What fraction of its electrons have been removed?

Question

A copper ball with a diameter of 2.20 mm is charged to +90.0 nC. What fraction of its electrons have been removed? I have attempted to answer this question four times, but I have not been able to arrive at the correct solution. I followed the instructions provided in these two links: http://answers.yahoo.com/question/index?qid=201007… and http://answers.yahoo.com/question/index?qid=200808… The answers I have received are 1.3 x 10^-9, 4.1 x 10^-8, and 4.1 x 10^-9. I believed the last answer was correct, so I adjusted my significant figures to 4.09 x 10^-9, but that was also incorrect. I would appreciate any assistance you can provide.

Anonymous · Accepted Answer

e = charge on electron = 1.602*10^-19 C9010^-9 C represents the charge of 9010^-9/1.602*10^-19 removed electrons.= 56.18010^10 electrons = 5.618010^11 electrons.Mass of copper ball = volumedensity = (4/3)pir^3density r = 1.1 mm = 1.1*10^-3 mr^3 = 1.331*10^-9 m^3density = 8,940 kg/m^3mass of ball = (4/3)pi1.33110^-98,940 = 4984310^-9 kg = 4.984*10^-5 kgmolar mass of copper = 63.546 g/mol = 63.546*10^-3 kg/mol.Number of mols of copper in 4.98410^-5 kg = 4.98410^-5 kg/(63.54610^-3) = 7.84310^-4.1 mol of a substance contains Avogadros number of particles = 6.02214*10^237.84310^-4 mols contains 6.0221410^237.84310^-4 = 47.231*10^19 atoms.A neutral copper atom has 29 electrons.Number of electrons in ball = 2947.23110^19 = 1368.7110^19 = 1.368710^22 electrons.Fraction of electrons removed = 5.618010^11/1.368710^22  = 4.1046*10^-11Answer is 4.10*10^-11 to 3 sig figs

Anonymous

What fraction of its electrons have been removed?

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