Đáp án:
a)
\({{\bar M}_{Cl}} = 35,5\)
b)
\({{\bar M}_{Br}} = 79,92\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
\% {}_{17}^{35}Cl = \dfrac{3}{{3 + 1}} \times 100\% = 75\% \\
\Rightarrow \% {}_{17}^{37}Cl = 100\% - 75\% = 25\% \\
{{\bar M}_{Cl}} = \dfrac{{35 \times 75 + 37 \times 25}}{{100}} = 35,5\\
b)\\
\% {}_{35}^{79}Br = \dfrac{{27}}{{27 + 23}} \times 100\% = 54\% \\
\% {}_{35}^{81}Br = 100 - 54 = 46\% \\
{{\bar M}_{Br}} = \dfrac{{79 \times 54 + 81 \times 46}}{{100}} = 79,92
\end{array}\)
