Ответ:

Рассчитаем относительные молекулярные массы веществ:

  • $$M_r(H_2S) = 2 \cdot A_r(H) + A_r(S) = 2 \cdot 1 + 32 = 34$$
  • $$M_r(SO_3) = A_r(S) + 3 \cdot A_r(O) = 32 + 3 \cdot 16 = 32 + 48 = 80$$
  • $$M_r(Na_2CO_3) = 2 \cdot A_r(Na) + A_r(C) + 3 \cdot A_r(O) = 2 \cdot 23 + 12 + 3 \cdot 16 = 46 + 12 + 48 = 106$$
  • $$M_r(Cu(OH)_2) = A_r(Cu) + 2 \cdot A_r(O) + 2 \cdot A_r(H) = 64 + 2 \cdot 16 + 2 \cdot 1 = 64 + 32 + 2 = 98$$
  • $$M_r(P_2O_5) = 2 \cdot A_r(P) + 5 \cdot A_r(O) = 2 \cdot 31 + 5 \cdot 16 = 62 + 80 = 142$$
  • $$M_r(AlCl_3) = A_r(Al) + 3 \cdot A_r(Cl) = 27 + 3 \cdot 35.5 = 27 + 106.5 = 133.5$$
  • $$M_r(Ba(NO_3)_2) = A_r(Ba) + 2 \cdot A_r(N) + 6 \cdot A_r(O) = 137 + 2 \cdot 14 + 6 \cdot 16 = 137 + 28 + 96 = 261$$

Ответ: $$M_r(H_2S) = 34$$, $$M_r(SO_3) = 80$$, $$M_r(Na_2CO_3) = 106$$, $$M_r(Cu(OH)_2) = 98$$, $$M_r(P_2O_5) = 142$$, $$M_r(AlCl_3) = 133.5$$, $$M_r(Ba(NO_3)_2) = 261$$.