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a) Calculate the number of electrons in a small, electrically neutral silver pin that has a mass of 6.3 g. Silver has 47 electrons per atom, and its molar mass is 107.87 g/mole.

b) Imagine adding electrons to the pin until the negative charge has the very large value 1.10mC. How many electrons are added for every 10^9 electrons already present?
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Added Wed, 26 Aug '15
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a) Moles of silver = Mass/Molar mass = 6.3/107.87 = 0.058403634

Atom of silver = 0.058403634 X 6.023 X 10^23 = 3.51765 X 10^22 atoms

So No of electrons = 47 X 3.51765 X 10^22 = 1.6533 X 10^24 electrons

b) Charge of one electron = 1.6 X 10^-19 C

No of electrons added = Charge of pin/ Charge of one electron = 1.10 X 10^-3/1.6 X 10^-19

= 6.875 X 10^15 electrons

No of electrons before addition in silver (X10^9) = 1.6533 X 10^24/10^9 = 1.6533 X 10^15

Electrons added for every 10^9 electrons already present = 6.875 X 10^15/1.6533 X 10^15

=4.158
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Added Wed, 26 Aug '15
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