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Sever21 [200]
3 years ago
9

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.)
Chemistry
1 answer:
Otrada [13]3 years ago
4 0
First, find how many copper atoms make up the ball: 
<span>moles of atoms = (46.6 g) / (63.5 g per mol of atoms) = 0.73385 mol </span>
<span># of atoms = (0.73385 mol) (6.02 × 10^23 atoms per mol) = 4.4178 × 10^23 atoms </span>

<span>There is normally one electron for every proton in copper. This means there are normally 29 electrons per atom: </span>
<span>normal # electrons = (4.4178 × 10^23 atoms) (29 electrons per atom) = 1.2811 × 10^25 electrons </span>

<span>Currently, the charge in the ball is 2.5 µC, which means -2.5 µC worth of electrons have been removed. </span>
<span># removed electrons = (-2.5 µC) / (1.602 × 10^-13 µC per electron) = 1.5604 × 10^13 electrons removed </span>

<span># removed electrons / normal # electrons = </span>
<span>(1.5604 × 10^13 electrons removed) / (1.2811 × 10^25 electrons) = 1.22 × 10^-12 </span>

<span>That's 1 / 1.22 × 10^12 </span>
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