[drainedbrainz999]

I need HELP with this problem IMMEDIATELY so PLEASE help me..! We r supposed 2 try 2 figure out the thickness of a piece of foil in ATOMS (I know, ridiculous? When is this going 2 help me in life?! ) we were given a paragraph on background info, and our teacher said not all of it was able to be utilized, it was just thrown in there to confuse us (like he doesn't do that already). So here's the problem and what I have so far... PLEASE help me... I don't WANT TO FAIL chemistry..!

Given a rectangular piece of aluminum foil, your problem is to find the number of atoms that make up the thickness of the foil. Record all measured datat w/ units and labels. Use Significant Figures (what r those? )

BG info: (I streamlined it 2 what seemed more important) Al's melting point: 660 C
atomic weight-26.98 grams per 1 mole of atoms; 1 mole=6.02 x 10 to the 23rd power particles; 1 atom of Al's radius =1.43 angstroms; 1 angstrom= ten to the negtaive tenth meters; density of Al=2.7 g/cubic cm

PLEASE HELP ME... I would reallllllyyyy appreciate it... o and help me 2 see how to show my work... chemistry is sooooooo tough. This is what I have so far:

Foil mass--3.6 grams
Foil width--20.6 cm
Foil length--16.7 cm
I did the D = m/v

v = w*l*h = 20.6*16.7*h
the density is given as 2.70 g/cm^3
so then you make it: 2.70 = 3.6/(20.6*16.7)(h)
so then 2.70* (20.6*16.7)(h)=3.6
and then you simplify that...
next, since (according to the bg info paragraph) radius of one Al atom is 1.43 Ångströms, the diameter is 2r, and therefore 2.86 Ångströms

I don't know what I'm doing and y I'm doing it but I'm doing this and I don't know where 2 go next I have about an hour and a half to get this done and go to bed so if sum1 doesn't help me by 12 30 am EST time in the US then I'm DOOMED. WAT Should I DO?!

[bignaked101]

Man, that sounds tough... but luckily there is always the bright idea to copy someone's work?

Seriously.. ask a friend..

[Lorsung23647]

Hate to almost just give you the answer, but you might want to get your height right. Height would be your volume (2.70g/cm^3/3.6g=.75cm^3) divided by your length (20.6cm), then width (16.7cm). You take .75cm^3/20.6/16.7=(h). So your atom diameter (2.86) * (h). Now you take and divide it by 1*10^-9 (what 1 Ångström equals). Now since that would be for meters you need to divide that by 100. And there you have your answer, I'd be happy to check your answer and work if you post it. It's relatively simple if you can remember the numbers.

[BobJ]

The thickness of aluminum foil is about 0.002 cm.
A mole of aluminum weighs 27 grams
The density of aluminum is 2.7 grams per cubic cm
So one mole occupies a cube with volume of 10 cc.
Or 1 cc contains 0.1 mols or 6.02E22 atoms.
The cube root gives 3.92E7 atoms/cm on a side
Multiply this by the thickness of the foil gives ~75,000 atoms thick

[supraman]

I need HELP with this problem IMMEDIATELY so PLEASE help me...!!!! We r supposed 2 try 2 figure out the thickness of a piece of foil in ATOMS (i knw, ridiculous?! wen is this going 2 help me in life?!!) we were given a paragraph on background info, and our teacher said not all of it was able to be utilized, it was just thrown in there to confuse us (like he doesn't do that already). So here's the problem and what i have so far...PLEASE help me.....i don't WANT TO FAIL chemistry.....!!!

Given a rectangular piece of aluminum foil, your problem is to find the number of atoms that make up the thickness of the foil. Record all measured datat w/ units and labels. Use Significant Figures (what r those?!)

BG info: (i streamlined it 2 what seemed more important) Al's melting point: 660 C
atomic weight-26.98 grams per 1 mole of atoms; 1 mole=6.02 x 10 to the 23rd power particles; 1 atom of Al's radius =1.43 angstroms; 1 angstrom= ten to the negtaive tenth meters; density of Al=2.7 g/cubic cm

PLEASE HELP ME.....i wud reallllllyyyy appreciate it.....o and help me 2 see how to show my work.......chemistry is sooooooo tough. This is wat i hav so far:

Foil mass--3.6 grams
Foil width--20.6 cm
Foil length--16.7 cm
i did the D = m/v

v = w*l*h = 20.6*16.7*h
the density is given as 2.70 g/cm^3
so then you make it: 2.70 = 3.6/(20.6*16.7)(h)
so then 2.70* (20.6*16.7)(h)=3.6
and then you simplify that...
next, since (according to the bg info paragraph) radius of one Al atom is 1.43 angstroms, the diameter is 2r, and therefore 2.86 angstroms

I don't knw wat im doing and y im doing it but im doing this and i dont knw where 2 go next i hav about an hour and a half to get this done and go to bed so if sum1 doesnt help me by 12 30 am EST time in the US then im DOOMED. WAT SHUD I DO?!!

We took squares of aluminum foil, measured the surface area and mass. By dividing the mass by a known density value, we obtained the volume of aluminum foil in our sample. By dividing that by the surface area, we found the thickness of the aluminum foil in cm. (The value was 2.86 *10^-3 cm.) The value converted to be 2.86*10^5 Ångströms. Each aluminum atom is about 1.48 Ångströms, so we can divide. Assuming that each aluminum atom is stacked directly on another one (which they aren't, but we can't measure without making this assumption), we come up with 1.93 * 10^5 atoms thick.