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

I think it'll start at 2 p.m.

我想大概是下午2点吧。

When 实战 - SpeakingMax英语口语达人

So if I start off with a list of length n, how many times can I divide it by 2, until I get to something no more than two left?

我能够除以多少次2呢？,直到我得到的长度不超过2么？,对数次，对吧？就像刚才那位同学说的那样？

麻省理工公开课 - 计算机科学及编程导论课程节选

And I start by taking my first path here.

路径2本来。

麻省理工公开课 - 热力学与动力学课程节选

So, in particular, let's start working out what share of the votes you'd get if you chose position 1 or position 2, against different positions the other guy can choose.

先来看看如果你选择立场1和2,在你对手选择不同立场时,你分别能获得多少选票

耶鲁公开课 - 博弈论课程节选

Now we're going to start in with that pi 2 p orbitals, which gives us 1 each, and then two each in those, we'll go up to our sigma 2 p z orbital.

现在我们要填π2p轨道，每个1个，然后每个2个，我们我们填sigma2pz轨道。

麻省理工公开课 - 化学原理课程节选

2 6 8 1 3 7 5 If I start off with fou, two, six, eight, one, three, seven, 8 five, so my list is of size N equals 8 at the moment.

顺序如下：,现在列表的大小N等于。

哈佛公开课 - 计算机科学课程节选

If you start with only one, you have two pieces of DNA, then you'll get 2 to the Nth fragments after N cycles because each cycle you're doubling the number.

如果你从仅仅一个DNA开始,你有两条DNA链,经过N次循环后,就得到二的N次方个DNA片段,因为每次循环都使其数量翻倍

耶鲁公开课 - 生物医学工程探索课程节选

Again I should have said first, index 0, the first one. I can similarly go in and say I'd like all the things between index 2 and index 4. And again, remember what a b c that does. Index 2 says start a 0. 1, 2. So a, b, c.

我还是要说一遍，索引为0的元素，是第一个元素，我可以要求返回索引,2和4之前的所有元素，请记住2,是从0开始的,那么0，1，2对应的是。

麻省理工公开课 - 计算机科学及编程导论课程节选

Log n Log n, because at each stage I'm cutting the problem in half. So I start off with n then it's n n/2 n/4 n/8 over two n over four n over eight.

因为总共有多少层？,因为在每一层，我都是把问题分解成两半,因此以n开始，然后是。

麻省理工公开课 - 计算机科学及编程导论课程节选

So what we can say is look at each of these separately, so if we start with looking at the 2 p z orbital, the highest probability of finding an electron in the 2 p z orbital, is going to be along this z-axis.

我们可以来分别看看这些图，首先来看看2pz轨道，在2pz轨道里,找到电子的最大概率，是沿着ｚ轴。

麻省理工公开课 - 化学原理课程节选

So what we'll do is this problem here, which is let's calculate out what the wavelength of radiation n would be emitted from a hydrogen atom if we start at the n equals 3 level and we go down to the n equals 2 level.

我们来做这个问题,让我们来计算一下,从n等于3到,等于2能级氢原子辐射的波长是多少。

麻省理工公开课 - 化学原理课程节选

So essentially, we have two ethene or ethylene molecules here to start with where these blue are our 2 s p 2 hybrid orbitals, so you can see that for each carbon atom, one is already used up binding to another carbon atom.

本质上，我们从两个乙烯分子开始,蓝色的是2sp2杂化轨道，你可以看到，对于每一个碳原子，其中一个已经用来和另外一个碳原子成键。

麻省理工公开课 - 化学原理课程节选

So I guess we'll start with helium 2.

我们从He2开始。

麻省理工公开课 - 化学原理课程节选

So the payoff from choosing 1, let's start against 2.

我们先看看我选立场1,我们先从选立场2开始吧

耶鲁公开课 - 博弈论课程节选

first So what do I do? I find the midpoint by taking last minus first, divide by 2, and add to start.

要怎么做？我把last减去,除以二，加到起点上去。

麻省理工公开课 - 计算机科学及编程导论课程节选

So, our two glitches we see when we go from the 2 p, or from 2 s to start filling the 2 p, and then we also get another glitch when we've half-filled the 2 p, and now we're adding and having to double up in one of those p orbitals.

因此，我们的看到的两个小偏差，一个是在开始进入，2，p，轨道，或者说在填满，2，s，轨道之后,开始填，2，p，轨道的时候出现的，另一个则是在，2，p，轨道半满之后,开始继续加电子，使得其中一个，p，轨道上的。

麻省理工公开课 - 化学原理课程节选

So carbon has four valence electrons, so if we talk about c 2, again we're going to start filling in our molecular orbitals, and now we're going to have eight electrons to fill into our molecular orbitals.

碳有4个价电子，所以如果我们考虑C2的话，同样，我们开始填分子轨道，现在我们有8个电子,要填入分子轨道。

麻省理工公开课 - 化学原理课程节选

We'll start with Week 2 talking about Genetic Engineering; what's DNA, how can it be manipulated, how is our ability to manipulate DNA led to things like gene therapy which can now be done in people.

从第二周开始我们会讲基因工程,什么是DNA,怎么样才能操控它们,我们如何操作DNA,并将它应用于人类的基因疗法

耶鲁公开课 - 生物医学工程探索课程节选

So essentially, each of these orbitals come from linear combinations of all of the original orbitals, and it's hard to picture exactly how that happens, but one that you can at least start to get an idea is if you think about combining the 2 s and the 2 p z here, which is not quite accurate because of course, we're combining all of them.

本质上，这些轨道每个都来,自原来所有轨道的线性组合，我们很难想象这是怎么发生的，但你们可以至少有个概念,如果你们考虑2s和2pz轨道的结合，这当然是不太准确的，因为我们要把所有的都组合起来。

麻省理工公开课 - 化学原理课程节选