Question

Answer 1

\(n_{XO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ M_{XO_2}=\dfrac{12,8}{0,2}=64\left(\dfrac{g}{mol}\right)\\ M\text{à}:M_{XO_2}=M_X+32\left(\dfrac{g}{mol}\right)\\ \Rightarrow M_X+32=64\\ \Leftrightarrow M_X=32\left(\dfrac{g}{mol}\right)\\ \Rightarrow X:L\text{ưu}.hu\text{ỳ}nh\left(S=32\right)\\ \Rightarrow Ch\text{ọn}.A\)

Answer 2

\(nXO_2=\dfrac{4,48}{22,4}=0,2mol\)

\(MXO_2=\dfrac{12,8}{0,2}=64g/mol\)

Mà \(MXO_2=MX+2MO\)

\(\rightarrow MX=MXO_2-2MO=64-16\times2=32g/mol\)

\(\rightarrow X\) là \(S\) ( lưu huỳnh ) 

\(\Rightarrow ChọnA\)