Решение:
Для расчета используем атомные массы элементов: Ca = 40, O = 16, C = 12, Cu = 64, N = 14.
- CaO: \( M_r(\text{CaO}) = M_r(\text{Ca}) + M_r(\text{O}) = 40 + 16 = 56 \)
- CaCO3: \( M_r(\text{CaCO}_3) = M_r(\text{Ca}) + M_r(\text{C}) + 3 \cdot M_r(\text{O}) = 40 + 12 + 3 \cdot 16 = 52 + 48 = 100 \)
- Cu(NO3)2: \( M_r(\text{Cu(NO}_3\text{)}_2) = M_r(\text{Cu}) + 2 \cdot (M_r(\text{N}) + 3 \cdot M_r(\text{O})) = 64 + 2 \cdot (14 + 3 \cdot 16) = 64 + 2 \cdot (14 + 48) = 64 + 2 \cdot 62 = 64 + 124 = 188 \)
Ответ: Mr(CaO) = 56; Mr(CaCO3) = 100; Mr(Cu(NO3)2) = 188.