• Even with the rise, the latest figure remains nearly 50 percent below the consumer confidence number reported two years ago.

    VOA: standard.2009.08.25

  • And it takes the set of stocks to simulate, a fig, figure number, this is going to print a bunch of graphs, and I want to say what graph it is, fig mo and whether or not I believe in big mo.

    我们需要一些股票才能进行模拟,很多很多的股票,我们会打印一些图,然后我会告诉你们这些图是什么意思,或者我们看到这个。

    麻省理工公开课 - 计算机科学及编程导论课程节选

  • All right so I probably should have chosen a shorter number 71 because now I can't figure it in here, but let's call it 71.

    好的,我可能应该选择一个更短的数字,因为我不能指出它,我把它叫做。

    哈佛公开课 - 计算机科学课程节选

  • The way that we can figure this out is using something called bond order, and bond order is equal to 1/2 times the number of bonding electrons, minus the number of anti-bonding electrons.

    我们可以用叫做,键序的概念来弄明白它,键序等于1/2乘以成键电子,数目减去反键电子数目。

    麻省理工公开课 - 化学原理课程节选

  • And for any of the other compounds. So in other words, by defining that reference state, we can then figure out or measure heats of formation of a vast number of compounds.

    对吧?对于任何其他的,化合物也是一样,所以换句话讲,定义了这个参考状态,我们就能算出或测量许多。

    麻省理工公开课 - 热力学与动力学课程节选

  • Well, the figure I saw last was a number too small to be mentioned; I mean, to say you just can't really say how much it is, but all right, two percent.

    这个数字简直,小的可忽略了,我的意思是说你实在说不出精确地数字,但实际上只有2%左右

    耶鲁公开课 - 古希腊历史简介课程节选

  • You can go ahead and use that equation, or you could figure it out every time, because if you know the total number of nodes, and you know the angular node number, then you know how many nodes you're going to have left.

    你们可以直接用这个方程,或者每次都自己算出来,因为如果你们知道了总的节点数,又知道角向节点数,就知道剩下的节点数是多少。

    麻省理工公开课 - 化学原理课程节选

  • So then in order to figure out the complete number of valence electrons in our molecule, we just add 5 plus 4 plus 1.

    那么接下来为了得到,这个分子中价电子的总个数,我们只需要将五加上四,再加上一。

    麻省理工公开课 - 化学原理课程节选

  • So, if we want to figure out the formal charge on the carbon, we need to take the number of valence electrons, so that's 4. We need to subtract the lone pair, what number is that? It's 2.

    如果我们想算出碳原子的形式电荷,我们需要将价电子的个数,也就是四,减去孤对电子的个数,它是多少?是二。

    麻省理工公开课 - 化学原理课程节选

  • So to figure out bonding electrons, -- what we take is that number 18, which is our total number of electrons we need to fill valence shells, and we subtract it from our number of valence electrons, which is 10.

    那么为了找出成键电子,我们将十八,也就是填满所有价壳层,所需要电子的总个数,减去我们所有的价电子的个数,也就是十。

    麻省理工公开课 - 化学原理课程节选

  • And there is a relationship here, and that is that the number of photons absorbed by the metal are related to the number of electrons ejected from the metal. So, in this figure here what I'm actually showing is these little sunshines, which let's say are each one individual photon.

    这里有一个关系是,被金属吸收的光子,和被金属逐出的电子是相关的,所以,我展示的这个图,是这些小太阳,我们这里代指每一个独立的光子。

    麻省理工公开课 - 化学原理课程节选

  • We know we need to divide by 266, 266 but what we need you to help us with is to figure out this top number here and see how many particles are going to backscatter. So if the TAs can come up and quickly hand out 1 particle to everyone.

    知道背散射的概率就可以了,我们知道要除以,但还需要你们来搞清楚,分子上的数是多少,有多少个发生背散射的粒子,助教们请过来,把这些球分给同学们。

    麻省理工公开课 - 化学原理课程节选

  • psi So we're going to for psi, and before that, we're going to figure out that instead of n just that one quantum number n, we're going to have a few other quantum numbers that fall out of solving the Schrodinger equation for what psi is.

    我们要讲到,但在这之前,我们已经知道了,主量子数,现在我们需要知道,其他一些,解psi的薛定谔方程,所需要的量子数。

    麻省理工公开课 - 化学原理课程节选

  • And we can generalize to figure out, based on any principle quantum number n, how many orbitals we have of the same energy, n and what we can say is that for any shell n, there are n squared degenerate orbitals.

    我们可以总结出来,在,主量子数为n的情况下,同一个能量上,有多少个轨道,我们可以说,对任何壳层,有n平方个简并轨道。

    麻省理工公开课 - 化学原理课程节选

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