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{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Ng\u01b0\u1eddi ta c\u00f3 th\u1ec3 d\u00e1t m\u1ecfng \u0111\u01b0\u01a1\u0323c nh\u00f4m th\u00e0nh th\u00eca, xoong, ch\u1eadu, gi\u1ea5y g\u00f3i b\u00e1nh k\u1eb9o l\u00e0 do nh\u00f4m c\u00f3 t\u00ednh:","select":["A. d\u1ebbo.","B. d\u1eabn \u0111i\u1ec7n.","C . d\u1eabn nhi\u1ec7t.","D . \u00e1nh kim."],"hint":"","explain":"<span class='basic_left'>Nh\u00f4m c\u00f3 t\u00ednh d\u1ebbo n\u00ean c\u00f3 th\u1ec3 d\u00e1t m\u1ecfng v\u00e0 k\u00e9o th\u00e0nh s\u1ee3i, \u0111\u01b0\u1ee3c \u1ee9ng d\u1ee5ng r\u1ed9ng r\u00e3i trong s\u1ea3n xu\u1ea5t c\u00e1c \u0111\u1ed3 gia d\u1ee5ng: th\u00eca, xoong,....<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2051},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Kim lo\u1ea1i nh\u00f4m c\u00f3 \u0111\u1ed9 d\u1eabn \u0111i\u1ec7n y\u1ebfu h\u01a1n kim lo\u1ea1i:","select":["A. Cu, Ag","B. Ag, Fe","C. Fe, Au ","D. Fe, Cu"],"hint":"","explain":"<span class='basic_left'>\u0110\u1ed9 d\u1eabn \u0111i\u1ec7n c\u1ee7a c\u00e1c kim lo\u1ea1i \u0111\u01b0\u1ee3c s\u1eafp x\u1ebfp nh\u01b0 sau: Ag > Cu > Au > Al > Fe<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2052},{"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 n\u00f3ng m\u1ed9t t\u1edd gi\u1ea5y l\u00e0m b\u1eb1ng nh\u00f4m th\u1ea5y ph\u1ea7n kh\u00f4ng ti\u1ebfp x\u00fac v\u1edbi ng\u1ecdn l\u1eeda c\u0169ng b\u1ecb n\u00f3ng l\u00ean. Th\u00ed nghi\u1ec7m ch\u1ee9ng t\u1ecf nh\u00f4m c\u00f3 t\u00ednh ch\u1ea5t","select":["A. d\u1ebbo. ","B. d\u1eabn \u0111i\u1ec7n.","C . d\u1eabn nhi\u1ec7t. ","D . \u00e1nh kim."],"hint":"","explain":"<span class='basic_left'>Nh\u00f4m d\u1eabn nhi\u1ec7t t\u1ed1t n\u00ean ph\u1ea7n ti\u1ebfp x\u00fac tr\u1ef1c ti\u1ebfp v\u1edbi ng\u1ecdn l\u1eeda c\u0169ng b\u1ecb n\u00f3ng l\u00ean<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2053},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Nguy\u00ean li\u1ec7u s\u1ea3n xu\u1ea5t nh\u00f4m l\u00e0 qu\u1eb7ng:","select":["A. Hematit ","B. Manhetit ","C. B\u00f4xit ","D. Pirit."],"hint":"","explain":"<span class='basic_left'>qu\u1eb7ng boxit d\u00f9ng \u0111\u1ec3 s\u1ea3n xu\u1ea5t nh\u00f4m<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2054},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Cho l\u00e1 nh\u00f4m v\u00e0o dung d\u1ecbch axit HCl c\u00f3 d\u01b0 thu \u0111\u01b0\u1ee3c 3,36 l\u00edt kh\u00ed hi\u0111ro (\u1edf \u0111ktc). Kh\u1ed1i l\u01b0\u1ee3ng nh\u00f4m \u0111\u00e3 ph\u1ea3n \u1ee9ng l\u00e0:","select":["A. 1,8 g ","B. 2,7 g","C. 4,05 g ","D. 5,4 g"],"hint":"","explain":"<span class='basic_left'>${{n}_{{{H}_{2}}}}=\\dfrac{3,36}{22,4}=0,15mol$ <br\/> $2Al+6HCl\\to 2AlC{{l}_{3}}+3{{H}_{2}}$ <br\/> $\\begin{aligned}& {{n}_{Al}}=\\dfrac{2}{3}{{n}_{{{H}_{2}}}}=\\dfrac{2}{3}.0,15=0,1mol \\ & {{m}_{Al}}=0,1.27=2,7gam \\ \\end{aligned}$. \\\\ \\end{aligned}$. <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2055},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Nh\u00f4m kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi ch\u1ea5t n\u00e0o sau \u0111\u00e2y","select":["A. $Cl_2$","B. Dung d\u1ecbch $HCl$","C. Dung d\u1ecbch $NaOH$","D. $H_2SO_4$\u0111\u1eb7c ngu\u1ed9i"],"hint":"","explain":"<span class='basic_left'>Al, Fe kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi $HNO_3$\u0111\u1eb7c, ngu\u1ed9i; $H_2SO_4$ \u0111\u1eb7c, ngu\u1ed9i.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2056},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" C\u1eb7p ch\u1ea5t n\u00e0o d\u01b0\u1edbi \u0111\u00e2y c\u00f3 ph\u1ea3n \u1ee9ng ?","select":["A. $Al + HNO_3$ \u0111\u1eb7c , ngu\u1ed9i ","B. $Al + HCl$","C. $Fe + HNO_3$ \u0111\u1eb7c , ngu\u1ed9i ","D. $Fe + Al_2(SO4)_3$"],"hint":"","explain":"<span class='basic_left'>Al \u0111\u1ee9ng tr\u01b0\u1edbc H n\u00ean c\u00f3 ph\u1ea3n \u1ee9ng v\u1edbi dung d\u1ecbch HCl<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2057},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" \u0110\u1ec3 ph\u00e2n bi\u1ec7t \u0111\u01b0\u1ee3c kim lo\u1ea1i nh\u00f4m v\u1edbi s\u1eaft, c\u00f3 th\u1ec3 s\u1eed d\u1ee5ng","select":["A. dung d\u1ecbch ki\u1ec1m ","B. dung d\u1ecbch mu\u1ed1i magie","C. dung d\u1ecbch mu\u1ed1i k\u1ebdm ","D. dung d\u1ecbch axit sunfuric"],"hint":"","explain":"<span class='basic_left'>Kim lo\u1ea1i n\u00e0o c\u00f3 ph\u1ea3n \u1ee9ng v\u1edbi dung d\u1ecbch ki\u1ec1m t\u1ea1o kh\u00ed kh\u00f4ng m\u00e0u l\u00e0 Al $2Al+2NaOH+2{H}_{2}O\\to 2NaAl{{O}_{2}}+3{H}_{2}$ Kim lo\u1ea1i c\u00f2n l\u1ea1i kh\u00f4ng ph\u1ea3n \u1ee9ng l\u00e0 Fe <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2058},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Nh\u1eefng kim lo\u1ea1i t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi dung d\u1ecbch HCl l\u00e0","select":["A. Fe;Al","B. Cu; Fe ","C. Ag; Al ","D. Cu; Al"],"hint":"","explain":"<span class='basic_left'>C\u00e1c kim lo\u1ea1i \u0111\u1ee9ng sau H nh\u01b0 Cu, Ag kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi $HCl, H_2SO_4$ lo\u00e3ng <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2059},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Th\u1ea3 m\u1ed9t m\u1ea3nh nh\u00f4m v\u00e0o \u1ed1ng nghi\u1ec7m ch\u1ee9a dung d\u1ecbch $CuSO_4$. X\u1ea3y ra hi\u1ec7n t\u01b0\u1ee3ng:","select":["A. Kh\u00f4ng c\u00f3 d\u1ea5u hi\u1ec7u ph\u1ea3n \u1ee9ng ","B. C\u00f3 ch\u1ea5t r\u1eafn m\u00e0u tr\u1eafng b\u00e1m ngo\u00e0i l\u00e1 nh\u00f4m, m\u00e0u xanh c\u1ee7a dung d\u1ecbch $CuSO_4$ nh\u1ea1t d\u1ea7n.","C. C\u00f3 ch\u1ea5t r\u1eafn m\u00e0u \u0111\u1ecf b\u00e1m ngo\u00e0i l\u00e1 nh\u00f4m, m\u00e0u xanh c\u1ee7a dung d\u1ecbch $CuSO_4$ nh\u1ea1t d\u1ea7n. ","D. C\u00f3 ch\u1ea5t kh\u00ed bay ra, dung d\u1ecbch kh\u00f4ng \u0111\u1ed5i m\u00e0u"],"hint":"","explain":"<span class='basic_left'>Al t\u00e1c d\u1ee5ng v\u1edbi $CuSO_4$ theo ph\u1ea3n \u1ee9ng $2Al+3CuS{{O}_{4}}\\to A{{l}_{2}}{{(S{{O}_{4}})}_{3}}+3Cu$ Cu m\u00e0u \u0111\u1ecf b\u00e1m v\u00e0o l\u00e1 Al, dung d\u1ecbch $CuSO_4$ ban \u0111\u1ea7u c\u00f3 m\u00e0u xanh, khi tham gia ph\u1ea3n \u1ee9ng m\u00e0u c\u1ee7a dung d\u1ecbch s\u1ebd nh\u1ea1t d\u1ea7n v\u00e0 m\u1ea5t m\u00e0u khi $CuSO_4$ ph\u1ea3n \u1ee9ng h\u1ebft.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2060},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Cho 5,4g b\u1ed9t nh\u00f4m v\u00e0o dung d\u1ecbch b\u1ea1c nitrat d\u01b0. Sau ph\u1ea3n \u1ee9ng thu \u0111\u01b0\u1ee3c bao nhi\u00eau gam k\u1ebft t\u1ee7a b\u1ea1c ","select":["A. 648 ","B. 64,8","C. 6,48 ","D.0, 648"],"hint":"","explain":"<span class='basic_left'> $Al$ \u0111\u1ea9y $Ag$ ra kh\u1ecfi mu\u1ed1i theo ph\u1ea3n \u1ee9ng<br\/>$Al+3AgN{{O}_{3}}\\to Al{{(N{{O}_{3}})}_{3}}+3Ag$<br\/>${{n}_{Al}}=\\dfrac{5,4}{27}=0,2mol$<br\/> $n_Ag = 3n_Al = 0,2.3 = 0,6 mol $<br\/>$m_Ag= 0,6.108 = 64,8 gam$ <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2061},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Kh\u00f4ng \u0111\u01b0\u1ee3c d\u00f9ng ch\u1eadu nh\u00f4m \u0111\u1ec3 ch\u1ee9a n\u01b0\u1edbc v\u00f4i trong do","select":["A. nh\u00f4m t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi dung d\u1ecbch axit.","B. nh\u00f4m t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi dung d\u1ecbch baz\u01a1.","C. nh\u00f4m \u0111\u1ea9y \u0111\u01b0\u1ee3c kim lo\u1ea1i y\u1ebfu h\u01a1n n\u00f3 ra kh\u1ecfi dung d\u1ecbch mu\u1ed1i. ","D. nh\u00f4m l\u00e0 kim lo\u1ea1i ho\u1ea1t \u0111\u1ed9ng h\u00f3a h\u1ecdc m\u1ea1nh"],"hint":"","explain":"<span class='basic_left'>N\u01b0\u1edbc v\u00f4i trong ch\u00ednh l\u00e0 dung d\u1ecbch $Ca(OH)_2$. Al t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi dung d\u1ecbch baz\u01a1 $Ca(OH)_2$ n\u00ean n\u1ebfu \u0111\u1ef1ng dung d\u1ecbch n\u01b0\u1edbc v\u00f4i trong b\u1eb1ng ch\u1eadu Al b\u1ecb m\u00f2n d\u1ea7n v\u00e0 h\u1ecfng. $2Al+Ca{{(OH)}_{2}}+2{{H}_{2}}O\\to Ca{{(Al{{O}_{2}})}_{2}}+3{{H}_{2}}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":2062},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" \u0110\u1ed1t nh\u00f4m trong b\u00ecnh kh\u00ed Clo . Sau ph\u1ea3n \u1ee9ng th\u1ea5y kh\u1ed1i l\u01b0\u1ee3ng ch\u1ea5t r\u1eafn trong b\u00ecnh t\u0103ng 7,1g . Kh\u1ed1i l\u01b0\u1ee3ng nh\u00f4m \u0111\u00e3 tham gia ph\u1ea3n \u1ee9ng l\u00e0 :","select":["A. 2,7g. ","B. 1,8g.","C. 4,1g ","D. 5,4g."],"hint":"","explain":"<span class='basic_left'>Kh\u1ed1i l\u01b0\u1ee3ng ch\u1ea5t r\u1eafn t\u0103ng ch\u00ednh l\u00e0 kh\u1ed1i l\u01b0\u1ee3ng $Cl_2$ ph\u1ea3n \u1ee9ng$ \\begin{aligned}& {{n}_{C{{l}_{2}}}}=\\dfrac{7,1}{71}=0,1mol \\\\ & 2Al+3C{{l}_{2}}\\to 2AlC{{l}_{3}} \\\\ & {{n}_{Al}}=\\dfrac{2}{3}{{n}_{C{{l}_{2}}}}=\\dfrac{2}{3}.0,1=\\dfrac{1}{15}mol \\\\ & {{m}_{Al}}=\\dfrac{1}{15}.27=1,8gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2063},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" C\u00f3 dung d\u1ecbch mu\u1ed1i $AlCl_3$ l\u1eabn t\u1ea1p ch\u1ea5t $CuCl_2$ , d\u00f9ng ch\u1ea5t n\u00e0o sau \u0111\u00e2y \u0111\u1ec3 l\u00e0m s\u1ea1ch mu\u1ed1i nh\u00f4m?","select":["A. $AgNO_3$ ","B. $HCl$","C. $Mg$","D. $Al$"],"hint":"","explain":"<span class='basic_left'>\u0110\u1ec3 l\u00e0m s\u1ea1ch mu\u1ed1i Al l\u1eabn $CuCl_2$ ph\u1ea3i cho h\u1ed7n h\u1ee3p ph\u1ea3n \u1ee9ng v\u1edbi ch\u1ea5t ch\u1ec9 ph\u1ea3n \u1ee9ng v\u1edbi $CuCl_2$, kh\u00f4ng ph\u1ea3n \u1ee9ng v\u1edbi $AlCl_3$. S\u1ea3n ph\u1ea9m sinh ra ph\u1ea3i d\u1ec5 t\u00e1ch kh\u1ecfi dung d\u1ecbch ho\u1eb7c sinh ra lu\u00f4n mu\u1ed1i $AlCl_3$. <br\/> Ch\u1ec9 c\u00f3 Al th\u1ecfa m\u00e3n: $2Al+3CuC{{l}_{2}}\\to 2AlC{{l}_{3}}+3Cu$.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2064},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Cho 1,35g Al v\u00e0o dung d\u1ecbch ch\u1ee9a 9,8g $H_2SO_4$ lo\u00e3ng. L\u01b0\u1ee3ng kh\u00ed tho\u00e1t ra \u1edf \u0111ktc l\u00e0","select":["A. 1,68 l\u00edt ","B. 2,24 l\u00edt","C. 3,36 l\u00edt ","D. 4,48 l\u00edt"],"hint":"","explain":"<span class='basic_left'>${{n}_{Al}}=\\dfrac{1,35}{27}=0,05mol$ <br\/>$\\begin{aligned}& {{n}_{{{H}_{2}}S{{O}_{4}}}}=\\dfrac{9,8}{98}=0,1mol \\\\ & 2Al+3{{H}_{2}}S{{O}_{4}}\\to A{{l}_{2}}{{(S{{O}_{4}})}_{3}}+3{{H}_{2}} \\\\ \\end{aligned} $<br\/> X\u00e9t t\u1ef7 l\u1ec7: $\\dfrac{n{}_{Al}}{2}<\\dfrac{{{n}_{{{H}_{2}}S{{O}_{4}}}}}{3}$ => Al ph\u1ea3n \u1ee9ng h\u1ebft, $H_2SO_4$ d\u01b0, t\u00ednh theo $Al$ <br\/>$\\begin{aligned}& {{n}_{{{H}_{2}}}}=\\dfrac{3}{2}{{n}_{Al}}=\\dfrac{3}{2}.0,05=0,075 mol\\\\ & {{V}_{{{H}_{2}}}}=0,075.22,4=1,68l \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2065},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Cho tan ho\u00e0n to\u00e0n 10g h\u1ed7n h\u1ee3p g\u1ed3m Al v\u00e0 $Al_2O_3$ trong dung d\u1ecbch NaOH d\u01b0, thu \u0111\u01b0\u1ee3c 6,72 l\u00edt $H_2$ (\u0111ktc). Ph\u1ea7n tr\u0103m kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a Al trong h\u1ed7n h\u1ee3p l\u00e0:","select":["A. 48% ","B. 50%","C. 52% ","D. 54%"],"hint":"","explain":"<span class='basic_left'>:${{n}_{{{H}_{2}}}}=\\dfrac{6,72}{22,4}=0,3mol$ $\\begin{aligned}& 2Al+2NaOH+2{{H}_{2}}O\\to 2NaAl{{O}_{2}}+3{{H}_{2}} \\\\ & A{{l}_{2}}{{O}_{3}}+2NaOH+\\to 2NaAl{{O}_{2}}+{{H}_{2}}O \\\\ & {{n}_{Al}}=\\dfrac{2}{3}{{n}_{{{H}_{2}}}}=\\dfrac{2}{3}.0,3=0,2mol \\\\ & {{m}_{Al}}=0,2.27=5,4gam \\\\ & \\%{{m}_{Al}}=\\dfrac{5,4}{10}.100\\%=54\\% \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2066},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Khi cho t\u1eeb t\u1eeb (\u0111\u1ebfn d\u01b0) dung d\u1ecbch ki\u1ec1m v\u00e0o dung d\u1ecbch nh\u00f4m clorua ta th\u1ea5y:","select":["A. \u0111\u1ea7u ti\u00ean xu\u1ea5t hi\u1ec7n k\u1ebft t\u1ee7a keo v\u00e0 kh\u00f4ng tan trong ki\u1ec1m d\u01b0 ","B. \u0111\u1ea7u ti\u00ean xu\u1ea5t hi\u1ec7n k\u1ebft t\u1ee7a keo tr\u1eafng, k\u1ebft t\u1ee7a tan d\u1ea7n v\u00e0 dung d\u1ecbch tr\u1edf l\u1ea1i trong su\u1ed1t.","C. kh\u00f4ng xu\u1ea5t hi\u1ec7n k\u1ebft t\u1ee7a v\u00e0 dung d\u1ecbch c\u00f3 m\u00e0u xanh ","D. kh\u00f4ng c\u00f3 hi\u1ec7n t\u01b0\u1ee3ng n\u00e0o x\u1ea3y ra."],"hint":"","explain":"<span class='basic_left'>Ban \u0111\u1ea7u NaOH ph\u1ea3n \u1ee9ng v\u1edbi $AlCl_3$ theo ph\u1ea3n \u1ee9ng <br\/> $3NaOH+AlC{{l}_{3}}\\to 3NaCl+Al{{(OH)}_{3}}\\downarrow <br\/> $Sau \u0111\u00f3 NaOH d\u01b0 th\u00ec NaOH ph\u1ea3n \u1ee9ng h\u00f2a tan k\u1ebft t\u1ee7a keo $Al(OH)_3$ theo ph\u1ea3n \u1ee9ng <br\/> $Al{{(OH)}_{3}}+NaOH\\to NaAl{{O}_{2}}+2{{H}_{2}}O$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":2067},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Ph\u01b0\u01a1ng tr\u00ecnh ph\u1ea3n \u1ee9ng : $4A{{l}_{R}}+3{{O}_{2(K)}}\\to 2A{{l}_{2}}{{O}_{3(R)}}$cho ph\u00e9p d\u1ef1 \u0111o\u00e1n: ","select":["A. 4 g Al ph\u1ea3n \u1ee9ng ho\u00e0n to\u00e0n v\u1edbi 3 g kh\u00ed $O_2$ cho 2g $Al_2O_3$ ","B. 108g Al ph\u1ea3n \u1ee9ng ho\u00e0n to\u00e0n v\u1edbi 3 l\u00edt kh\u00ed $O_2$ cho 2g $Al_2O_3$.","C. 2,7 g Al ph\u1ea3n \u1ee9ng ho\u00e0n to\u00e0n v\u1edbi 2,4 g kh\u00ed $O_2$ cho 5,1g $Al_2O_3$.","D. 4 mol Al ph\u1ea3n \u1ee9ng ho\u00e0n to\u00e0n v\u1edbi 3 l\u00edt kh\u00ed $O_2$ cho 2 mol $Al_2O_3$."],"hint":"","explain":"<span class='basic_left'>C\u00e1c h\u1ec7 s\u1ed1 \u0111\u1ee9ng tr\u01b0\u1edbc c\u00e1c ch\u1ea5t ph\u1ea3n \u1ee9ng cho bi\u1ebft t\u1ec9 l\u1ec7 v\u1ec1 s\u1ed1 mol c\u1ee7a c\u00e1c ch\u1ea5t trong ph\u01b0\u01a1ng tr\u00ecnh ph\u1ea3n \u1ee9ng. <br\/> C\u1ee9 2 mol Al ph\u1ea3n \u1ee9ng ho\u00e0n to\u00e0n v\u1edbi 3 mol $O_2$t\u1ea1o ra 2 mol $Al_2O_3$ <br\/> T\u1eeb \u0111\u00f3 ta c\u00f3 th\u1ec3 \u0111\u1ed5i ra kh\u1ed1i l\u01b0\u1ee3ng ho\u1eb7c th\u1ec3 t\u00edch t\u01b0\u01a1ng \u1ee9ng.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2068},{"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 lo\u1ea1i qu\u1eb7ng boxit ch\u1ee9a 60% $Al_2O_3$ . T\u1eeb 2,125 t\u1ea5n qu\u1eb7ng boxit \u0111\u00f3 s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c 0,54 t\u1ea5n Al. Hi\u1ec7u su\u1ea5t c\u1ee7a qu\u00e1 tr\u00ecnh l\u00e0 : ","select":["A. 70%","B. 80%","C. 85%","D. 90%"],"hint":"","explain":"<span class='basic_left'>${{m}_{A{{l}_{2}}{{O}_{3}}}}=2,125.60\\%=1,275$ (t\u1ea5n)<br\/> $\\begin{aligned}& A{{l}_{2}}{{O}_{3}}\\,\\,\\underrightarrow{\u0111pnc}\\,\\,4Al+3{{O}_{2}} \\\\ & 2,102\\,\\,\\text{t\u1ea5n}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,4,27\\,\\text{t\u1ea5n} \\\\ & \\text{1,275}\\,\\text{t\u1ea5n}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x\\,\\,\\text{t\u1ea5n} \\\\ \\end{aligned}$ <br\/>Suy ra:$x=\\dfrac{1,275.4,27}{2,102}=0,675$ (t\u1ea5n)<br\/> $H=\\dfrac{0,54}{0,675}.100\\%=80\\%$ .<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2069},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" H\u00f2a tan 11g h\u1ed7n h\u1ee3p g\u1ed3m Al, Fe v\u00e0o dung d\u1ecbch $H_2SO_4$ lo\u00e3ng, d\u01b0. Ph\u1ea3n \u1ee9ng k\u1ebft th\u00fac thu \u0111\u01b0\u1ee3c 8,96 l\u00edt kh\u00ed $H_2$ (\u0111ktc). Th\u00e0nh ph\u1ea7n % kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a Al, Fe trong h\u1ed7n h\u1ee3p l\u1ea7n l\u01b0\u1ee3t l\u00e0 ","select":["A. 70% v\u00e0 30% ","B. 49% v\u00e0 51%","C. 51% v\u00e0 49%","D. 30% v\u00e0 70%"],"hint":"","explain":"<span class='basic_left'>\u0110\u1eb7t s\u1ed1 mol c\u1ee7a $Al, Fe$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 $x, y$ <br\/> Kh\u1ed1i l\u01b0\u1ee3ng h\u1ed7n h\u1ee3p b\u1eb1ng 11 gam n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh :<br\/>$27x + 56y = 11 $ (1) <br\/> C\u00e1c ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc <br\/>$\\begin{aligned} & 2Al+3{{H}_{2}}S{{O}_{4}}\\to A{{l}_{2}}{{\\left( S{{O}_{4}} \\right)}_{3}}+3{{H}_{2}} \\\\ & x\\,mol\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,1,5x\\,mol \\\\ & Fe+{{H}_{2}}S{{O}_{4}}\\to FeS{{O}_{4}}+{{H}_{2}} \\\\ & y\\,mol\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,y\\,mol \\\\ \\end{aligned}$<br\/> ${{n}_{{{H}_{2}}}}=\\dfrac{8,96}{22,4}=0,4mol \\Rightarrow 1,5x + y = 0,4$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) ta c\u00f3 $x = 0,2 mol; y = 0,1 mol$ <br\/> $\\begin{aligned}& \\%{{m}_{Al}}=\\dfrac{0,2.27}{11}.100\\%=49\\% \\\\ & <br\/> \\%{{m}_{Fe}}=100\\%-49\\%=51\\% \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2070}],"lesson":{"save":0,"level":2}}

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Câu hỏi này theo dạng chọn đáp án đúng, sau khi đọc xong câu hỏi, bạn bấm vào một trong số các đáp án mà chương trình đưa ra bên dưới, sau đó bấm vào nút gửi để kiểm tra đáp án và sẵn sàng chuyển sang câu hỏi kế tiếp

Trả lời đúng trong khoảng thời gian quy định bạn sẽ được + số điểm như sau:
Trong khoảng 1 phút đầu tiên + 5 điểm
Trong khoảng 1 phút -> 2 phút + 4 điểm
Trong khoảng 2 phút -> 3 phút + 3 điểm
Trong khoảng 3 phút -> 4 phút + 2 điểm
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