Chemistry questions for junior high school, use it tonight, ask for masters!! Find the reason!! Find

1.c cuso4 colored. DFECL3 and CUSO4 are colored.

agno3 ====agcl + nano3 .AGCL precipitation.

So first b2A can't. pH=12 means that it contains -OH hydrogen ions that react with -OH to form water.

b No, you can't. Hydroxide ions, Al3+ reaction to form Al(OH)3 precipitate.

c Yes. D ag+, Cl-, CO32- precipitate.

3.After adding purple litmus solution dropwise to the solution in A, the solution turns blue, indicating that there is -OH, CO32-K+

then B contains: H+ Ba2+ Cl- No3-

The following substances can coexist in large quantities in water and obtain a colorless and transparent solution.

A NaCl, AgNO3, and K2CO3 form AGCL precipitates, and B H2SO4, KNO3, and NaCl can coexist.

C Ba(OH)2, Na2CO3, CuSO4 to generate BaSO4 precipitate D FeCl3, CuSO4, HNo3 colored solution.

The following ions can coexist in large quantities in aqueous solutions with pH = 12 (the original questions are all chemical formulas, but I don't know how to write chemical formulas, so I use kanji).

A sulfate ions, nitrate ions, potassium ions, hydrogen ions H+ do not do B sodium ions, Cl-, hydroxide ions, Al3+ to generate metaaluminate.

C Cl-, NO-3 (-in the upper 3 in the bottom), potassium ions, Na+ can coexist in D Ag+, Cl-, CO2-3 (2- in the upper 3 below) K+ to form silver chloride precipitates.

Dissolve some soluble compounds in water and configure two bottles of solution A and B, which contain H+, K+, Ba2+ (2+ on top) OH-, Cl-, CO2-3 (2-on 3 on top) NO-2 (-on 2 on bottom) eight ions, each bottle of solution contains four different ions, it is known that after adding purple litmus test solution dropwise to the solution in A, the solution turns blue, and the ions in bottle B are (four).

What is the problem, it seems that there is one less ion?

A, AGCL white precipitate.

C Cuso4 Blue.

d fecl3 red, cuso4 blue.

pH = 12 is alkaline.

a Hydrogen ions are not allowed.

b Al3+ and hydroxide ion precipitation.

Ag(OH) precipitation.

After the purple litmus test, the solution turns blue and is alkaline.

There are K+, OH-, CO2-3 (2- in the top, 3 in the bottom), NO-2 (- in the top, 2 in the bottom) that are already left.

1.CD is excluded first because the copper ions are blue and the iron ions are yellow. Because silver chloride ions will form silver chloride precipitates with silver ions, A is incorrect, so B is chosen

2.First exclude D for the same reason as in question 1. b Similarly, hydroxide and aluminum ions react to form aluminum chloride precipitates. Because pH = 12 is a base, A is excluded, and because a neutralization reaction occurs, B is chosen

3.See, the purple stone Lei turned blue, indicating that it is alkaline, and the chloride ions cannot be in the same bottle as the silver ions, so A should be: potassium hydroxide, barium hydroxide, and barium chloride. So b is carbonic acid, nitric acid, silver nitrate. So it's hydrogen ions, nitrates, silver ions, carbonates.

I'm also in the third year of junior high school, and I'm not sure about the answers and explanations.

Because A generates AgCl precipitates, CuSO4 in CD is blue.

That is, an alkaline solution with hydroxide ions, combined with hydrogen ions in A to form water, combined with Al3+ in B for precipitation, and Ag+ and Cl- in D to react, so C is selected

There are only 7 kinds of 3 questions, and the questions are incomplete, so I'm sorry!

The first question, A will generate AGCL, B is correct, C will generate barium carbonate barium sulfate copper hydroxide d ferric ions are brown and yellow copper ions are blue The second question is alkaline, A does not coexist with hydrogen ions B Aluminum ions do not coexist with hydroxide c D Silver ions do not coexist with hydroxide Question 3, there is a question to know that B has hydrogen ions, then there is hydroxide carbonate in A (hydroxide carbonate and hydrogen ions do not coexist), because there are carbonate in A and barium ions do not coexist, then there are barium ions in B, Since the solution is not electrically invisible, it is known that A has potassium ions, and B has chloride ions and nitrates.

1. Select B, A: NaCl and AgNO3 will produce white precipitate C: Ba(OH)2 and Na2CO3 will produce white precipitate.

Ba(OH)2 and CuSO4 produce white and blue precipitates D:FeCl3 is a yellow solution and CuSO4 is a blue solution.

2. The aqueous solution of pH=12 indicates that there is a large amount of hydroxide in the solution, so it must be an ion that can coexist with hydroxide, so C is selected

Hydrogen ions in A are not good, aluminum ions and hydroxides in B are not, and AG+, Cl-, and not in D.

3. It is known that after adding purple litmus test solution dropwise to the solution in A, the solution turns blue, so the solution is alkaline, so there is OH-, then the ion in A must be able to coexist with OH-, CO2-3 (2- in the upper 3 below) NO-2, K+, then B is H+, Ba2+ (2+ on the top) OH-, CL-, (Note: there must be cations in a bottle of solution, and the whole system can be relatively stable).

1. a, sodium chloride reacts with silver nitrate to form a precipitate b, if it is potassium nitrate, it is correct c, barium hydroxide and carbonate ions and sulfate ions all form precipitates d, iron chloride is yellow.

2. pH=12 means that it is an alkaline solution. a. Hydrogen ions do not coexist with alkali b. Aluminum ions and hydroxide form aluminum hydroxide precipitates.

c. Correct D. Silver ions cannot coexist with carbonate ions, nor can they exist in alkaline solutions.

3. Anyway, you only wrote about seven ions...

Choose b a will have silver chloride precipitation, c has copper ions are blue, d has copper ions and trivalent iron ions are all colored.

Choose c pH = 12 means that there are a lot of hydroxide ions, hydrogen ions can react with hydroxides, aluminum ions and hydroxide ions will precipitate, silver ions and chloride ions carbonate will precipitate.

The third question is obviously only 7 ions...

m(s) = 400 * 1% = 4kg

m(so2) = 4 * 64 / 32 = 8kg

About 8 kg of sulfur dioxide is emitted from the atmosphere.

One percent of the sulfur content is 4 kilograms.

s---so2

4kg xx= 12kg.

That's probably it, I hope it helps.

8kg Analysis:

Mass fraction of sulfur in sulfur dioxide = 32 64 =;

The mass of sulfur in coal = 400 * 1% = 4kg;

Because "all the sulfur elements in it are converted into sulfur dioxide", there is a sulfur dioxide mass = 4

Boss, you can't do such a simple question, it's incredible, 400kg contains 1% is 4kg, and the mass fraction of sulfur dioxide in sulfur dioxide is 50%, so the mass of sulfur dioxide is naturally 8kg, give it to me.

Because the S content is 1%, that is, the S is 4kg

s+o2=so2

4 x when 8kg

1. a (calculates the masses of carbon and hydrogen, which add up to equal to.)

2. A (the law of conservation of mass, cd mass and (conservation of mass).

The carbon ratio to oxygen is 12 to 32

5. The limit case when the remaining gas is all oxygen and all hydrogen).

Is polyvinyl chloride a polymer material? Yes.

Is polyester a synthetic fiber? Yes, polyester fibers are synthetic fibers, commonly known as polyester.

Can you tell the difference between wool and polyester by combustion? Yes, wool has a smell when burned, and polyester melts without a smell when burned.

Is PVC a composite? No, it is obtained by polymerization of vinyl chloride.

1. Compared with A and B, B has more marble, reflecting more residual quality, indicating that B's marble is excessive, compared with A and C, C has 20g more hydrochloric acid, reflecting the remaining 20g, indicating that there is excess hydrochloric acid, in summary, A happens to have a complete reaction.

1 Methyl-C generation.

Come to Ethyl D. b 4: (since 42-6) = 4:9

Answer: B2 solution fades.

Bai color. du solution fading The former is a physical change, and the latter is a chemical change.

3 caco3---co2100 44x 66gx=150g

Question 1 is selected b. In the second question, both of them fade, the reason is that the addition of charcoal is a physical change, which adsorbs the red substance generated, and the heating is the escape of CO2 by heating. The answer to the third question is 150g

m (sodium carbonate) =

co2+2naoh=na2co3+h2o

44 80 106 184g

There are 4g of caustic soda involved in the reaction, and the mass of the original solution is.

The mass fraction of the solute of the original solution is 4

Carbon dioxide reacts with caustic soda to form sodium carbonate, calculate the mass of sodium carbonate multiplied by Decker, then, the carbon dioxide added is, the mass is gram, and the mass of NaOH is 4g, minus Deke (the mass of the original caustic soda solution), 4 divided by the mass fraction of the caustic soda solution used is about.