- #1

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I was looking through my notes and came across this (above) I understand the calculation etc but I don't see where the number of atoms N was calculated?

I don't see where it has come from

Any help would be great thanks in advanced

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- Thread starter KiNGGeexD
- Start date

- #1

- 317

- 1

I was looking through my notes and came across this (above) I understand the calculation etc but I don't see where the number of atoms N was calculated?

I don't see where it has come from

Any help would be great thanks in advanced

- #2

BvU

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28 gram is one mole

density is 0.81 g/cm3

so 28 gram is 28/0.81 = 35 cm3, right ? i.e. 35 cm3/mole

one mole is N_avogadro molecules and 35/(6 e23) = 5.8 e-23 cm3/molecule.

Take the ##\root 3 \of \ ## and you get something like 3.9 e-8 cm or 0.39 nm

- #3

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- #4

BvU

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Would N in the top equation not be just Avergadros number?

- #6

BvU

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Yes. Since we are talking J/**mol**, N=N_{A}.

- #7

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1.95E-21 J

Thanks

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