Final answer:
The number of electrons in a 10.0 g silver pin is approximately 2.62 x 10^24. To achieve a charge of 1.00 µC, approximately 6.25 x 10^12 electrons are added. This equates to about 6250 electrons added for every 10^9 electrons already present.
Step-by-step explanation:
To calculate the number of electrons in a 10.0 g silver pin, we first need to find out how many silver atoms there are in it. This can be done by using the molar mass of silver and Avogadro's number. Since the atomic mass of silver is 107.87 g/mol, we can calculate the number of moles of silver in the pin as follows:
Number of moles = mass (g) / molar mass (g/mol) = 10.0 g / 107.87 g/mol ≈ 0.0927 mol
Knowing that each mole contains Avogadro's number of particles (approximately 6.022 x 1023 particles/mol), we multiply the number of moles by Avogadro's number to get the number of silver atoms:
Number of atoms = number of moles x Avogadro's number ≈ 0.0927 mol x 6.022 x 1023 atoms/mol ≈ 5.58 x 1022 atoms
Since each silver atom has 47 electrons, the total number of electrons in the pin is:
Total electrons = number of atoms x electrons per atom ≈ 5.58 x 1022 atoms x 47 electrons/atom ≈ 2.62 x 1024 electrons
For part (b), when adding enough electrons to the pin to result in a charge of 1.00 µC (or 1.00 x 10-6 C), we first determine the number of electrons associated with this charge:
Number of electrons = charge / charge per electron = 1.00 x 10-6 C / 1.60 x 10-19 C/electron ≈ 6.25 x 1012 electrons
To find out how many electrons are added for every 109 electrons already present, we set up a ratio:
Added electrons per existing electrons = 6.25 x 1012 / 109 ≈ 6250
Therefore, approximately 6250 electrons are added for every 109 electrons already present.