(b) Calculate the standard free energy change of the reaction: p =-145,000 Cal/mole and JG1000, M20 > = -55,000 Cal/mole. Also comment on the possibility of reducing Moo, by H2 at 727 at 1 atm pressure.

i think is conclusions.i think so

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