{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" \u0110\u01a1n ch\u1ea5t t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch $H_2SO_4$ lo\u00e3ng gi\u1ea3i ph\u00f3ng kh\u00ed hi\u0111r\u00f4 l\u00e0:","select":["A. \u0110\u1ed3ng ","B. L\u01b0u hu\u1ef3nh","C. K\u1ebdm ","D. Thu\u1ef7 ng\u00e2n"],"hint":"","explain":"<span class='basic_left'>C\u00e1c kim lo\u1ea1i m\u1ea1nh nh\u01b0 Zn, Mg, Al, Fe t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi axit $HCl, H_2SO_4$ lo\u00e3ng gi\u1ea3i ph\u00f3ng kh\u00ed hi\u0111r\u00f4<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2011},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" \u0110\u1ec3 l\u00e0m s\u1ea1ch m\u1eabu ch\u00ec b\u1ecb l\u1eabn k\u1ebdm, ng\u01b0\u01a1\u00ec ta ng\u00e2m m\u1eabu ch\u00ec n\u00e0y v\u00e0o m\u1ed9t l\u01b0\u1ee3ng d\u01b0 dung d\u1ecbch","select":["A. $ZnSO_4$ ","B. $Pb(NO_3)_2$","C. $CuCl_2$","D. $Na_2CO_3$"],"hint":"","explain":"<span class='basic_left'>Ta cho m\u1eabu n\u00e0y t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch $Pb(NO_3)_2$ \u0111\u1ec3 h\u00f2a tan h\u1ebft $Zn$ theo ph\u1ea3n \u1ee9ng:<br\/>$Zn+Pb{{(N{{O}_{3}})}_{2}}\\to Zn{{(N{{O}_{3}})}_{2}}+Pb$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2012},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" D\u00e3y c\u00e1c kim lo\u1ea1i c\u00f3 \u0111\u1ed9 ho\u1ea1t \u0111\u1ed9ng h\u00f3a h\u1ecdc m\u1ea1nh h\u01a1n Cu l\u00e0:","select":["A. Al, Zn, Mg ","B. Zn, Fe, Ag","C. Al, Ag, Au ","D. Al, Mg, Hg"],"hint":"","explain":"<span class='basic_left'>C\u00e1c kim lo\u1ea1i m\u1ea1nh h\u01a1n Cu (tr\u1eeb kim lo\u1ea1i ki\u1ec1m, ki\u1ec1n th\u1ed5 nh\u01b0: Zn, Fe, Mg, Al,....)s\u1ebd t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi dung d\u1ecbch $Cu(NO_3)_2$ t\u1ea1o th\u00e0nh kim lo\u1ea1i \u0111\u1ed3ng<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2013},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Kim lo\u1ea1i v\u1eeba t\u00e1c d\u1ee5ng v\u1edbi dd HCl v\u1eeba t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi dung d\u1ecbch KOH","select":["A. Fe, Al","B. Ag, Zn","C. Al, Cu ","D. Al, Zn"],"hint":"","explain":"<span class='basic_left'>C\u00e1c kim lo\u1ea1i c\u00f3 t\u00ednh l\u01b0\u1ee1ng t\u00ednh nh\u01b0 Al, Zn v\u1eeba t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi dung d\u1ecbch axit, v\u1eeba t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi dung d\u1ecbch baz\u01a1.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2014},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Khi cho s\u1eaft t\u00e1c d\u1ee5ng v\u1edbi kh\u00ed clo mu\u1ed1i thu \u0111\u01b0\u1ee3c c\u00f3 c\u00f4ng th\u1ee9c l\u00e0:","select":["A. $FeCl$ ","B. $FeCl_2$","C. $FeCl_3$ ","D. $FeCl_2$ v\u00e0 $FeCl_3$"],"hint":"","explain":"<span class='basic_left'>$2Fe+3C{{l}_{2}}\\xrightarrow{{{t}^{0}}}2FeC{{l}_{3}}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2015},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Cho ph\u1ea3n \u1ee9ng: $Cu+2AgN{{O}_{3}}\\to Cu{{(N{{O}_{3}})}_{2}}+2Ag$. Ph\u1ea3n \u1ee9ng tr\u00ean c\u00f3 th\u1ec3 k\u1ebft lu\u1eadn \u0111\u01b0\u1ee3c l\u00e0","select":["A. Ag ho\u1ea1t \u0111\u1ed9ng h\u00f3a h\u1ecdc y\u1ebfu h\u01a1n Cu ","B. Ag ho\u1ea1t \u0111\u1ed9ng h\u00f3a h\u1ecdc m\u1ea1nh h\u01a1n Cu","C. Cu v\u00e0 Ag c\u00f3 \u0111\u1ed9 ho\u1ea1t \u0111\u1ed9ng h\u00f3a h\u1ecdc m\u1ea1nh nh\u01b0 nhau ","D. Kh\u00f4ng th\u1ec3 k\u1ebft lu\u1eadn \u0111\u01b0\u1ee3c \u0111\u1ed9 ho\u1ea1t \u0111\u1ed9ng h\u00f3a h\u1ecdc m\u1ea1nh y\u1ebfu c\u1ee7a 2 kim lo\u1ea1i."],"hint":"","explain":"<span class='basic_left'>Cu \u0111\u1ea9y \u0111\u01b0\u1ee3c Ag ra kh\u1ecfi mu\u1ed1i n\u00ean c\u00f3 th\u1ec3 k\u1ebft lu\u1eadn Cu ho\u1ea1t \u0111\u1ed9ng h\u00f3a h\u1ecdc m\u1ea1nh h\u01a1n Ag.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":1}]}],"id_ques":2016},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Kh\u1ed1i l\u01b0\u1ee3ng kh\u00ed t\u1ea1o th\u00e0nh khi cho Zn t\u00e1c d\u1ee5ng v\u1eeba \u0111\u1ee7 v\u1edbi 200 ml dung d\u1ecbch HCl 1M l\u00e0:","select":["A. 0,2 gam","B. 0,4 gam ","C. 0,6 gam ","D. 0,8 gam."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& Zn+2HCl\\to ZnC{{l}_{2}}+{{H}_{2}}\\uparrow \\\\ & {{n}_{{{H}_{2}}}}=\\dfrac{1}{2}{{n}_{HCl}}=\\dfrac{1}{2}.0,2\\,=0,1\\,mol \\\\ & {{m}_{{{H}_{2}}}}=0,2\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2017},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Khi th\u1ea3 m\u1ed9t c\u00e2y \u0111inh s\u1eaft s\u1ea1ch v\u00e0o dung d\u1ecbch $Cu{{(N{{O}_{3}})}_{2}}$ lo\u00e3ng, c\u00f3 hi\u1ec7n t\u01b0\u1ee3ng sau:","select":["A. S\u1ee7i b\u1ecdt kh\u00ed, m\u00e0u xanh c\u1ee7a dung d\u1ecbch nh\u1ea1t d\u1ea7n. ","B. C\u00f3 m\u1ed9t l\u1edbp \u0111\u1ed3ng m\u00e0u \u0111\u1ecf ph\u1ee7 l\u00ean \u0111inh s\u1eaft, m\u00e0u xanh c\u1ee7a dung d\u1ecbch \u0111\u1eadm d\u1ea7n.","C. C\u00f3 m\u1ed9t l\u1edbp \u0111\u1ed3ng m\u00e0u \u0111\u1ecf ph\u1ee7 l\u00ean \u0111inh s\u1eaft, dung d\u1ecbch kh\u00f4ng \u0111\u1ed5i m\u00e0u. ","D. C\u00f3 m\u1ed9t l\u1edbp \u0111\u1ed3ng m\u00e0u \u0111\u1ecf ph\u1ee7 l\u00ean \u0111inh s\u1eaft, m\u00e0u xanh c\u1ee7a dung d\u1ecbch nh\u1ea1t d\u1ea7n "],"hint":"","explain":"<span class='basic_left'>$Fe+Cu{{(N{{O}_{3}})}_{2}}\\to Fe{{(N{{O}_{3}})}_{2}}+Cu$ <br\/> Ph\u1ea3n \u1ee9ng t\u1ea1o ra Cu m\u00e0u \u0111\u1ecf b\u00e1m v\u00e0o \u0111inh s\u1eaft. Dung d\u1ecbch $Cu(NO_3)_2$ ph\u1ea3n \u1ee9ng n\u00ean m\u00e0u xanh nh\u1ea1t d\u1ea7n<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":2018},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" H\u00e0m l\u01b0\u1ee3ng s\u1eaft trong $Fe_3O_4$:","select":["A. 70% ","B. 72,41%","C. 46,66%","D. 48,27%"],"hint":"","explain":"<span class='basic_left'>$\\%{{m}_{Fe}}=\\dfrac{56.3}{232}.100\\%\\approx 72,41\\%$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2019},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Hi\u1ec7n t\u01b0\u1ee3ng x\u1ea3y ra khi \u0111\u1ed1t natri trong b\u00ecnh kh\u00ed clo l\u00e0:","select":["A. Kh\u00f3i m\u00e0u tr\u1eafng sinh ra. ","B. Xu\u1ea5t hi\u1ec7n nh\u1eefng tia s\u00e1ng ch\u00f3i.","C. T\u1ea1o ch\u1ea5t b\u1ed9t xanh b\u00e1m xung quanh th\u00e0nh b\u00ecnh. ","D. C\u00f3 kh\u00f3i m\u00e0u n\u00e2u \u0111\u1ecf t\u1ea1o th\u00e0nh."],"hint":"","explain":"<span class='basic_left'>$2Na+C{{l}_{2}}\\xrightarrow{{{t}^{0}}}2NaCl$<br\/>Natri ch\u00e1y trong kh\u00ed clo t\u1ea1o th\u00e0nh kh\u00f3i tr\u1eafng. <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":1}]}],"id_ques":2020},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" C\u00e1c kim lo\u1ea1i t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi dung d\u1ecbch $Cu(NO_3)_2$ t\u1ea1o th\u00e0nh kim lo\u1ea1i \u0111\u1ed3ng:","select":["A. Al, Zn, Fe ","B. Mg, Fe, Ag","C. Zn, Pb, Au","D. Na, Mg, Al"],"hint":"","explain":"<span class='basic_left'>C\u00e1c kim lo\u1ea1i m\u1ea1nh h\u01a1n Cu (tr\u1eeb kim lo\u1ea1i ki\u1ec1m, ki\u1ec1n th\u1ed5 nh\u01b0: Zn, Fe, Mg, Al,....)s\u1ebd t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi dung d\u1ecbch $Cu(NO_3)_2$ t\u1ea1o th\u00e0nh kim lo\u1ea1i \u0111\u1ed3ng<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2021},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Khi cho 1 thanh Cu v\u00e0o dung d\u1ecbch $H_2SO_4$ lo\u00e3ng hi\u1ec7n t\u01b0\u1ee3ng quan s\u00e1t \u0111\u01b0\u1ee3c l\u00e0:","select":["A. C\u00f3 kh\u00ed kh\u00f4ng m\u00e0u tho\u00e1t ra, dung d\u1ecbch m\u00e0u xanh ","B. C\u00f3 kh\u00ed kh\u00f4ng m\u00e0u tho\u00e1t ra, dung d\u1ecbch kh\u00f4ng m\u00e0u","C. Kh\u00f4ng c\u00f3 kh\u00ed tho\u00e1t ra, dung d\u1ecbch m\u00e0u xanh ","D. Kh\u00f4ng c\u00f3 hi\u1ec7n t\u01b0\u1ee3ng g\u00ec."],"hint":"","explain":"<span class='basic_left'>Cu kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi axit $HCl, H_2SO_4$ lo\u00e3ng n\u00ean kh\u00f4ng c\u00f3 hi\u1ec7n t\u01b0\u1ee3ng g\u00ec x\u1ea3y ra.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2022},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" L\u1ea5y m\u1ed9t \u00edt b\u1ed9t Fe cho v\u00e0o dung d\u1ecbch HCl v\u1eeba \u0111\u1ee7 r\u1ed3i nh\u1ecf t\u1eeb t\u1eeb dung d\u1ecbch NaOH \u0111\u1ebfn d\u01b0 v\u00e0o dung d\u1ecbch . Hi\u1ec7n t\u01b0\u1ee3ng x\u1ea3y ra l\u00e0","select":["A. C\u00f3 kh\u00ed bay ra v\u00e0 dung d\u1ecbch c\u00f3 m\u00e0u xanh lam. ","B. Kh\u00f4ng th\u1ea5y hi\u1ec7n t\u01b0\u1ee3ng g\u00ec.","C. Ban \u0111\u1ea7u c\u00f3 kh\u00ed tho\u00e1t ra v\u00e0 dd c\u00f3 k\u1ebft t\u1ee7a tr\u1eafng xanh r\u1ed3i chuy\u1ec3n d\u1ea7n th\u00e0nh m\u00e0u n\u00e2u \u0111\u1ecf. ","D. C\u00f3 kh\u00ed tho\u00e1t ra v\u00e0 t\u1ea1o k\u1ebft t\u1ee7a m\u00e0u tr\u1eafng xanh \u0111\u1ebfn khi k\u1ebft th\u00fac ."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& Fe+2HCl\\to FeC{{l}_{2}}+{{H}_{2}}\\uparrow \\\\ & FeC{{l}_{2}}+2NaOH\\to 2NaCl+Fe{{(OH)}_{2}}\\downarrow \\, \\\\ \\end{aligned}$ <br\/>$Fe(OH)_2$ c\u00f3 m\u00e0u tr\u1eafng xanh.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":2023},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Nung 6,4g Cu ngo\u00e0i kh\u00f4ng kh\u00ed thu \u0111\u01b0\u1ee3c 6,4g CuO. Hi\u1ec7u su\u1ea5t ph\u1ea3n \u1ee9ng l\u00e0:","select":["A. 100%. ","B. 80%.","C. 70%.","D. 60%."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& 2Cu+{{O}_{2}}\\xrightarrow{{{t}^{0}}}2CuO \\\\ & {{n}_{CuO}}={{n}_{Cu}}=0,1\\,mol \\\\ & {{m}_{CuO\\,(LT)}}=8\\,gam \\\\ & H=\\dfrac{{{m}_{CuO\\,(TT)}}}{{{m}_{CuO\\,(LT)}}}=\\dfrac{6,4}{8}.100\\%=80\\% \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2024},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Ho\u00e0 tan ho\u00e0n to\u00e0n 3,25g m\u1ed9t kim lo\u1ea1i X (ho\u00e1 tr\u1ecb II) b\u1eb1ng dung d\u1ecbch $HCl$ lo\u00e3ng thu \u0111\u01b0\u1ee3c 1,12 l\u00edt kh\u00ed $H_2$ \u1edf \u0111ktc. V\u1eady X l\u00e0 kim lo\u1ea1i n\u00e0o sau \u0111\u00e2y","select":["A. Fe ","B. Mg","C. Ca","D. Zn"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& X+2HCl\\to XC{{l}_{2}}+{{H}_{2}}\\uparrow \\\\ & {{n}_{{{H}_{2}}}}={{n}_{X}}=0,05\\,mol \\\\ & {{M}_{X}}=\\dfrac{3,25}{0,05}=65\\,\\,(Zn) \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2025},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Cho v\u00e0o dung d\u1ecbch HCl m\u1ed9t c\u00e2y \u0111inh s\u1eaft , sau m\u1ed9t th\u1eddi gian thu \u0111\u01b0\u1ee3c 11,2 l\u00edt kh\u00ed hi\u0111r\u00f4 (\u0111ktc). Kh\u1ed1i l\u01b0\u1ee3ng s\u1eaft \u0111\u00e3 ph\u1ea3n \u1ee9ng l\u00e0 ","select":["A. 28 gam ","B. 12,5 gam","C. 8 gam ","D. 36 gam"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& Fe+2HCl\\to FeC{{l}_{2}}+{{H}_{2}}\\uparrow \\\\ & {{n}_{Fe}}={{n}_{{{H}_{2}}}}=0,5\\,mol \\\\ & {{m}_{Fe}}=28\\,gam \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2026},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Cho 10,5g h\u1ed7n h\u1ee3p 2 kim lo\u1ea1i $Cu$ v\u00e0 $Zn$ v\u00e0o dd $H_2SO_4$ lo\u00e3ng d\u01b0, ng\u01b0\u1eddi ta thu \u0111\u01b0\u1ee3c 2,24 l\u00edt kh\u00ed (\u0111ktc). Th\u00e0nh ph\u1ea7n % theo kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a $Cu$ v\u00e0 $Zn$ l\u1ea7n l\u01b0\u1ee3t l\u00e0:","select":["A. 61,9% v\u00e0 38,1% ","B. 38,1 % v\u00e0 61,9%","C. 65% v\u00e0 35% ","D. 35% v\u00e0 65%"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& Zn+{{H}_{2}}S{{O}_{4}}\\to ZnS{{O}_{4}}+{{H}_{2}}\\uparrow \\\\ & {{n}_{Zn}}={{n}_{{{H}_{2}}}}=0,1\\,mol \\\\ & {{m}_{Zn}}=6,5\\,gam \\\\ & \\%{{m}_{Zn}}=\\dfrac{6,5}{10,5}.100\\%\\approx 61,9\\% \\\\ & \\%{{m}_{Cu}}=100\\%-61,9\\%=38,1\\% \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2027},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Ho\u00e0 tan h\u1ebft 2,3g Na kim lo\u1ea1i v\u00e0o 97,8g n\u01b0\u1edbc thu \u0111\u01b0\u1ee3c dung d\u1ecbch c\u00f3 n\u1ed3ng \u0111\u1ed9:","select":["A. 3,99%. ","B. 4,0%.","C. 23,0%.","D. 5,8%."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{Na}}=0,1\\,mol \\\\ & 2Na+2{{H}_{2}}O\\to 2NaOH+{{H}_{2}}\\uparrow \\\\ & {{n}_{{{H}_{2}}}}=\\dfrac{1}{2}{{n}_{Na}}=0,05\\,mol \\\\ & {{m}_{{{H}_{2}}}}=0,1\\,gam \\\\ & {{m}_{\\text{dd}}}={{m}_{Na}}+{{m}_{{{H}_{2}}O}}-{{m}_{{{H}_{2}}}}=100\\,gam \\\\ & {{n}_{NaOH}}={{n}_{Na}}=0,1\\,mol \\\\ & {{m}_{NaOH}}=4\\,gam \\\\ & C\\%=\\dfrac{4}{100}.100\\%=4\\% \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2028},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho 100 gam h\u1ed7n h\u1ee3p g\u1ed3m 2 kim lo\u1ea1i $Fe, Cu$ v\u00e0o dung d\u1ecbch $CuSO_4$ d\u01b0 sau ph\u1ea3n \u1ee9ng th\u1ea5y kh\u1ed1i l\u01b0\u1ee3ng ch\u1ea5t r\u1eafn thu \u0111\u01b0\u1ee3c t\u0103ng th\u00eam 4 gam so v\u1edbi ban \u0111\u1ea7u . V\u1eady % kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a $Cu$ trong h\u1ed7n h\u1ee3p ban \u0111\u1ea7u l\u00e0 ","select":["A. 100% ","B. 32%","C. 72% ","D. 28%"],"hint":"","explain":"<span class='basic_left'>G\u1ecdi x l\u00e0 s\u1ed1 mol c\u1ee7a Fe c\u00f3 trong h\u1ed7n h\u1ee3p<br\/>$\\begin{aligned}& Fe+CuS{{O}_{4}}\\to FeS{{O}_{4}}+Cu \\\\ & x\\,mol\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x\\,mol \\\\ \\end{aligned}$<br\/> Kh\u1ed1i l\u01b0\u1ee3ng ch\u1ea5t r\u1eafn t\u0103ng b\u1eb1ng kh\u1ed1i l\u01b0\u1ee3ng Cu b\u00e1m v\u00e0o tr\u1eeb kh\u1ed1i l\u01b0\u1ee3ng Fe ph\u1ea3n \u1ee9ng.<br\/>$\\begin{aligned}& 64x-56x=4 \\\\ & =>\\,x=0,5\\,mol \\\\ & {{m}_{Fe}}=0,5.56=28\\,gam \\\\ & \\%{{m}_{Fe}}=\\dfrac{28}{100}.100\\%=28\\% \\\\ & \\%{{m}_{Cu}}=100\\%-28\\%=72\\% \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2029},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Cho 8,1g m\u1ed9t kim lo\u1ea1i X (ho\u00e1 tr\u1ecb III) t\u00e1c d\u1ee5ng v\u1edbi kh\u00ed clo c\u00f3 d\u01b0 thu \u0111\u01b0\u1ee3c 40,05g mu\u1ed1i. X\u00e1c \u0111\u1ecbnh kim lo\u1ea1i \u0111em ph\u1ea3n \u1ee9ng","select":["A. Cr ","B. Al","C. Fe","D. Au"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& 2X+3C{{l}_{2}}\\xrightarrow{{{t}^{0}}}2XC{{l}_{3}} \\\\ & {{m}_{C{{l}_{2}}}}=40,05-8,1=31,95\\,gam \\\\ & {{n}_{C{{l}_{2}}}}=0,45\\,mol \\\\ & {{n}_{X}}=\\dfrac{2}{3}{{n}_{C{{l}_{2}}}}=0,3\\,\\,mol \\\\ & {{M}_{X}}=\\dfrac{8,1}{0,3}=27\\,(Al) \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2030}],"lesson":{"save":0,"level":2}}