В 100 мл воды-12 г соли, в 20 000 мл-х г соли, отсюда х=2400 г
если принять плотность раствора=1 г/мл, m=плотность*v=20 000 г
<span>m(H2O)=20 000-2400=17600 г</span>
2СO + O2 → 2CO2↑
V(CO) : V(O2) = 2 : 1
Кислород взяли в избытке.
V1(O2) = 100/2 = 50 л
V(CO2) = 100 л
V2 = 100 л + 50 л = 150 л
ω(X) = 100\%*V(X)/V
ω(CO2) = 100\%*100/150 = 67\%
ω(O2) = 100\% - 67\% = 33\%
Ответ: 67\% СО2, 33\% О2
NH3+NO->N2+H2O
4 NH3 + 6 NO → 5 N20 + 6 H2O
2H2SO4+C=2SO2+CO2+2H2<span>O </span>→ 2+1=2+1+2 → 3+5=8 (Ответ 2)
1.
![K_{2}[Cu(CN)_{4}] = 2K^{+} + [Cu(CN)_{4}]^{2-}](https://tex.z-dn.net/?f=K_%7B2%7D%5BCu%28CN%29_%7B4%7D%5D%20%3D%202K%5E%7B%2B%7D%20%2B%20%5BCu%28CN%29_%7B4%7D%5D%5E%7B2-%7D)
![[Cu(CN)_{4}]^{2-} = Cu^{2+} + 4CN^{-}](https://tex.z-dn.net/?f=%5BCu%28CN%29_%7B4%7D%5D%5E%7B2-%7D%20%3D%20Cu%5E%7B2%2B%7D%20%2B%204CN%5E%7B-%7D)
![K = \frac{[Cu^{2+}][CN^{-}]^4}{[[Cu(CN)_{4}]^{2-}]}](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B%5BCu%5E%7B2%2B%7D%5D%5BCN%5E%7B-%7D%5D%5E4%7D%7B%5B%5BCu%28CN%29_%7B4%7D%5D%5E%7B2-%7D%5D%7D)
2.
![Na[PdI_{4}] = Na^{+} + [PdI_{4}]^{-}](https://tex.z-dn.net/?f=Na%5BPdI_%7B4%7D%5D%20%3D%20Na%5E%7B%2B%7D%20%2B%20%5BPdI_%7B4%7D%5D%5E%7B-%7D)
![[PdI_{4}]^{-} = Pd^{+3} + 4I^{-}](https://tex.z-dn.net/?f=%5BPdI_%7B4%7D%5D%5E%7B-%7D%20%3D%20Pd%5E%7B%2B3%7D%20%2B%204I%5E%7B-%7D)
![K = \frac{[Pd^{3+}][I^{-}]^4}{[[PdI_{4}]^{-}]}](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B%5BPd%5E%7B3%2B%7D%5D%5BI%5E%7B-%7D%5D%5E4%7D%7B%5B%5BPdI_%7B4%7D%5D%5E%7B-%7D%5D%7D)
3.
Bd - неизвестный химический элемент; скорее всего вы допустили опечатку; замените этот символ на нужный в дальнейшем; ответ пишу в исходном виде.
![[Bd(H_{2}O)(NH_{3})_{2}Cl]Cl = Bd(H_{2}O)(NH_{3})_{2}Cl]^{+} + Cl^{-}](https://tex.z-dn.net/?f=%5BBd%28H_%7B2%7DO%29%28NH_%7B3%7D%29_%7B2%7DCl%5DCl%20%3D%20Bd%28H_%7B2%7DO%29%28NH_%7B3%7D%29_%7B2%7DCl%5D%5E%7B%2B%7D%20%2B%20Cl%5E%7B-%7D)
![[Bd(H_{2}O)(NH_{3})_{2}Cl]^{+} = Bd^{2+} + H_{2}O + 2NH_{3} + Cl^{-}](https://tex.z-dn.net/?f=%20%5BBd%28H_%7B2%7DO%29%28NH_%7B3%7D%29_%7B2%7DCl%5D%5E%7B%2B%7D%20%3D%20Bd%5E%7B2%2B%7D%20%2B%20H_%7B2%7DO%20%2B%202NH_%7B3%7D%20%2B%20Cl%5E%7B-%7D)
![K = \frac{[Bd^{2+}][H_{2}O][NH_{3}]^2[Cl^{-}]}{[[Bd(H_{2}O)(NH_{3})_{2}Cl]^{+}]}](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B%5BBd%5E%7B2%2B%7D%5D%5BH_%7B2%7DO%5D%5BNH_%7B3%7D%5D%5E2%5BCl%5E%7B-%7D%5D%7D%7B%5B%5BBd%28H_%7B2%7DO%29%28NH_%7B3%7D%29_%7B2%7DCl%5D%5E%7B%2B%7D%5D%7D)
4.
![[Pt(NH_{3})_{2}Cl_{4}] = Pt^{4+} + 2NH_{3} + 4Cl^{-}](https://tex.z-dn.net/?f=%5BPt%28NH_%7B3%7D%29_%7B2%7DCl_%7B4%7D%5D%20%3D%20Pt%5E%7B4%2B%7D%20%2B%202NH_%7B3%7D%20%2B%204Cl%5E%7B-%7D)
![K = \frac{[Pt^{4+}][NH_{3}]^2[Cl^{-}]^4}{[Pt(NH_{3})_{2}Cl_{4}]}](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B%5BPt%5E%7B4%2B%7D%5D%5BNH_%7B3%7D%5D%5E2%5BCl%5E%7B-%7D%5D%5E4%7D%7B%5BPt%28NH_%7B3%7D%29_%7B2%7DCl_%7B4%7D%5D%7D)
