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- anonymous

Anyone know know how I can find the % of the piece I have cut (20.0 cm by 20.0 cm or 8 inch by 8 inch) in relation to the whole roll (area is 6.96 m^2)?

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- anonymous

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- JFraser

Find the area of the individual piece (in m2), divide by the total area. Percentage of the total area.

- anonymous

I need to find the cost PER ATOM of my sample aluminum foil (20.0 cm by 20.0 cm), which has a mass of 1.76 g. I found its number of moles: 0.0652 mol, and its number of atoms in the sample: 3.925 x 10^22. Is it good so far?

- anonymous

ohh.. so how do I find the total area of the sample? o.o

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- JFraser

try length x width, in meters

- anonymous

oh. I am so stupid.

- anonymous

so basically convert 20.0 cm into metres, which is 0.2 m.

- JFraser

So the area of the sample is 0.4m^2, and the area of the whole roll is 6.96m^2. Not a very big percentage. Multiply that percentage by the cost per roll and you have the cost per sample

- JFraser

Divide the cost of the sample by the number of atoms, and you have the cost per atom.

- anonymous

:O THANK YOU SO MUCH JFRASER!!

- anonymous

So.. I got Area % = Area of sample/area of whole roll = 0.4 m^2/6.96 m^2 = 0.00574 m^2.
Cost per sample = Area % x Cost per roll = 0.0575 m^2 x $1.59 = $ 0.0914/m^2, and
cost per atom = Cost per sample/# of atoms = 0.0914 m^2/3.925 x 10^22 atom = 2.29299363 x 10^20...
So the cost per atom is 2.29 x 10^22? o.O ahh!!

- anonymous

please help!!

- JFraser

when you divide\[\frac{$0.0914}{3.925*10{^2}{^2}}\] you should gets some incredibly small #, not 10^20.If the whole roll only costs $1.59, a small fraction of that roll can't cost billions of times more money. You probably divided upside down.

- anonymous

ohh.. so wait is it going to be divided:
$0.0914
--------
3.925*10^22
= 2.29299363*10^-24... I'm so confused D; I'm so sorry!

- JFraser

that's the cost per atom of the aluminum.

- anonymous

it is? How? $2.29? o.o

- JFraser

don't ignore the exponent at the tail end of that number. it's not $2.29, it's $0.0000000000000000000000229

- anonymous

oh. is it really the cost? It looks weird for it to be cost.. o.o

- JFraser

that's the point. It's an incredibly small cost per atom, because you have an incredibly large number of atoms in the aluminum foil.

- anonymous

ohh.. so it's correct or wrong? D: ahh.

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