There s also a chemistry question, a chemistry question

A. When the solution is a single salt (CH3coona) solution, the solution is alkaline due to acetate hydrolysis.

Concentration-size relationship: C(Na) C(CH3COO) C(OH) C(H).

Conservation relation: conservation of charge: C(CH3COO)+C(OH)=C(Na)+C(H);

Conservation of materials: C(CH3COO) C(CH3COOH)=C(NA);

Conservation of protons: C(OH) = C(CH3CoOH) C(H).

B. When the solution is neutral, the solute is a mixture of CH3COONA and CH3COOH, which is equivalent to CH3COONA is not hydrolyzed and CH3COOH is not ionized.

Concentration-size relationship: C(Na) C(CH3COO) C(OH) C(H).

Conservation relation: conservation of charge: C(CH3COO)+C(OH)=C(Na)+C(H);

Conservation of materials: c(ch3coo) = c(na);

Conservation of protons: c(oh) = c(h).

C. When the solution is acidic, the solute is a mixture of CH3COONA and CH3COOH, which is equivalent to the mixed solution of CH3COONA and CH3COOH and then added acetic acid solution on the basis of neutral.

Concentration magnitude relationship: C(CH3COO)>C(Na)>C(H)C(OH).

d. In the case of excess of strong alkali and weak acid and alkali, the solute is a mixture of CH3Coona and NaOH.

Concentration magnitude relationship: c(na) c(ch3coo) c(oh) c(h) or c(na) c(ch3coo) = c(oh) c(h) or c(na) c(oh) c(ch3coo) c(h).

First of all, there can be no acetic acid, once there is acetic acid, there is sodium hydroxide, which will react, it should be H+, OH- phase combination, CH3COO, Na+, phase combination, but there are also separate individual ions.

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But pay attention to the pressure of 0 in the beaker T and the pressure of d large m gas 0, if you put the beaker in the place where water 7 is very deep, 1 will have less 7 amounts of water 2 into e g) If the mouth of the cup q is not y is perpendicular to the water 7 facing down f, then o will have air 4 bubbles, and at the same time, it can only prove that there is air 5 in 1 in the beaker, but not o can prove that it is filled with air 6 r

It is impossible to coexist.

H+, OH- form H20

CH3COO H+ forms acetic acid.

Summary. 1.According to the law of conservation of mass, the total mass of the reactants should be equal to the total mass of the product, the reactant is, the product now has grams, and the extra grams are the mass of carbon dioxide.

2.This question requires the mass fraction of sodium chloride, which should come from two parts, one part is the original sodium chloride impurity, and the other part is the newly produced sodium chloride, and then divided by the total mass gram of the solution, which is the mass fraction of sodium chloride. 3.

Then write the equation, let the mass of sodium carbonate be x, the mass of sodium chloride produced is y, calculate the equation x=g, is the mass of sodium carbonate in the original solid, use, is the mass of sodium chloride contained in the original solid. Calculate y=, which is the mass of the newly produced sodium chloride. 4.

The mass of the two parts of sodium chloride is added, and it is obtained, divided by the mass of the solution, multiplied by 100%, and you can get the final number.

Okay, is that the math problem<>

I don't understand the last question.

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Let's see if the process is clear.

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1.According to the law of conservation of mass, the total mass of the reactants should be equal to the total mass of the product, the reactant is, the product now has grams, and the extra grams are the mass of carbon dioxide. 2.

This question requires the mass fraction of sodium chloride, which should come from two parts, one part is the original sodium chloride impurity, and the other part is the newly produced sodium chloride, and then divided by the total mass gram of the solution, which is the mass fraction of sodium chloride. 3.Then write the equation, let the mass of sodium carbonate be x, the mass of sodium chloride produced is y, calculate the equation x=g, is the mass of sodium carbonate in the original solid, use, is the mass of sodium chloride contained in the original solid.

Calculate y=, which is the mass of the newly produced sodium chloride. 4.The mass of the two parts of sodium chloride is added, and it is obtained, divided by the mass of the solution, multiplied by 100%, and you can get the final number.

This question is indeed difficult, but if you sort out the whole process, it should be relatively clear.