一个高二化学问题O2(g)=O2 +(g)+e- H1=1175.7kJ/molPtF6(g)+e-=PtF6-(g)
化学H2(g)+2/1O2(g)==H2O(g)△H1=-241.8kj/mol
高二化学2H2(g)+O2(g)=2H2O(1) △H=571、6kj、mol-1次方 C3H8(g)+5O2(g)=3
热化学方程式的比较C(g)+O2(g)=CO2(g) △H1 C(g)+1/2O2(g)=CO(g) △H2这两条化学反
N2(g)+2O2(g)=2NO2……① △H1=+67.7kJ/mol
(1)CH4(g)+2O2(g)=CO2(g)+2H2O(l),△H1=-890.3kj/mol
,已知下列反应 :H2(g)=2H(g) ΔH1=+Q1 kJ/mol O2(g)=2O(g) ΔH2=+Q2 kJ/m
已知 ①C2H5OH(g)+3O2(g)==2CO2(g)+3H2O(g);△H1=-a kJ/mol; ②H2O(g)
H2(g)+O2(g)=H2O(g) △H1
根据下列热化学方程式(1)C(s)+O2(g)=CO2(g)△H1=-393.5kJ/mol(2)H2(g)+12
化学反映热的计算已知A(g)+B(g)=C(g) H1D(g)+E(g)=E(g) H2,且H1
已知2H2(g)+O2(g)=2H2O(l);△H=-571.6kJ/mol,C3H8(g)+5O2(g)=3CO2(g
已知:2H2(g)+O2(g)=2H2O(l) △H=-571.6kj/mol 2h2(g)+O2(g)=2H2O(g)