How many atoms are there in a piece of aluminum foil, 7.3 inches in length and 12.0 inches in width, and 0.141 mm thick? According to this problem, aluminum foil is a flat-rolled product, rectangular in cross section, of thickness from 0.006(0.15mm) to 0.00025(0.006mm). The density of Al is 2.699 g/cm3.
the nice and cozy button is Avogadro' sort that's 6.0226 x 10^23 atoms consistent with mole So if the cost of the Al roll is $2.fifty two for one lb Then proceed as follows: a million lb = 454 gms At Wt Al = 26.ninety 8 gms consistent with mole Moles of Al in a million lb = 454/26.ninety 8 = sixteen.80 two moles sort of atms of Al = sixteen.80 two x 6.0226 x 10^23 = a million.013 x 10^25 cost consistent with atom = $2.fifty two / a million.013 x 10^25 = $2.2847 x 10^-25 consistent with atom or 0.0000000000000000000000228 cents consistent with atom. BTW, you probably did no longer finished specify the situation - how thick is teh Al foil? In any journey Al atoms are enormously inexpensive. desire this helps
What you want to do is determine the mass of the piece of aluminum foil and then convert that to number of atoms. I'll give some pointers and you can fill in the rest: 1. Calculate the mass of the foil: a. Convert the length and width of the foil into centimeters. b. Convert the thickness of the foil into centimeters. Use your 0.141 mm figure. c. Now multiply the length times width times the height to give you the volume in centimeters. d. Now multiply this volume calculated in step c times the density of aluminum. That will give you the mass of your piece of foil. 2. Convert this mass into number of atoms. a. From the periodic table you know that there are 26.982 grams per mole of Al b. There are 6.02 x 10^23 atoms in a mole. c. So, take the mass calculated in Part 1 and divide it by 26.982 grams per mole of Al. That will give you the number of moles of Al in your piece of foil. d. Now take this number of moles and multiply it by 6.02 x 10^23 atoms in a mole, and that will give you the number of atoms in your piece of foil.
