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Hello, everybody.
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Today we're talking about aluminum.
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So we're given in the problem that diameter of the nucleus is equal to 8 times 10 to the negative 15th meters.
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And the diameter of the atom is equal to 1 .2 times 10 to the negative 10th meters.
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The density, when you've got a bunch of aluminum together, is equal to 2 ,700.
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Kilograms per meter cubed.
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So part a asks us to get the density of just a single atom of aluminum.
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So the density is equal to mass per volume, or in the case of a sphere, mass is equal to four, thirds pi r cubed but since we have density we're going to do density or since we have diameter we're going to do diameter we're too cute so plugging in our numbers we get 4 .4 8 times 10 to the negative 26 pillar grams that's the mass of a single atom of aluminum and we'll multiply that by 3 from the bottom and then pi times 4 times 1 .2 times times times times times times times the negatives divided by 2.
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That value cubed.
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And that is going to equal 49 ,500 kilograms per meter cubed.
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So as we can see, that's larger than the density of just regular aluminum.
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We can conclude that there are spaces between the aluminum atoms.
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They're all packed as close as they can possibly be.
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So let's figure out what the average amount of spaces with these aluminum atoms get.
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So volume per atom is going to be equal to mass divided by density...