d) \(CaO+2HCl\rightarrow CaCl_2+H_2O\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(K_2O+2HCl\rightarrow2KCl+H_2O\)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
\(ZnO+2HCl\rightarrow ZnCl_3+H_2O\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

a) \(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)
\(SO_3+H_2O\rightarrow H_2SO_4\)
\(P_2O_5+3H_2O\rightarrow2H_3PO_4\)
\(N_2O_5+H_2O\rightarrow2HNO_3\)
\(K_2O+H_2O\rightarrow2KOH\)
b) \(SO_3+2KOH\rightarrow K_2SO_4+H_2O\)
\(P_2O_5+6KOH\rightarrow2K_3PO_4+3H_2O\)
\(N_2O_5+2KOH\rightarrow2KNO_3+H_2O\)
\(Al_2O_3+2KOH\rightarrow2KAlO_2+H_2O\)
c) \(SO_3+Ba\left(OH\right)_2\rightarrow BaSO_4+H_2O\)
\(P_2O_5+3Ba\left(OH\right)_2\rightarrow Ba_3\left(PO_4\right)_2+3H_2O\)
\(N_2O_5+Ba\left(OH\right)_2\rightarrow Ba\left(NO_3\right)_2+H_2O\)
\(SiO_2+Ba\left(OH\right)_2\rightarrow BaSiO_3+H_2O\)
\(ZnO+Ba\left(OH\right)_2\rightarrow BaZnO_2+H_2O\)

Bài 7
a) \(\dfrac{3}{4}-\dfrac{5}{6}.\left(\dfrac{1}{6}+\dfrac{1}{8}\right):\dfrac{7}{12}\)
\(=\dfrac{3}{4}-\dfrac{5}{6}.\dfrac{7}{24}:\dfrac{7}{12}\)
\(=\dfrac{3}{4}-\dfrac{5}{12}=\dfrac{9}{12}-\dfrac{5}{12}=\dfrac{4}{12}=\dfrac{1}{3}\)
b) \(\dfrac{15}{8}:\dfrac{45}{6}-\dfrac{1}{13}.\left(\dfrac{4}{9}-\dfrac{1}{12}\right)\)
\(=\dfrac{15}{8}:\dfrac{45}{6}-\dfrac{1}{13}.\dfrac{13}{36}\)
\(=\dfrac{1}{4}-\dfrac{1}{36}=\dfrac{9}{36}-\dfrac{1}{36}=\dfrac{8}{36}=\dfrac{2}{9}\)

Bài 7 

Bài 6 
Chu vi tấm bìa là \(4.\dfrac{31}{7}=\dfrac{124}{7}dm\)
DT tấm bìa là \(\dfrac{31}{7}.\dfrac{31}{7}=\dfrac{961}{49}dm\)

Câu 16
\(n_{Cu\left(OH\right)_2}=\dfrac{m_{Cu\left(OH\right)_2}}{M_{Cu\left(OH\right)_2}}=\dfrac{19,6}{98}=0,2\left(mol\right)\)
\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)
Tỉ lệ: 1           1          1
\(0,2\)    \(\rightarrow\)       \(0,2\)        \(0,2\left(mol\right)\)
\(m_{CuO}=n_{CuO}.m_{CuO}=0,2.80=16g\)
=> Chọn A

15. A
16. A
17. C
18. A

19. C
20. D