Odpowiedź:
[tex]\huge\boxed{\boxed{m_{\text{Ca}(\text{OH})_2}=74 \ u}}[/tex]
Wyjaśnienie:
[tex]m_\text{Ca}=40 \ u\\\\m_\text{H}=1 \ u\\\\m_\text{O}=16 \ u\\\\m_{\text{Ca}(\text{OH})_2}=40 \ u+2\cdot1 \ u+2\cdot16 \ u=40 \ u+2 \ u+32 \ u=74 \ u[/tex]
mCa(OH)2= 40u+ 2×16u+ 2×1u= 40u+ 32u+2u= 74u
