Question

Giúp mình vs ạ, huhu

Answer 1

Bài 8:

a) PTK(hc)= 2.NTK(X) + 3.NTK(Y)

<=> 4,25. NTK(Mg)= 2.NTK(X) + 3.NTK(Y)

<=> 2.NTK(X) + 3.NTK(Y)= 4,25. 24=102(đ.v.C)

=> PTK(hc)=102(đ.v.C)

b) Ta có:

\(\dfrac{2.NTK_X}{102}.100\%=52,94\%\\ \Leftrightarrow NTK_X=27\left(\dfrac{g}{mol}\right)\)

=> X là nhôm (Al=27)

2.27+3.NTK(Y)=102

<=>NTK(Y)=16(đ.v.C)

=>Y là Oxi (O=16)