All the chemical equations of the sophomore year, what is the chemical equation of the sophomore yea

The chemistry equation for the second year of high school is as follows:

1. Aluminum and hydrochloric acid: 2Al+6HCl=2AlCl3+3H2 gas.

2. Aluminum and sulfuric acid: 2Al+3H2SO4=Al2(SO4)3+3H2 gas.

3. Aluminum and sodium hydroxide: 2Al+2NaOH+H2O=2Naalo2+3H2 gas.

4. Aluminum hydroxide heating is divided into 2AL (OH) 3 = (heating) AL2O3 + H2O.

5. Aluminum sulfate and ammonia: Al2(SO4)3 6NH3, H2O 2AL(OH)3 precipitation 3(NH4)2SO4.

6. Alumina and hydrochloric acid: Al2O3 6HCl=2ALCL3+3H2O.

7. Alumina and sodium hydroxide: Al2O3+2NaOH=2Naalo2+H2O.

8. Aluminum chloride and sodium hydroxide: AlCl3 + 3NaOH = Al(OH)3 precipitation + 3NaCl.

9. Aluminum hydroxide and sodium hydroxide: Al(OH)3 NaOH=Naalo2+2H2O.

10. Aluminum chloride and sodium hydroxide: AlCl3+4NaOH=Naalo2+2H2O.

Electrolysis = 2NaOH + H2 + Cl2

cl2+cu=cucl2;3Cl2+2Fe=2FeCl3 (These 3 are more classic, and the whole writing is incomplete.) )

11.Tested with agno3, if there is cl-, there is a white precipitate. If there is br-, it has a light yellow color. If there is cl-, there is a yellow precipitate.

The ionic equations are: ag+

Cl-AGCL (white precipitate);

ag+br-

AGBR (light yellow precipitate); ag+

i-AGI (yellow precipitate).

12.Total equation: Zn+H2SO4=ZnSO4+H2 The electrons flow from the negative electrode to the positive electrode, then the current flows from the positive electrode to the negative electrode.

Negative electrode: Zn as negative electrode, Zn2E-

Zn2+ cathode: Cu is the cathode, 2H+2E-

Ignition = SO2, SO2 + H2O = H2SO3, 2H2SO3 + O2 = 2H2SO4

Concentrate) + 2NaCl = Heating = Na2SO4 + 2HCl

Concentrate) = Heating = CuSO4 + 2H2O + SO2

2H2SO4 (concentrated) + C = CO2 +2SO2 +2H2O

16.The chemical equation is: Na2SO4 + BACL2 = NaCl2 + BaSO4

Ion equation: SO42-

ba2+baso4↓

High temperature and high pressure catalyst = 2NH3; N2 + O2 = Discharge = 2NO

High temperature and high pressure catalyst = 2NH3;

Reversible symbol) NH4+H2O

H+ Since ammonium ions are hydrolyzed, the concentration of ammonium ions is less than that of chloride ions.

Electrolysis = Cu + Cl2

High temperature = F3O4 + 4H2

High temperature = 2Fe + Al2O3

Heating = 2al(OH)3+3h2

h+alo2-

Heating = Fe3O4 + 4H2

High temperature = 2Fe + Al2O3

Upstairs, it's thin, not thick.

Fe3O4 + 10Hno3 concentrate = 3Fe (No3) 3 + No2 (arrow) + 5H2O

3Fe3O4+28Hno3 (dilute) = 9Fe(NO3)3+NO+14H20

Fe3O4+8Hi==3FE2+4H2O+I2 (Fe3O4 can be regarded as Fe2O3·FEO, Fe3+ oxidizable I- is I2), condition: no special conditions (can be regarded as no condition).

Upstairs trim error!

3Fe3O4+28Hno3 (concentrated) = 9Fe(NO3)3+NO+14H20

fe3o4+8hi==3fei2+4h2o+i2

Bringing it to the exam room? You're cheating! It's better not to do that!

It's easy to spot! Finding out is 0 points... It's the final exam, and you don't want your family to scold you, if you are discovered, you can't get by!

You should take the test honestly, even if you fail the test, it is better than cheating on the exam and being discovered!

The high one is mainly a redox problem, and it is necessary to clarify the elemental valency of reactants and the resulting species, and the same is true for the ion equation.

High oxygen loss and low also.

The valency becomes higher and loses electrons, is oxidized, and the reducing agent undergoes an oxidation reaction to produce oxidation products.

The valency becomes lower to obtain electrons, which are reduced, and the oxidant undergoes a reduction reaction to produce a reduction product.

The oxidation of oxidation products is weak by oxidants, and the reducing properties of reducing products are weak by reducing agents.

The representation can be expressed by a two-wire bridge and a single-line bridge, the two-wire bridge is easy to express and is better used in reactions with many valence changes, and the single-wire bridge is faster in the redox simpler reaction formula.

The total reaction equation redox gains and losses should be balanced, so the relative numbers in the reactants and products can be derived from this.

I don't think I need to answer this question professionally, because it would be very difficult to get started. First, the number of each element in the equation should be conserved; Secondly, the transfer of electrons should be conserved (this is very important in the second and third years of high school, and it is basically used). In the first year of high school, it is necessary to follow the current order, but in the second or third year of high school, it should be used upside down, that is, first transfer the conservation of electricity and then the conservation of numbers.

Redox trimming.

Three steps: 1. Conservation of electrons.

2. Conservation of electric charge.

3. Conservation of atoms.