# Calculate the number of electron

+1 vote

(a) Calculate the number of electrons in a small, electrically neutral silver pin that has a mass of 13.0 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 3.00 mC. How many electrons are added for every 109 electrons already present?

asked Jan 15, 2012 in Physics

+1 vote

For part A you can approach this problem by first setting up a relation as follows:

(47 electrons/ 1 atom) *13 g *(1mol / 107.87 g) *(6.022*10^23 / 1 mol)  your goal is to arrange it so the units all cancel out and you are left with only electrons in your final answer which would be 3.41*10^24.

Part B can be solved similarly by once again setting up a relation as follows:

*note the charge of one electron is 1.602*10^-19 C

3 mC * (1 C / 1000 mC) = 0.003 C  * (1electron / 1.602*10^-19 C) =

once again you are trying to get an answer in terms on electron, so upon simplfying you get 1.87*10^16 electrons as the number of electrons added.

The next step is to divide by the number of electrons already present which we calculated in part A.

so:  (1.87*10^16) / (3.41*10^24) = 5.49 * 10^-9  and hence there are 5.49 electrons added for every 10^9 electron present.
answered Jan 17, 2012 by ~Expert~ (3,856 points)
selected Jan 17, 2012 by Roar3215