Hicc :)))) đợi đợt sau

\(HNO_3 \to H^{+}+NO_3^{-}\\ [H^{+}]=CM_{HNO_3}=0,00125M\\ pH=-log(0,00125)=2,9\)

\(a/ Na_2SO_3+2HCl \to 2NaCl+SO_2+H_2O\\ b/ Ba(OH)_2+SO_2 \to BaSO_3+H_2O\\ c/ N_2O_5+H_2O \to 2HNO_3\\ d// K_2O+H_2O \to 2KOH\)

\(\text{Cho quỳ tím ẩm vào 3 mâu:}\\ \text{- Hoá xanh: } CaO\\ \text{- Hoá đỏ: } P_2O_5\\ \text{- Không hiện tượng: } CuO\\ CaO+H_2O \to Ca(OH)_2\\ P_2O_5+3H_2O \to 2H_3PO_4 \)

\(Fe+H_2SO_4 \to FeSO_4+H_2\\ n_{H_2}=0,15(mol)\\ a/\\ n_{Fe}=n_{H_2}=0,15(mol)\\ m_{Fe}=0,15.56=8,4(g)\\ b/\\ n_{H_2SO_4}=n_{H_2}=0,15(mol)\\ CM_{H_2SO_4}=\dfrac{0,15}{2}=0,75M c/\\ n_{FeSO_4}=n_{H_2}=0,15(mol)\\ CM_{FeSO_4}=\dfrac{0,15}{0,2}=0,75M\\\)

\(\text{Chất tác dụng đuọc với axit: ZnO}\\ ZnO+2HCl \to ZnCl_2+H_2O\)

\(n_{H_2}=0,1(mol)\\ \text{Bảo toàn H}\\ n_{H_SO_2}=n_{H_2}=0,1(mol)\\ BTKL:\\ m_{hh}+m_{HCl}=m_{muối}+m_{H_2}\\ 3,28+0,1.36,5=m_{muối}+0,1.2\\ m_{muối}=6,73(g)\)

\(Al_2O_3+6HCl \to 2AlCl_3+3H_2\)

\(Tổng: 2p+n=28(1)\\ MĐ > KMĐ: 2p-n=8(2)\\ (1)(2)\\ p=e=9\\ n=10\\ A=10+9=19 (F)\)