475 Multiply by 1.7475 787kJ/mol and you end up with minus 787 kilojoules per mole.

乘以1。,最终得到。

麻省理工公开课 - 固态化学导论课程节选

For the carbon, we start with 4 valence electrons, we have 0 lone pair electrons minus 4, and we end up with a formal charge of 0.

对于碳，我们从四个价电子开始，我们有零个孤对电子，再减去四，最终我们有零个形式电荷。

麻省理工公开课 - 化学原理课程节选

So I'm first going to look for something that's not in the list, I'm going to see, is minus 1 in this list, so it's going to be at the far end, and if I do that in the basic case, bam.

如果我试试第一种最基本的方法,噢，一下就完成了对不对？,因为这种方法查了下第一个元素，然后发现目标数比较下，因为目标数小于第一个元素。

麻省理工公开课 - 计算机科学及编程导论课程节选

And we plug in our values and end up with mv squared mv^2/r-Ze^2/ over r minus Ze squared over And I am going to call this equation two.

我们最后的结果,就是,我把这称为方程式二。

麻省理工公开课 - 固态化学导论课程节选

That's equivalent to doing the integral, and so, what we end up getting is that the reversible work v2 pdv is equal to minus integral V1, V2, p dV.

这与刚才的积分过程效果相同，最后，我们得到的结论是可逆过程的功,是负的积分，从v1到。

麻省理工公开课 - 热力学与动力学课程节选

For the sulfur, we start off with 6 valence electrons, minus 4 lone pair electrons, minus 2, taking in account our bonding electrons, so we end up with a formal charge of 0.

对于硫，我们从六个价电子开始，减去四个孤对电子，再减去二，算上我们的成键电子，因此最终我们有零个形式电荷。

麻省理工公开课 - 化学原理课程节选

So for the bond order we want to take 1/2 of the total number of bonding electrons, so that's going to be 4 minus anti-bonding is 4, so we end up getting a bond order that's equal to 0.

键序等于1/2乘以,总的成键电子数，也就是4，减去反键电子数，也就是4。,所以最后得到键序为0。

麻省理工公开课 - 化学原理课程节选

If we look at our last structure here where we have nitrogen the middle, we can also figure out all those formal charges, and in this case we have plus 1 on the nitrogen, we have minus 2 on the carbon, and then we end up with a 0 on the sulfur there.

如果我们来看看最后一个结构,在中间的原子是氮，我们同样可以计算出所有的形式电荷，而在这种情况下，氮为正一，碳为负二，而最后硫为零。

麻省理工公开课 - 化学原理课程节选