A junior high school chemistry question is urgent A junior high school chemistry question online, e

Note: Insoluble solids and gases do not count in the mass of the solution.

According to the title, we only look at the quality of the solution.

A: Originally agno3 and water; Cu(NO3)2 and water after the reaction.

The mass of water and NO3- remains unchanged for Cu6H, so the mass of the solution increases.

C: The original Ca(OH)2 and water, after the reaction of water (CO2 dissolved part is not considered) according to the chemical equation Ca(OH)2 + CO2 = CaCO3 + H2O. The mass of the resulting water is less than the mass of the original Ca(OH)2.

And the quality of the water in the original solution remains unchanged. So the mass of the solution is reduced.

D: original HCl and water; After the reaction, NaCl and water (the dissolved part of CO2 is not considered) Na2CO3+2HCl=2NaCl+CO2+H2O, 2NaCl+H2O-CO2>2HCl, so the mass of the solution increases.

a.Copper displaces silver in solution, and the relative atomic weight of copper is less than that of silver, and the mass of the solution is reduced.

b.Aluminum displaces hydrogen, and according to the chemical equation of the reaction, the mass of the 2al>6h solution increases.

Introduced into Ca(OH)2, the reaction CO2 + Ca(OH)2=CaCO3 + H2O, CaCO3 is a substance that is insoluble in water, which is equivalent to the mass of the solution reducing the mass of CaO.

Na2CO3 dissolves and carbon dioxide gas is generated at the same time, and the relative molecular weight of Na2CO3 is greater than that of CO2, and the mass of the solution increases.

So you should choose A, C

Option A explains the same as above.

For B, 2Al+3H2SO4=Al2(SO4)3+3H2, the mass becomes larger.

For C, when CO2 is continuously added, CaCO3 becomes Ca(HCO3)2 and dissolves, so the mass increases.

For D, when there is an excess of Na2CO3, Na2CO3 and HCl produce NaHCO3, no CO2 gas is produced, and the mass of the solution becomes larger.

The answer is a parsing:

CU is added to agno3.

Cu+2agNO3=Cu(NO3)2+2AGog solution silver is replaced by copper, and the relative atomic mass of copper is less than silver.

Al is added to H2SO4.

Al replaces H, and the relative atomic mass of Al is greater than H

CO2 is passed into Ca(OH)2.

CaCO3 precipitation.

Na2CO3 solids are added to HCl.

CO2 gas is produced.

The answer is: A A in the Cu displacement is relatively reduced, 2 * 108-64 = 152 B is the increase of 1 aluminum for 3 hydrogen plus 27-3 = 24

Not necessarily. When CO2 is small, the CaCO3 solution is generated. When CO2 is excessively generated, dicalcium carbonate (soluble in water) increases.

d increase.

A cu is added to agno3.

cu+2agno3=cu(no3)2+2ag

Silver is replaced by copper in solution, and the relative atomic mass of copper is less than that of silver.

After evaporating 80 grams of water, 640 grams of saturated solution remains, and 10 grams of a precipitated at this time, so 720 grams of a solution can be divided into 640 grams of saturated solution and an unsaturated solution containing 80 grams of water dissolved in 10 grams of a substance for analysis. Thereafter, 60 g of substance A is added, which can be considered as the addition of 60 g of A to an unsaturated solution containing 80 g of water. It is easy to know that only 54 grams are dissolved at this time, and the solution containing 80 grams of water is also saturated at this time, plus the original 10 grams of a, a total of 64 grams of a is dissolved, that is, 64 grams of a is dissolved in 80 grams of water in 80 grams of water, and the solubility of the substance is 80 grams (that is, 80 grams of this substance can be dissolved in 100 grams of water).

The original solution must be an unsaturated solution, and the solubility is calculated according to the equation "If it evaporates 80 grams of water and returns to 20, 10 grams of a substance will be precipitated".

Set 720 grams of solution containing AM grams.

Finally, the last two times all reach the saturated state, and the quality score of the saturated state is the same.

m-20)/(720-10-20)=(m+60-6)/(720+60-6)

m=(690*54+20*774)/(774-690)m=

Solubility of the substance at 20 = (grams.)

Suppose 720 grams dissolve x, 640 grams in x-10 grams, 720 grams to saturation is x+54, 80 grams of water can dissolve 64 grams, solubility is 100 grams of water dissolved 80 grams.

I don't remember the unit of solubility, it's like the number of grams of a substance dissolved when it reaches saturation in 100 grams of solvent. Just checked the information

Let's talk about the idea: In the solution of a substance A, the solute, that is, the mass of substance A, is unchanged, and the equation is listed according to this condition.

When the solution is saturated, the percentage of solute A in the solution is x%, and the equation is:

720-80-10）*x+10=（720+60-6）*x-（60-6）

When the solution is saturated, the percentage of solute is calculated and converted to solubility.

The equation can be solved according to the equation of the saturation column after evaporation and the equation of the saturated column after the addition of solute

Because there is crystal precipitation after evaporating 80 grams of water, so the remaining solution is saturated solution, I can mix the evaporated 80 grams of water, 10 grams of a, and 60 grams of a later, there should be 6 grams of a leftover, so 80 grams of water can dissolve 64 grams of a, then the solubility is 80 grams.

Let the solute mass in the original solution be x, then: 720-80-10 720+54x-10 x

Solve x, x = grams.

Solubility: s=(.

Solubility, the key is that the solution is saturated.

After selecting DA, after dissolving in water, the solute is only CuSO4, which is less than 5%B, which is 5%.

C, calcium hydroxide produced by the reaction of calcium oxide and water is slightly soluble in water, less than 5%d, SO3 + H2O = H2SO4, greater than 5%.

Selecting D and D will consume water by reacting with H2O.

a.Less than. b.Amount.

Calcium hydroxide is slightly soluble in water.

【Analysis】This question is an experimental problem about the reaction between acids and alkalis. (1) According to the experimental operation and phenomenon, it can be judged that Xiaoli's scheme is correct, the pH of the solution after the reaction = 7, and the solution is neutral, indicating that sodium hydroxide and dilute hydrochloric acid happen to be completely reactive; When the hydrochloric acid is excessive, the phenolphthalein test solution does not change color, and Xiaoliang's scheme is wrong; During the reaction, sodium chloride is generated, and dropwise addition of dilute nitric acid and silver nitrate solution will definitely produce a white precipitate, and Kobayashi's plan is wrong. (2) After the mixed reaction of sodium hydroxide and dilute hydrochloric acid, there are three possibilities for the solute of the obtained solution, one is that it happens to be completely reacted, and the solute is NaCl, the second is that there is a surplus of hydrochloric acid, and the solute is NaCl and HCl, and the third is that there is a surplus of sodium hydroxide, and the solute is NaCl and Naoh, and sodium hydroxide and hydrochloric acid cannot coexist, so Xiaoliang's conjecture is wrong; If Kobayashi's conjecture is correct, the solution contains hydrochloric acid, and the addition of sodium carbonate solution will produce carbon dioxide gas; If your guess is correct, the solution contains sodium hydroxide, and drops of copper sulfate or copper chloride solutions will produce a blue precipitate.

Answer: (1) Xiaoli.

The reason for the error of Xiaoliang's scheme: if the solution after the reaction is acidic, the phenolphthalein test solution does not change color, so it cannot be judged whether the hydrochloric acid has been just neutralized.

The reason for the error of Kobayashi's scheme: because the product of the neutralization reaction has NaCl (or contains Cl-), adding AgNO3 solution will also produce a white precipitate, which does not indicate whether the hydrochloric acid is excessive.

2)nacl、naoh

Hydrochloric acid and NaOH solution to react (or H+ and OH- can not coexist) and bubbles are generated.

Take the reaction solution in a test tube, and add CuSO4 solution (soluble copper salt solution is acceptable).

1.Xiaoli did the right thing.

Xiaoliang error reason: the discoloration range of phenolphthalein is, but the discoloration does not mean that the reaction happens to be completely carried out Kobayashi error reason: chloride ions have always existed in the solution, and silver chloride only proves that there are chloride ions, which cannot explain the degree of reaction progress.

2.Conjectures: NACL, NAOH

As long as the conjecture is logical, it can be the same as what they guessed, but there are only four situations in this question, and it needs to be consistent with the following, to generate blue precipitated copper hydroxide, so there must be sodium hydroxide.

False reason: NaOH and HCl cannot coexist, so it is impossible to have experimental phenomena in solution at the same time:1There are bubbles generated.

2.Add a small amount of copper sulfate solution.

Manganese dioxide is used as a catalyst, and the mass does not change before and after the reaction. The mass of the reduced solids after the reaction is the quality of the oxygen produced.

Suppose that the mass of potassium chlorate in the original mixture is x

2kClO3 = Heating = 2kCl + 3O2

2* 3*32x

The proportional solution of the column formula yields x==

Potassium chlorate is present in the original mixture.

2kclo3=(mno2) 2kcl + 3o2245 149 96

If the meaning of the x case is less than before the reaction, then the two reductions are oxygen, and x= is calculated

The question should be how many grams of potassium chlorate are in the original mixture, and how many grams are grams.

The reaction of manganese dioxide is used as a catalyst, does not participate in the reaction, 2kclo3--2kCl + 3O2, the oxygen mass fraction is 32, the mass decreases after the evolution of the generated grams of oxygen, and the consumption of kclo3 is 2, 25 grams.

The pre-reaction and post-reaction masses should be equal, why is there less? That's because oxygen is generated, oxygen runs, so the mass decreases, that is, the mass of oxygen is. Now that the mass of oxygen is known, then this question is simple, so according to the calculation of the chemical equation, the mass of potassium chloride is, and potassium chlorate is.