## anonymous 5 years ago 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)?

1. JFraser

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

2. 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?

3. anonymous

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

4. JFraser

try length x width, in meters

5. anonymous

oh. I am so stupid.

6. anonymous

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

7. 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

8. JFraser

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

9. anonymous

:O THANK YOU SO MUCH JFRASER!!

10. 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!!

11. anonymous

12. 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. 13. 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!

14. JFraser

that's the cost per atom of the aluminum.

15. anonymous

it is? How? $2.29? o.o 16. JFraser don't ignore the exponent at the tail end of that number. it's not$2.29, it's \$0.0000000000000000000000229

17. anonymous

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

18. 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.

19. anonymous

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