A simple high school chemistry problem, but I just couldn t figure it out

1. Possible reactions of CO2 introduction: CO2 + 2NaOH = Na2CO3 + H2O

co2 + na2co3 +h2o = 2nahco3

2. Add enough lime water: Na2CO3 + Ca(OH)2 = CaCO3 + 2NaOH

nahco3 + ca（oh）2 = caco3 + naoh + h2o

3. Add a sufficient amount of CaCl2 Na2CO3 + CaCl2 = CaCO3 + 2NaCl

From the above reaction, it can be seen that if there is a solution in the amount of CaCO3 in Nahco is not the same, so when A=B, only the right is in the solution.

The two chemical formulas are strung together and written out, and it's good to do.

Here's how it is simplified: CO2+2NaOH=Na2CO3+H2O+Ca(OH)2=CaCO3+2NaOH

ag co2+2naoh=na2co3~~+cacl2=caco3+2nacl

The two different products generated by BG are the key to solving the problem.

If A=B is the flower, since the carbonate ions are equal, it means that no calcium bicarbonate is generated, i.e., only sodium carbonate, and the answer is C

This should be a multiple-choice question.

Let's start with the case of purity.

CaCO3 + 2HCl = H2O + CaCl2 + CO2100g calcium carbonate produces carbon dioxide.

10 g calcium carbonate carbon dioxide.

NCO2=So CAC3=

So when it's pure, exactly 10g of calcium carbonate produces CO2, but the title says it's impurities.

So we can think of it this way.

The average amount of CO2 produced by impurity A and impurity B is the same as that of calcium carbonate, and the amount of carbon dioxide produced by magnesium carbonate is 84g, and silica is not produced (greater than calcium carbonate).

So they can produce the same average value as calcium carbonate.

It is estimated that several other options are either greater than calcium carbonate or all are smaller than calcium carbonate.

Actually, there is an error in the titration experiment, and there is a half-drop principle, which is similar to the meaning of this topic.

Here's the analysis.

Since it is neutralized by a strong acid and a strong alkali in the amount and concentration of the same substance, the required solution volume ratio is 1:1, right, it is 100 ml, that is, the solution l unit is converted.

naoh+hcl=nacl+h2o

The answer is that the titration error is within .

In case (1), the amount of NaOH drops is insufficient, and the amount of NaOH drops is .

H+ ions remaining in the solution.

The amount of its substance is:

The volume of the solution after mixing is +

The concentration of H+ in the solution is mol l

pH = i.e., the lowest pH after the reaction.

Situation (2).

The same can be obtained by adding an excessive droplet of NaOH.

NaOH is added to the volume.

The concentration of OH- in the solution is:

The method of calculating pH by mol l 1 can be calculated according to the ion product conversion kw=10 (-14) h+ and calculate it yourself.

2 You can also use the definition of poh.

POH = pH + POH=14 in aqueous solution at room temperature (based on KW=[H+]*OH-]=10 (-14)).

pH = is exactly the same as the previous method.

Therefore, when titrating hydrochloric acid with a solution, if the titration error is within plus or minus, after the reaction is completed, the pH range of the solution is.

If the titration error is negative, there will be more hydrochloric acid.

Hydrochloric acid concentration: , pH = positive error of titration, then AOH is more, pH =

1. When the error is +, the HCl solution is neutralized, and the pH value of the mixed solution is determined by the unreacted, just find the concentration of H ions at this time The concentration of H at this time is C=, which is converted to pH=

2. When the error is, the HCl solution is completely neutralized, and the pH value of the mixed solution is determined by the remainder, and the OH concentration at this time [OH]=, that is, POH= then PH= , is solved

KSP = [S2-][Fe2+] = when [Fe2+] = 1 mol L, substituting the above equation gives [S2-] = mol L

Ka = [H+] 2 [S2-] = substituting [S2-] = mol for H2S into the above formula.

The solution yields [H+] = 4 10 -3 mol lph = - lg[h+] = 3 - lg4 = 3 - 2 lg2 =

C(S2-)= in solution can be obtained by Ksp= of Fes and C(Fe2+) in solution reaching 1mol L

Then it can be seen from: c2(h)c(s2-)=, c(h)*c(h)=

c(h)=, so ph=3-2lg2=

It has nothing to do with it, the question asks about the concentration, and by the knowledge of dissolution equilibrium, the concentration of "CO3 2-" ions in Na2CO3 solution is already very large, and Ksp is a fixed value at a certain temperature, so the concentration of Ba2+ in Na2CO3 solution is the smallest, otherwise which one do you choose?

Unrelated, sodium carbonate in a has a homoionic effect, which inhibits the dissolution of barium carbonate.

It doesn't matter at all, this question asks you the concentration of barium ions, not how much barium ions are dissolved, and the less the volume of the solution, the less the melting point, but the concentration after dissolution can be different, because the solubility product of this barium carbonate is related to Ksp(BaCO3)=C(Ba2+)XC(CO32-)=C(Ba2+) 2=C(CO32-) 2For the B and D options, the concentration is the same, for A and C options, just look at the concentration of carbonate or barium ions, and whoever has a larger concentration will have a smaller concentration of barium ions, because Ksp is a fixed value.

It had nothing to do with the volume of the solution added, he asked about the concentration, not the amount of the substance. A, the presence of CO32 shifts the equilibrium to the left and the concentration of Ba2+ decreases.

Because the addition is a solid powder, BaCO3 is an intolerant electrolyte, so the concentration of Ba2+ should be calculated according to the concentration of KSP CO32-, of course, the greater the concentration of carbonate ions, the smaller the concentration of Ba2+, and the concentration of Ba2+ in A and B are very small, which actually ignores the dissolution of CO32-.

BaCO3 is insoluble in water and base, according to the principle of ionization equilibrium, the concentration of H d c b a, the more H dissolves, the more Ba2+ dissolves.

1.Yes, sulfite oxidizes to sulfate.

2 Ions with the same number of protons, the same number of neutrons, and different numbers of electrons do not belong to the same kind of ions, but belong to the ions formed by the same element.

3 Metallic elements are not necessarily metals, and elements that are on the dividing line between metal-only and non-metals are both metallic and non-metallic.

4 mistakes, the outermost 8 are rare and difficult to lose.

5 The number of electrons in the outermost shell.

a. no!nn triple bond.

3molh to participate in the reaction; 3mole-

NH2 has one single electron, 9NaE-

Colloidal particles are a collection of several iron hydroxides. <

The question is simple. Option a contains electron pairs and should be constant. Option b should be right. The c option should be a 9 constant. Option d This colloidal reaction is reversible and cannot be completely reacted, so there is no constant.

a false, the number of neutrons is unequal;

b right. O2+ and Na+ in Na2O are 1:2;In Na2O2, peroxide ions and sodium ions are also 1:2;Therefore, no matter how you mix it, it will always be 1:2.

c false, there is also cl2 in the water, this part of the chlorine atom is not considered;

d false, there is no standard condition.