Calculate the number of electrons in a small, electrically neutral silver pin that has a mass of 10.0 g. silver has 47 electrons per atom, and its molar mass is 107.87 g/mol. (b) imagine adding electrons to the pin until the negative charge has the very large value 1.00 mc. how many electrons are added for every 109 electrons already present?

part A

moles of silver=10.0g/107.8g/mol=0.093moles

calculate number silver atom using the Avogadro law

=0.093 x 6.023 x 10^23 = 5.58 x 10 ^22atoms

number of electron is therefore number of atom x atomic number

that is 5.58 x 10^22 x 47=2.62 x 10^24 electrons

part B

we well know electron charge is 1.60 x 10^-19c

convert 1.00mc to coulombs = 1/1000=1.0 x 10^-3c

Number of electron in 1.0 x 10^-3 is therefore

{ (1.0 x 10^-3 / 1.60 x10^-19) } = 6.25 x 10^15

number of (10^9) electrons { (2.62 x10^24 / 10^9) }=2.62 x 10^15

number of electron added per 10^9 is therefore = { (6.25 x 10^25 / 2.62 x10^15) } = 2.3per (10^9)