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{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" C\u00f4ng th\u1ee9c c\u1ea5u t\u1ea1o c\u1ee7a axit axetic ($C_2H_4O_2$) l\u00e0:<br\/><table><tr><th>(1) $O = CH \u2013 O \u2013 CH_3$<\/th><th>(2) <br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/hoahoc/bai45/lv2/img\/Capture1.PNG' \/><\/center> <\/th><th>(3) <br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/hoahoc/bai45/lv2/img\/Capture1,1.PNG' \/><\/center> <\/th><th>(4) $CH_2 \u2013 O \u2013 O \u2013 CH_2$<\/th><\/tr><\/table>","select":["A. 1","B. 2 ","C. 3 ","D. 4 "],"hint":"","explain":"<span class='basic_left'>C\u00f4ng th\u1ee9c c\u1ea5u t\u1ea1o c\u1ee7a axit axetic ($C_2H_4O_2$) l\u00e0:<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/hoahoc/bai45/lv2/img\/Capture1.PNG' \/><\/center><br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2511},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Gi\u1ea5m \u0103n l\u00e0 dung d\u1ecbch axit axetic c\u00f3 n\u1ed3ng \u0111\u1ed9 ","select":["A. tr\u00ean 5%.","B. d\u01b0\u1edbi 2%.","C. t\u1eeb 2% - 5%.","D. t\u1eeb 3% - 6%."],"hint":"","explain":"<span class='basic_left'>Gi\u1ea5m \u0103n l\u00e0 dung d\u1ecbch axit axetic c\u00f3 n\u1ed3ng \u0111\u1ed9 t\u1eeb 2% - 5%.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2512},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" T\u00ednh ch\u1ea5t v\u1eadt l\u00fd c\u1ee7a axit axetic l\u00e0 ","select":["A. ch\u1ea5t l\u1ecfng, kh\u00f4ng m\u00e0u, v\u1ecb chua, tan v\u00f4 h\u1ea1n trong n\u01b0\u1edbc. ","B. ch\u1ea5t l\u1ecfng, m\u00e0u tr\u1eafng, v\u1ecb chua, tan v\u00f4 h\u1ea1n trong n\u01b0\u1edbc.","C. ch\u1ea5t l\u1ecfng, kh\u00f4ng m\u00e0u, v\u1ecb \u0111\u1eafng, tan v\u00f4 h\u1ea1n trong n\u01b0\u1edbc.","D. ch\u1ea5t l\u1ecfng, kh\u00f4ng m\u00e0u, v\u1ecb chua, kh\u00f4ng tan trong n\u01b0\u1edbc."],"hint":"","explain":"<span class='basic_left'>Axit axetic l\u00e0 ch\u1ea5t l\u1ecfng, kh\u00f4ng m\u00e0u, v\u1ecb chua, tan v\u00f4 h\u1ea1n trong n\u01b0\u1edbc.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":1}]}],"id_ques":2513},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Ph\u1ea3n \u1ee9ng gi\u1eefa axit axetic v\u1edbi dung d\u1ecbch baz\u01a1 thu\u1ed9c lo\u1ea1i ","select":["A. ph\u1ea3n \u1ee9ng oxi h\u00f3a - kh\u1eed.","B. ph\u1ea3n \u1ee9ng h\u00f3a h\u1ee3p.","C. ph\u1ea3n \u1ee9ng h\u00f3a h\u1ee3p.","D. ph\u1ea3n \u1ee9ng trung h\u00f2a."],"hint":"","explain":"<span class='basic_left'>Ph\u1ea3n \u1ee9ng gi\u1eefa axit axetic v\u1edbi dung d\u1ecbch baz\u01a1 thu\u1ed9c lo\u1ea1i ph\u1ea3n \u1ee9ng trung h\u00f2a.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2514},{"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 m\u1ed9t l\u01b0\u1ee3ng l\u1edbn axit axetic \u0111\u01b0\u1ee3c \u0111i\u1ec1u ch\u1ebf b\u1eb1ng c\u00e1ch ","select":["A. oxi h\u00f3a metan c\u00f3 x\u00fac t\u00e1c v\u00e0 nhi\u1ec7t \u0111\u1ed9 th\u00edch h\u1ee3p.","B. oxi h\u00f3a etilen c\u00f3 x\u00fac t\u00e1c v\u00e0 nhi\u1ec7t \u0111\u1ed9 th\u00edch h\u1ee3p.","C. oxi h\u00f3a etan c\u00f3 x\u00fac t\u00e1c v\u00e0 nhi\u1ec7t \u0111\u1ed9 th\u00edch h\u1ee3p.","D. oxi h\u00f3a butan c\u00f3 x\u00fac t\u00e1c v\u00e0 nhi\u1ec7t \u0111\u1ed9 th\u00edch h\u1ee3p."],"hint":"","explain":"<span class='basic_left'>Trong c\u00f4ng nghi\u1ec7p m\u1ed9t l\u01b0\u1ee3ng l\u1edbn axit axetic \u0111\u01b0\u1ee3c \u0111i\u1ec1u ch\u1ebf theo ph\u1ea3n \u1ee9ng sau: <br\/> $\\begin{aligned} & 2{{C}_{4}}{{H}_{10}}+5{{O}_{2}}\\xrightarrow[{{t}^{0}}]{XT}4C{{H}_{3}}\\text{COO}H+2{{H}_{2}}O \\\\ & (Bu\\tan )\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(Axit\\,\\text{ax}etic) \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":2515},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Axit axetic t\u00e1c d\u1ee5ng v\u1edbi k\u1ebdm gi\u1ea3i ph\u00f3ng kh\u00ed ","select":["A. hi\u0111ro","B. hi\u0111ro clorua","C. hi\u0111ro sunfua","D. amoniac"],"hint":"","explain":"<span class='basic_left'>Axit axetic t\u00e1c d\u1ee5ng v\u1edbi c\u00e1c kim lo\u1ea1i \u0111\u1ee9ng tr\u01b0\u1edbc H trong d\u00e3y ho\u1ea1t \u0111\u1ed9ng h\u00f3a h\u1ecdc gi\u1ea3i ph\u00f3ng kh\u00ed hi\u0111ro<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2516},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Axit axetic t\u00e1c d\u1ee5ng v\u1edbi mu\u1ed1i cacbonat gi\u1ea3i ph\u00f3ng kh\u00ed ","select":["A. cacbon \u0111ioxit.","B. l\u01b0u hu\u1ef3nh \u0111ioxit. ","C. l\u01b0u hu\u1ef3nh trioxit.","D. cacbon monooxit. "],"hint":"","explain":"<span class='basic_left'>V\u00ed d\u1ee5: $2C{{H}_{3}}COOH+N{{a}_{2}}C{{O}_{3}}\\to 2C{{H}_{3}}C\\text{OO}Na+C{{O}_{2}}\\uparrow +{{H}_{2}}O$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2517},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"T\u00ednh ch\u1ea5t v\u1eadt l\u00fd c\u1ee7a etyl axetat l\u00e0 ","select":["A. ch\u1ea5t l\u1ecfng, m\u00f9i th\u01a1m, \u00edt tan trong n\u01b0\u1edbc, d\u00f9ng l\u00e0m dung m\u00f4i trong c\u00f4ng nghi\u1ec7p.","B. ch\u1ea5t kh\u00ed m\u00f9i th\u01a1m, \u00edt tan trong n\u01b0\u1edbc, d\u00f9ng l\u00e0m dung m\u00f4i trong c\u00f4ng nghi\u1ec7p.","C. ch\u1ea5t l\u1ecfng kh\u00f4ng m\u00f9i, \u00edt tan trong n\u01b0\u1edbc, d\u00f9ng l\u00e0m dung m\u00f4i trong c\u00f4ng nghi\u1ec7p. ","D. ch\u1ea5t l\u1ecfng tan v\u00f4 h\u1ea1n trong n\u01b0\u1edbc, d\u00f9ng l\u00e0m dung m\u00f4i trong c\u00f4ng nghi\u1ec7p."],"hint":"","explain":"<span class='basic_left'>Etyl axetat l\u00e0 ch\u1ea5t l\u1ecfng, m\u00f9i th\u01a1m, \u00edt tan trong n\u01b0\u1edbc, d\u00f9ng l\u00e0m dung m\u00f4i trong c\u00f4ng nghi\u1ec7p.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":1}]}],"id_ques":2518},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Axit axetic c\u00f3 t\u00ednh axit v\u00ec:","select":["A. Ph\u00e2n t\u1eed c\u00f3 ch\u1ee9a nh\u00f3m \u2013 OH ","B. Ph\u00e2n t\u1eed c\u00f3 ch\u1ee9a nh\u00f3m \u2013 OH v\u00e0 \u2013 COOH ","C. Ph\u00e2n t\u1eed c\u00f3 ch\u1ee9a nh\u00f3m \u2013 COOH","D. Ph\u00e2n t\u1eed c\u00f3 ch\u1ee9a C, H, O."],"hint":"","explain":"<span class='basic_left'>Axit axetic c\u00f3 t\u00ednh axit v\u00ec trong ph\u00e2n t\u1eed c\u00f3 ch\u1ee9a nh\u00f3m \u2013 COOH <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2519},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Ph\u1ea3n \u1ee9ng este h\u00f3a l\u00e0 ph\u1ea3n \u1ee9ng x\u1ea3y ra gi\u1eefa:","select":["A. Gi\u1eefa axit v\u00e0 baz\u01a1","B. Gi\u1eefa 2 dung d\u1ecbch mu\u1ed1i","C. Gi\u1eefa axit h\u1eefu c\u01a1 v\u00e0 r\u01b0\u1ee3u ","D. Gi\u1eefa baz\u01a1 v\u00e0 r\u01b0\u1ee3u"],"hint":"","explain":"<span class='basic_left'>Ph\u1ea3n \u1ee9ng gi\u1eefa axit h\u1eefu c\u01a1 v\u00e0 r\u01b0\u1ee3u sinh ra este n\u00ean \u0111\u01b0\u1ee3c g\u1ecdi l\u00e0 ph\u1ea3n \u1ee9ng este h\u00f3a<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2520},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" H\u00f2a tan ho\u00e0n to\u00e0n 6,5 gam $Zn$ v\u00e0o dung d\u1ecbch $CH_3COOH$. Th\u1ec3 t\u00edch kh\u00ed $H_2$ tho\u00e1t ra ( \u0111ktc) l\u00e0 ","select":["A. 0,56 l\u00edt.","B. 1,12 l\u00edt","C. 2,24 l\u00edt.","D. 3,36 l\u00edt."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & 2C{{H}_{3}}COOH+Zn\\to {{(C{{H}_{3}}COO)}_{2}}Zn+{{H}_{2}}\\uparrow \\\\ & {{n}_{{{H}_{2}}}}={{n}_{Zn}}=0,1\\,mol\\Rightarrow {{V}_{\\uparrow }}=2,24\\,L \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2521},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" C\u1eb7p ch\u1ea5t n\u00e0o sau \u0111\u00e2y c\u00f3 ph\u1ea3n \u1ee9ng x\u1ea3y ra v\u00e0 sinh ra kh\u00ed $CO_2$ ","select":["A. $CH_3COOH$ v\u00e0 $ZnO$. ","B. $CH_3COOH$ v\u00e0 $Zn(OH)_2$.","C. $CH_3COOH$ v\u00e0 $ZnCO_3$.","D. $CH_3COONa$ v\u00e0 $K_2CO_3$."],"hint":"","explain":"<span class='basic_left'>$2C{{H}_{3}}COOH\\text{ + }ZnC{{O}_{3}}\\to {{(C{{H}_{3}}COO)}_{2}}Zn+C{{O}_{2}}\\uparrow +{{H}_{2}}O$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2522},{"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 t\u1ed3n t\u1ea1i \u0111\u01b0\u1ee3c trong m\u1ed9t dung d\u1ecbch l\u00e0 ( kh\u00f4ng x\u1ea3y ra ph\u1ea3n \u1ee9ng h\u00f3a h\u1ecdc v\u1edbi nhau) ","select":["A. $CH_3COOH$v\u00e0 $NaOH$. ","B. $CH_3COOH$ v\u00e0 $H_3PO_4$.","C. $CH_3COOH$ v\u00e0 $Ca(OH)_2$.","D. $CH_3COOH$ v\u00e0 $Na_2CO_3$."],"hint":"","explain":"<span class='basic_left'>C\u00e1c ch\u1ea5t t\u1ed3n t\u1ea1i \u0111\u01b0\u1ee3c trong 1 dung d\u1ecbch khi ch\u00fang kh\u00f4ng ph\u1ea3n \u00fang v\u1edbi nhau. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n B th\u1ecfa m\u00e3n<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2523},{"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 ph\u00e2n bi\u1ec7t dung d\u1ecbch $CH_3COOH$ v\u00e0 $C_2H_5OH$ ta d\u00f9ng ","select":["A. Na.","B. Zn. ","C. K. ","D. Cu."],"hint":"","explain":"<span class='basic_left'>Cho 2 dung d\u1ecbch l\u1ea7n l\u01b0\u1ee3t ph\u1ea3n \u1ee9ng v\u1edbi Zn, ch\u1ea5t n\u00e0o ph\u1ea3n \u1ee9ng t\u1ea1o kh\u00ed kh\u00f4ng m\u00e0u l\u00e0 axit axetic. Ch\u1ea5t c\u00f2n l\u1ea1i kh\u00f4ng c\u00f3 hi\u1ec7n t\u01b0\u1ee3ng g\u00ec l\u00e0 r\u01b0\u1ee3u etylic. <br\/> $2C{{H}_{3}}COOH+Zn\\to {{(C{{H}_{3}}COO)}_{2}}Zn+{{H}_{2}}\\uparrow $ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2524},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Cho th\u00eam $Cu(OH)_2$ v\u00e0o hai \u1ed1ng nghi\u1ec7m \u0111\u1ef1ng $CH_3COOH$v\u00e0 $C_2H_5OH$. Hi\u1ec7n t\u01b0\u1ee3ng n\u00e0o sau \u0111\u00e2y ch\u1ee9ng t\u1ecf l\u00e0 axit $CH_3COOH$ ","select":["A. Dung d\u1ecbch c\u00f3 m\u00e0u xanh.","B. Dung d\u1ecbch m\u00e0u v\u00e0ng n\u00e2u.","C. C\u00f3 k\u1ebft t\u1ee7a tr\u1eafng. ","D. C\u00f3 k\u1ebft t\u1ee7a n\u00e2u \u0111\u1ecf."],"hint":"","explain":"<span class='basic_left'>$CH_3COOH$ ph\u1ea3n \u1ee9ng v\u1edbi $Cu(OH)_2$ t\u1ea1o th\u00e0nh mu\u1ed1i c\u1ee7a \u0111\u1ed3ng c\u00f3 m\u00e0u xanh theo ph\u1ea3n \u1ee9ng: <br\/> $2C{{H}_{3}}COOH\\text{ + Cu(OH}{{\\text{)}}_{2}}\\to {{(C{{H}_{3}}COO)}_{2}}Cu+2{{H}_{2}}O$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2525},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Cho 100 ml dung d\u1ecbch $CH_3COOH$ 0,1M v\u00e0o 100 ml dung d\u1ecbch $Ca(OH)_2$ 0,1M. Dung d\u1ecbch sau ph\u1ea3n \u1ee9ng c\u00f3 kh\u1ea3 n\u0103ng ","select":["A. l\u00e0m qu\u1ef3 t\u00edm h\u00f3a xanh","B. l\u00e0m qu\u1ef3 t\u00edm h\u00f3a \u0111\u1ecf.","C. kh\u00f4ng l\u00e0m qu\u1ef3 t\u00edm \u0111\u1ed5i m\u00e0u.","D. t\u00e1c d\u1ee5ng v\u1edbi $Mg$ gi\u1ea3i ph\u00f3ng kh\u00ed $H_2$. "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{C{{H}_{3}}COOH}}=0,01\\,mol;\\,\\,{{n}_{Ca{{(OH)}_{2}}}}=0,01\\,mol \\\\ & 2C{{H}_{3}}COOH+Ca{{(OH)}_{2}}\\to {{(C{{H}_{3}}COO)}_{2}}Ca+2{{H}_{2}}O \\\\ \\end{aligned}$ <br\/> X\u00e9t t\u1ec9 l\u1ec7: $\\dfrac{{{n}_{C{{H}_{3}}COOH}}}{2}<\\dfrac{\\,{{n}_{Ca{{(OH)}_{2}}}}}{1}$ <br\/>V\u1eady $CH_3COOH$ h\u1ebft, $Ca(OH)_2$ d\u01b0 n\u00ean dung d\u1ecbch sau ph\u1ea3n \u1ee9ng l\u00e0m qu\u1ef3 t\u00edm h\u00f3a xanh.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":1}]}],"id_ques":2526},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Cho 30 ml dung d\u1ecbch $CH_3COOH$ 1M v\u00e0o \u1ed1ng nghi\u1ec7m ch\u1ee9a 0,36 gam $Mg$, sau khi ph\u1ea3n \u1ee9ng k\u1ebft th\u00fac thu \u0111\u01b0\u1ee3c ","select":["A. dung d\u1ecbch c\u00f3 m\u00e0u xanh, c\u00f3 kh\u00ed bay l\u00ean.","B. dung d\u1ecbch kh\u00f4ng m\u00e0u, c\u00f3 m\u1ed9t ph\u1ea7n ch\u1ea5t r\u1eafn m\u00e0u tr\u1eafng kh\u00f4ng tan, c\u00f3 kh\u00ed bay l\u00ean","C. dung d\u1ecbch m\u00e0u xanh, c\u00f3 m\u1ed9t ph\u1ea7n ch\u1ea5t r\u1eafn m\u00e0u tr\u1eafng kh\u00f4ng tan, c\u00f3 kh\u00ed bay l\u00ean","D. dung d\u1ecbch kh\u00f4ng m\u00e0u, c\u00f3 kh\u00ed bay l\u00ean."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & 2C{{H}_{3}}COOH+Mg\\to {{(C{{H}_{3}}COO)}_{2}}Mg+{{H}_{2}}\\uparrow \\\\ & {{n}_{C{{H}_{3}}COOH}}=0,03\\,mol;\\,\\,{{n}_{Mg}}=0,015\\,mol \\\\ \\end{aligned}$ <br\/> X\u00e9t t\u1ec9 l\u1ec7: $\\dfrac{{{n}_{C{{H}_{3}}COOH}}}{2}=\\dfrac{{{n}_{Mg}}}{1}$ <br\/> V\u1eady c\u1ea3 2 ch\u1ea5t ph\u1ea3n \u1ee9ng v\u1eeba \u0111\u1ee7 n\u00ean thu \u0111\u01b0\u1ee3c dung d\u1ecbch kh\u00f4ng m\u00e0u sau ph\u1ea3n \u1ee9ng v\u00e0 c\u00f3 kh\u00ed kh\u00f4ng m\u00e0u bay l\u00ean.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":2527},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Cho axit axetic t\u00e1c d\u1ee5ng v\u1edbi r\u01b0\u1ee3u etylic c\u00f3 m\u1eb7t $H_2SO_4$ \u0111\u1eb7c l\u00e0m x\u00fac t\u00e1c v\u00e0 \u0111un n\u00f3ng. Sau ph\u1ea3n \u1ee9ng thu \u0111\u01b0\u1ee3c 44 gam etyl axetat. Kh\u1ed1i l\u01b0\u1ee3ng $CH_3COOH$ v\u00e0 $C_2H_5OH$ \u0111\u00e3 ph\u1ea3n \u1ee9ng l\u00e0:","select":["A. 60 gam v\u00e0 46 gam.","B. 30 gam v\u00e0 23 gam. ","C. 15 gam v\u00e0 11,5 gam. ","D. 45 gam v\u00e0 34,5 gam."],"hint":"","explain":"<span class='basic_left'> <br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/hoahoc/bai45/lv2/img\/Capture18.PNG' \/><\/center> $\\begin{aligned} & {{n}_{C{{H}_{3}}COOH}}={{n}_{{{C}_{2}}{{H}_{5}}OH}}={{n}_{C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}}}=\\frac{44}{88}=0,5\\,mol \\\\ & \\Rightarrow {{m}_{C{{H}_{3}}COOH}}=30\\,gam;\\,\\,\\,{{m}_{{{C}_{2}}{{H}_{5}}OH}}=23\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2528},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Cho dung d\u1ecbch ch\u1ee9a 10 gam h\u1ed7n h\u1ee3p $C_2H_5OH$ v\u00e0 $CH_3COOH$ t\u00e1c d\u1ee5ng v\u1edbi $Zn$ d\u01b0 thu \u0111\u01b0\u1ee3c 1,12 l\u00edt kh\u00ed $H_2$ (\u0111ktc) . Th\u00e0nh ph\u1ea7n ph\u1ea7n tr\u0103m theo kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a r\u01b0\u1ee3u etylic v\u00e0 axit axetic l\u1ea7n l\u01b0\u1ee3t l\u00e0 ","select":["A. 30% v\u00e0 70%.","B. 40% v\u00e0 60%.","C. 70% v\u00e0 30%.","D. 60% v\u00e0 40%."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & 2C{{H}_{3}}COOH+Zn\\to {{(C{{H}_{3}}COO)}_{2}}Zn+{{H}_{2}}\\uparrow \\\\ & {{n}_{C{{H}_{3}}COOH}}=2{{n}_{{{H}_{2}}}}=2.0,05\\,=0,1\\,mol \\\\ & {{m}_{C{{H}_{3}}COOH}}=6\\,gam \\\\ & \\%{{m}_{C{{H}_{3}}COOH}}=\\dfrac{6}{10}.100\\%=60\\%\\Rightarrow \\%{{m}_{{{C}_{2}}{{H}_{5}}OH}}=40\\% \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2529},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho dung d\u1ecbch axit axetic c\u00f3 n\u1ed3ng \u0111\u1ed9 a% t\u00e1c d\u1ee5ng v\u1eeba \u0111\u1ee7 v\u1edbi dung d\u1ecbch NaOH n\u1ed3ng \u0111\u1ed9 10% thu \u0111\u01b0\u1ee3c dung d\u1ecbch c\u00f3 n\u1ed3ng \u0111\u1ed9 10,25%. Gi\u00e1 tr\u1ecb c\u1ee7a a l\u00e0:","select":["A. 5","B. 10","C. 15","D. 20"],"hint":"","explain":"<span class='basic_left'>Gi\u1ea3 s\u1eed c\u00f3 1 mol axit axetic. <br\/> $\\begin{aligned} & \\Rightarrow {{m}_{C{{H}_{3}}COOH}}=60\\,gam.\\Rightarrow {{m}_{\\text{dd}\\,C{{H}_{3}}COOH\\,}}=\\dfrac{60.100\\%}{a\\%}=\\dfrac{6000}{a}\\,gam \\\\ & C{{H}_{3}}COOH+NaOH\\to C{{H}_{3}}COONa+{{H}_{2}}O \\\\ & {{n}_{NaOH}}={{n}_{C{{H}_{3}}COOH}}=1\\,mol\\Rightarrow {{m}_{NaOH}}=40\\,gam \\\\ & \\Rightarrow {{m}_{\\text{dd}\\,NaOH}}=\\dfrac{40.100\\%}{10\\%}=400\\,gam \\\\ & {{m}_{\\text{dd}}}=400+\\dfrac{6000}{a} \\\\ & {{n}_{C{{H}_{3}}COONa}}={{n}_{C{{H}_{3}}COOH}}=1\\,mol\\Rightarrow {{m}_{C{{H}_{3}}COONa}}=82\\,gam \\\\ & C{{\\%}_{C{{H}_{3}}COONa}}=10,25\\%\\Rightarrow \\dfrac{82}{400+\\dfrac{6000}{a}}.100\\%=10,25\\%\\Rightarrow a=15 \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2530}],"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
Trong khoảng 4 phút -> 5 phút + 1 điểm

Quá 5 phút: không được cộng điểm

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