) + S (turbidity) + 2H2O (the volume ratio of H2S consumed to SO2 generated is 1:1);

Through soda lime, the volume is reduced to 10%, indicating that it contains CO2, and the total content of H2S and CO2 is 90%;

The mixture of gases may be **, which means that H2 and O2 exist at the same time;

Ignition: 2H2+O2=2H2O

Two parts of H2 and one part of O2 are accounted for, then H2 accounts for 5% and O2 accounts for O2;

Through the hot copper mesh, the copper mesh turns black, indicating that there is excess oxygen, then O2 accounts for a total of 5%;

The raw gas consumes 10% of the gas through Cuso4:

cuso4+h2s=cus+h2so4

If it contains 10% H2S, it contains 80% CO2;

Then the gas composition is: 5% H% O % H2S, 80% CO2;

In summary, the average relative molecular mass of the raw gas is:

From (1), no chlorine and nitrogen dioxide can be obtained.

From (2) the available gas is free of nitric oxide and sulfur dioxide.

From (3), there is hydrogen sulfide in the raw gas, but there is no ammonia. Hydrogen sulfide and concentrated sulfuric acid produce sulfur, so it is turbid; The same amount of sulfur dioxide is generated, so the volume does not change.

From the reaction of (4), hydrogen and oxygen can be obtained in the gas. A blackening of the copper mesh indicates an excess of oxygen.

Hydrogen sulfide volume fraction is stated by (5) 10%.

In summary, hydrogen and oxygen have an overall integral number of 10%. Only these two gases will not be absorbed by soda lime, but how can the remaining gas volume after these two gases ignite to be 25% of the original?

This proves that there is something wrong with the question.

(1) No Cl2 NO2

2) No SO2 NO

3) There are H2S (possibly CO2).

4) There is H2 O2 (possibly NH3).

5) There is H2S 10% (no CO2).

(1) It can be seen that the mixed gas is colorless and free of Cl2 and NO2; The density is greater than that of air and has a molecular weight greater than 29.

2) It can be seen that the magenta test solution that cannot be wetted by the mixed gas fades without SO2, and the gas has no color change after opening the cork.

No NO(3) is turbid when passing through concentrated sulfuric acid, but the gas volume remains unchanged; NH3 without H2S;

4) Ignition may have H2, O2, and O2 remaining, (5) combined with 3 can know that H2S accounts for 10%, while there is no CO2, comprehensive can know that there are H2S, H2, O2 three gases, of which H2S 10%, let the reaction O2 is A, so there is:

3a+25%+10%=100% to get a=

Average molecular weight = 32 * (25+.)

According to (1), there is no Cl2, and O2 is at most one. According to (2), there is no SO2, and there is no color change when the plug is opened. According to (3), since there is no change in the volume of concentrated sulfuric acid to know that there is no NH3, then what makes it turbid must be H2S, which reacts with concentrated sulfuric acid to generate the same amount of SO2 and generate S precipitation, and the equation you can match yourself, only H2 and O2 can survive after passing the alkali lime, and account for 10% of the total volume.

According to (4), knowing that there must be H2 and O2, and O2 remaining after ignition, accounting for the total volume, then some H2 and O2 react away, according to the equation of H2 combustion, knowing that H2 accounts for 5% and O2 accounts for O2, then O2 accounts for 5% in total. According to (5), H2S accounts for 10% (the reaction proceeds in the direction of acidic weakening, so CO2 does not react with CuSO4, as you should know), so CO2 accounts for 80%, then the average relative molecular weight is.

Fainting, how to learn chemistry in the rest of the buildings, CO2 and CuSO4 do not react, because H2SO4 + CUCO3 = CuSO4 + CO2 + H2O, this reaction is irreversible.

10% H2S353O21753H2 with an average molecular weight is.

This question is based on the basic principle of "strong acid to weak acid", or it can also be understood as "the ionization ability of large ionization capacity is weak".

Based on the known ionization constants, we can derive the acidic strength (ionization capacity) of these ions: H2SO3 > H2CO3 > HSO3- >HCO3-

Now let's analyze the options:

A: CO32- and HSO3-: In this group, HSO3- is more acidic than HCO3-, so a reaction occurs:

HSO3- +CO32- = HCO3- +SO32- (HCO3- is made from HSO3-, which can be understood as a strong acid to make a weak acid, or it can be understood as a weak acid with strong ionization ability), so it is impossible for a large number of ions to coexist in this group of ions.

B: HCO3- and HSO3-, the former corresponds to H2CO3, but H2CO3 is more acidic than HSO3-, so HSO3- will not react with HCO3- to obtain H2CO3; In the same way, because the acid corresponding to Hso3- is H2SO3, and HCO3- is much less acidic than H2SO3, it is even more unlikely that H2SO3 can be obtained by the reaction of HCO3- and HSO3-So this group can coexist.

c: SO32- and HSO3-: This group of ions coexists in equilibrium: HSO3- = SO32- +H+, which can coexist in large quantities;

The group of D:H2SO3 and HSO3- cannot coexist in large quantities because there is another equilibrium in the solution in addition to the equilibrium one: H2SO3 = H+ +HSO3-

H2SO3 = SO2 + H2O, when both are present in large quantities, SO2 gas will escape from the solution. It's easier to understand that way.

To sum up, it is BC that can coexist

It can be known from the conditions given.

2x + y = az (g) initial (mol) 2 2 0 reaction (mol) 2 m am equilibrium (mol) 2-2m 2-m am Since the ratio of the amount of matter x y z at equilibrium is 1:3:2 (2-2m):

2-m）：am=1：3：

2 yields m= a=1

A: If x is solid and y is gaseous, it can be seen that 2moly is a gas, and the sum after equilibrium is still a gas, that is to say, this is a reaction with constant volume, then the pressure is also constant, and the temperature is also unchanged under the same conditions, and the time taken to reach equilibrium is of course the same.

B: This option contradicts A. "x is solid and y is gaseous"This is a case where x and y are not both gaseous, and the average molar masses of i and ii are the same.

C: This option contradicts A. Under the condition of a, the volume of the container at equilibrium vld:

If xy is gaseous, the amount of total matter at equilibrium will be smaller than at the beginning. Because 2 1 3 gas molecules before the reaction and 1 gas molecule after the reaction, the amount of matter during the reaction decreases. And since the reactants and products are gases, the pressure of the gas will decrease during the reaction, and the average molar mass will decrease.

While the pressure of remains constant, the average molar mass should remain constant. The result should be:

a When x is a solid and y is a gas, the reaction can be seen from the equation as an equal volume reaction, and there is no difference between isochoric and isobaric reactions. So the time for the reaction to reach equilibrium is the same.

b (a), if only x is a gas, then this is a reaction with a smaller volume, the pressure of the container will become smaller when isochoric (i.e., less than the starting value), isobaric pressure as the name suggests is the same as the starting value, so the conversion rate is higher than , that is to say, the mass ratio of the gas is smaller, maybe the volume you notice has also become smaller, but there is a main factor here, that is, the change in the amount of the substance causes the change in volume, so the volume can be ignored, and the volume is considered to be unchanged, so the relative molar mass of the gas is less than , 2) If only y is a gas, Then the analysis is the analysis of option a, at this time the relative molar mass of the gas is equal to (3) If xy is not a gas then it is basically similar to the analysis of (1), and the conclusion is that the relative molar mass of the gas is less than . To sum up, if x and y are not both gaseous, then the average molar mass of the gas at equilibrium:

It is easy to see from B's analysis that the volume is closely related to the state of xy, and C's statement is obviously wrong.

d can be known by b.

This is a redox reaction, the title means that there is a lot of sulfate in the solution, and the amount of hydrogen ions generated in the reaction is required, and the sulfur in Fes2 is -1

The equation is valence s-+7fe3++4h2o=so42-+7fe2++8h+

So 8mol of sulfuric acid is generated.