Who has a high school chemistry beginner s exercise

The first question should be D,25, the ionization equilibrium constant of nitrite is about 10 (. The ionization equilibrium constant of nitrite can be calculated according to the data of point A in the figure, the solute of the solution at point A is all nitrite acid, so the concentration of nitrite acid is, according to the pH of a, the concentration of hydrogen ions in the solution can be calculated to be 10 (, the hydrogen ions in the solution are ionized by water and nitrite respectively, so the hydrogen ion concentration is equal to the sum of the concentration of nitrite ions and hydroxide ions, because the concentration of hydroxide ions is small, it can be ignored, Therefore, it can be calculated that the nitrite ion concentration in the solution is approximately equal to the hydrogen ion concentration, that is, 10 (, and finally the data is substituted into the ionization equilibrium constant calculation formula, and the ionization equilibrium constant can be calculated to be about 10 (.

2NaHCO3=NA2CO3 + H2O + CO2 The mass of weight loss is the mass of CO2 + H2O.

xx= g sodium bicarbonate is mol

Sodium carbonate = (mol

2na + 2h2o = 2naoh + h2↑1 mol 1 mol mol

The concentration of sodium hydroxide is 1 mol l

So the gas volume is l

One. 2NaHCO3 ==Na2CO3 + CO2(gas) +H2O, Na2CO3 does not react. The reduced mass in the reaction is the CO2 and H2O produced by the NaHCO3 reaction.

So it's the quality of CO2 and H2O.

n=m÷m= 18)=

n nahco3= , m nahco3=nxm=so m na2co3=, n na2co3=m m=!! Enough detail!

If the weight of an S atom is 32 grams, and the weight of an O atom is 16 grams, then the weight of an SO2 molecule is 64 grams, and the weight of an SO3 molecule is 80 grams, and now there are 320 grams of SO2 and SO3 each.

Then there are 5 SO2 molecules, 4 SO3 molecules.

The ratio of the number of oxygen atoms = (5*2) :(4*3) = 5:6, a is not the ratio of the mass of the element containing s, that is, the ratio of the number of atoms of s = 5:

4, b is not the ratio of the mass of the element containing o, that is, the ratio of the number of oxygen atoms = (5 * 2) :( 4 * 3) = 5:6, c is correct.

The ratio of the number of atoms = 5:4, d is not true.

Calculate the ratio of the number of molecules = 5:4 first, and then it's simple, answer c

The molecular weight of SO2 is 64 and the molecular weight of SO3 is 80

With the same mass, with 5molSO2 and 4molS03, the oxygen atom is 10:12

The sulfur element is 5:4

The mass ratio of oxygen is equal to the mass ratio of matter: 5:6

So choose C

1) Cu+4Hno3 (concentrated) ==Cu(No3)2+2NO2+2H20 Cu valency increased, reducing agent, oxidized; n is restored.

2) 2kmNO4+5H2S+6HCl==2kCl+5S+2MnCl2+8H2O S are oxidized; MN is restored.

3) 6FeSO4+2HNO3+3H2SO4==3Fe2(SO4)3+2NO+4H2O Fe is oxidized; n is restored.

4) 2FeCl3+Cu==2FeCl2+CuCl2 Cu is oxidized; FE is restored.

1) 10kmNO4 + 6I2 + 4H2O==10mNO2 + 6KIO3 + 4KOH oxidant KMNO4 reducing agent I2

2) 2kmNO4 + 16HCl==2KCl + 2mnCl2 + 5Cl2 + 8H2O oxidant KMNo4 reducing agent HCl

3) HNO (very rare) + ZN (NO3) + NH4NO3 + H2O ? Didn't write it clearly.

4) kmno4+naso4+h2so4==mnso4+k2so4+naso4+h2o I copied it wrong.

Understanding first ...

1) Cu + Hno3 (Thick) - Cu (No3) 2 + No3 + H20 (no answer, the question is questionable).

2) 2kmNO4+5H2S+6HCl——2KCl+5S+2MnCl2+8H2O (S is oxidized, Mn is reduced, and the electron movement is 10).

3) 6FeSO4 + 2HNO3 + 3H2SO4 - 3Fe2 (SO4) + 2NO+4H2O (Fe is oxidized N is reduced and electrons move 6).

4) 2FeCl3+Cu——2FeCl2+CuCl2 (Cu is oxidized, Fe is reduced and electrons move 2).

Second. 1) 10kmNO4+3I2+2H2O——10mNO2+6KIO3+4KOH (KMN04 oxidant I2 reducing agent).

2) 2kmNO4+16HCl——2KCl+2mnCl2+5Cl2+8H2O (KMno4 oxidant, HCl reducing agent).

3) Hno (extremely rare) + Zn (no3) + NH4NO3 + H2O (no solution).

4) KMno4+NaSO4+H2SO4 - MnSO4+K2SO4+NaSO4+H2O (no solution).

cu + hno3 = cu(no3)2 + h2o + nocu + hno3 = cu(no3)2 + h2o + no22 kmno4 + 5 h2s + 6 hcl = 2 kcl + 5 s + 2 mncl2 + 8 h2o

6 feso4 + 2 hno3 + 3 h2so4 = 3 fe2(so4)3 + 2 no + 4 h2o

Add me the rest and tell you.

It must contain BA (NO3)2, Na2CO3 (produce precipitation) and must not contain CUSO4 (colored, blue).

May contain NaCl

1. BA2+ +CO32- ==2BAC3 (precipitation symbol) 2, BAC3+2H+==BA2+ +CO2 (gas symbol) +H2O3, BA2+ +SO42-==BASO4 precipitation symbol AG+ +CO32-==AG2CO3 (precipitation symbol, silver carbonate is yellow, high sensitivity, insoluble in water, looks white).

Must contain barium nitrate, sodium carbonate; There must be no copper sulphate; May contain sodium chloride.

CO32- ====BaCO3 (precipitation symbol) + 2H+ =====Ba2+ +CO2 (gas symbol) +H2O

SO4 2- ==== BASO4 (precipitation symbol) 4It's possible.

SO4 2- +AG+ ====AGo4 (precipitation symbol) indicates that there is no sodium chloride.

Cl- +AG+ ==== AGCL (precipitation symbol) indicates the presence of sodium chloride.

(1)ba(no3)2、na2co3 cuso4 nacl

2）ba+co3=baco3

baco3+2h=h2o+co2+ba

ba+so4=baso4

2ag+co3=ag2co3

The valence of the owner's ions is marked by himself, and it is not easy to mark here. There is also the precipitation symbol.