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Problem 129

Calculate the weighted average atomic mass of zi…

Problem 128

Based on the relative abundances of each isotope,
predict to which isotope's mass the average atomic
mass of zirconium is going to be closest.





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Video Transcript

So now we'll work on problem 128 from chapter four in this problem, we're asked to protect relative abundance are based on the relative abundance for the isotopes of zirconium. Protect which isotopes mass will be closest to the average atomic mass. So we can go ahead and write in the elements. And they're, uh they're isotopes, and their abundance is so we can, uh, compare for each one. So we have zirconium. 90 91 92 94 the last is 90 six. So this is the isotope, and the abundances are the 1st 2 is 51 point for an 11.2, and we have 17.2, 17.4 and 2.8 for the last one. So these is this is isotope and abundance. So the question is which, for the average atomic mass of, uh, zirconium, which will be closest, Which mess will it be closest to? So the most abundant is, or Coney, um 90. We can see that, but we also do have a little abundance in zirconium 92 94 which is significant. Um, each is about 1/3 of the abundance of zirconium. 90 so it won't. The average will be close to 90 because it's at the end, even though it's the most effective. But it would probably be around, um, zirconium 91 because the most abundant is 90. But because we have a little bit more abundance of 92 94 we bring the average a little bit above 90 to around zirconium 91 as the closest to the average topic mass, and this can be checked to see if we're correct by looking at the period.

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