早教吧作业答案频道 -->化学-->
已知:(1)2Fe(s)+O2(g)=2FeO(s)△H=-544kJ•mol-1(2)2Al(s)+32O2(g)=Al2O3(s)△H=-1675kJ•mol-1则2Al(s)+3FeO(s)=Al2O3(s)+3Fe(s)的△H为()A.-2491kJ•mol-1B.+859kJ•mol-1C.-1403
题目详情
已知:
(1)2Fe(s)+O2(g)=2FeO(s)△H=-544kJ•mol-1
(2)2Al(s)+
O2(g)=Al2O3(s)△H=-1675kJ•mol-1
则 2Al(s)+3FeO(s)=Al2O3(s)+3Fe(s)的△H为( )
A.-2491 kJ•mol-1
B.+859 kJ•mol-1
C.-1403 kJ•mol-1
D.-859 kJ•mol-1
2-1
O2(g)=Al2O3(s)△H=-1675kJ•mol-1
则 2Al(s)+3FeO(s)=Al2O3(s)+3Fe(s)的△H为( )
A.-2491 kJ•mol-1
B.+859 kJ•mol-1
C.-1403 kJ•mol-1
D.-859 kJ•mol-1
3 3 2 2 223-1
23
-1
-1
-1
-1
(1)2Fe(s)+O2(g)=2FeO(s)△H=-544kJ•mol-1
(2)2Al(s)+
3 |
2 |
则 2Al(s)+3FeO(s)=Al2O3(s)+3Fe(s)的△H为( )
A.-2491 kJ•mol-1
B.+859 kJ•mol-1
C.-1403 kJ•mol-1
D.-859 kJ•mol-1
2-1
3 |
2 |
则 2Al(s)+3FeO(s)=Al2O3(s)+3Fe(s)的△H为( )
A.-2491 kJ•mol-1
B.+859 kJ•mol-1
C.-1403 kJ•mol-1
D.-859 kJ•mol-1
3 |
2 |
23
-1
-1
-1
-1
▼优质解答
答案和解析
已知:(1)2Fe(s)+O22(g)=2FeO(s)△H=-544kJ•mol-1-1
(2)2Al(s)+
O2(g)=Al2O3(s)△H=-1675kJ•mol-1
据盖斯定律,(2)-(1)×
得:2Al(s)+3FeO(s)=Al2O3(s)+3Fe(s)△H=-859KJ/mol
故选:D
3 3 32 2 2O22(g)=Al22O33(s)△H=-1675kJ•mol-1-1
据盖斯定律,(2)-(1)×
得:2Al(s)+3FeO(s)=Al2O3(s)+3Fe(s)△H=-859KJ/mol
故选:D
3 3 32 2 2得:2Al(s)+3FeO(s)=Al22O33(s)+3Fe(s)△H=-859KJ/mol
故选:D
(2)2Al(s)+
3 |
2 |
据盖斯定律,(2)-(1)×
3 |
2 |
故选:D
3 |
2 |
据盖斯定律,(2)-(1)×
3 |
2 |
故选:D
3 |
2 |
故选:D
看了 已知:(1)2Fe(s)+O...的网友还看了以下:
已知下列反应的反应热(1)CH3COOH(l)+2O2(g)=2CO2(g)+2H2O(l)△H1 2020-05-13 …
氨气的燃烧热的热化学方程式NH3(g)燃烧生成NO2和H2O,已知(1)H2(g)+1/2O2(g 2020-05-14 …
已知1/3≤a≤1若函数f(x)=ax²-2x+1在区间[1,3]上的最大值为Ma最小值为Na令g 2020-05-16 …
1.设函数f(x)是定义在R上的周期为2的偶函数,当x∈[0,1]时,f(x)=x+1,则f(3/ 2020-06-03 …
已知函数f(x)=ax3-3ax,g(x)=bx2+clnx,且g(x)在点(1,g(1))处的切 2020-06-16 …
已知f(x)、g(x)都是定义在R上的函数,g(x)≠0,f(x)g(x)=ax,且f′(x)g( 2020-06-16 …
已知函数f(x-1)的图像与函数g(x)的图像关于直线y=x对称,且g(1)=2则:A,f(1)= 2020-06-27 …
甲醇是一种优质燃料,在工业上常用CO和H2合成甲醇,反应方程式为CO(g)+H2(g)CH3OH( 2020-07-19 …
已知f(x),g(x)都是定义在R上的函数,g(x)≠0,f(x)g′(x)>f′(x)g(x), 2020-08-02 …
(1)求g(x)的表达式.(2)若存在x>0使f(x)<=0成立,求实数m的取值范围已知二次函数g( 2020-12-08 …