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If I get you down to a cluster of about 30 water molecules, the boiling point is a function of the size of the water droplet.
如果对于一个由30个水分子,组成的簇，沸点是,水珠尺寸的函数。
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For me... I mean eventually I would like to, hopefully get a post doc and
对于我……我是说最终我想，希望能拿个博士后然后
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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。
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So, there will be a whole -- I'm looking ahead now, projecting into some topics that we'll explore more closely later.
所以对于这一部分,大家先有个印象,我们以后将会更细致的讨论
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Often, for a particular state change, it is easy to calculate two of these, but not the third.
对于一些特定的状态变化，它们中的两个很好计算，但是第三个往往不容易计算。
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Let's see if we can do that, and it's a little bit of a challenge for the performer.
让我们看看能不能做到这个,对于表演者来说,还真是个挑战呢
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And, here, I've got 17 protons in the nucleus.
对于氯，它的原子核里有17个质子。
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But you know, you kind of wonder if that's a good example for fire insurance because if the whole city burns down, then insurance companies would go bankrupt anyway, right?
但是你也许会疑惑,对于阐述火险来说,这是不是个好例子,因为如果整个城市被烧毁,那么保险公司就会破产,对吧
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We might have a term for this sort of thing--possession.
对于这种现象我们有个名词，夺舍。
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Remember last time, I said that there's different kinds of complexity in our code, and I suggested for simple branching programs, the amount of time it takes to run that program is, in essence, bounded by the number of instructions, because you only execute each instruction at most once.
但是这里有个很重要的点,记得上节课，我提过在我们的代码中，有不同种类的复杂度，而且我还说了对于简单的分支程序,运行这种程序需要的总体时间,大体上，是和指令的数目相关的,因为每个指令只会被执行最多一次。
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You can brute force it and unfortunately, the so-called Caesar cipher is not very secure because if you assume, for our purposes in English alphabet with 26 characters, say all lowercase for all uppercase, my God, you only have to try like 24, 25, 26 possible rotations until you can figure out what his secret message is.
你可以强行解密但是不幸地是，所谓的凯撒密码不是很安全的,因为如果假设，对于我们的26个字符的字母表,指明所有的小写为大写，我的天，你只要试24，25次，26次就可以解答出,密文的内容。
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I suggested that there was a critical response to those issues that was somewhat negative, and I want to sort of remind you of that from W. E. B. Du Bois's review of Black Boy when it came out.
对于负面问题,有个关键评述，大家看一下,关于《黑孩子》的评论。
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There may be a little bit of uncertainty about that.
对于你能否进食还有个不确定因素
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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。
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For some of you that haven't I'll just say that it's a protein, it's 238 amino acids, which means that it's about 1,000, actually more than 1,000 atoms in size, and this protein is fluorescent.
对于那些没听说过的人,我要说它是一种蛋白质，它有238个氨基酸，也就意味着它有约1000个，实际上是超过1000个原子，这种蛋白质是有萤光的。
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That is a very important difference that was, you had to act in such a way as to make life possible and decent, and for the Greeks that always meant being part of a decent community, the polis.
这也是个非常重要的区别,你得活出体面来,对于古希腊人而言就是,成为体面的社区即城邦的一部分
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s1 So for the fourth period, now we're into the 4 s 1 3d for potassium here. And what we notice when we get to the third element in 4s2 and the fourth period is 3d that we go 4 s 2 and then we're back to the 3 d's.
对于第四周期到现在我们来到钾的1,然后我们返回到,我们注意到当我们看到第三个元素,第四周期我们来到,然后我们返回到。
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So when we talk about p orbitals, it's similar to talking about s orbitals, and the difference lies, and now we have a different value for l, so l equals 1 for a p orbital, and we know if we have l equal 1, we can have three different total orbitals that have sub-shell of l equalling 1.
当我们考虑ｐ轨道时，这和ｓ轨道的情形和相似，不同之处在于ｌ的值不一样，对于ｐ轨道，ｌ等于１,我们知道如果ｌ等于１,我们有３个,不同的轨道。
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It is. It's drawn like an array, it effectively is an array so bracket zero means go to the zero's location in that array, which happens to be F and do what with it?
它看起来像个数组，实际上也是一个数组,所以【0】就是定位到那个数组的0的存储单元，对于f发生了什么？用它做什么？
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/2 So this would mean the bond order is equal to 1/2, and in terms of valence electrons, how many bonding valence electrons do we have?
这意味着键序等于，对于价电子，有多少个成键价电子？
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So for a 2 s orbital, how many total nodes will we have?
对于２ｓ轨道，有多少个节点？
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And in terms of radial nodes, we expect to see one node.
对于径向节点，应该有１个。
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F Cl There is fluorine and chlorine where you have seven electrons.
而对于,有7个价电子。
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So for some you that are less interested in maybe the physical structure of an individual atom, now some more exciting material for you might be coming up if you like to think about how, instead, molecules behave, either within bonding, within themselves, or with other molecules, that's what we're going to be heading to in this next unit.
那么对于某些同学，你们或许不感兴趣,对于单个原子的物理结构，现在可能有令你感到兴奋的内容,要出现了，如果你更喜欢思考,分子的行为，或者是关于成键的，或者是关于它们本身的，又或者与其它分子之间的行为，这些将是我们下个单元要讲的内容。
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So, this should be pretty straight forward, 100% let's see if we can get close to a 100% on this one, which is how many radial nodes does a 4 p orbital have?
很简单,让我们来看看这题，我们是不是可以接近%,对于一个4p轨道，它有多少个节点？,给你们10秒钟？
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So, remember, boron was one of those exceptions to our Lewis structure rules where it was perfectly happy not having a full octet.
所以我们可以加上3个氢原子,对于BH3，这是个稳定结构,记住，B是Lewis结构的一个特例。
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So before long, we'll see, especially for problems that we can't allocate an int every time we want to store something, because what if we want to store 140,000 English words.
所以不久后，我们将看到，特别是对于,我们不能为int分配内存，每次我们想要,存储一些东西时，因为当我们想要,存储140，000个英语单词时。
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So what we end up with is one radial node for the 2 s orbital of hydrogen, and we can apply that for argon or any other multi-electron atom here, we also have one radial node for the 2 s orbital of argon.
那意味着它们都是径向节点,所以我们得出的结论是，氢的2s轨道是1个径向节点,我们可以将它应用，到氩或者任意一个多电子原子,对于氩的2s轨道。
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And similarly, particularly for those less comfortable, do we offer prior to the start of each week for a new piece at what we dub a walkthrough whereby one of the teaching fellows will literally start on the first page, give you a mental framework for the week's problems and talk about the ideas that it covers, and then walk through the piece set so that you have a roadmap of sorts that you can follow.
同样，特别对于那些不熟悉编程跟不上进度的同学，每周上课前都可以预习新的知识点,一名教师从第一页开始,对于每周的问题,都给你们简历一个心理框架,并讨论它所涵盖的思想，然后了解了每个部分之后,就可以有个学习的路线图。
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All right, so in terms of s p 3 hybrid orbitals, let's combine all four together on one axis, because this is what's going to happen in an s p 3 carbon atom.
对于sp3杂化轨道，让我们把4个结合起来，因为这是sp3碳原子中发生的情况。
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