\(a,NTK_X=2.NTK_O=2.16=32\left(\text{đ}.v.C\right)\\ \Rightarrow X:L\text{ưu}.hu\text{ỳ}nh\\ b,PTK_{Na_2SO_4}=2.NTK_{Na}+NTK_S+4.NTK_O=23.2+32+4.16=142\left(\text{đ}.v.C\right)\\ PTK_{K_2O}=2.NTK_K+NTK_O=2.39+16=94\left(\text{đ}.v.C\right)\\ PTK_{MgCl_2}=NTK_{Mg}+2.NTK_{Cl}=24+35,5.2=95\left(\text{đ}.v.C\right)\)
a) Nguyên tử X nặng gấp 2 lần nguyên tử Oxi
⇒ X = 2 . O = 2 . 16 = 32
⇒ X là nguyên tố Lưu Huỳnh (S)
b) Na2SO4: 23.2 + 32.1 + 16.4 = 142 đvC
K2O: 31.2 + 16.1 = 78 đvC
MgCl2: 24.1 + 35,5.2 = 95 đvC