Đây là đề cương ôn tập ah, help me

`a)` `\overline{M_A}={90.50+91.12+92.18+94.18+96.2}/{100}=91,32` amu.

`b)`

Đặt \(\%_{29}^{63}Cu=x\%\)

\(\to\%_{29}^{65}Cu=(100-x)\%\)

`->\overline{M_{Cu}}={63x+65(100-x)}/{100}=63,546`

`->x=72,7`

\(\to\%_{29}^{63}Cu=72,7\%;\%_{29}^{65}Cu=27,3\%\)

Xét 1 mol `CuSO_4:`

`m_{CuSO_4}=63,546+96=159,546(g)`

`n_{Cu}=n_{CuSO_4}=1(mol)`

\(\to n_{_{29}^{63}Cu}=1.72,7\%=0,727(mol)\)

\(\to\%m_{_{29}^{63}Cu}=\dfrac{0,727.63}{159,546}.100\%\approx 28,707\%\)