Question

Answer 1

\(m_{hh}=\dfrac{110g}{mol}\)

\(=>M_{O2}=\dfrac{110.43,46\%}{100}=48g\)

=>\(M_P=62\)g

\(m_{O2}=\dfrac{48}{16}=3mol\)

\(m_P=\dfrac{62}{31}=2mol\)

\(=>CTHH:P_2O_3\)

Answer 2

c2 : 
\(m_O=\dfrac{110.43,46}{100}=48\left(g\right)\\ m_P=110-48=62\left(g\right)\\ n_O=\dfrac{48}{16}=3\left(mol\right)\\ n_P=\dfrac{62}{31}=2\left(mol\right)\) 
-> CTHH là P₂O₃

Answer 3

hỏi 1 lần thôi bn =v