Summary of chemical equations by family .

I can send you the equation that must be taken one or two.

2o3=3o2

Fe+S (heating) = Fes

s + h2 (heat) = h2s

2H2O2 (MNO2) = 2H20 + O2 (arrow) SO2 + H20 (mutual inversion) = H2SO4

2SO2 + O2 (mutual inversion, catalyst, heating) = 2SO3SO3 + H2O = H2SO4

2H2SO4 (concentrated) + Cu (heated) = CuSO4 + 2H2O + SO2 (arrow).

2H2SO4 (concentrated) + C (heated) = CO2 (arrow) + 2H2O + 2SO2 (arrow).

SiO2 + 2C (high temperature) = Si + 2Co (arrow) SiO2 + Cao (high temperature) = Casio3

SiO2+2NaOH = Na2SiO3 + H2ONa2CO3 + SiO2 (high temperature) = Na2SiO2 + CO2 (arrow) CaCO3 + SiO2 (high temperature) = Casio3 + CO2 (arrow).

S + H2 ==H2S (heated) Fe + S ==Feds (heated).

2o3 ==3o2 3o2 ==2o3 (discharge).

2H2O2 ==2H2O + O2 (MnO2 catalysis) H2S ==H2 + S (high temperature).

2H2S + 3O2 ==2H2O +2So2 2H2S + O2 ==2H2O +2S are all ignited.

fes + 2hcl ==fecl2 + h2s↑ fes + h2so4 ==feso4 + h2s↑

SO2 + H2O ==H2SO4 (reversible) SO3 + H2O ==H2SO4

Cu + S ==Cu2S (Black Solid) H2O2 + H2S ==S + H2O

NaSO3 + H2SO4 ==NaSO4 + SO2 + H2O Laboratory-made sulfur dioxide.

2H2SO4 concentrate + Cu ==CuSO4 +2H2O + SO2 heating.

2H2SO4 concentrate + C ==CO2 +2H2O+ 2SO2 heating.

2SO2 + O2 ==2SO3 (needs to be heated, catalyst and reversible).

H2S + H2SO4 (concentrated) ==S +SO2 + 2H2O

H2S +2Hno3 (Concentrated) ==3S +2No + 4H2O

Mutual conversion of O2 and O3.

3o2 = discharge = 2o3

2o3=3o2

SO2 is soluble in water.

so2+h2o=h2so3

The pH of acid rain becomes smaller.

so2+h2o=h2so3

2h2so3+o2=2h2so4

SO2 vs. H2S

so2+2h2s=3s2↓+2h2o

H2S deteriorates in the air.

2h2s+o2=2s↓+2h2o

Cl2 is passed into H2S

cl2+h2s=s+2hcl

Industrial SO3, H2SO4,

2SO2 + O2 = Reversible = 2SO3 (catalyst V2O5) SO3 + H2O = H2SO4

Na2SO3 deteriorates.

2na2so3+o2=2na2so4

NaOH passes less or more SO2

so2+2naoh=na2so3+h2o

so2+naoh=nahso3

Lab-produced H2S, SO2

fes+h2so4=h2s↑+feso4

H2SO4 (concentrated) + Na2SO3 = SO2 + H2O+Na2SO4

1 3O2 = 2O3 conditional discharge.

2`so2+h2o=h2so3

3 Because 2H2SO3 + O2 = H2SO4

5`h2s+cl2=2hcl+s

6 2SO2+O2=3SO3 condition high temperature, catalyst SO2+H2O=H2SO3

8. Less Na2SO3 is generated, and more Na2SO4 is generated

(1) Group A element C has strong reducibility, so it can replace with silica to form silicon and carbon monoxide, so the answer is: SiO2

2C high temperature. si+2co；

2) The greater the bond energy of the chemical bond, the greater the energy required to break the chemical bond, and the higher the melting point of the substance, as you can see from the chart, the bond energy of the Si-O bond is greater than that of the Si-Cl bond, so SiCl4

The melting point is higher than that of SiO2

The answer is: ;

sicl4（g）+h2（g）=1

si（s）+2hcl（g）△h=+118 kj?mol-1, h = bond energy of reactants - bond energy of products = (1

360 4+h-h bond energy -1

176×2-2×431）kj?mol-1=+118 kj?mol-1

The H-H bond energy is 436 kJ?mol-1, so the answer is: 436

(1) Group A element C has strong reducibility, so it can replace with silica to generate silicon and carbon monoxide, so the answer is: SiO2 +2C high temperature

si+2co；

2) The greater the bond energy of the chemical bond, the greater the energy required to break the chemical bond, and the higher the melting point of the substance, as we know from the chart, the bond energy of the Si-O bond is greater than that of the Si-Cl bond, so the melting point of SiCl4 is lower than that of SiO2

So the answer is: ;

sicl4 （g）+h2 （g）=1 2

si（s）+2hcl（g）△h=+118 kj?mol-1, h = bond energy of reactant - bond energy of product = (1 2 360 4 + bond energy of h-h -1 2

176×2-2×431）kj?mol-1 =+118 kj?mol-1

The H-H bond energy is 436 kJ?mol-1, so the answer is: 436

The expansion and contraction vibration of the c=c of the aromatic hydrocarbon compound has 3 characteristic absorption frequencies cm-1 if you agree with my answer,. Yours is the motivation for me to answer the question, o( o thank you!!)