with the final conclusion that the result of the textbook is not congruent to the calculation parameters of the same book! In case we really imply that the descriptions must mean the identical as the “Simple Cubic Packing” we have to end in, conform to Sev (Posting #3), that V A consumes the same Volume as V U and that must result in: I was not able to detect during writing my posting (Replay #8), that you edited your Replay #7 in the same moment in such a significant way, that my answer is not really fitting your posting anymore. Could anyone please help me to see where I may have erred? However, the answer in my textbook gives 143 pm. Since the volume of a sphere is (4/3)pi(r^3), the radius of one Al atom would be 1.57E-8 cm = 15.7 pm. Therefore, one atom would have a volume of 1.633E-23 cm^3. Since 1 mol = 26.98 g, this foil would be 0.01557 mol. State your answer for the radius of an Al atom in picometers.ĭimension of Al foil = 10.1 cm x 10.1 cm x 1.5E-3 cm
To make it easier, think of the sphere sitting inside a cube. Based on your previous determination of the thickness of a piece of Aluminum foil, you will need the measurements and the calculated thickness of the foil (among other things) to determine the radius of one aluminium atom.Īssume that an aluminium atom is spherical. I am having some trouble with the following activity.ġ.
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