anonymous
  • anonymous
Hi everyone! I had to do an 'Avogadro Goes to Court-- Case Study,' which is to find the cost PER ATOM. I got my answer but I asked several of my classmates and they got a different answer: 2.29299363 x 10-23. Help?
Chemistry
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
I got 2.29299363 x 10^-24/atom in the aluminum foil. Aluminum foil roll costs $ 1.59 Area of the roll is 6.96 m^2 Sample of the aluminum foil: 8 inch by 8 inch or 20.0 cm by 20.0 cm Mass of my aluminum foil sample: 1.76 g Avogadro's constant: 6.02 x 10^23 Aluminum symbol: Al Aluminum's atomic number: 13 Aluminum's atomic mass: 26. 98154 So after calculating the mass of my aluminum foil sample to its molar mass and moles I got: Al molar mass: 26.98154 g/mol Mole: 0.0652 mol of aluminum After calculating the number of atoms in the sample I have, there are 3.925 x 10^22 atoms in the aluminum foil sample. Length & width of aluminum foil sample: 20.0 cm by 20.0 cm Area of the roll: 6.96 m^2 Aluminum foil roll costs $1.59 # of atoms of the aluminum foil sample: 3.925 x 10^22 atoms. Lastly, after calculating the area of my aluminum foil sample: 0.4 m^2, percentage of the total area: 0.0575 m^2 (is it?), and cost of per atom: 2.29299363 x 10^-24? Ahhhh.
Xishem
  • Xishem
Alright, I'll go ahead and work it out using your original numbers but ignoring your work and see what I come up with. \[1.76g\ Al∗\frac{1mol\ Al}{26.98154g\ Al}=6.52_{298}∗10^{-2}mol\ Al\](subscript numbers just represent insignificant figures) Given that there are this many moles of aluminum in a 0.2*0.2m piece, we need to find out how many moles are in the entire sheet. We can do this by setting up a proportion. \[\frac{6.52298∗10^{-2}mol\ Al}{(0.200m)^2}=\frac{x}{6.96m^2}\] Solving for x, we will get... \[x=11.3_{500}mol\ Al\] Now, we need to find how many atoms of aluminum this is... \[11.3_{500}mol\ Al∗\frac{6.022∗10^{23}atoms\ Al}{1mol\ Al}=6.83_{496}*10^{24}atoms\ Al\] Now, we need to find COST per ATOM... \[\frac{$1.59}{6.83_{496}*10^{24}atoms\ Al}=\frac{$2.33*10^{-25}}{atom\ Al}\]
Xishem
  • Xishem
I can tell you that YOUR answer differs from mine because you miscalculated the area of the original piece.\[A=l^2=(20.0cm)^2=(400cm^2)\]\[400cm^2*\frac{1m^2}{10,000cm^2}=0.04m^2\]That would explain why you were only 1 order of magnitude off of my answer. Your classmates' answers, however, I cannot explain. It sounds like they calculated the area of the cut piece like this...\[A=l^2=(20.0cm)^2=400cm^2*\frac{1m^2}{100cm^2}=4m^2\]Which would explain their difference in order of magnitude. I don't know if this is the error they made, but it is certainly one possible explanation.

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Xishem
  • Xishem
The reason the conversion factor is 1:10,000 from square meters to square centimeters is... Given two squares with equal lengths (1m and 100cm)...\[A=l^2 \rightarrow A=(1m)^2=1m^2\]Now the square with the same length, but in cm...\[A=l^2 \rightarrow A=(100cm)^2=10,000cm^2\]
anonymous
  • anonymous
A = l^2 -> A = (100cm)^2? How do I..
anonymous
  • anonymous
why is it .200 m^2? please answer :$
anonymous
  • anonymous
when .2 m by .2 m is .04 m^2???
Xishem
  • Xishem
Yes. 20cm is 0.2m. So the area is...\[A=(0.200m)^2=0.0400m^2\]

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