[Co(H2O)6](NO3)2 + NH3 + (NH4)2CO3 + H2O2 = [Co(NH3)4CO3]NO3 + NH4NO3 + H2O Calculate the theoretical yield of your product assuming you used: 4.8g (NH4)CO3 in 10ml H2O; 10ml 15 M NH3; 2.5g of Co(NO3)2 * 6H2O in 5ml H2O; 1.5 ml of 30% H2O2 (the density of 30% H2O2 is 1.11g/ml)

the answer for sodium is 0.186

the answer for rubidium is 0.251

explanation:

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