Urgent need! Summary of chemistry in the third year of junior high school, urgent and urgent!! Probl

You enter it in the library.

Summary of the third chemistry solution acid-base salts.

That's it. You try.

Answer: (1) A, potassium chlorate (KCLO3), B, manganese dioxide (mNO2) C, oxygen (O2) D, carbon (C) E,

Carbon dioxide (CO2).

2) 2kClO3== ==(mNO2)== 2kCl+3O2 C+O2==Ignition==CO2

Analysis: First review the question, and then we push it back and forward according to the conditions in the question. First of all, according to the white turbidity phenomenon of the colorless and odorless gas E and the colorless liquid F, it can be judged that the colorless liquid F may be clarified lime water, because this is its unique property!

In fact, there are many questions to do, and there must be a sensitivity to this special phenomenon. Then the gas E may be (carbon dioxide)CO2 , and then, according to the condition "the black solid D is ignited and put into a gas collector cylinder containing C, D burns violently to produce another colorless and odorless gas E" We already know that E is (carbon dioxide) CO2, then we can judge that C may be (oxygen) O2, because supporting combustion is a property of (oxygen) O2, then D that can be violently burned in (oxygen) O2 and produce (carbon dioxide) CO2 may be carbon (C), and that's it, we already know that C is oxygen (O2), so let's look at the first condition in the title, we naturally think of the reactions of oxygen production mentioned in the book, and then according to the physical properties of A and B given in the title, we can judge that A is potassium chlorate (KCLO3) B is manganese dioxide (Mno2). The answer can be found from the above analysis.

Note: The most important thing to do this kind of question is to be careful, don't feel big when you see the question, feel that the conditions given in the question are unknown, how to push it! In fact, in this type of question, there are often some conditions that clearly tell you what they are, and as long as we grasp these conditions, we can easily introduce which ambiguous substances we don't know what they are.

The most important thing is to practice more and memorize the physical and chemical properties of some commonly tested substances.

a Potassium chlorate.

b Manganese dioxide.

c oxygen d carbon.

e carbon dioxide.

Potassium chlorate == manganese dioxide (catalyst) == potassium chloride + oxygen carbon + oxygen === combustion === carbon dioxide

If you think you can, you can give it satisfaction.

a, b, c, d, e are KCLO3, MNO2, O2, C, CO2, potassium chlorate is thermally decomposed under the catalytic action of manganese dioxide to produce potassium chloride and oxygen, and charcoal is violently burned in oxygen to produce carbon dioxide gas.

2kClO3 = (mNO2, heated) 2kCl + 3O2 (gas) C + O2 = (ignited) CO2

(1) A: Potassium chlorate B: Manganese dioxide C: Oxygen D: Carbon E: Carbon dioxide F: Calcium hydroxide solution.

2) 2kClO3 ==mnO2 (catalyst) == 2kCl + 3O2

c + o2 = co2

The breakthrough of this problem is C, a piece of black solid D is ignited and put into a gas collector cylinder containing C, and D burns violently, indicating that C has the effect of combustion, or can react violently with D. From this, it is very likely that C is oxygen.

The order of analysis is as follows:

d:co2f:ca(oh)2

d:cc:o2

A:B is sodium hypochlorite and manganese dioxide.

2kClO3 = (mNO2, ) = 2kCl + 3O2 (gas).

This is a typical extrapolation question, and the answer is: a is potassium chlorate, b is manganese dioxide, c is oxygen, d is charcoal, and e is carbon dioxide.

Manganese dioxide. Potassium chlorate ——— potassium chloride + oxygen.

Heat ignition. Charcoal + oxygen ——— carbon dioxide.

Breakthrough in problem solving: D can burn violently to indicate that C is oxygen, and the generated gas can make the colorless liquid turbid, which is just in line with carbon dioxide to make the clarified lime water turbid, which proves that D is charcoal. C is oxygen, then A and B can be determined according to the color characteristics.

I'll try to ignore the opening sentence for now, and see if I can introduce something later.

The black solid D can ignite and burn violently in C gas. According to our intuition when studying chemistry, d may be carbon, while c may be oxygen. Then, e is carbon dioxide gas.

Further down, pour a colorless liquid with white turbidity, according to what we have learned, only calcium hydroxide, that is, lime water can react with carbon dioxide to produce white calcium carbonate precipitation. Then, f may be a calcium hydroxide solution.

Looking back, if C is oxygen, then the white solid (don't let the crystals confuse the eyes) and the black solid are substances that can produce gas, we know that the only solid oxygen we learn now is potassium chlorate, manganese dioxide and potassium permanganate, and only potassium chlorate and manganese dioxide are heated to produce oxygen. Manganese dioxide is black, so potassium chlorate is white (I don't know the color of potassium chlorate).

Okay, so the answer to the first question is.

a. Potassium chlorate, b. manganese dioxide, c. oxygen, d. carbon e, carbon dioxide.

The second question is now easy.

A and B: Potassium chlorate = = manganese dioxide Heating = = potassium chloride + oxygen (manganese dioxide is written above the equal sign, and heating is written below the equal sign).

Carbon + oxygen == ignition == carbon dioxide (don't write combustion, combustion is not a condition, just like electrolysis of water, only electricity can be written, not electrolysis).

A is potassium chlorate, B is manganese dioxide, and C is oxygen. D is carbon, which is burned in oxygen to produce E is carbon dioxide, and poured into a colorless liquid F is clarified lime water to produce white precipitated calcium carbonate.

Potassium chlorate is heated under the catalysis of manganese dioxide to produce oxygen and potassium chloride.

Carbon is burned in oxygen to produce carbon dioxide.

(1) a. potassium chlorate b manganese dioxide c oxygen d charcoal e carbon dioxide f clarified lime water.

2) 1. Potassium chlorate, manganese dioxide, potassium chloride + oxygen heating.

2. Charcoal + oxygen - ignition - carbon dioxide.

The more detailed, the better. Thank you Answer: BC Using the principle of conservation of charge, since the nitrate ion is not told, the total charge number of cations is required to be greater than the total charge number of anions 3*2+1*3<10*2 in A, 3*2+ in B

A: Potassium chlorate B: Manganese dioxide C: Oxygen D: Carbon E: Carbon dioxide F: Clarified lime water.

2kClO3 (white solid) Mno2 (black solid).

2kclo3)==2kCl + 3O2(gas) conditional heating manganese dioxide catalyst.

1 A potassium chlorate.

b Manganese dioxide.

c Oxygen. dCarbon or charcoal.

e carbon dioxide.

2 Potassium chloride (on the arrow: manganese dioxide under the arrow: heating) potassium chloride + oxygen carbon + oxygen (on the arrow: ignited) carbon dioxide.

a. potassium chlorate, b manganese dioxide, c, c, oxygen, d, charcoal, e carbon dioxide.

A reacts with B, B acts as a catalyst, and A undergoes a decomposition reaction to generate potassium chloride and oxygen.

Charcoal is burned in oxygen to produce carbon dioxide.

Let O2 be xkg

ch4 + 2o2 == co2 + 2h2o16 64

16kg x

The solution is x=64kg

So 64kg of oxygen is needed.

The volume is 64*1000

Because O2 accounts for about 21% of the air

So the volume of air is.

CH4 + 2O2 = ignition = CO2 + 2H2O

16 xx=64kg

The volume of oxygen consumed is 64*1000

Because the amount of oxygen in the air is about 1 5

So the volume of air required is 44755*5=223775L

CH4 + 2 O2 = CO2 + 2 H2O Relative molecular weight of methane: 16 mass 16 000 g

1 2 Amount of substance: 1000mol

1000 x

Obtain: x=500mol

So the mass of oxygen: 500x32=16000 g= kg volume of oxygen: v=m density=16000 l (approximately) oxygen makes up 21% of the air so to the volume of air:

11189 l (approximately equal to).

The concept of solubility in C is how many grams of solute melted in 100 grams, and A and B both took out 10 grams, and B precipitated 3 grams, and A precipitated 2 grams, so the solubility of A is less than the solubility of B.

If you choose B, it means that substance A dissolves a lot in the same water, so the solubility is large.

But I think it should be d, saturation plus saturation or saturation.

Reply upstairs, 3g precipitation indicates that 7g is dissolved, 2g precipitation indicates that 8g is dissolved, so that A is dissolved more.

It is not said that the solute and the solvent are the same, but that the quality of the solution is the same, which is probably wrong.

The answer should be b:a 2b=c 2d

If you are proficient in the question, you will find that the relative molecular mass given in the question is a substance that we are very familiar with, and the easiest to judge is B-32-O2; c-44-co2；d-18-h2o。Because the mass of C and D increases, it means that they are products, and this reaction is the reaction of a substance with oxygen to produce carbon dioxide and water. The presence of carbon in the product means that A is a carbonaceous compound, and because the relative atomic mass of carbon is 12 and the relative molecular mass of A is 16, it means that it contains 1 carbon.

From this, it is easy to conclude that A is CH4. Finally, trimming is sufficient.

b reacts 1mola and 2molb from the question to produce 1molc

n=m/m

According to the law of conservation of mass, the mass before and after the reaction of each substance is found, and then the mass is divided by the relative molecular mass (molar mass) to obtain the relationship between them.

In the reaction, A consumes 16-12 = 4 g

B consumes 16g

c generates 27-16 = 11g

So x: (4 16) = y: (16 32) = z: (11 44), that is, 4x = 2y = 4z is the smallest.

x=1, y=2,z=1

So choose B

Alas, obviously, this is methane reacting with oxygen.

(1) The higher the position of the metal, the easier it is for the metal to (lose) electrons and become ions in the aqueous solution, and the more active it is

2) The metal in front of (hydrogen) can replace the hydrogen in the acid, and the reaction of the metal with concentrated sulfuric acid, nitric acid, etc. (not) "fill yes or no" replacement reaction.

3) Only the metal in the front can replace the metal in the back of it from their salt solution, and k, ca, and na generally (can) "fill in or can't" replace the reaction, but instead of displacing the metal elements in the salt, they replace the hydrogen elements in the water.

4) Use the order of metal activity (can) to "fill in or can't" to judge the displacement reaction in aqueous solution (except for metals such as K, Ca, Na and other metals that are easy to react with water).

1: Weakness.

2: h No.

3: The front and the back can't.

4: No, you can't.

(1) Lose strong.

2: h No.

3: The front and the back can't.

4) Yes.