HOC24
Lớp học
Môn học
Chủ đề / Chương
Bài học
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 7a) \(\dfrac{3}{4}-\dfrac{5}{6}.\left(\dfrac{1}{6}+\dfrac{1}{8}\right):\dfrac{7}{12}\)\(=\dfrac{3}{4}-\dfrac{5}{6}.\dfrac{7}{24}:\dfrac{7}{12}\)\(=\dfrac{3}{4}-\dfrac{5}{12}=\dfrac{9}{12}-\dfrac{5}{12}=\dfrac{4}{12}=\dfrac{1}{3}\)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. A16. A17. C18. A
19. C20. D
Dạ vg;-;
bn đăng từng bài 1 thôi ạBài 3a) ĐK \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)Thay \(x=9\) vào Q ta đc \(Q=\dfrac{\sqrt{9}+2}{\sqrt{9}-2}=\dfrac{3+2}{3-2}=5\)b) \(P=\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{x-2\sqrt{x}}{x-4}\)\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)+\sqrt{x}\left(\sqrt{x}-2\right)-x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)\(=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}-x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)\(=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)c) \(M=\dfrac{P}{Q}=\dfrac{\sqrt{x}}{\sqrt{x}-2}.\dfrac{\sqrt{x}-2}{\sqrt{x}+2}=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)\(M< \dfrac{1}{2}< =>\dfrac{\sqrt{x}}{\sqrt{x}+2}< \dfrac{1}{2}< =>2\sqrt{x}< \sqrt{x}+2\)\(< =>\sqrt{x}< 2< =>x< 4\)\(=>0\le x< 4\)