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)

Kinetic energy = 1/2mv^2. since the mass is constant we can form a ratio of it's kinetic energy at 60 divided by that at 30 to see ke(60)/ke(30) = +(60^2)/30^2). this may be said to 300% increase.