试题详情
常温下N
2H
4为液体,可作为火箭发动机的燃料,与氧化剂N
2O
4发生反应:2N
2H
4 + N
2O
4 ![](https://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmrow%3E%3Cmunder+accentunder%3D%22true%22%3E%3Cmunder+accentunder%3D%22true%22%3E%3Cmrow%3E%3Cmi%3E%E7%82%B9%3C%2Fmi%3E%3Cmi%3E%E7%87%83%3C%2Fmi%3E%3C%2Fmrow%3E%3Cmo+stretchy%3D%22true%22%3E_%3C%2Fmo%3E%3C%2Fmunder%3E%3Cmo+stretchy%3D%22true%22%3E_%3C%2Fmo%3E%3C%2Fmunder%3E%3C%2Fmrow%3E%3C%2Fmath%3E)
3N
2 + 4H
2O。分子球棍模型如下图所示,N
A为阿伏加德罗常数的值,下列叙述正确的是( )
![](http://tikupic.21cnjy.com/2e/b9/2eb99edf04c0ad78f200d8310989b57f.png)
- A、32 g N2H4中含有共用电子对数为6 NA
- B、标准状况下,22.4L N2H4中含有的分子数为NA
- C、N2H4和N2O4混合的总物质的量为1mol时,含氮原子数为4 NA
- D、上述反应消耗1mol N2H4 , 转移电子数为4 NA
知识点
参考答案