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19 March, 16:20

A 49.3 g ball of copper has a net charge of 2.0 µc. what fraction of the copper's electrons has been removed? (each copper atom has 29 protons, and copper has an atomic mass of 63.5.)

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  1. 19 March, 16:31
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    First, find how many copper atoms make up the ball:

    moles of atoms = (49.3 g) / (63.5 g per mol of atoms) = 0. 77638 mol

    # of atoms = (0.77638 mol) (6.02 * 10^23 atoms per mol) = 4.6738*10^23 atoms

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

    normal # electrons = (4.6738 * 10^23 atoms) (29 electrons per atom) = 1.3554 * 10^25 electrons

    Currently, the charge in the ball is 2.0 µC, which means - 2.0 µC worth of electrons have been removed.

    # removed electrons = (-2.0 µC) / (1.602 * 10^-13 µC per electron) = 1.2484 * 10^13 electrons removed

    # removed electrons / normal # electrons =

    (1.2484 * 10^13 electrons removed) / (1.3554 * 10^25 electrons) = 9.21 * 10^-13

    That's 1 / 9.21 * 10^13
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