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tuggler
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Homework Statement
A 48.7 g ball of copper has a net charge of 2.2 µC. What fraction of the copper's electrons have been removed? (Each copper atom has 29 protons, and copper has an atomic mass of 63.5.).
Homework Equations
The Attempt at a Solution
Number of electrons normally present ..
(48.7/63.5)mol x 6.022^23(atoms per mol) x 29(electrons per atom) = 1.34^24 electrons
Electrons removed = n
ne = 2.20^-6C .. .. n = 2.20^-6C / 1.60^-19C/e .. .. n = 1.38^13 electrons removed
Fraction removed = 1.38^13 removed / 1.34^24 total
But the answer is wrong. I don't know what I am doing wrong?