Lab Question (Based on your aluminum foil sample) If the population of the world is 5.6 x 10^9 individuals, how many aluminum atoms could you distribute to each person from your sample of aluminum foil?
Do a very accurate massing of your sample. Convert this value to 'moles' (refer to the atomic weight value) Convert the mole value to atoms. (standard conversion, 6.022 x 10^23) Divide the # of atoms value by the population.
RE: Copper (II) Chloride Aluminum Lab! ? whilst watching the filtrate of this lab, that's the Aluminum Chloride, what actual observations could be made approximately it? additionally, once you first pour the Copper (II) Chloride crystals into the water, is it heterogeneous or homogeneous? How approximately in case you enable the beaker sit down undisturbed for some...
You have to start with the mass of your aluminum foil so weigh it out, let's call it m. Then find the molecular weight in g/mol from the periodic table, we'll call this MW. Next, calculate how many mols of Al are in your foil: # mols = m / MW Then calculate the number of atoms using Avogadro's Number. # atoms = # mols * Avogadro's Number Finally, divide the # atoms by the population of the world to see how many atoms each person could get from your sample. -trickshottim
