Ireally need ! calculate the number of moles of al2 o3 that are produced when 0.60 mol of fe is produced in the following reaction. 2al(s) + 3feo(s) arrow 3fe(s) + al 2 o3 (s)a.0.20 molc.0.60 molb.0.40 mold.0.90 mol 

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