1,

Áp dụng sơ đồ đường chéo:

$\dfrac{^{79}Br}{^{81}Br}=\dfrac{81-79,904}{79,904-79}=\dfrac{137}{113}$

Giả sử $n_{KBrO_4}=1\ (mol)$

$\to n_{Br}=n_{KBrO_4}=1\ (mol)$

$\to n_{^{79}Br}=1.\dfrac{137}{137+113}=0,548\ (mol)$

$\to \%m_{^{79}Br}=\dfrac{0,548.79}{39+79,904+16.4}.100\%=23,67\%$

2,

$\overline{M}_{Cr}=\dfrac{50.4,4+52.83,8+53.9,5+54.2,4}{100}=52,107$