Junior 3 chemistry questions, help me pull, a chemistry problem for junior 3, help me

H2CO3 = CO2 + H2O (= plus heating symbol) CaCO3 + H2O + CO2 = CA (HCO3) 2 Ca(HCO3) 2 = CaCO3 + H2O + CO2 (= plus heating symbol) Cao + H2O = Ca(OH)2

2naoh+ co2=na2co3+h2o

Phosphorus and oxygen are spontaneous combustion options

The R element is +6 valence and the oxygen element is -2 valence, so the chemical formula of the oxide is Ro3 and the relative atomic mass of the R element is x

Then the mass fraction of r in the oxide is x (x+3*16)=40%, and the solution is x=32, the number of protons = neutron number = 16, and r is the sulfur element.

The oxide chemical formula is SO3

Trioxide r16 * 3 = 48, 18 divided by 60% in multiplication by 40 = 32

Look up the table, it's SSO3

Precipitate barium carbonate, according to the equation, the mass of sodium chloride and sodium carbonate are generated, the mass of primary sodium chloride is 100+, and the solute mass fraction in the solution after the reaction is (

Barium chloride and sodium carbonate produce barium carbonate precipitates. This, the relative molecular mass of barium carbonate is 197, then precipitated into a molar mixture of sodium carbonate, that is, gram of original sodium chloride gram of newly generated gram solute cogram. The quality of the solution is.

The precipitate is barium carbonate, the relative molecular weight is 197, the precipitate mol, the sodium carbonate gram in the original mixture, and the sodium carbonate of the original sodium chloride gram reacts into sodium chloride gram. The mass of the solute total gram solution is 100+, and the solute mass fraction is (

The precipitate is barium carbonate, and the relative molecular weight of barium carbonate is 197, then the precipitate is mol, that is, the carbonate is mol, and in the original mixture, sodium carbonate mol, that is, grams, then the original sodium chloride grams. Later, moles of sodium carbonate reacted into sodium chloride, i.e., grams. The solute totaled 20 grams.

1.Drip water on the absorbent cotton wrapped with sodium peroxide powder, and the absorbent cotton burns, please analyze the cause of the burning of the absorbent cotton.

Teacher's answer: Sodium peroxide reacts with water to release heat, and the oxygen produced can support combustion.

I think: sodium peroxide and water produce sodium hydroxide and oxygen. The resulting sodium hydroxide dissolves in water and releases a large amount of heat to bring the temperature to the ignition point, while oxygen supports combustion.

Which one is better?

2.Add water to the test tube containing sodium peroxide powder, and after the full reaction, take a small amount of solution in the test tube and put it in another small test tube, add a few drops of colorless phenolphthalein test solution, the solution turns red, and after a while, the red color fades again.

Ask the question: Why does the solution turn red first, and then fade after a while?

Conjecture: Student A thinks that it is because the sodium hydroxide solution reacts with carbon dioxide in the air. Student B thinks that Student A's conjecture is incorrect, on the grounds that the aqueous solution of sodium carbonate after the reaction of sodium hydroxide and carbon dioxide is alkaline, which can make phenolphthalein red).

Access to information:1When the solution with phenolphthalein showing red dropwise encounters hydrogen peroxide in alkaline solution, hydrogen peroxide can destroy the structure of phenolphthalein, so that phenolphthalein can no longer appear red in alkaline solution; 2.

Phenolphthalein appears stable red in dilute alkaline solutions, while red and faded in concentrated sodium hydroxide and other solutions.

Conjecture: After going through the tea data, the interest group made a conjecture about the main reason for the red fading of the alkaline solution.

Conjecture A: Hydrogen peroxide may be formed by the reaction of sodium hydroxide with water; Conjecture B: The resulting sodium hydroxide solution may be too concentrated.

Experiment**: Please complete the following protocol.

Conjecture A Experimental scheme: take the solution after the reaction of hydrogen peroxide and water into the test tube, (add a small amount of manganese dioxide, and extend a small wooden strip with sparks into the test tube) Experimental phenomenon: (there are colorless bubbles generated, and the small wooden strip rekindles) Conclusion: Conjecture A is correct.

Conjecture B Experimental scheme: take the solution after the reaction of sodium peroxide with water and put it into a test tube, ( mm ) Experimental phenomenon: (nn ) Conclusion: Conjecture B is wrong.

It is concluded that according to the experimental analysis, the reaction of sodium peroxide with water has another reaction in addition to sodium hydroxide and oxygen. The chemical formula of this reaction is: sodium peroxide + water - sodium hydroxide + hydrogen peroxide.

That's what our teacher said.

Empty mm: Add a few drops of colorless phenolphthalein test solution dropwise.

Empty nn: The solution turns red and does not fade.

I don't think it's right, the original experiment was to add phenolphthalein test solution to the solution after the reaction, and the phenolphthalein test solution was red on the edge, and the red faded after a while. When you do the experiment, you will take the solution after the previous reaction and do the same experiment, and the red color will not fade.

1. Water in nature is ( ).

A, mixture B, pure matter C, compound D, elemental 2, alum can be used for water purification, belongs to the following methods ( ) method A, adsorption B, precipitation C, filtration D, distillation 3, the following groups of substances can be separated by filtration is ( ) A, salt, sediment B, alcohol, water.

c. Copper oxide, carbon d, water, sucrose.

a、a、a

2hcl+caco3=cacl2+co2+h2o100 44x=8

2: First of all, gram gas is generated when 40 grams of dilute hydrochloric acid are added, and gram gas is also generated when 50 grams of dilute hydrochloric acid is added, which proves that when 40 grams of dilute hydrochloric acid is added, it is exactly a complete reaction.

Therefore, the quality of the solution only needs to be 40+

According to the first 3 steps, we can know that 10g HCl is formed into gas, then the fourth step is just completely reflected, which is CO2, according to the reaction equation m(CaCO3)=

2) The amount of solution in beaker 4 is the sample mass + 40g hydrochloric acid - the mass of carbon dioxide generated = 8+, I hope it will help you.

Add the mass g of dilute hydrochloric acid

The mass g of the gas is formed after sufficient reaction

It can be seen that the addition of 40 g of dilute hydrochloric acid is a complete reaction, so it should be used as a calculation data, assuming that the mass of calcium carbonate in the sample is x.

i.e. 44 x = 100

The solution is x=8g, so the mass fraction of calcium carbonate in the sample is 8g 10g*100%=80%.

The mass of the solution obtained after the full reaction of the substances in the beaker is 8g+ Answer: ··

This question should pay attention to the analysis of the data in ** to judge the degree of reaction in different beakers, it is obvious that the reaction sample in the beaker is excessive, 4 beakers are just completely reacting, and 5 beakers are overdose of hydrochloric acid. Therefore, the calculations should be made using the reaction in the 4 beaker.

Solution: Let the mass of calcium carbonate in 10g of sample be x

caco3+2hcl===cacl2+h2o+co2100 44

x100：44=x： x=8g

The mass fraction of calcium carbonate in the sample is: 8g 10g x 100% = 80%4 The mass of the solution obtained after the full reaction of the substances in the beaker is; 8g+ A: Omitted.