Question

Giúp mình với

Answer 1

a) 

\(m_O=\dfrac{110.43,64}{100}=48\left(g\right)=>n_O=\dfrac{48}{16}=3\left(mol\right)\)

\(m_P=110-48=62\left(g\right)=>n_P=\dfrac{62}{31}=2\left(mol\right)\)

=> CTHH: P₂O₃

b)

 \(m_{Na}=\dfrac{43,4.106}{100}=46\left(g\right)=>n_{Na}=\dfrac{46}{23}=2\left(mol\right)\)

\(m_C=\dfrac{11,3.106}{100}=12\left(g\right)=>n_C=\dfrac{12}{12}=1\left(mol\right)\)

\(m_O=\dfrac{45,3.106}{100}=48\left(g\right)=>n_O=\dfrac{48}{16}=3\left(mol\right)\)

> CTHH: Na₂CO₃

c) 

M = 8,5.2 = 17 (g/mol)

\(m_N=\dfrac{82,35.17}{100}=14\left(g\right)=>n_N=\dfrac{14}{14}=1\left(mol\right)\)

\(m_H=\dfrac{17,65.17}{100}=3\left(g\right)>n_H=\dfrac{3}{1}=3\left(mol\right)\)

=> CTHH: NH₃