{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" C\u00f4ng th\u1ee9c ph\u00e2n t\u1eed c\u1ee7a glucoz\u01a1 l\u00e0:","select":["A. $C_2H_5OH$","B. $CH_3COOH$ ","C. $C_6H_{12}O_6$ ","D. $C_{12}H_{22}O_{11}$"],"hint":"","explain":"<span class='basic_left'>C\u00f4ng th\u1ee9c ph\u00e2n t\u1eed c\u1ee7a glucoz\u01a1 l\u00e0: $C_6H_{12}O_6$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2591},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" \u1ee8ng d\u1ee5ng n\u00e0o KH\u00d4NG ph\u1ea3i l\u00e0 \u1ee9ng d\u1ee5ng c\u1ee7a glucoz\u01a1","select":["A. L\u00e0m th\u1ef1c ph\u1ea9m dinh d\u01b0\u1ee1ng v\u00e0 thu\u1ed1c t\u0103ng l\u1ef1c ","B. Tr\u00e1ng g\u01b0\u01a1ng, tr\u00e1ng ph\u00edch","C. Nguy\u00ean li\u1ec7u s\u1ea3n xu\u1ea5t ancol etylic","D. Nguy\u00ean li\u1ec7u s\u1ea3n xu\u1ea5t PVC "],"hint":"","explain":"<span class='basic_left'>\u1ee8ng d\u1ee5ng c\u1ee7a glucoz\u01a1 l\u00e0: <br\/> L\u00e0m th\u1ef1c ph\u1ea9m dinh d\u01b0\u1ee1ng v\u00e0 thu\u1ed1c t\u0103ng l\u1ef1c.<br\/> Tr\u00e1ng g\u01b0\u01a1ng, tr\u00e1ng ph\u00edch. <br\/> Nguy\u00ean li\u1ec7u s\u1ea3n xu\u1ea5t ancoletylic<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":2592},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Trong ph\u1ea3n \u1ee9ng tr\u00e1ng g\u01b0\u01a1ng t\u1ea1o th\u00e0nh b\u1ea1c v\u00e0:","select":["A. Axit gluconic","B. R\u01b0\u1ee3u etylic","C. Axit axetic","D. Kh\u00ed cacbonic"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{C}_{6}}{{H}_{12}}{{O}_{6}}+A{{g}_{2}}O\\xrightarrow{N{{H}_{3}}}{{C}_{6}}{{H}_{12}}{{O}_{7}}+2Ag \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(Axit\\,gluconic) \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2593},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Trong c\u00f4ng nghi\u1ec7p ch\u1ebf t\u1ea1o ru\u1ed9t ph\u00edch, ng\u01b0\u1eddi ta th\u01b0\u1eddng s\u1eed d\u1ee5ng ph\u1ea3n \u1ee9ng ho\u00e1 h\u1ecdc n\u00e0o sau \u0111\u00e2y","select":["A. Cho axetilen t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch $AgNO_3\/NH_3$.","B. Cho an\u0111ehit fomic t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch $AgNO_3\/NH_3$.","C. Cho axit fomic t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch $AgNO_3\/NH_3$.","D. Cho glucoz\u01a1 t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch $AgNO_3\/NH_3$."],"hint":"","explain":"<span class='basic_left'>Trong c\u00f4ng nghi\u1ec7p ch\u1ebf t\u1ea1o ru\u1ed9t ph\u00edch, ng\u01b0\u1eddi ta th\u01b0\u1eddng s\u1eed d\u1ee5ng ph\u1ea3n \u1ee9ng ho\u00e1 h\u1ecdc: <br\/>${{C}_{6}}{{H}_{12}}{{O}_{6}}+A{{g}_{2}}O\\xrightarrow{N{{H}_{3}}}{{C}_{6}}{{H}_{12}}{{O}_{7}}+2Ag$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":2594},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Glucoz\u01a1 l\u00e0 ch\u1ea5t r\u1eafn:","select":["A. M\u00e0u tr\u1eafng, v\u1ecb ng\u1ecdt, kh\u00f4ng tan trong n\u01b0\u1edbc.","B. Kh\u00f4ng m\u00e0u, kh\u00f4ng v\u1ecb, tan nhi\u1ec1u trong n\u01b0\u1edbc","C. Kh\u00f4ng m\u00e0u, v\u1ecb ng\u1ecdt, tan nhi\u1ec1u trong n\u01b0\u1edbc ","D. M\u00e0u tr\u1eafng, kh\u00f4ng v\u1ecb, kh\u00f4ng tan trong n\u01b0\u1edbc"],"hint":"","explain":"<span class='basic_left'>Glucoz\u01a1 l\u00e0 ch\u1ea5t r\u1eafn k\u1ebft tinh kh\u00f4ng m\u00e0u, v\u1ecb ng\u1ecdt, tan nhi\u1ec1u trong n\u01b0\u1edbc.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2595},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Nh\u1ecf v\u00e0i gi\u1ecdt dung d\u1ecbch b\u1ea1c nitrat v\u00e0o \u1ed1ng nghi\u1ec7m \u0111\u1ef1ng dung d\u1ecbch amoniac, th\u00eam v\u00e0i gi\u1ecdt dung d\u1ecbch glucoz\u01a1, \u0111\u1eb7t v\u00e0o c\u1ed1c n\u01b0\u1edbc n\u00f3ng hi\u1ec7n t\u01b0\u1ee3ng quan s\u00e1t \u0111\u01b0\u1ee3c l\u00e0:","select":["A. C\u00f3 ch\u1ea5t m\u00e0u v\u00e0ng xu\u1ea5t hi\u1ec7n tr\u00ean th\u00e0nh \u1ed1ng nghi\u1ec7m","B. Dung d\u1ecbch v\u1ea9n \u0111\u1ee5c, c\u00f3 kh\u00ed bay l\u00ean","C. C\u00f3 ch\u1ea5t m\u00e0u tr\u1eafng s\u00e1ng b\u00e1m tr\u00ean th\u00e0nh \u1ed1ng nghi\u1ec7m ","D. C\u00f3 ch\u1ea5t m\u00e0u \u0111en b\u00e1m tr\u00ean th\u00e0nh \u1ed1ng nghi\u1ec7m."],"hint":"","explain":"<span class='basic_left'>Nh\u1ecf v\u00e0i gi\u1ecdt dung d\u1ecbch b\u1ea1c nitrat v\u00e0o \u1ed1ng nghi\u1ec7m \u0111\u1ef1ng dung d\u1ecbch amoniac,th\u00eam v\u00e0i gi\u1ecdt dung d\u1ecbch glucoz\u01a1 \u0111\u1eb7t v\u00e0o c\u1ed1c n\u01b0\u1edbc n\u00f3ng sau m\u1ed9t th\u1eddi gian c\u00f3 ch\u1ea5t m\u00e0u tr\u1eafng s\u00e1ng b\u00e1m tr\u00ean th\u00e0nh \u1ed1ng nghi\u1ec7m: <br\/> ${{C}_{6}}{{H}_{12}}{{O}_{6}}+A{{g}_{2}}O\\xrightarrow{N{{H}_{3}}}{{C}_{6}}{{H}_{12}}{{O}_{7}}+2Ag$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2596},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Glucoz\u01a1 c\u00f3 nhi\u1ec1u nh\u1ea5t trong lo\u1ea1i qu\u1ea3 n\u00e0o sau \u0111\u00e2y:","select":["A. Qu\u1ea3 h\u1ed3ng xanh","B. Qu\u1ea3 nh\u00f3t ch\u00edn","C. Qu\u1ea3 nho ch\u00edn","D. Qu\u1ea3 d\u01b0a h\u1ea5u"],"hint":"","explain":"<span class='basic_left'>Glucoz\u01a1 c\u00f3 nhi\u1ec1u nh\u1ea5t trong qu\u1ea3 nho ch\u00edn<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2597},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Trong m\u00e1u ng\u01b0\u1eddi lu\u00f4n ch\u1ee9a m\u1ed9t t\u1ec9 l\u1ec7 glucoz\u01a1 kh\u00f4ng \u0111\u1ed5i l\u00e0 :","select":["A. 0,01% ","B. 0,1% ","C. 1% ","D. 10%"],"hint":"","explain":"<span class='basic_left'>Trong m\u00e1u ng\u01b0\u1eddi lu\u00f4n ch\u1ee9a m\u1ed9t t\u1ec9 l\u1ec7 glucoz\u01a1 kh\u00f4ng \u0111\u1ed5i l\u00e0 kho\u1ea3ng 0,1% <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2598},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" T\u1ed5ng h\u1ec7 s\u1ed1 c\u00e2n b\u1eb1ng c\u1ee7a ph\u1ea3n \u1ee9ng l\u00ean men r\u01b0\u1ee3u t\u1eeb glucoz\u01a1 l\u00e0:","select":["A. 3","B. 4","C. 5 ","D. 6"],"hint":"","explain":"<span class='basic_left'>${{C}_{6}}{{H}_{12}}{{O}_{6}}\\xrightarrow[{{30}^{0}}-{{32}^{0}}]{men}2{{C}_{2}}{{H}_{5}}OH+2C{{O}_{2}}$. <br\/> T\u1ed5ng h\u1ec7 s\u1ed1 c\u00e2n b\u1eb1ng l\u00e0 1 + 2 + 2 = 5 <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2599},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" 1 mol glucoz\u01a1 ph\u1ea3n \u1ee9ng tr\u00e1ng g\u01b0\u01a1ng t\u1ea1o t\u1ed1i \u0111a bao nhi\u00eau mol b\u1ea1c","select":["A. 1 ","B. 2","C. 3","D. 4"],"hint":"","explain":"<span class='basic_left'>${{C}_{6}}{{H}_{12}}{{O}_{6}}+A{{g}_{2}}O\\xrightarrow{N{{H}_{3}}}{{C}_{6}}{{H}_{12}}{{O}_{7}}+2Ag$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2600},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Trong \u0111i\u1ec1u ki\u1ec7n th\u00edch h\u1ee3p glucoz\u01a1 l\u00ean men t\u1ea1o th\u00e0nh kh\u00ed $CO_2$ v\u00e0","select":["A. $C_2H_5OH$ ","B. $CH_3COOH$. ","C. $HCOOH$. ","D. $CH_3CHO$."],"hint":"","explain":"<span class='basic_left'>${{C}_{6}}{{H}_{12}}{{O}_{6}}\\xrightarrow[{{30}^{0}}-{{32}^{0}}]{men}2{{C}_{2}}{{H}_{5}}OH+2C{{O}_{2}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2601},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" \u0110un n\u00f3ng 50g dung d\u1ecbch glucoz\u01a1 v\u1edbi l\u01b0\u1ee3ng dung d\u1ecbch $AgNO_3\/NH_3$ d\u01b0 thu \u0111\u01b0\u1ee3c 4,32g b\u1ea1c. N\u1ed3ng \u0111\u1ed9 % c\u1ee7a dung d\u1ecbch glucoz\u01a1 l\u00e0: ","select":["A. 13,4% ","B. 7,2% ","C. 12,4%","D. 14,4%"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{C}_{6}}{{H}_{12}}{{O}_{6}}+A{{g}_{2}}O\\xrightarrow{N{{H}_{3}}}{{C}_{2}}{{H}_{12}}{{O}_{7}}+2Ag \\\\ & {{n}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=\\dfrac{1}{2}{{n}_{Ag}}=\\dfrac{1}{2}0,04=0,02\\,mol \\\\ & \\Rightarrow {{m}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=0,02.180=3,6\\,gam \\\\ & C{{\\%}_{\\text{dd}\\,\\,{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=\\dfrac{3,6}{50}.100\\%=7,2\\% \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2602},{"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 c\u00e1c dung d\u1ecbch sau: r\u01b0\u1ee3u etylic, glucoz\u01a1 v\u00e0 axit axetic. Ta c\u00f3 th\u1ec3 d\u00f9ng ","select":["A. Gi\u1ea5y qu\u1ef3 t\u00edm v\u00e0 dung d\u1ecbch $AgNO_3\/NH_3$.","B. Gi\u1ea5y qu\u1ef3 t\u00edm v\u00e0 $Na$.","C. $Na$ v\u00e0 dung d\u1ecbch $AgNO_3\/NH_3$. ","D. $Na$ v\u00e0 dung d\u1ecbch $HCl$."],"hint":"","explain":"<span class='basic_left'>Nh\u00fang qu\u1ef3 t\u00edm v\u00e0o 3 dung d\u1ecbch, dung d\u1ecbch n\u00e0o l\u00e0m qu\u1ef3 chuy\u1ec3n \u0111\u1ecf nh\u1ea1t l\u00e0 dung d\u1ecbch axit axetic. 2 dung d\u1ecbch c\u00f2n l\u1ea1i qu\u1ef3 kh\u00f4ng chuy\u1ec3n m\u00e0u. Cho 2 dung d\u1ecbch c\u00f2n l\u1ea1i ph\u1ea3n \u1ee9ng v\u1edbi dung d\u1ecbch AgNO3\/NH3 dung d\u1ecbch n\u00e0o t\u1ea1o th\u00e0nh ch\u1ea5t m\u00e0u tr\u1eafng s\u00e1ng do t\u1ea1o b\u1ea1c l\u00e0 dung d\u1ecbch glucoz\u01a1, r\u01b0\u1ee3u etylic kh\u00f4ng c\u00f3 hi\u1ec7n t\u01b0\u1ee3ng g\u00ec. <br\/> ${{C}_{6}}{{H}_{12}}{{O}_{6}}+A{{g}_{2}}O\\xrightarrow{N{{H}_{3}}}{{C}_{6}}{{H}_{12}}{{O}_{7}}+2Ag$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":1}]}],"id_ques":2603},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho 25,2 gam glucoz\u01a1 l\u00ean men r\u01b0\u1ee3u. Kh\u1ed1i l\u01b0\u1ee3ng r\u01b0\u1ee3u v\u00e0 th\u1ec3 t\u00edch kh\u00ed $CO_2$ (\u1edf \u0111ktc) thu \u0111\u01b0\u1ee3c l\u1ea7n l\u01b0\u1ee3t l\u00e0 (hi\u1ec7u su\u1ea5t 100%) ","select":["A. 12,88 gam v\u00e0 12,32 l\u00edt.","B. 128,8 gam v\u00e0 62,72 l\u00edt.","C. 12,88 gam v\u00e0 6,272 l\u00edt.","D. 12,88 gam v\u00e0 62,72 l\u00edt."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{C}_{6}}{{H}_{12}}{{O}_{6}}\\xrightarrow[{{30}^{0}}-{{32}^{0}}]{men}2{{C}_{2}}{{H}_{5}}OH+2C{{O}_{2}} \\\\ & {{n}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=0,14\\,mol \\\\ & {{n}_{{{C}_{2}}{{H}_{5}}OH}}={{n}_{C{{O}_{2}}}}=2{{n}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=0,28\\,mol \\\\ & \\Rightarrow {{m}_{{{C}_{2}}{{H}_{5}}OH}}=12,88\\,gam;\\,\\,{{V}_{C{{O}_{2}}}}=6,272\\,L \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2604},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" \u0110un 100 ml dung d\u1ecbch glucoz\u01a1 v\u1edbi m\u1ed9t l\u01b0\u1ee3ng d\u01b0 dung d\u1ecbch $AgNO_3\/NH_3$ thu \u0111\u01b0\u1ee3c 5,4 gam b\u1ea1c. N\u1ed3ng \u0111\u1ed9 mol c\u1ee7a dung d\u1ecbch glucoz\u01a1 l\u00e0 ","select":["A. 0,025 M.","B. 0,05 M.","C. 0,25 M.","D. 0,725 M."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{C}_{6}}{{H}_{12}}{{O}_{6}}+A{{g}_{2}}O\\xrightarrow{N{{H}_{3}}}{{C}_{6}}{{H}_{12}}{{O}_{7}}+2Ag \\\\ & {{n}_{Ag}}=0,05\\,\\,mol \\\\ & {{n}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=\\dfrac{1}{2}{{n}_{Ag}}=0,025\\,mol \\\\ & \\Rightarrow {{C}_{M\\,\\,{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=\\dfrac{0,025}{0,1}=0,25M \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2605},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" \u0110un n\u00f3ng 25 gam dung d\u1ecbch glucoz\u01a1 v\u1edbi l\u01b0\u1ee3ng d\u01b0 dung d\u1ecbch $AgNO_3\/NH_3$ thu \u0111\u01b0\u1ee3c 2,16 gam b\u1ea1c. N\u1ed3ng \u0111\u1ed9 % c\u1ee7a dung d\u1ecbch glucoz\u01a1 l\u00e0 ","select":["A. 7,2 %.","B. 11,4 %.","C. 14,4 %. ","D. 17,2 %."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{C}_{6}}{{H}_{12}}{{O}_{6}}+A{{g}_{2}}O\\xrightarrow{N{{H}_{3}}}{{C}_{6}}{{H}_{12}}{{O}_{7}}+2Ag \\\\ & {{n}_{Ag}}=0,02\\,\\,mol \\\\ & {{n}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=\\dfrac{1}{2}{{n}_{Ag}}=0,01\\,mol\\Rightarrow {{m}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=1,8\\,gam \\\\ & \\Rightarrow C{{\\%}_{\\text{dd}\\,\\,{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=\\dfrac{1,8}{25}.100\\%=7,2\\% \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2606},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" T\u00ednh kh\u1ed1i l\u01b0\u1ee3ng glucoz\u01a1 c\u1ea7n l\u1ea5y \u0111\u1ec3 pha \u0111\u01b0\u1ee3c 200 ml dung d\u1ecbch glucozo 5% c\u00f3 D = 1$g\/cm^3$","select":["A. 12,5 g","B. 10,0 g","C. 15,6 g ","D. 17,6 g"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{m}_{\\text{dd}}}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}=200.1=200\\,gam \\\\ & \\Rightarrow {{m}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=\\dfrac{200.5\\%}{100\\%}=10\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2607},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" L\u00ean men r\u01b0\u1ee3u t\u1eeb glucoz\u01a1 sinh ra 4,48 l\u00edt kh\u00ed cacbonic \u1edf \u0111ktc. L\u01b0\u1ee3ng natri c\u1ea7n l\u1ea5y \u0111\u1ec3 t\u00e1c d\u1ee5ng h\u1ebft v\u1edbi l\u01b0\u1ee3ng r\u01b0\u1ee3u sinh ra l\u00e0:","select":["A. 46 gam.","B. 2,3 gam","C. 6,4 gam. ","D. 4,6 gam."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{C}_{6}}{{H}_{12}}{{O}_{6}}\\xrightarrow[{{30}^{0}}-{{32}^{0}}]{men}2{{C}_{2}}{{H}_{5}}OH+2C{{O}_{2}} \\\\ & {{n}_{{{C}_{2}}{{H}_{5}}OH}}={{n}_{C{{O}_{2}}}}=0,2\\,mol \\\\ & 2{{C}_{2}}{{H}_{5}}OH+2Na\\to 2{{C}_{2}}{{H}_{5}}ONa+{{H}_{2}}\\uparrow \\\\ & {{n}_{Na}}={{n}_{{{C}_{2}}{{H}_{5}}OH}}=0,2\\,mol\\Rightarrow {{m}_{Na}}=4,6\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2608},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Cho m gam glucoz\u01a1 tham gia ph\u1ea3n \u1ee9ng tr\u00e1ng b\u1ea1c trong dung d\u1ecbch b\u1ea1c nitrat trong m\u00f4i tr\u01b0\u1eddng amoniac. Bi\u1ebft l\u01b0\u1ee3ng b\u1ea1c t\u1ea1o th\u00e0nh ph\u1ea3n \u1ee9ng v\u1eeba \u0111\u1ee7 v\u1edbi axit sunfuric \u0111\u1eb7c t\u1ea1o ra 2,24 l\u00edt kh\u00ed $SO_2$ duy nh\u1ea5t. Gi\u00e1 tr\u1ecb c\u1ee7a m l\u00e0","select":["A. 9 gam","B. 12 gam","C. 18 gam. ","D. 24 gam"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{C}_{6}}{{H}_{12}}{{O}_{6}}+A{{g}_{2}}O\\xrightarrow{N{{H}_{3}}}{{C}_{6}}{{H}_{12}}{{O}_{7}}+2Ag \\\\ & 2Ag+2{{H}_{2}}S{{O}_{4\\,\\text{\u0111}}}\\to A{{g}_{2}}S{{O}_{4}}+S{{O}_{2}}\\uparrow +2{{H}_{2}}O \\\\ & {{n}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=\\dfrac{1}{2}{{n}_{Ag}}=\\dfrac{1}{2}.2{{n}_{S{{O}_{2}}}}={{n}_{S{{O}_{2}}}}=0,1\\,mol \\\\ & \\Rightarrow {{m}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=18\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2609},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho 2,5 kg glucoz\u01a1 ch\u1ee9a 20% t\u1ea1p ch\u1ea5t l\u00ean men th\u00e0nh ancol etylic. T\u00ednh th\u1ec3 t\u00edch ancol etylic $40^0$ thu \u0111\u01b0\u1ee3c bi\u1ebft ancol etylic c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang l\u00e0 0,8 g\/ml v\u00e0 qu\u00e1 tr\u00ecnh ch\u1ebf bi\u1ebfn anol etylic hao h\u1ee5t 10%.","select":["A. 3194,4 ml","B. 27850 ml","C. 2875 ml ","D. 23000 ml"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{m}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=\\dfrac{2,5.80\\%}{100\\%}=2\\,kg=2000\\,gam \\\\ & {{n}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=\\dfrac{100}{9}\\,mol \\\\ & {{C}_{6}}{{H}_{12}}{{O}_{6}}\\xrightarrow[{{30}^{0}}-{{32}^{0}}]{men}2{{C}_{2}}{{H}_{5}}OH+2C{{O}_{2}} \\\\ & {{n}_{{{C}_{2}}{{H}_{5}}OH}}=2{{n}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=\\dfrac{200}{9}mol\\Rightarrow {{m}_{{{C}_{2}}{{H}_{5}}OH}}\\approx 1022\\,,22gam \\\\ & {{V}_{{{C}_{2}}{{H}_{5}}OH}}=\\dfrac{1022}{0,8}=1277,8\\,ml \\\\ & \\Rightarrow {{V}_{\\text{dd}\\,{{C}_{2}}{{H}_{5}}OH}}=\\dfrac{1277,5.100}{40}\\approx 3194\\,ml \\\\ \\end{aligned}$ <br\/> V\u00ec qu\u00e1 tr\u00ecnh ch\u1ebf bi\u1ebfn hao h\u1ee5t 10% => H = 90% <br\/> $\\Rightarrow {{V}_{\\text{dd}\\,\\,{{C}_{2}}{{H}_{5}}OH\\,\\,(TT)}}=\\dfrac{3194.90\\%}{100\\%}=2875\\,ml$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2610}],"lesson":{"save":0,"level":2}}