Giải:
a) \(PTK_{CaCO3}=NTK_{Ca}+NTK_C+3NTK_O=40+12+3.16=100\left(đvC\right)\)
\(PTK_{CuO}=NTK_{Cu}+NTK_O=64+16=80\left(đvC\right)\)
b) \(2PTK_{CuO}=2\left(NTK_{Cu}+NTK_O\right)=2.\left(64+16\right)=2.80=160\left(đvC\right)\)
CaCO3= 40.1 + 12.1+16.3= 100(đvC)
CuO= 64.1 + 16.1= 80(đvC)
(CuO)2 = 2.64 + 2.16 = 160(đvC)