{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Clo l\u00e0 ch\u1ea5t kh\u00ed c\u00f3 m\u00e0u ","select":["A. n\u00e2u \u0111\u1ecf ","B. v\u00e0ng l\u1ee5c.","C. l\u1ee5c nh\u1ea1t ","D. tr\u1eafng xanh"],"hint":"","explain":"<span class='basic_left'>Clo l\u00e0 ch\u1ea5t kh\u00ed c\u00f3 m\u00e0u v\u00e0ng l\u1ee5c<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2191},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" T\u00ednh ch\u1ea5t n\u00e0o sau \u0111\u00e2y l\u00e0 c\u1ee7a kh\u00ed clo ","select":["A. T\u00e1c d\u1ee5ng v\u1edbi n\u01b0\u1edbc t\u1ea1o th\u00e0nh dung d\u1ecbch baz\u01a1.","B. T\u00e1c d\u1ee5ng v\u1edbi n\u01b0\u1edbc t\u1ea1o th\u00e0nh axit clor\u01a1 ($HClO_2$).","C. T\u00e1c d\u1ee5ng v\u1edbi oxi t\u1ea1o th\u00e0nh oxit.","D. C\u00f3 t\u00ednh t\u1ea9y m\u00e0u trong kh\u00f4ng kh\u00ed \u1ea9m."],"hint":"","explain":"<span class='basic_left'>Kh\u00ed clo trong kh\u00f4ng kh\u00ed \u1ea9m t\u1ea1o th\u00e0nh HClO c\u00f3 t\u00ednh t\u1ea9y m\u00e0u.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2192},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Clo l\u00e0 phi kim c\u00f3 \u0111\u1ed9 ho\u1ea1t \u0111\u1ed9ng ho\u00e1 h\u1ecdc ","select":["A. m\u1ea1nh h\u01a1n photpho, l\u01b0u hu\u1ef3nh nh\u01b0ng y\u1ebfu h\u01a1n flo.","B. m\u1ea1nh h\u01a1n photpho, l\u01b0u hu\u1ef3nh v\u00e0 flo","C. y\u1ebfu h\u01a1n flo, l\u01b0u hu\u1ef3nh nh\u01b0ng m\u1ea1nh h\u01a1n photpho.","D. y\u1ebfu h\u01a1n flo, photpho v\u00e0 l\u01b0u hu\u1ef3nh"],"hint":"","explain":"<span class='basic_left'>Clo ho\u1ea1t \u0111\u1ed9ng h\u00f3a h\u1ecdc m\u1ea1nh h\u01a1n photpho, l\u01b0u hu\u1ef3nh nh\u01b0ng y\u1ebfu h\u01a1n flo.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2193},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Clo t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch KOH","select":["A. t\u1ea1o ra h\u1ed7n h\u1ee3p hai axit. ","B. t\u1ea1o ra h\u1ed7n h\u1ee3p hai baz\u01a1","C. t\u1ea1o ra h\u1ed7n h\u1ee3p mu\u1ed1i","D. t\u1ea1o ra m\u1ed9t axit hipoclor\u01a1"],"hint":"","explain":"<span class='basic_left'>$C{{l}_{2}}+2KOH\\to KCl+KClO+{{H}_{2}}O$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2194},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Ch\u1ea5t d\u00f9ng \u0111\u1ec3 \u0111i\u1ec1u ch\u1ebf clo trong ph\u00f2ng th\u00ed nghi\u1ec7m l\u00e0 ","select":["A. mangan \u0111ioxit v\u00e0 axit clohi\u0111ric \u0111\u1eb7c. ","B. mangan \u0111ioxit v\u00e0 axit sunfuric \u0111\u1eb7c","C. mangan \u0111ioxit v\u00e0 axit nitric \u0111\u1eb7c. ","D. mangan \u0111ioxit v\u00e0 mu\u1ed1i natri clorua."],"hint":"","explain":"<span class='basic_left'>Trong ph\u00f2ng th\u00ed nghi\u1ec7m kh\u00ed clo \u0111\u01b0\u1ee3c \u0111i\u1ec1u ch\u1ebf b\u1eb1ng c\u00e1ch \u0111un n\u00f3ng dung d\u1ecbch $HCl$ \u0111\u1eadm \u0111\u1eb7c v\u1edbi c\u00e1c ch\u1ea5t oxi h\u00f3a m\u1ea1nh nh\u01b0 $MnO_2$ ho\u1eb7c $KMnO_4$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":1}]}],"id_ques":2195},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Trong c\u00f4ng nghi\u1ec7p ng\u01b0\u1eddi ta \u0111i\u1ec1u ch\u1ebf clo b\u1eb1ng c\u00e1ch ","select":["A. \u0111i\u1ec7n ph\u00e2n dung d\u1ecbch mu\u1ed1i \u0103n b\u00e3o ho\u00e0 ","B. \u0111i\u1ec7n ph\u00e2n dung d\u1ecbch mu\u1ed1i \u0103n b\u00e3o ho\u00e0 trong b\u00ecnh \u0111i\u1ec7n ph\u00e2n c\u00f3 m\u00e0ng ng\u0103n.","C. nung n\u00f3ng mu\u1ed1i \u0103n. ","D. \u0111un nh\u1eb9 mangan \u0111ioxit v\u1edbi axit clohi\u0111ric \u0111\u1eb7c."],"hint":"","explain":"<span class='basic_left'>Trong c\u00f4ng nghi\u1ec7p kh\u00ed clo \u0111\u01b0\u1ee3c \u0111i\u1ec1u ch\u1ebf b\u1eb1ng c\u00e1ch \u0111i\u1ec7n ph\u00e2n dung d\u1ecbch mu\u1ed1i \u0103n b\u00e3o h\u00f2a c\u00f3 m\u00e0ng ng\u0103n<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":2196},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" C\u00e1c kh\u00ed c\u00f3 th\u1ec3 t\u1ed3n t\u1ea1i trong m\u1ed9t h\u1ed7n h\u1ee3p \u1edf b\u1ea5t k\u00ec \u0111i\u1ec1u ki\u1ec7n n\u00e0o ","select":["A. $H_2$ v\u00e0 $O_2$. ","B. $Cl_2$ v\u00e0 $H_2$.","C. $Cl_2$ v\u00e0 $O_2$. ","D. $O_2$ v\u00e0 $SO_2$."],"hint":"","explain":"<span class='basic_left'>C\u00e1c kh\u00ed c\u00f3 th\u1ec3 t\u1ed3n t\u1ea1i c\u00f9ng nhau trong 1 h\u1ed7n h\u1ee3p khi ch\u00fang kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi nhau. Kh\u00ed clo kh\u00f4ng t\u00e1c d\u1ee5ng tr\u1ef1c ti\u1ebfp v\u1edbi oxi.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2197},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Kh\u1ed1i l\u01b0\u1ee3ng \u0111\u1ed3ng ph\u1ea3n \u1ee9ng v\u1eeba \u0111\u1ee7 v\u1edbi 5,04 l\u00edt kh\u00ed clo l\u00e0:","select":["A. 6,4 gam","B. 12,8 gam","C. 14,4 gam ","D. 19,2 gam"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& Cu+C{{l}_{2}}\\xrightarrow{{{t}^{0}}}CuC{{l}_{2}} \\\\ & {{n}_{Cu}}={{n}_{C{{l}_{2}}}}=0,025\\,mol \\\\ & {{m}_{Cu}}=14,4\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2198},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Khi \u0111i\u1ec7n ph\u00e2n c\u00f3 m\u00e0ng ng\u0103n dung d\u1ecbch b\u00e3o h\u00f2a mu\u1ed1i \u0103n trong n\u01b0\u1edbc c\u00f3 m\u00e0ng ng\u0103n x\u1ed1p th\u00ec x\u1ea3y ra hi\u1ec7n t\u01b0\u1ee3ng n\u00e0o trong s\u1ed1 c\u00e1c hi\u1ec7n t\u01b0\u1ee3ng cho d\u01b0\u1edbi \u0111\u00e2y","select":["A. Kh\u00ed $O_2$ tho\u00e1t ra \u1edf c\u1ef1c d\u01b0\u01a1ng v\u00e0 kh\u00ed $Cl_2$ tho\u00e1t ra \u1edf c\u1ef1c \u00e2m ","B. Kh\u00ed $H_2$ tho\u00e1t ra \u1edf c\u1ef1c \u00e2m v\u00e0 kh\u00ed $Cl_2$ tho\u00e1t ra \u1edf c\u1ef1c d\u01b0\u01a1ng","C. Kim lo\u1ea1i $Na$ tho\u00e1t ra \u1edf catot v\u00e0 kh\u00ed $Cl_2$ tho\u00e1t ra \u1edf anot","D. N\u01b0\u1edbc gia ven \u0111\u01b0\u1ee3c t\u1ea1o ra trong b\u00ecnh \u0111i\u1ec7n ph\u00e2n"],"hint":"","explain":"<span class='basic_left'>Khi \u0111i\u1ec7n ph\u00e2n dung d\u1ecbch b\u00e3o h\u00f2a mu\u1ed1i \u0103n c\u00f3 m\u00e0ng ng\u0103n x\u1ed1p thu \u0111\u01b0\u1ee3c kh\u00ed clo \u1edf c\u1ef1c d\u01b0\u01a1ng , hi\u0111ro \u1edf c\u1ef1c \u00e2m v\u00e0 dung d\u1ecbch NaOH<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":2199},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" H\u1ee3p ch\u1ea5t n\u00e0o sau \u0111\u00e2y ph\u1ea3n \u1ee9ng \u0111\u01b0\u1ee3c v\u1edbi n\u01b0\u1edbc clo ","select":["A. $NaOH$","B. $NaCl$","C. $CaSO_4$ ","D. $Cu(NO_3)_2$"],"hint":"","explain":"<span class='basic_left'>$2NaOH+C{{l}_{2}}\\to NaCl+NaClO+{{H}_{2}}O$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2200},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Trong 4 h\u1ed7n h\u1ee3p d\u01b0\u1edbi \u0111\u00e2y, h\u1ed7n h\u1ee3p n\u00e0o l\u00e0 n\u01b0\u1edbc Gia-ven","select":["A. $NaCl + NaClO + H_2O$","B. $NaCl + NaClO_3 + H_2O$ ","C. $NaCl + NaClO_2 + H_2O$","D. $NaCl + NaClO_4 + H_2O$"],"hint":"","explain":"<span class='basic_left'>$2NaOH+C{{l}_{2}}\\to NaCl+NaClO+{{H}_{2}}O$<br\/>H\u1ed7n h\u1ee3p s\u1ea3n ph\u1ea9m thu \u0111\u01b0\u1ee3c l\u00e0 n\u01b0\u1edbc Gia \u2013 ven.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2201},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Kh\u00ed clo thu \u0111\u01b0\u1ee3c trong ph\u00f2ng th\u00ed nghi\u1ec7m \u0111\u01b0\u1ee3c l\u00e0 kh\u00f4 b\u1eb1ng","select":["A. $NaOH$ ","B. $P$","C. $H_2SO_4$ \u0111\u1eb7c ","D. $CaO$."],"hint":"","explain":"<span class='basic_left'>Kh\u00ed clo \u0111\u01b0\u1ee3c l\u00e0m kh\u00f4 b\u1eb1ng ch\u1ea5t kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi n\u00f3. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n C th\u1ecfa m\u00e3n<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2202},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" N\u01b0\u1edbc clo c\u00f3 t\u00ednh t\u1ea9y m\u00e0u v\u00ec ","select":["A. clo t\u00e1c d\u1ee5ng v\u1edbi n\u01b0\u1edbc t\u1ea1o n\u00ean axit HCl c\u00f3 t\u00ednh t\u1ea9y m\u00e0u.","B. clo h\u1ea5p ph\u1ee5 \u0111\u01b0\u1ee3c m\u00e0u","C. clo t\u00e1c d\u1ee5ng n\u01b0\u1edbc t\u1ea1o n\u00ean axit HClO c\u00f3 t\u00ednh t\u1ea9y m\u00e0u. ","D. khi d\u1eabn kh\u00ed clo v\u00e0o n\u01b0\u1edbc kh\u00f4ng x\u1ea3y ra ph\u1ea3n \u1ee9ng ho\u00e1 h\u1ecdc."],"hint":"","explain":"<span class='basic_left'>$C{{l}_{2}}+{{H}_{2}}O\\rightleftharpoons HCl+HClO$ <br\/>N\u01b0\u1edbc clo c\u00f3 t\u00ednh t\u1ea9y m\u00e0u do t\u00ednh t\u1ea9y m\u00e0u c\u1ee7a HClO<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2203},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" \u0110\u1ed1t $Al$ trong b\u00ecnh kh\u00ed $Cl_2$ , sau ph\u1ea3n \u1ee9ng th\u1ea5y kh\u1ed1i l\u01b0\u1ee3ng ch\u1ea5t r\u1eafn trong b\u00ecnh t\u0103ng 3,55 g. Kh\u1ed1i l\u01b0\u1ee3ng $Al$ \u0111\u00e3 tham gia ph\u1ea3n \u1ee9ng l\u00e0","select":["A. 2,7 g ","B. 1,8 g ","C. 0,9 g ","D. 5,4 g"],"hint":"","explain":"<span class='basic_left'>Kh\u1ed1i l\u01b0\u1ee3ng ch\u1ea5t r\u1eafn trong b\u00ecnh t\u0103ng ch\u00ednh l\u00e0 kh\u1ed1i l\u01b0\u1ee3ng clo tham gia ph\u1ea3n \u1ee9ng<br\/>$\\begin{aligned}& 2Al+3C{{l}_{2}}\\xrightarrow{{{t}^{0}}}2AlC{{l}_{3}} \\\\ & {{n}_{Cl2}} = \\dfrac{3,55}{71}=0,05 mol \\\\ &{{n}_{Al}}=\\dfrac{2}{3}{{n}_{C{{l}_{2}}}}=\\dfrac{2}{3}.0,05=\\dfrac{1}{30}\\,mol \\\\ & {{m}_{Al}}=\\dfrac{1}{30}.27=0,9\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2204},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Cho 11,2 gam b\u1ed9t s\u1eaft t\u00e1c d\u1ee5ng v\u1edbi kh\u00ed clo d\u01b0. Sau ph\u1ea3n \u1ee9ng thu \u0111\u01b0\u1ee3c 32,5 gam mu\u1ed1i s\u1eaft. Kh\u1ed1i l\u01b0\u1ee3ng kh\u00ed clo tham gia ph\u1ea3n \u1ee9ng l\u00e0 ","select":["A. 21,3 gam. ","B. 20,50 gam","C. 20,50 gam ","D. 10,65 gam."],"hint":"","explain":"<span class='basic_left'>\u00c1p d\u1ee5ng \u0111\u1ecbnh lu\u1eadt b\u1ea3o to\u00e0n kh\u1ed1i l\u01b0\u1ee3ng:<br\/>$\\begin{aligned}& \\,\\,\\,\\,\\,\\,{{m}_{Fe}}+{{m}_{C{{l}_{2}}}}={{m}_{\\text{mu\u1ed1i}}} \\\\ & \\Rightarrow 11,2+{{m}_{C{{l}_{2}}}}=32,5 \\\\ & \\Rightarrow {{m}_{C{{l}_{2}}}}=21,3\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2205},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Cho m\u1ed9t lu\u1ed3ng kh\u00ed clo d\u01b0 t\u00e1c d\u1ee5ng v\u1edbi 9,2 gam kim lo\u1ea1i X sinh ra 23,4 gam mu\u1ed1i kim lo\u1ea1i ho\u00e1 tr\u1ecb I. Kim lo\u1ea1i \u0111\u00f3 l\u00e0 ","select":["A. K","B. Na.","C. Li ","D. Rb"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& 2X+C{{l}_{2}}\\xrightarrow{{{t}^{0}}}2XCl \\\\ & {{n}_{X}}=\\dfrac{9,2}{X} \\\\ & {{n}_{XCl}}=\\dfrac{23,4}{X+35,5} \\\\ & {{n}_{X}}={{n}_{XCl}}\\Rightarrow \\dfrac{9,2}{X}=\\dfrac{23,4}{X+35,5}\\Rightarrow X=23\\,(Na) \\\\ & \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2206},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Cho dung d\u1ecbch axit c\u00f3 ch\u1ee9a 7,3 gam $HCl$ \u0111\u1eb7c t\u00e1c d\u1ee5ng v\u1edbi $MnO_2$ d\u01b0. Th\u1ec3 t\u00edch kh\u00ed clo sinh ra (\u0111ktc)","select":["A. 1,12 l\u00edt ","B. 2,24 l\u00edt.","C. 11,2 l\u00edt ","D. 22,4 l\u00edt."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& Mn{{O}_{2}}+4HC{{l}_{\\text{\u0111}}}\\xrightarrow{{{t}^{0}}}MnC{{l}_{2}}+C{{l}_{2}}+2{{H}_{2}}O \\\\ & {{n}_{C{{l}_{2}}}}=\\dfrac{1}{4}{{n}_{HCl}}=\\dfrac{1}{4}.0,2=0,05\\,mol \\\\ & {{V}_{C{{l}_{2}}}}=1,12\\,l \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2207},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho 1,12 l\u00edt kh\u00ed clo (\u0111ktc) v\u00e0o dung d\u1ecbch NaOH 0,5M . Sau khi ph\u1ea3n \u1ee9ng k\u1ebft th\u00fac, th\u1ec3 t\u00edch dung d\u1ecbch NaOH c\u1ea7n d\u00f9ng l\u00e0 ","select":["A. 0,1 l\u00edt ","B. 0,15 l\u00edt.","C. 0,2 l\u00edt","D. 0,25 l\u00edt."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& C{{l}_{2}}+2NaOH\\to NaCl+NaClO+{{H}_{2}}O \\\\ & {{n}_{NaOH}}=2{{n}_{C{{l}_{2}}}}=0,1\\,mol \\\\ & {{V}_{NaOH}}=\\dfrac{0,1}{0,5}=0,2\\,l \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2208},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" \u0110i\u1ec1u ch\u1ebf kh\u00ed clo trong ph\u00f2ng th\u00ed nghi\u1ec7m b\u1eb1ng c\u00e1ch cho $KMnO_4$ t\u00e1c d\u1ee5ng v\u1edbi axit $HCl$ \u0111\u1eb7c. Kh\u1ed1i l\u01b0\u1ee3ng $KMnO_4$ c\u1ea7n d\u00f9ng l\u00e0 bao nhi\u00eau gam \u0111\u1ec3 thu \u0111\u01b0\u1ee3c l\u01b0\u1ee3ng kh\u00ed clo \u0111\u1ee7 \u0111\u1ec3 ph\u1ea3n \u1ee9ng v\u1edbi 5,6 gam s\u1eaft. Bi\u1ebft hi\u1ec7u su\u1ea5t c\u1ee7a qu\u00e1 tr\u00ecnh \u0111i\u1ec1u ch\u1ebf \u0111\u1ea1t 80%","select":["A. 9,48 gam","B. 10,05 gam","C. 11,85 gam ","D. 7,584 gam"],"hint":"","explain":"<span class='basic_left'>G\u1ecdi x l\u00e0 s\u1ed1 mol kh\u00ed clo thu \u0111\u01b0\u1ee3c.<br\/>$\\begin{aligned} & 2KMn{{O}_{4}}+16HCl\\xrightarrow{{{t}^{0}}}2KCl+2MnC{{l}_{2}}+5C{{l}_{2}}+8{{H}_{2}}O \\\\ & 2Fe+3C{{l}_{2}}\\xrightarrow{{{t}^{0}}}2FeC{{l}_{3}} \\\\ & {{n}_{Fe}}=0,1\\,mol \\\\ & {{n}_{KMn{{O}_{4}}}}=\\dfrac{2}{5\\,}{{n}_{C{{l}_{2}}}}=\\dfrac{2}{5}.\\dfrac{3}{2}{{n}_{Fe}}=0,06\\,mol \\\\ & {{m}_{KMn{{O}_{4}}\\,(LT)}}=9,48\\,gam \\\\ & H=80\\%\\Rightarrow {{m}_{KMn{{O}_{4}}\\,(TT)}}=\\dfrac{9,48.100\\%}{80\\%}=11,85\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2209},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" R\u1eafc b\u1ed9t s\u1eaft \u0111un n\u00f3ng v\u00e0o l\u1ecd ch\u1ee9a kh\u00ed Clo d\u01b0. H\u1ed7n h\u1ee3p sau ph\u1ea3n \u1ee9ng cho t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch $HCl$ d\u01b0 th\u1ea5y t\u1ea1o ra 2,24 l\u00edt $H_2$ (\u0111ktc).N\u1ebfu cho h\u1ed7n h\u1ee3p sau ph\u1ea3n \u1ee9ng t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch $NaOH$ th\u00ec t\u1ea1o ra 0,03 mol ch\u1ea5t k\u1ebft t\u1ee7a m\u00e0u n\u00e2u \u0111\u1ecf. Hi\u1ec7u su\u1ea5t c\u1ee7a ph\u1ea3n \u1ee9ng $Fe$ t\u00e1c d\u1ee5ng v\u1edbi Clo l\u00e0","select":["A. 30 %","B. 50 %","C. 47 % ","D. 23 %"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& 2Fe+3C{{l}_{2}}\\xrightarrow{{{t}^{o}}}2FeC{{l}_{3}}\\,\\,\\,\\,\\,\\,\\,(1) \\\\ & F{{e}_{\\text{d\u01b0}}}+2HCl\\to FeC{{l}_{2}}+{{H}_{2}}\\uparrow \\\\ & FeC{{l}_{3}}+3NaOH\\to Fe{{(OH)}_{3}}\\downarrow +3NaCl \\\\ & {{n}_{FeC{{l}_{3}}}}={{n}_{\\downarrow }}=0,03\\,mol \\\\ & {{n}_{Fe\\,\\text{d\u01b0}}}={{n}_{{{H}_{2}}}}=0,1\\,mol \\\\ & {{n}_{Fe}}={{n}_{Fe\\,\\text{d\u01b0}}}+{{n}_{FeC{{l}_{3}}}}=0,13\\,mol \\\\ & H=\\dfrac{{{n}_{Fe\\,(\\text{\u1edf}\\,1)}}}{{{n}_{Fe}}}=\\dfrac{0,03}{0,13}.100\\%\\approx 23\\% \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2210}],"lesson":{"save":0,"level":2}}