`a)`

Đặt \(\%_{35}^{79}Br=x\%\)

\(\to \%_{35}^{81}Br=(100-x)\%\)

`->M_{Br}={79x+81(100-x)}/{100}=79,88`

`->x=56`

\(\to \%_{35}^{79}Br=56\%;\%_{35}^{81}Br=44\%\)

`b)`

Cho `n_{Br}=1(mol)`

\(\to n_{_{35}^{79}Br}=1.56\%=0,56(mol)\)

\(\to\%m_{_{35}^{79}Br}=\dfrac{79.0,56}{79,88}.100\%\approx 55,38\%\)

\(\to\%m_{_{35}^{81}Br}=100-55,38=44,62\%\)

`c)` \(\%m_{_{35}^{79}Br}=\dfrac{3.79.56\%}{27+79,88.3}.100\%\approx 49,77\%\)

`d)`

`n_{Br}=3n_{FeBr_3}=6(mol)`

\(\to n_{_{35}^{81}Br}=6.44\%=2,64(mol)\)

Số nguyên tử \(_{35}^{81}Br\) là `2,64.6,022.10^{23}=15,89808.10^{23}`