Calculate the Gibbs Free-Energy and the standard Gibbs Free- Energy by the cell in the given cell notation: Zn(s) |Zn2+(aq) || H+(aq)| H2(g) |Pt (s)
When [Zn+2] = 4.0 x 10 -3 M, [H2] = 0.010 M, [H+] = 1.0 M

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally

it would only be the stuff in module eight so you need to study all the lessons in that you also could check the flvs

hope this : )