1. 2
2. 2
3. 4
4. 3
5. 4
6. 2
7. 4
8. 235
40 %, 48%.................
В VI группе................................................
А)Cu(OH)₂+2HCl=CuCl₂+2H2O
Cu(OH)₂+2H⁺ + 2Cl⁻=Cu²⁺ + 2Cl⁻ +2H2O
Cu(OH)₂+2H⁺ =Cu²⁺ +2H2O
б)Na₂CO₃+H₂SO₄=Na₂SO₄+H₂O+CO₂
2Na⁺ + CO₃²⁻ +2H⁺ + SO₄²⁻ =2Na⁺ + SO₄²⁻ +H₂O+CO₂
CO₃²⁻ +2H⁺ =H₂O+CO₂
в)Fe(OH)₂ +2HCl = FeCl₂ + 2H₂O
Fe(OH)₂ +2H⁺ + 2Cl⁻ = Fe²⁺ + 2Cl⁻ + 2H₂O
Fe(OH)₂ +2H⁺ + = Fe²⁺ + 2H₂O
Cu+2H2SO4=CuSO4+SO2+2H2O
Cu(0)-2e=Cu(+2). 1. ок-ие
в-ль
S(+6)+2e=S(+4). 1. в-ие
ок-ль