{"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 n\u00e0o sau \u0111\u00e2y l\u00e0 c\u00f4ng th\u1ee9c c\u1ee7a ch\u1ea5t b\u00e9o","select":["A. $C_{17}H_{35}COOH$ ","B. $C_3H_5(OH)_3$","C. $(C_{17}H_{35}COO)_3C_3H_5$","D. $(CH_3COO)_3C_3H_5$"],"hint":"","explain":"<span class='basic_left'>Ch\u1ea5t b\u00e9o l\u00e0 este c\u1ee7a axit b\u00e9o v\u00e0 glixerol.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2571},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Ch\u1ea5t t\u1ecfa ra n\u0103ng l\u01b0\u1ee3ng nhi\u1ec1u nh\u1ea5t khi oxi h\u00f3a th\u1ee9c \u0103n l\u00e0:","select":["A. Ch\u1ea5t \u0111\u1ea1m ","B. Ch\u1ea5t b\u1ed9t ","C. Ch\u1ea5t b\u00e9o ","D. Ch\u1ea5t x\u01a1"],"hint":"","explain":"<span class='basic_left'>Ch\u1ea5t b\u00e9o \u0111\u01b0\u1ee3c t\u1ea1o th\u00e0nh khi oxi h\u00f3a th\u1ee9c \u0103n t\u1ecfa ra nhi\u1ec1u n\u0103ng l\u01b0\u1ee3ng<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2572},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Axit axetic KH\u00d4NG t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi:","select":["A. Mg ","B. CuO","C. NaOH ","D. Ag"],"hint":"","explain":"<span class='basic_left'>Ag l\u00e0 kim lo\u1ea1i y\u1ebfu kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi axit axetic.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2573},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Th\u1ee7y ph\u00e2n ho\u00e0n to\u00e0n etyl axetat trong m\u00f4i tr\u01b0\u1eddng ki\u1ec1m thu \u0111\u01b0\u1ee3c:","select":["A. Mu\u1ed1i axetat c\u1ee7a kim lo\u1ea1i ki\u1ec1m v\u00e0 ch\u1ea5t b\u00e9o","B. Mu\u1ed1i axetat c\u1ee7a kim lo\u1ea1i ki\u1ec1m v\u00e0 glixerol","C. Mu\u1ed1i axetat c\u1ee7a kim lo\u1ea1i ki\u1ec1m v\u00e0 r\u01b0\u1ee3u etylic","D. Mu\u1ed1i axetat c\u1ee7a kim lo\u1ea1i ki\u1ec1m v\u00e0 axit axetic."],"hint":"","explain":"<span class='basic_left'>V\u00ed d\u1ee5: $C{{H}_{3}}C\\text{OO}{{\\text{C}}_{2}}{{H}_{5}}+NaOH\\to C{{H}_{3}}\\text{COONa}\\,\\text{+}\\,{{\\text{C}}_{2}}{{H}_{5}}OH$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2574},{"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 r\u01b0\u1ee3u v\u00e0 r\u01b0\u1ee3u etylic v\u00e0 d\u1ea7u \u0103n tan trong r\u01b0\u1ee3u etylic ta d\u00f9ng:","select":["A. Qu\u1ef3 t\u00edm","B. Gi\u1ea5m \u0103n ","C. N\u01b0\u1edbc ","D. CaO "],"hint":"","explain":"<span class='basic_left'>H\u00f2a n\u01b0\u1edbc v\u00e0o 2 dung d\u1ecbch. R\u01b0\u1ee3u tan nhi\u1ec1u trong n\u01b0\u1edbc n\u00ean t\u1ea1o th\u00e0nh dung d\u1ecbch \u0111\u1ed3ng nh\u1ea5t. C\u00f2n h\u1ed7n h\u1ee3p d\u1ea7u \u0103n tan trong r\u01b0\u1ee3u etylic kh\u00f4ng tan kh\u00f4ng tan trong n\u01b0\u1edbc ph\u00e2n th\u00e0nh 2 l\u1edbp.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2575},{"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 ml dung d\u1ecbch axit axetic t\u00e1c d\u1ee5ng ho\u00e0n to\u00e0n v\u1edbi kim lo\u1ea1i magie. C\u00f4 c\u1ea1n dung d\u1ecbch sau ph\u1ea3n \u1ee9ng \u0111\u01b0\u1ee3c 0,71 g mu\u1ed1i khan. N\u1ed3ng \u0111\u1ed9 mol c\u1ee7a dung d\u1ecbch axit axetic l\u00e0:","select":["A. 0,2 M","B. 0,3 M","C. 0,4 M","D. 0,5 M"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & 2C{{H}_{3}}\\text{COOH}\\,\\text{+}\\,\\text{Mg}\\to {{(C{{H}_{3}}\\text{COO})}_{2}}Mg+{{H}_{2}}\\uparrow \\\\ & {{n}_{C{{H}_{3}}\\text{COOH}\\,}}=2{{n}_{{{(C{{H}_{3}}\\text{COO})}_{2}}Mg}}={{2.5.10}^{-3}}=0,01 mol \\\\ & {{C}_{M\\,\\,C{{H}_{3}}\\text{COOH}\\,}}=\\dfrac{2,{{5.10}^{-3}}}{0,025}=0,1M \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2576},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Bi\u1ebft kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a r\u01b0\u1ee3u l\u00e0 0,8 g\/ml. H\u1ecfi 225ml r\u01b0\u1ee3u nguy\u00ean ch\u1ea5t n\u1eb7ng bao nhi\u00eau gam?","select":["A. 150 g. ","B. 180g.","C. 120 g. ","D. 110 g."],"hint":"","explain":"<span class='basic_left'>${{m}_{\\text{dd}}}=V.D=225.0,8=180\\,gam$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2577},{"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 axit axetic v\u1edbi r\u01b0\u1ee3u etylic c\u00f3 axit sunfuric l\u00e0m x\u00fac t\u00e1c th\u00ec ng\u01b0\u1eddi ta thu \u0111\u01b0\u1ee3c m\u1ed9t ch\u1ea5t l\u1ecfng kh\u00f4ng m\u00e0u, m\u00f9i th\u01a1m, kh\u00f4ng tan trong n\u01b0\u1edbc v\u00e0 n\u1ed5i tr\u00ean m\u1eb7t n\u01b0\u1edbc. S\u1ea3n ph\u1ea9m \u0111\u00f3 l\u00e0","select":["A. \u0111imetyl ete","B. etyl axetat","C. r\u01b0\u1ee3u etylic","D. metan"],"hint":"","explain":"<span class='basic_left'> <br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/hoahoc/bai48/lv2/img\/Capture18.PNG' \/><\/center> <br\/> $CH_3COOC_2H_5$: Etyl axetat <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2578},{"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\u00e0i","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'>$\\begin{aligned} & {{C}_{2}}{{H}_{5}}OH+3{{O}_{2}}\\xrightarrow{{{t}^{0}}}2C{{O}_{2}}+3{{H}_{2}}O \\\\ & {{n}_{{{O}_{2}}}}=3{{n}_{{{C}_{2}}{{H}_{5}}OH}}=3.0,3=0,9\\,\\,mol \\\\ & \\Rightarrow {{V}_{{{O}_{2}}}}=20,16\\,L \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2579},{"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 t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi natri v\u00ec","select":["A. trong ph\u00e2n t\u1eed c\u00f3 nguy\u00ean t\u1eed oxi.","B. trong ph\u00e2n t\u1eed c\u00f3 nguy\u00ean t\u1eed hi\u0111ro v\u00e0 nguy\u00ean t\u1eed oxi.","C. trong ph\u00e2n t\u1eed c\u00f3 nguy\u00ean t\u1eed cacbon, hi\u0111ro v\u00e0 nguy\u00ean t\u1eed oxi ","D. trong ph\u00e2n t\u1eed c\u00f3 nh\u00f3m \u2013 OH."],"hint":"","explain":"<span class='basic_left'>Trong ph\u00e2n t\u1eed r\u01b0\u1ee3u etylic c\u00f3 nh\u00f3m \u2013 OH n\u00ean r\u01b0\u1ee3u etylic c\u00f3 kh\u1ea3 n\u0103ng ph\u1ea3n \u1ee9ng v\u1edbi Na gi\u1ea3i ph\u00f3ng kh\u00ed hi\u0111ro<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":2580},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Tr\u00ean nh\u00e3n c\u1ee7a m\u1ed9t chai r\u01b0\u1ee3u ghi $18^0$ c\u00f3 ngh\u0129a l\u00e0","select":["A. Nhi\u1ec7t \u0111\u1ed9 s\u00f4i c\u1ee7a r\u01b0\u1ee3u etylic l\u00e0 $18^0C$.","B. Nhi\u1ec7t \u0111\u1ed9 \u0111\u00f4ng \u0111\u1eb7c c\u1ee7a r\u01b0\u1ee3u etylic l\u00e0 $18^0C$.","C. Trong 100 ml r\u01b0\u1ee3u c\u00f3 18 ml r\u01b0\u1ee3u etylic nguy\u00ean ch\u1ea5t v\u00e0 82 ml n\u01b0\u1edbc ","D. Trong 100 ml r\u01b0\u1ee3u c\u00f3 18 ml n\u01b0\u1edbc v\u00e0 82 ml r\u01b0\u1ee3u etylic nguy\u00ean Ch\u1ea5t"],"hint":"","explain":"<span class='basic_left'>$18^0$ l\u00e0 \u0111\u1ed9 r\u01b0\u1ee3u. V\u1eady n\u00ean $18^0$: trong 100 ml r\u01b0\u1ee3u c\u00f3 18 ml r\u01b0\u1ee3u etylic nguy\u00ean ch\u1ea5t v\u00e0 82 ml n\u01b0\u1edbc.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2581},{"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 ch\u00e1y ho\u00e0n to\u00e0n 18,4 gam glixerol. D\u1eabn kh\u00ed thu \u0111\u01b0\u1ee3c v\u00e0o dung d\u1ecbch n\u01b0\u1edbc v\u00f4i trong d\u01b0 thu \u0111\u01b0\u1ee3c kh\u1ed1i l\u01b0\u1ee3ng k\u1ebft t\u1ee7a l\u00e0:","select":["A. 40 gam","B. 50 gam ","C. 60 gam ","D. 70 gam"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & 2{{C}_{3}}{{H}_{5}}{{(OH)}_{3}}+7{{O}_{2}}\\xrightarrow{{{t}^{0}}}6C{{O}_{2}}+8{{H}_{2}}O \\\\ & C{{O}_{2}}+Ca{{(OH)}_{2}}\\to CaC{{O}_{3}}\\downarrow +{{H}_{2}}O \\\\ & {{n}_{\\downarrow }}={{n}_{C{{O}_{2}}}}=3{{n}_{{{C}_{3}}{{H}_{5}}{{(OH)}_{3}}}}=3.0,2=0,6\\,mol \\\\ & \\Rightarrow {{m}_{\\downarrow }}=60\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2582},{"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'>$\\begin{aligned} & {{C}_{2}}{{H}_{4}}+{{H}_{2}}O\\xrightarrow{Axit}{{C}_{2}}{{H}_{5}}OH\\,\\,(1) \\\\ & {{n}_{{{C}_{2}}{{H}_{4}}\\,(1)}}={{n}_{{{C}_{2}}{{H}_{5}}OH}}=0,2\\,mol \\\\ & {{n}_{{{C}_{2}}{{H}_{4}}\\,(\\text{b\u0111)}}}=0,5\\,mol \\\\ & H=\\dfrac{{{n}_{{{C}_{2}}{{H}_{4}}\\,(1)}}}{{{n}_{{{C}_{2}}{{H}_{4}}\\,(\\text{b\u0111})\\,}}}.100\\%=\\dfrac{0,2}{0,5}.100\\%=40\\% \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2583},{"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 $90^0$ t\u00e1c d\u1ee5ng v\u1edbi natri. S\u1ed1 ph\u1ea3n \u1ee9ng h\u00f3a h\u1ecdc c\u00f3 th\u1ec3 x\u1ea3y ra l\u00e0","select":["A. 1 ","B. 2","C. 3","D. 4"],"hint":"","explain":"<span class='basic_left'>Trong r\u01b0\u1ee3u $90^0$ c\u00f3 c\u1ea3 r\u01b0\u1ee3u v\u00e0 n\u01b0\u1edbc n\u00ean c\u00f3 th\u1ec3 x\u1ea3y ra 2 ph\u1ea3n \u1ee9ng: <br\/> $\\begin{aligned} & 2{{C}_{2}}{{H}_{5}}OH+2Na\\to 2{{C}_{2}}{{H}_{5}}ONa+{{H}_{2}}\\uparrow \\\\ & 2Na+2{{H}_{2}}O\\to 2NaOH+{{H}_{2}}\\uparrow \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2584},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"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'>$\\begin{aligned} & 2{{C}_{2}}{{H}_{5}}OH+2Na\\to 2{{C}_{2}}{{H}_{5}}ONa+{{H}_{2}}\\uparrow \\\\ & {{n}_{\\uparrow }}=\\dfrac{1}{2}{{n}_{{{C}_{2}}{{H}_{5}}OH}}=\\dfrac{1}{2}.0,5=0,25\\,mol \\\\ & {{V}_{\\uparrow }}=5,6\\,L \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2585},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Trung h\u00f2a 400 ml dung d\u1ecbch axit axetic 0,5M b\u1eb1ng dung d\u1ecbch NaOH 0,5M. Th\u1ec3 t\u00edch dung d\u1ecbch NaOH c\u1ea7n d\u00f9ng l\u00e0","select":["A. 100 ml ","B. 200 ml.","C. 300 ml. ","D. 400 ml."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & C{{H}_{3}}COOH+NaOH\\to C{{H}_{3}}COONa+{{H}_{2}}O \\\\ & {{n}_{NaOH}}={{n}_{C{{H}_{3}}COOH}}=0,2\\,mol \\\\ & \\Rightarrow {{V}_{NaOH}}=\\dfrac{0,2}{0,5}=0,4\\,L=400\\,ml \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2586},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Cho 200 gam dung d\u1ecbch $CH_3COOH$ 9% t\u00e1c d\u1ee5ng v\u1eeba \u0111\u1ee7 v\u1edbi $Na_2CO_3$. Th\u1ec3 t\u00edch kh\u00ed $CO_2$ sinh ra \u1edf \u0111ktc l\u00e0 ","select":["A. 4,48 l\u00edt ","B. 2,8 l\u00edt.","C. 3,36 l\u00edt. ","D. 2,24 l\u00edt."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{m}_{C{{H}_{3}}COOH}}=\\dfrac{200.9\\%}{100\\%}=18\\,gam\\Rightarrow {{n}_{C{{H}_{3}}COOH}}=0,3\\,mol \\\\ & 2C{{H}_{3}}COOH+N{{a}_{2}}C{{O}_{3}}\\to 2C{{H}_{3}}COONa+C{{O}_{2}}\\uparrow +{{H}_{2}}O \\\\ & {{n}_{\\uparrow }}=\\dfrac{1}{2}{{n}_{C{{H}_{3}}COOH}}=0,15\\,mol\\Rightarrow {{V}_{\\uparrow }}=3,36\\,L \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2587},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Cho 18 g axit axetic t\u00e1c d\u1ee5ng v\u1edbi 11,5 g r\u01b0\u1ee3u etylic c\u00f3 $H_2SO_4$ \u0111\u1eb7c l\u00e0m x\u00fac t\u00e1c. Kh\u1ed1i l\u01b0\u1ee3ng este thu \u0111\u01b0\u1ee3c l\u00e0","select":["A. 23 g. ","B. 22 g.","C. 24 g. ","D. 25 g."],"hint":"","explain":"<span class='basic_left'>$ {{n}_{C{{H}_{3}}COOH}}=0,3\\,mol;\\,\\,{{n}_{{{C}_{2}}{{H}_{5}}OH}}=0,25\\,mol$ <br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/hoahoc/bai48/lv2/img\/Capture18.PNG' \/><\/center><br\/> X\u00e9t t\u1ec9 l\u1ec7 <br\/> $\\dfrac{{{n}_{C{{H}_{3}}COOH}}}{1}>\\dfrac{{{n}_{{{C}_{2}}{{H}_{5}}OH}}}{1}$ => CH3COOH d\u01b0, t\u00ednh theo C2H5OH <br\/> $\\begin{aligned} & {{n}_{C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}}}={{n}_{{{C}_{2}}{{H}_{5}}OH}}=0,25\\,mol \\\\ & \\Rightarrow {{m}_{C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}}}=22\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2588},{"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 dung d\u1ecbch axit axetic 12% t\u00e1c d\u1ee5ng v\u1eeba \u0111\u1ee7 v\u1edbi dung d\u1ecbch $NaHCO_3$ 8,4%. N\u1ed3ng \u0111\u1ed9 % c\u1ee7a dung d\u1ecbch mu\u1ed1i thu \u0111\u01b0\u1ee3c l\u00e0:","select":["A. 4,68%","B. 4,47%","C. 5,63% ","D. 6,81%"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{m}_{C{{H}_{3}}COOH}}=\\dfrac{100.12\\%}{100\\%}=12\\,gam\\Rightarrow {{n}_{C{{H}_{3}}COOH}}=0,2\\,mol \\\\ & C{{H}_{3}}COOH+NaHC{{O}_{3}}\\to C{{H}_{3}}COONa+C{{O}_{2}}\\uparrow +{{H}_{2}}O \\\\ & {{n}_{C{{H}_{3}}COONa}}={{n}_{NaHC{{O}_{3}}}}={{n}_{C{{O}_{2}}}}={{n}_{C{{H}_{3}}COOH}}=0,2\\,mol \\\\ & {{m}_{C{{H}_{3}}COONa}}=16,4\\,gam \\\\ & {{m}_{NaHC{{O}_{3}}}}=16,8\\,gam\\Rightarrow {{m}_{\\text{dd}\\,\\,NaHC{{O}_{3}}}}=\\dfrac{16,8.100\\%}{8,4\\%}=200\\,gam \\\\ & {{m}_{\\text{dd}\\,C{{H}_{3}}COONa}}=100+200-{{m}_{C{{O}_{2}}}}=300-0,2.44=291,2\\,gam \\\\ & C{{\\%}_{\\text{dd}\\,C{{H}_{3}}COONa}}=\\dfrac{16,4}{291,3}.100\\%\\approx 5,63\\% \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2589},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" T\u00ednh kh\u1ed1i l\u01b0\u1ee3ng x\u00e0 ph\u00f2ng t\u1ea1o th\u00e0nh khi x\u00e0 ph\u00f2ng h\u00f3a 89 gam ch\u1ea5t b\u00e9o. Bi\u1ebft ph\u1ea3n \u1ee9ng t\u1ea1o ra 9,2 gam glixerol.","select":["A. 98,2 gam ","B. 95,4 gam","C. 91,8 gam","D. 100,3 gam"],"hint":"","explain":"<span class='basic_left'>G\u1ecdi x l\u00e0 kh\u1ed1i l\u01b0\u1ee3ng x\u00e0 ph\u00f2ng thu \u0111\u01b0\u1ee3c <br\/> $\\begin{aligned} & {{(RCOO)}_{3}}{{C}_{3}}{{H}_{5}}+3NaOH\\to 3RCOONa+{{C}_{3}}{{H}_{5}}{{(OH)}_{3}} \\\\ & {{n}_{{{C}_{3}}{{H}_{5}}{{(OH)}_{3}}}}=0,1\\,mol\\Rightarrow {{n}_{NaOH}}=3{{n}_{{{C}_{3}}{{H}_{5}}{{(OH)}_{3}}}}=0,3\\,mol \\\\ \\end{aligned}$ <br\/> \u00c1p d\u1ee5ng \u0111\u1ecbnh lu\u1eadt b\u1ea3o to\u00e0n kh\u1ed1i l\u01b0\u1ee3ng ta c\u00f3: <br\/> $\\begin{aligned} & {{m}_{{{(RCOO)}_{3}}{{C}_{3}}{{H}_{5}}}}+{{m}_{NaOH}}={{m}_{xp}}+{{m}_{{{C}_{3}}{{H}_{5}}{{(OH)}_{3}}}} \\\\ & \\Rightarrow 89\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+0,3.40=x\\,\\,\\,\\,\\,+9,2 \\\\ & \\Rightarrow x=91,8\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2590}],"lesson":{"save":0,"level":2}}