已知CaCO3(s)=CaO(s)+CO2(g);ΔH=+175.7kj/mol C(s)+O2(g)=CO2(g);Δ
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已知CaCO3(s)=CaO(s)+CO2(g);ΔH=+175.7kj/mol C(s)+O2(g)=CO2(g);Δ=-393.5kj/mol,求1t碳酸钙煅烧生成石灰,理论上需要含杂质10%的焦炭多少千克?这个题我没看明白什么意思,烧碳酸钙要焦炭干什么
![已知CaCO3(s)=CaO(s)+CO2(g);ΔH=+175.7kj/mol C(s)+O2(g)=CO2(g);Δ](/uploads/image/z/4044765-21-5.jpg?t=%E5%B7%B2%E7%9F%A5CaCO3%EF%BC%88s%29%3DCaO%28s%29%2BCO2%28g%29%3B%CE%94H%3D%2B175.7kj%2Fmol+C%28s%29%2BO2%28g%29%3DCO2%28g%29%3B%CE%94)
碳酸钙完全分解需吸热,所以需要焦炭提供热量
已知CaCO3(s)=CaO(s)+CO2(g);ΔH=+175.7kj/mol C(s)+O2(g)=CO2(g);Δ
已知:C(s)+O2(g)=CO2(g) 变量H=-393.5kJ/mol,CaCO3(s)=CaO(s)+CO2(g)
已知:C(S)+O2(g)=CO2(g) △H=-393.5kj/mol CACO3(S)=CAO(S)+CO2(g)
已知热化学方程式CaCO3(S)=CaO(s)+CO2(g)-178.3kj,C(s)+O2(g)=CO2+395kj
CaCO3(s)==CaO(s)+CO2(g) ΔH=+178.5 KJ·mol-1 的意义是
C(s)+O2(g)=CO2(g) ΔH=-393.5 kJ/mol
CaCO3(s)→CO2(g)+CaO(s)-157KJ;C(s)+O2(g)+393.5KJ
已知:C(s)+1/2O2(g)=CO(s)▲H=-110.35kj/mol C(s)+O2(g)=CO2(g)▲H=-
1.c(s)+o2(g)=co2(g);△H=-393.5kj/mol
已知:C(s)+O2(g)=CO2(g);ΔH=-393.5KJ/mol.H2(g)+
已知 (1) 标况下 2C(s)+O2(g)==2CO2(g) △H=-221KJ/mol
9.已知热化学方程式:C(s)+O2(g)=CO2(g); △H=-393.3kJ/mol