Question concerning Electromagnetism

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To solve the problem of determining the fraction of electrons removed from a charged copper ball, first calculate the total number of electrons in the uncharged state using the atomic mass and the number of protons in copper. The ball weighs 47.0 g, and with 29 protons per atom, find the number of atoms and multiply by the number of electrons per atom to get the total electron count. Next, convert the net charge of 1.7 µC into elementary charges to find out how many electrons have been removed. Finally, divide the number of removed electrons by the total number of electrons to find the fraction. Understanding these concepts of charge, moles, and atomic mass is crucial for solving the problem accurately.
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Here is a homework question that I have for my general physics class; our professor ran out of time during lecture and did not get to explain the material very well so I am a little lost.

A 47.0 g ball of copper has a net charge of 1.7 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.)

I know that it is going to involve some stoichiometry-type solving, but I just can't seem to figure it out
 
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How many electrons does the ball have when not charged?
 
Yes, you will have to check out what a coulomb is in terms of 'elementary charges' (which is best to look on Wiki because of the relations to other things you'll find there, and look also for ampere) and revise what mole and atomic mass mean.

In my day mnyah mynah the calculation would have been a bit more direct because then the coulomb was defined in terms of electrochemical measurements, the amount of silver deposited by the flow of current in silver nitrate, something very acceptably precisely measurable, so the given number of coulombs \equiv a number of moles of Ag \equiv half that number of moles of Cu considering the atomic charges.

That old definition has been superseded because even more precisely measurable standards have been evolved but it would still be good enough for the calculation asked here.
 
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