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5 June, 17:29

A 48.6 g ball of copper has a net charge of 2.4 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.)

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  1. 5 June, 20:32
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    First, find how many copper atoms make up the ball:

    moles of atoms = (46.6 g) / (63.5 g per mol of atoms) = 0.73385 mol

    # of atoms = (0.73385 mol) (6.02 * 10^23 atoms per mol) = 4.4178 * 10^23 atoms

    There is normally one electron for every proton in copper. This means there are normally 29 electrons per atom:

    normal # electrons = (4.4178 * 10^23 atoms) (29 electrons per atom) = 1.2811 * 10^25 electrons

    Currently, the charge in the ball is 2.5 µC, which means - 2.5 µC worth of electrons have been removed.

    # removed electrons = (-2.5 µC) / (1.602 * 10^-13 µC per electron) = 1.5604 * 10^13 electrons removed

    # removed electrons / normal # electrons =

    (1.5604 * 10^13 electrons removed) / (1.2811 * 10^25 electrons) = 1.22 * 10^-12

    That's 1 / 1.22 * 10^12
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