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Unread 10 Nov 2005, 05:57   #1
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Molecule dispersion question

How the hell can there be a 99% chance that the glass of water on my desk has a molecule in it that was in Socrates cup of hemlock? Have you ever someone say that? I think it was my chemistry teacher ages ago. How can that be true?
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Unread 10 Nov 2005, 06:19   #2
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Re: Molecule dispersion question

eh?
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Unread 10 Nov 2005, 06:23   #3
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Re: Molecule dispersion question

If you breathe deeply there's better than a 99% chance that you'll inhale one of the molecules that was exhaled in caesar's dying breath.


PS There's a lot of molecules in a glass of water.
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Unread 10 Nov 2005, 06:51   #4
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Re: Molecule dispersion question

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Originally Posted by JonnyBGood
If you breathe deeply there's better than a 99% chance that you'll inhale one of the molecules that was exhaled in caesar's dying breath.


PS There's a lot of molecules in a glass of water.
How can that be possible?
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Unread 10 Nov 2005, 08:22   #5
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Re: Molecule dispersion question

OK... nitrogen is the main constituent of air (roughly 78%), so for simplicity's sake we'll assume that air has an identical molecular mass as nitrogen (because I can't be bothered working out an average). At room temperature, air has a density of something like 1.29 x 10-3 g/cm3 (i.e. a mass of 0.00129 g for every cubic cm of air). We'll assume that the breath you're talking about is about 2l, so 2000* 0.00129 = 2.58 g of "air" in the breath we're considering.

Now, to discover exactly how many molecules we have in our breath sample, we can use a clever little thing called the avogadro constant (remember though that I'm assuming air is comprised only of nitrogren, so these figures will turn out a little off, should give the idea though). Essentially the avogadro constant can tell us how many atoms of a compound we have. The atomic mass of nitrogen is roughly 14 (close enough for this explanation). If we have 14g of nitrogen, we have the avogadro constant number of nitrogen atoms (which is 6.023 x 10 to the 23 mol-1, a rather large number).

We're assuming we just have 2.58g of nitrogen though, about 18.43% of 14g. So: 0.1843 * 6.023 x 10 to the 23 mol-1 = ~ 1110038900000000000000000000(seeing as I don't have a calc that does standard form I just had to work it out, should be close though). We'll divide that number by two, as we used the atomic weight and not the molecular weight (nitrogen in the air comes in the gaseous N2 molecular form) . So if we ignore: the fact that I just assumed nitrogen, my poor maths, the fact that it's 07:10 and any other criticisms anyone cares to pull up, then the 2l breath of air we started with contains in the region of 555019450000000000000000000 molecules. That's a whole lot of molecules, looking at the size of the numbers makes it seem a little more plausible that you could currently be inhaling Caesar's death throes.
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Unread 10 Nov 2005, 08:39   #6
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Re: Molecule dispersion question

though if u divide that by all the modlecules of N2 in the world.....
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Unread 10 Nov 2005, 08:46   #7
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Re: Molecule dispersion question

How did the molecule get from Italy to Arizona? Let's all inhale deeply and inhale a molecule from Ceaser's dying breath. Infact let's all inhale deeply and inhale a molecule from the last breath of anyone who ever lived and died. No, it doesn't make sense to me. Sorry. Even with those large numbers.
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Unread 10 Nov 2005, 09:52   #8
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Re: Molecule dispersion question

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Originally Posted by Kal
though if u divide that by all the modlecules of N2 in the world.....
true, but that isn't the point. the point is that you have to look at the numbers of say square metres of air in the world. THis calculation is really hacky, but the surface area of the world is 4*pi*r^2

or say with r being 6x10^6m

we have a surface area of the earth as 4x20^14 m^2

3/4 of the atmosphere is in the first 11 km of altitude, so we will ignore the rest for now and just say that up to that point, the density of air is even, giving a total atmospheric colume of about

5x10^18 m^3

and there are 1000 liters in every m^3, so the total atmospheric volume is

5x^10^21 liters

prawn calculated what,

5x10^26 molecules of nitrogen in each 2 l breath (I am assuming that is right for simplicity and not bothering to check it)

so that means that in each breath, assuming that the air is distributed evenly, you should be breathing in some 55,000 molecules of ceasar's dying breath.
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Unread 10 Nov 2005, 09:56   #9
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Re: Molecule dispersion question

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Originally Posted by s|k
How did the molecule get from Italy to Arizona? Let's all inhale deeply and inhale a molecule from Ceaser's dying breath. Infact let's all inhale deeply and inhale a molecule from the last breath of anyone who ever lived and died. No, it doesn't make sense to me. Sorry. Even with those large numbers.
if you actually work the numbers out, it makes a reasonable amount of sense, because the numbers are vast. 55,000 molecules per breath constitutes some 10^-20th of a percent of every breath. which is a very tiddly little number. somewhere along the lines of the percentage of the earth that a family car weighs.

of course, the getting from italy to arizona bit might seem a problem, but look at how much the winds move. in the UK we regularly get wind coming in from siberia over winter. you can tell because it is damned cold. if you've seen hurricanes, you should appreciate how much wind those things shift around. so getting from italy to arizona isn't as difficult as one might imagine, though I would expect there will be differences in concentrations of caesar's dying breath with latitude because of the trade winds and such.
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Unread 10 Nov 2005, 10:06   #10
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Re: Molecule dispersion question

Quote:
Originally Posted by s|k
How the hell can there be a 99% chance that the glass of water on my desk has a molecule in it that was in Socrates cup of hemlock? Have you ever someone say that? I think it was my chemistry teacher ages ago. How can that be true?
Dude, you really nead to get out more if this is the sort of thing that worries you
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Unread 10 Nov 2005, 13:51   #11
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Re: Molecule dispersion question

Quote:
Originally Posted by s|k
No, it doesn't make sense to me. Sorry. Even with those large numbers.
What number of people is required such that the chance of there being at least one birthday shared by two people is greater than 50%? ie, that there is a greater than 1/2 chance for, within that group, two people to have the same birthday.

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Unread 10 Nov 2005, 15:25   #12
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Re: Molecule dispersion question

inhale my farts, bastards.
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Unread 10 Nov 2005, 16:11   #13
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Exclamation Re: Molecule dispersion question

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inhale my farts, bastards.
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Unread 10 Nov 2005, 17:11   #14
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Re: Molecule dispersion question

Quote:
Originally Posted by s|k
How the hell can there be a 99% chance that the glass of water on my desk has a molecule in it that was in Socrates cup of hemlock?
To be fair, there's probably a greater than 1% chance that our knowledge of Greek history is wrong and Socrates didnt really drink hemlock.
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Unread 10 Nov 2005, 17:13   #15
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Talking Re: Molecule dispersion question

Quote:
Originally Posted by Nodrog
To be fair, there's probably a greater than 1% chance that our knowledge of Greek history is wrong and Socrates didnt really drink hemlock.
To be fair, there's probably a greater than 1% chance that Nodrog doesn't exist and is actually a fat plumber from Morecambe.
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Unread 11 Nov 2005, 14:45   #16
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Re: Molecule dispersion question

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Originally Posted by NEWSBOT3
To be fair, there's probably a greater than 1% chance that Nodrog doesn't exist and is actually a fat plumber from Morecambe.
That's rather hard. He either doesn't exist, OR he's a fat plumber from Morecambe.

As for the rest of this, GAH, LARGE NUMBERS!
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Unread 11 Nov 2005, 16:06   #17
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Re: Molecule dispersion question

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Originally Posted by flapjack
As for the rest of this, GAH, LARGE NUMBERS!
Amen brother.
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