$@Heley$
$2.a)$
$PTK_{O_{3}}=3.NTK_{O}=3.16=48(đvC)$
$PTK_{H_{3}PO_{4}}=3.NTK_{H}+1.NTK_{P}+4.NTK_{O}=3.1+1.31+4.16=98(đvC)$
$PTK_{Al_{2}(SO_{4})_{3}}=2.NTK_{Al}+3.(1.NTK_{S}+4.NTK_{O})=2.27+3.(1.32+4.16)=342(đvC)$
$PTK_{FeSO_{4}}=1.NTK_{Fe}+1.NTK_{S}+4.NTK_{O}=1.56+1.32+4.16=152(đvC)$
$PTK_{7H_{2}O}=7.(2.NTK_{H}+1.NTK_{O})=7.(2.1+1.16)=126(đvC)$
$PTK_{Cl_{2}}=2.NTK_{Cl}=2.35,5=71(đvC)$
$PTK_{Ba(HCO_{3})_{2}}=1.NTK_{Ba}+2.(1.NTK_{H}+1.NTK_{C}+3.NTK_{O}) =1.137+2.(1.1+1.12+3.16)=259(đvC)$
$PTK_{Mg(H_{2}SO_{4})_{2}}=1.NTK_{Mg}+2.(2.NTK_{H}+1.NTK_{S}+4.NTK_{O})=1.24+2.(2.1+1.32+4.16)=220(đvC)$
$PTK_{N_{2}}=2.NTK_{N}=2.14=28(đvC)$
$b)$
Đơn chất: $O_{3},Cl_{2},N_{2}$
Vì những chất trên chỉ có 1 nguyên tố
Hợp chất: $H_{3}PO_{4},Al_{2}(SO_{4})_{3},FeSO_{4},7H_{2}O,Ba(HCO_{3})_{2},Mg(H_{2}SO_{4})_{2}$
Vì những chất trên có 2 nguyên tố trở lên
#$BTS$
