Question

Answer 1

M_A = 8,5.2 = 17 (g/mol)

\(m_N=\dfrac{17.82,35}{100}=14\left(g\right)=>n_N=\dfrac{14}{14}=1\left(mol\right)\)

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

=> CTHH: NH₃

Answer 2

nani Hóa lớp 7 :)

Answer 3

Ta có: \(d_{A/H_2}=8,5\)

\(\Rightarrow M_a=8,5\times H_2=8,5\times1\times2=17\left(g/mol\right)\)

\(CTHH\) dạng chung \(:N_xH_y\)

\(\%N=\dfrac{14x.100}{17}=82,35\%\)

\(\rightarrow x=1\)

\(\%H=\dfrac{1y.100}{17}=17,65\%\)

\(\rightarrow y=3\)

Vậy \(CTHH\) là \(:NH_3\)