{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" C\u00f4ng th\u1ee9c c\u1ea5u t\u1ea1o c\u1ee7a r\u01b0\u1ee3u etylic l\u00e0 ","select":["A. $CH_2 \u2013 CH_3 \u2013 OH$.","B. $CH_3 \u2013 O \u2013 CH_3$.","C. $CH_2 \u2013 CH_2 \u2013 OH_2$.","D. $CH_3 \u2013 CH_2 \u2013 OH$"],"hint":"","explain":"<span class='basic_left'>R\u01b0\u1ee3u etylic: $CH_3 \u2013 CH_2 \u2013 OH$.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2491},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Nhi\u1ec7t \u0111\u1ed9 s\u00f4i c\u1ee7a r\u01b0\u1ee3u etylic l\u00e0 ","select":["A. $78,3 ^0C$.","B. $87,3^0C$.","C. $73,8^0C$","D. $83,7^0C$."],"hint":"","explain":"<span class='basic_left'>R\u01b0\u1ee3u etylic s\u00f4i \u1edf $78,3 ^0C$. <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2492},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"R\u01b0\u1ee3u etylic l\u00e0 ","select":["A. ch\u1ea5t l\u1ecfng kh\u00f4ng m\u00e0u, nh\u1eb9 h\u01a1n n\u01b0\u1edbc, tan v\u00f4 h\u1ea1n trong n\u01b0\u1edbc, h\u00f2a tan \u0111\u01b0\u1ee3c nhi\u1ec1u ch\u1ea5t nh\u01b0 iot, benzen,\u2026 ","B. ch\u1ea5t l\u1ecfng m\u00e0u h\u1ed3ng , nh\u1eb9 h\u01a1n n\u01b0\u1edbc, tan v\u00f4 h\u1ea1n trong n\u01b0\u1edbc, h\u00f2a tan \u0111\u01b0\u1ee3c nhi\u1ec1u ch\u1ea5t nh\u01b0: iot, benzen,\u2026","C. ch\u1ea5t l\u1ecfng kh\u00f4ng m\u00e0u, kh\u00f4ng tan trong n\u01b0\u1edbc, h\u00f2a tan \u0111\u01b0\u1ee3c nhi\u1ec1u ch\u1ea5t nh\u01b0: iot, benzen,\u2026 ","D. ch\u1ea5t l\u1ecfng kh\u00f4ng m\u00e0u, n\u1eb7ng h\u01a1n n\u01b0\u1edbc, tan v\u00f4 h\u1ea1n trong n\u01b0\u1edbc, h\u00f2a tan \u0111\u01b0\u1ee3c nhi\u1ec1u ch\u1ea5t nh\u01b0: iot, benzen,\u2026"],"hint":"","explain":"<span class='basic_left'>R\u01b0\u1ee3u etylic l\u00e0 ch\u1ea5t l\u1ecfng kh\u00f4ng m\u00e0u, nh\u1eb9 h\u01a1n n\u01b0\u1edbc, tan v\u00f4 h\u1ea1n trong n\u01b0\u1edbc, h\u00f2a tan \u0111\u01b0\u1ee3c nhi\u1ec1u ch\u1ea5t nh\u01b0 iot, benzen,\u2026 <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":1}]}],"id_ques":2493},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" \u0110\u1ed9 r\u01b0\u1ee3u l\u00e0:","select":["A. s\u1ed1 ml r\u01b0\u1ee3u etylic c\u00f3 trong 100 ml h\u1ed7n h\u1ee3p r\u01b0\u1ee3u v\u1edbi n\u01b0\u1edbc.","B. s\u1ed1 ml n\u01b0\u1edbc c\u00f3 trong 100 ml h\u1ed7n h\u1ee3p r\u01b0\u1ee3u v\u1edbi n\u01b0\u1edbc.","C. s\u1ed1 gam r\u01b0\u1ee3u etylic c\u00f3 trong 100 ml h\u1ed7n h\u1ee3p r\u01b0\u1ee3u v\u1edbi n\u01b0\u1edbc.","D. s\u1ed1 gam n\u01b0\u1edbc c\u00f3 trong 100 gam h\u1ed7n h\u1ee3p r\u01b0\u1ee3u v\u1edbi n\u01b0\u1edbc."],"hint":"","explain":"<span class='basic_left'>\u0110\u1ed9 r\u01b0\u1ee3u l\u00e0 s\u1ed1 ml r\u01b0\u1ee3u nguy\u00ean ch\u1ea5t c\u00f3 trong 100 ml dung d\u1ecbch r\u01b0\u1ee3u.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":1}]}],"id_ques":2494},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Nh\u00f3m \u2013OH trong ph\u00e2n t\u1eed r\u01b0\u1ee3u etylic c\u00f3 t\u00ednh ch\u1ea5t h\u00f3a h\u1ecdc \u0111\u1eb7c tr\u01b0ng l\u00e0 ","select":["A. t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi kim lo\u1ea1i gi\u1ea3i ph\u00f3ng kh\u00ed hi\u0111ro.","B. t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi natri, kali gi\u1ea3i ph\u00f3ng kh\u00ed hi\u0111ro.","C. t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi magie, natri gi\u1ea3i ph\u00f3ng kh\u00ed hi\u0111ro.","D. t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi kali, k\u1ebdm gi\u1ea3i ph\u00f3ng kh\u00ed hi\u0111ro."],"hint":"","explain":"<span class='basic_left'>Nh\u00f3m \u2013OH trong ph\u00e2n t\u1eed r\u01b0\u1ee3u etylic c\u00f3 t\u00ednh ch\u1ea5t h\u00f3a h\u1ecdc \u0111\u1eb7c tr\u01b0ng l\u00e0 t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi c\u00e1c kim lo\u1ea1i ki\u1ec1m, ki\u1ec1m th\u1ed5 gi\u1ea3i ph\u00f3ng kh\u00ed hi\u0111ro<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":2495},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"R\u01b0\u1ee3u etylic ch\u00e1y trong kh\u00f4ng kh\u00ed, hi\u1ec7n t\u01b0\u1ee3ng quan s\u00e1t \u0111\u01b0\u1ee3c l\u00e0 ","select":["A. ng\u1ecdn l\u1eeda m\u00e0u \u0111\u1ecf, t\u1ecfa nhi\u1ec1u nhi\u1ec7t. ","B. ng\u1ecdn l\u1eeda m\u00e0u v\u00e0ng, t\u1ecfa nhi\u1ec1u nhi\u1ec7t.","C. ng\u1ecdn l\u1eeda m\u00e0u xanh, t\u1ecfa nhi\u1ec1u nhi\u1ec7t. ","D. ng\u1ecdn l\u1eeda m\u00e0u xanh, kh\u00f4ng t\u1ecfa nhi\u1ec7t."],"hint":"","explain":"<span class='basic_left'>R\u01b0\u1ee3u etylic ch\u00e1y trong kh\u00f4ng kh\u00ed v\u1edbi ng\u1ecdn l\u1eeda m\u00e0u xanh, t\u1ecfa nhi\u1ec1u nhi\u1ec7t<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2496},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" T\u00ean g\u1ecdi kh\u00e1c c\u1ee7a r\u01b0\u1ee3u etylic l\u00e0","select":["A. Etylat","B. Axetic","C. Etanol","D. Etanal"],"hint":"","explain":"<span class='basic_left'>R\u01b0\u1ee3u etylic hay c\u00f2n g\u1ecdi l\u00e0 etanol<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2497},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" R\u01b0\u1ee3u etylic c\u00f3 th\u1ec3 \u0111\u01b0\u1ee3c \u0111i\u1ec1u ch\u1ebf t\u1eeb:","select":["A. \u0110\u01b0\u1eddng","B. Etilen","C. Tinh b\u1ed9t ","D. C\u1ea3 3 \u0111\u00e1p \u00e1n tr\u00ean \u0111\u1ec1u \u0111\u00fang"],"hint":"","explain":"<span class='basic_left'>R\u01b0\u1ee3u etylic c\u00f3 th\u1ec3 \u0111\u01b0\u1ee3c \u0111i\u1ec1u ch\u1ebf theo 2 c\u00e1ch: <br\/> $\\begin{aligned} & \\text{Tinh}\\,\\text{b\u1ed9t}\\,\\text{ho\u1eb7c }\\,\\text{\u0111\u01b0\u1eddng}\\,\\xrightarrow{\\text{l\u00ean}\\,\\text{men}}\\text{R\u01b0\u1ee3u}\\,\\text{etylic} \\\\ & {{C}_{2}}{{H}_{4\\,(k)}}+{{H}_{2}}{{O}_{(l)}}\\xrightarrow{Axit}{{C}_{2}}{{H}_{5}}O{{H}_{(l)}} \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2498},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Cho r\u01b0\u1ee3u etylic nguy\u00ean ch\u1ea5t t\u00e1c d\u1ee5ng v\u1edbi K. S\u1ed1 ph\u1ea3n \u1ee9ng x\u1ea3y ra l\u00e0:","select":["A. 0","B. 1","C. 2 ","D. 3"],"hint":"","explain":"<span class='basic_left'>$2{{C}_{2}}{{H}_{5}}OH+2K\\to 2{{C}_{2}}{{H}_{5}}OK+{{H}_{2}}\\uparrow $ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2499},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Cho m\u1ed9t m\u1eabu natri v\u00e0o \u1ed1ng nghi\u1ec7m \u0111\u1ef1ng r\u01b0\u1ee3u etylic. Hi\u1ec7n t\u01b0\u1ee3ng quan s\u00e1t \u0111\u01b0\u1ee3c l\u00e0 ","select":["A. c\u00f3 b\u1ecdt kh\u00ed m\u00e0u n\u00e2u tho\u00e1t ra.","B. m\u1eabu natri tan d\u1ea7n kh\u00f4ng c\u00f3 b\u1ecdt kh\u00ed tho\u00e1t ra.","C. m\u1eabu natri n\u1eb1m d\u01b0\u1edbi b\u1ec1 m\u1eb7t ch\u1ea5t l\u1ecfng v\u00e0 kh\u00f4ng tan.","D. c\u00f3 b\u1ecdt kh\u00ed kh\u00f4ng m\u00e0u tho\u00e1t ra v\u00e0 natri tan d\u1ea7n."],"hint":"","explain":"<span class='basic_left'>Cho m\u1ed9t m\u1eabu natri v\u00e0o \u1ed1ng nghi\u1ec7m \u0111\u1ef1ng r\u01b0\u1ee3u etylic th\u00ec r\u01b0\u1ee3u etylic ph\u1ea3n \u1ee9ng v\u1edbi Na t\u1ea1o th\u00e0nh kh\u00ed hi\u0111ro theo ph\u1ea3n \u1ee9ng: <br\/> $2{{C}_{2}}{{H}_{5}}OH+2Na\\to 2{{C}_{2}}{{H}_{5}}ONa+{{H}_{2}}\\uparrow $ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":2500},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Th\u1ec3 t\u00edch kh\u00ed oxi ( \u0111ktc) c\u1ea7n d\u00f9ng \u0111\u1ec3 \u0111\u1ed1t ch\u00e1y ho\u00e0n to\u00e0n 13,8 gam r\u01b0\u1ee3u etylic nguy\u00ean ch\u1ea5t l\u00e0 ","select":["A. 16,20 l\u00edt.","B. 18,20 l\u00edt.","C. 20,16 l\u00edt.","D. 22,16 l\u00edt."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{C}_{2}}{{H}_{6}}O+3{{O}_{2}}\\xrightarrow{{{t}^{0}}}2C{{O}_{2}}+3{{H}_{2}}O \\\\ & {{n}_{{{O}_{2}}}}=3{{n}_{{{C}_{2}}{{H}_{6}}O}}=3.\\dfrac{13,8}{46}=0,9\\,mol \\\\ & \\Rightarrow {{V}_{{{O}_{2}}}}=20,16\\,L \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2501},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Cho 23 gam r\u01b0\u1ee3u etylic nguy\u00ean ch\u1ea5t t\u00e1c d\u1ee5ng v\u1edbi natri d\u01b0. Th\u1ec3 t\u00edch kh\u00ed $H_2$ tho\u00e1t ra ( \u0111ktc) l\u00e0 ","select":["A. 2,8 l\u00edt","B. 5,6 l\u00edt","C. 8,4 l\u00edt.","D. 11,2 l\u00edt."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & 2{{C}_{2}}{{H}_{5}}OH+2Na\\to 2{{C}_{2}}{{H}_{5}}ONa+{{H}_{2}}\\uparrow \\\\ & {{n}_{{{H}_{2}}}}=\\dfrac{1}{2}{{n}_{{C_{2}}{{H}_{6}}O}}=\\dfrac{1}{2}.1=0,5\\,\\,mol \\\\ & {{V}_{{{H}_{2}}}}=11,2\\,L \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2502},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Trong $100$ ml r\u01b0\u1ee3u $45^0$ c\u00f3 ch\u1ee9a ","select":["A. 45 ml n\u01b0\u1edbc v\u00e0 55 ml r\u01b0\u1ee3u nguy\u00ean ch\u1ea5t. ","B. 45 ml r\u01b0\u1ee3u nguy\u00ean ch\u1ea5t v\u00e0 55 ml n\u01b0\u1edbc.","C. 45 gam r\u01b0\u1ee3u nguy\u00ean ch\u1ea5t v\u00e0 55 gam n\u01b0\u1edbc.","D. 45 gam n\u01b0\u1edbc v\u00e0 55 gam r\u01b0\u1ee3u nguy\u00ean ch\u1ea5t."],"hint":"","explain":"<span class='basic_left'>Trong $100$ ml r\u01b0\u1ee3u $45^0$ c\u00f3 ch\u1ee9a 45 ml r\u01b0\u1ee3u nguy\u00ean ch\u1ea5t v\u00e0 55 ml n\u01b0\u1edbc<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":2503},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Mu\u1ed1n \u0111i\u1ec1u ch\u1ebf $100$ ml r\u01b0\u1ee3u etylic $65^0$ ta d\u00f9ng ","select":["A. 100 ml n\u01b0\u1edbc h\u00f2a v\u1edbi c\u00f3 65 ml r\u01b0\u1ee3u nguy\u00ean ch\u1ea5t.","B. 100 ml r\u01b0\u1ee3u etylic nguy\u00ean ch\u1ea5t c\u00f3 65 ml n\u01b0\u1edbc.","C. 65 ml r\u01b0\u1ee3u etylic nguy\u00ean ch\u1ea5t h\u00f2a v\u1edbi 35 ml n\u01b0\u1edbc.","D. 35 ml r\u01b0\u1ee3u nguy\u00ean ch\u1ea5t v\u1edbi 65 ml n\u01b0\u1edbc."],"hint":"","explain":"<span class='basic_left'>Trong $100$ ml r\u01b0\u1ee3u etylic $65^0$ c\u00f3 ch\u1ee9a 65 ml r\u01b0\u1ee3u nguy\u00ean ch\u1ea5t v\u00e0 35 ml n\u01b0\u1edbc. N\u00ean ta s\u1ebd pha 65 ml r\u01b0\u1ee3u etylic nguy\u00ean ch\u1ea5t h\u00f2a v\u1edbi 35 ml n\u01b0\u1edbc.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2504},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"\u0110\u1ed1t ch\u00e1y ho\u00e0n to\u00e0n 57,5 ml r\u01b0\u1ee3u etylic. Th\u1ec3 t\u00edch kh\u00ed $CO_2$ ( \u0111ktc) thu \u0111\u01b0\u1ee3c l\u00e0 ( bi\u1ebft D = 0,8g\/ml) ","select":["A. 2,24 l\u00edt.","B. 22,4 l\u00edt.","C. 4,48 l\u00edt","D. 44,8 l\u00edt."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{C}_{2}}{{H}_{6}}O+3{{O}_{2}}\\xrightarrow{{{t}^{0}}}2C{{O}_{2}}+3{{H}_{2}}O \\\\ & {{m}_{{{C}_{2}}{{H}_{6}}O}}=V.D= 57,5 . 0,8 = 46\\,gam \\\\ & {{n}_{{{C}_{2}}{{H}_{6}}O}}=\\dfrac{46}{46}=1\\,mol \\\\ & {{n}_{C{{O}_{2}}}}= 2.{{n}_{{{C}_{2}}{{H}_{6}}O}} = 2 mol \\Rightarrow {{V}_{C{{O}_{2}}}}={{n}_{C{{O}_{2}}}}.22,4 = 44,8\\,L \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2505},{"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 30 ml r\u01b0\u1ee3u etylic nguy\u00ean ch\u1ea5t v\u00e0o 90 ml n\u01b0\u1edbc c\u1ea5t thu \u0111\u01b0\u1ee3c ","select":["A. r\u01b0\u1ee3u etylic c\u00f3 \u0111\u1ed9 r\u01b0\u1ee3u l\u00e0 $20^0$. ","B. r\u01b0\u1ee3u etylic c\u00f3 \u0111\u1ed9 r\u01b0\u1ee3u l\u00e0 $25^0$.","C. r\u01b0\u1ee3u etylic c\u00f3 \u0111\u1ed9 r\u01b0\u1ee3u l\u00e0 $30^0$. ","D. r\u01b0\u1ee3u etylic c\u00f3 \u0111\u1ed9 r\u01b0\u1ee3u l\u00e0 $35^0$."],"hint":"","explain":"<span class='basic_left'>\u0110\u1ed9 r\u01b0\u1ee3u \u0111\u01b0\u1ee3c t\u00ednh theo c\u00f4ng th\u1ee9c: <br\/> $\\begin{aligned} & \\dfrac{{{V}_{{{C}_{2}}{{H}_{6}}O}}}{{{V}_{{{C}_{2}}{{H}_{6}}O}}+{{V}_{{{H}_{2}}O}}}.100 \\\\ & =\\dfrac{30}{30+90}={{25}^{0}} \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2506},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"\u0110\u1ec3 ph\u00e2n bi\u1ec7t hai ch\u1ea5t l\u1ecfng kh\u00f4ng m\u00e0u l\u00e0 benzen v\u00e0 r\u01b0\u1ee3u etylic ta d\u00f9ng ","select":["A. s\u1eaft","B. \u0111\u1ed3ng","C. natri. ","D. k\u1ebdm. "],"hint":"","explain":"<span class='basic_left'>Cho Na v\u00e0o 2 ch\u1ea5t l\u1ecfng. Ch\u1ea5t l\u1ecfng n\u00e0o c\u00f3 b\u1ecdt kh\u00ed kh\u00f4ng m\u00e0u bay ra l\u00e0 r\u01b0\u1ee3u etylic. Ch\u1ea5t l\u1ecfng c\u00f2n l\u1ea1i kh\u00f4ng c\u00f3 hi\u1ec7n t\u01b0\u1ee3ng g\u00ec l\u00e0 benzen.<br\/> $2{{C}_{2}}{{H}_{5}}OH+2Na\\to 2{{C}_{2}}{{H}_{5}}ONa+{{H}_{2}}\\uparrow $<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2507},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Cho 11,2 l\u00edt kh\u00ed etilen ( \u0111ktc) t\u00e1c d\u1ee5ng v\u1edbi n\u01b0\u1edbc c\u00f3 axit sunfuric ($H_2SO_4$) l\u00e0m x\u00fac t\u00e1c, thu \u0111\u01b0\u1ee3c 9,2 gam r\u01b0\u1ee3u etylic. Hi\u1ec7u su\u1ea5t ph\u1ea3n \u1ee9ng l\u00e0 ","select":["A. 40%.","B. 45%.","C. 50%.","D. 55%."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{C}_{2}}{{H}_{4}}+{{H}_{2}}O\\xrightarrow{{{H}_{2}}S{{O}_{4}}}{{C}_{2}}{{H}_{6}}O \\\\ & {{n}_{{{C}_{2}}{{H}_{6}}O\\,(LT)}}={{n}_{{{C}_{2}}{{H}_{4}}}}=0,5\\,mol \\\\ & {{n}_{{{C}_{2}}{{H}_{6}}O\\,(TT)}}=\\dfrac{9,2}{46}=0,2\\,mol \\\\ & H=\\dfrac{{{n}_{{{C}_{2}}{{H}_{6}}O\\,(TT)}}}{{{n}_{{{C}_{2}}{{H}_{6}}O\\,(LT)}}}.100\\%=\\dfrac{0,2}{0,5}.100\\%=40\\% \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2508},{"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 m\u1ed9t m\u1eabu kali d\u01b0 v\u00e0o r\u01b0\u1ee3u etylic nguy\u00ean ch\u1ea5t thu \u0111\u01b0\u1ee3c 2,24 l\u00edt kh\u00ed $H_2$ ( \u0111ktc). Th\u1ec3 t\u00edch r\u01b0\u1ee3u etylic \u0111\u00e3 d\u00f9ng l\u00e0 (Bi\u1ebft kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a r\u01b0\u1ee3u etylic l\u00e0 D= 0,8g\/ml) ","select":["A. 11,0 ml","B. 11,5 ml","C. 12,0 ml ","D. 12,5 ml."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & 2{{C}_{2}}{{H}_{5}}OH+2Na\\to 2{{C}_{2}}{{H}_{5}}ONa+{{H}_{2}}\\uparrow \\\\ & {{n}_{{C_{2}}{{H}_{6}}O}}=2{{n}_{{{H}_{2}}}}=2.0,1=0,2\\,\\,mol \\\\ & {{m}_{{{C}_{2}}{{H}_{6}}O}}=0,2.46=9,2\\,gam \\\\ & {{V}_{{{C}_{2}}{{H}_{6}}O}}=\\dfrac{m}{D}=\\dfrac{9,2}{0,8}=11,5\\,ml \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2509},{"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 ch\u00e1y ho\u00e0n to\u00e0n 20 ml r\u01b0\u1ee3u etylic $a^0$, d\u1eabn s\u1ea3n ph\u1ea9m kh\u00ed thu \u0111\u01b0\u1ee3c qua dung d\u1ecbch n\u01b0\u1edbc v\u00f4i trong d\u01b0 thu \u0111\u01b0\u1ee3c 60 gam k\u1ebft t\u1ee7a ( bi\u1ebft D = 0,8g\/ml). Gi\u00e1 tr\u1ecb c\u1ee7a a l\u00e0","select":["A. 68,25.","B. 86,25.","C. 25,86. ","D. 25,68"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{C}_{2}}{{H}_{6}}O+3{{O}_{2}}\\xrightarrow{{{t}^{0}}}2C{{O}_{2}}+3{{H}_{2}}O \\\\ & C{{O}_{2}}+Ca{{(OH)}_{2}}\\to CaC{{O}_{3}}\\downarrow +{{H}_{2}}O \\\\ & {{n}_{{{C}_{2}}{{H}_{6}}O}}=\\dfrac{1}{2}{{n}_{C{{O}_{2}}}}=\\dfrac{1}{2}{{n}_{\\downarrow }}=\\dfrac{1}{2}.0,6=0,3mol \\\\ & {{m}_{{{C}_{2}}{{H}_{6}}O}}=13,8\\,gam\\Rightarrow {{V}_{{{C}_{2}}{{H}_{6}}O}}=\\dfrac{m}{D}=17,25\\,ml \\\\ \\end{aligned}$ <br\/> => \u0110\u1ed9 r\u01b0\u1ee3u l\u00e0:$\\dfrac{{{V}_{{{C}_{2}}{{H}_{6}}O}}}{{{V}_{\\text{dd}}}}.100=\\dfrac{17,25}{20}.100=86,{{25}^{0}}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2510}],"lesson":{"save":0,"level":2}}