Excess electrons are placed on a small lead sphere with mass 8.00g so that it's net charge is -3.20x10^-9C (a) Find the number of excess electrons on the sphere. (b) how many excess electrons are there per lead atom? The atomic number of lead is 82, and its atomic mass is 207g/mol.
We have -3.20x10^-9C . All charge is "quantized" meaning it is an integer multiple of the charge "e" (-1.602x10^-19C/electron). To obtain the amount of "excess" (Neutrally charged atoms have the same amount of protons and electrons, so a negatively charged atom has more electrons, the charge is due to theses extra electrons) we divide our charge Q by e. Electrons= (-3.20x10^-9C)/(-1.602x10^-19C/electron)=1.997x10^10 which we round to 2x10^10 electrons. We then obtain the number of moles the lead ball has, multiply it by avodagro's number to obtain the amount of atoms, and then divide the amount of electrons by the amount of atoms.And we end up with 8.594x10^-13electrons/atom.[/B]
The Attempt at a Solution
I can't get my head around the fact that 8.59x10-13 electrons is not an integer multiple of 1 electron. What does it even mean to have https://www4c.wolframalpha.com/Calculate/MSP/MSP157620gi5hegf6iihfdf00005if81hhidie84ia2?MSPStoreType=image/gif&s=59 [Broken] extra electrons per atom? My understanding is that electrons are moved(transfered from one atom to the next) in integers, you can't really divide an electron to have 8.59x10^-13 of it. What does this actually mean?
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