What fraction of the copper's electrons has been removed? Each copper atom has 29 protons, and copper has an atomic mass of 63.5.

9.09*10^-13

Explanation:

Certain values of the problem where omitted, however the omitted values were captured in the solution.

Step1:

Avogadro's number (NA) I = 6.02*10^23 atoms/mole.

Step2:

To determine the number of moles of copper that are present, thus: Using the mass and atomic mass:

n = m/A

n = 50.0g/63.5g/mol

Therefore, since the are 29 protons per atom, I the number of protons can be determined as follows:

Np = nNA*29 protons / atom

Np = (50.0gm/63.5g/mol) (6.02*10^23 atoms/mol) * (29 protons / C u atom)

Np = 1.375*10^25 protons

Note that there are same number of electrons as protons in a neutral atom, I therefore the removal of electrons to give the copper a net change, hence the result is 1.375*10^25

Step3:

To determine the number electrons, removed to leave a net charge of 2.00Uc, then remove - 2.00Uc of charge, so that the number of electrons to be removed are as follows:

Ne (removed) =

Q/qe = - 2.00*10^-6c/-1.60 * 10^-19c

Ne (removed) = 1.25*10^13 electrons removed

Step4:

To calculate the fraction of copper's electron by taking the ratio of the number of electrons initially present:

Ne, removed/Ne, initially=1.25*10^13 / 1.37*10^25 = 9.09*10^-13