{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Ch\u1ea5t khi g\u1eb7p iot s\u1ebd bi\u1ebfn sang m\u00e0u xanh:","select":["A. L\u00f2ng tr\u1eafng tr\u1ee9ng ","B. H\u1ed3 tinh b\u1ed9t","C. Cao su","D. Ch\u1ea5t b\u00e9o"],"hint":"","explain":"<span class='basic_left'>H\u1ed3 tinh b\u1ed9t g\u1eb7p iot s\u1ebd chuy\u1ec3n th\u00e0nh m\u00e0u xanh<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2691},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" C\u1eb7p ch\u1ea5t t\u00e1c d\u1ee5ng v\u1edbi nhau \u0111\u1ec3 t\u1ea1o th\u00e0nh h\u1ee3p ch\u1ea5t kh\u00ed l\u00e0 ","select":["A. k\u1ebdm v\u1edbi axit clohi\u0111ric.","B. natri cacbonat v\u00e0 canxi clorua.","C. natri hi\u0111roxit v\u00e0 axit clohi\u0111ric.","D. natri cacbonat v\u00e0 axit clohi\u0111ric."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & Zn+2HCl\\to ZnC{{l}_{2}}+{{H}_{2}}\\uparrow \\\\ & N{{a}_{2}}C{{O}_{3}}+2HCl\\to 2NaCl+C{{O}_{2}}\\uparrow +{{H}_{2}}O \\\\ \\end{aligned}$ <br\/> $CO_2$ l\u00e0 h\u1ee3p ch\u1ea5t kh\u00ed l\u00e0 <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":2692},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Ch\u1ecdn c\u00e2u \u0111\u00fang. ","select":["A. Metan, etilen, axetilen \u0111\u1ec1u l\u00e0m m\u1ea5t m\u00e0u dung d\u1ecbch brom.","B. Etilen, axetilen, benzen \u0111\u1ec1u l\u00e0m m\u1ea5t m\u00e0u dung d\u1ecbch brom.","C. Etilen, axetilen \u0111\u1ec1u l\u00e0m m\u1ea5t m\u00e0u dung d\u1ecbch brom.","D. Metan, etilen, benzen \u0111\u1ec1u l\u00e0m m\u1ea5t m\u00e0u dung d\u1ecbch brom."],"hint":"","explain":"<span class='basic_left'>Etilen, axetilen \u0111\u1ec1u l\u00e0m m\u1ea5t m\u00e0u dung d\u1ecbch brom. <br\/> $\\begin{aligned} & {{C}_{2}}{{H}_{4}}+B{{r}_{2}}\\to {{C}_{2}}{{H}_{4}}B{{r}_{2}} \\\\ & {{C}_{2}}{{H}_{2}}+2B{{r}_{2}}\\to {{C}_{2}}{{H}_{2}}B{{r}_{4}} \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2693},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" D\u00e3y ch\u1ea5t \u0111\u1ec1u tan trong n\u01b0\u1edbc \u1edf nhi\u1ec7t \u0111\u1ed9 th\u01b0\u1eddng l\u00e0 ","select":["A. Saccaroz\u01a1 v\u00e0 tinh b\u1ed9t. ","B. Glucoz\u01a1 v\u00e0 xenluloz\u01a1.","C. Glucoz\u01a1 v\u00e0 saccaroz\u01a1. ","D. Saccaroz\u01a1 v\u00e0 xenluloz\u01a1."],"hint":"","explain":"<span class='basic_left'>Glucoz\u01a1 v\u00e0 saccaroz\u01a1 \u0111\u1ec1u tan trong n\u01b0\u1edbc \u1edf nhi\u1ec7t \u0111\u1ed9 th\u01b0\u1eddng.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2694},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" D\u00e3y c\u00e1c ch\u1ea5t \u0111\u1ec1u c\u00f3 ph\u1ea3n \u1ee9ng th\u1ee7y ph\u00e2n l\u00e0 ","select":["A. Tinh b\u1ed9t, xenluloz\u01a1, PVC, glucoz\u01a1.","B. Tinh b\u1ed9t, xenluloz\u01a1, saccaroz\u01a1, ch\u1ea5t b\u00e9o.","C. Tinh b\u1ed9t, xenluloz\u01a1, saccaroz\u01a1, glucoz\u01a1. ","D. Tinh b\u1ed9t, xenluloz\u01a1, saccaroz\u01a1, PE."],"hint":"","explain":"<span class='basic_left'>Tinh b\u1ed9t, xenluloz\u01a1, saccaroz\u01a1, ch\u1ea5t b\u00e9o \u0111\u1ec1u c\u00f3 ph\u1ea3n \u1ee9ng th\u1ee7y ph\u00e2n trong m\u00f4i tr\u01b0\u1eddng axit.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":2695},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Ch\u1ec9 ra c\u00e1c lo\u1ea1i ph\u00e2n \u0111\u1ea1m:","select":["A. $KCl, NH_4NO_3$ ","B. $(NH_2)_2CO, (NH_4)_2SO_4$ ","C. $Ca_3(PO_4)_2, KNO_3$ ","D. $(NH_4)_2HPO_4, Ca(H_2PO_4)_2$"],"hint":"","explain":"<span class='basic_left'>$(NH_2)_2CO, (NH_4)_2SO_4$: c\u00e1c lo\u1ea1i ph\u00e2n \u0111\u1ea1m (ch\u1ee9a h\u00e0m l\u01b0\u1ee3ng N cung c\u1ea5p dinh d\u01b0\u1ee1ng cho c\u00e2y) <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2696},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Kim lo\u1ea1i n\u00e0o d\u01b0\u1edbi \u0111\u00e2y v\u1eeba t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch HCl, v\u1eeba t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch NaOH:","select":["A. Fe","B. Cu","C. Al ","D. Ag"],"hint":"","explain":"<span class='basic_left'>Al v\u1eeba t\u00e1c d\u1ee5ng v\u1edbi axit, v\u1eeba t\u00e1c d\u1ee5ng v\u1edbi baz\u01a1 <br\/> $\\begin{aligned} & 2Al+6HCl\\to 2AlC{{l}_{3}}+3{{H}_{2}}\\uparrow \\\\ & 2Al+2NaOH+2{{H}_{2}}O\\to 2NaAl{{O}_{2}}+3{{H}_{2}}\\uparrow \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2697},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Kh\u00ed $CO_2$ d\u00f9ng \u0111\u1ec3 d\u1eadp t\u1eaft \u0111\u00e1m ch\u00e1y v\u00ec:","select":["A. $CO_2$ kh\u00f4ng ch\u00e1y \u0111\u01b0\u1ee3c.","B. $CO_2$ kh\u00f4ng duy tr\u00ec s\u1ef1 ch\u00e1y.","C. $CO_2$ n\u1eb7ng h\u01a1n kh\u00f4ng kh\u00ed v\u00e0 kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi oxi n\u00ean n\u00f3 c\u00f3 t\u00e1c d\u1ee5ng ng\u0103n kh\u00f4ng cho v\u1eadt ch\u00e1y ti\u1ebfp x\u00fac v\u1edbi oxi. ","D. $CO_2$ l\u00e0 s\u1ea3n ph\u1ea9m c\u1ee7a ph\u1ea3n \u1ee9ng ch\u00e1y n\u00ean kh\u00f4ng th\u1ec3 tham gia ph\u1ea3n \u1ee9ng ch\u00e1y n\u1eefa."],"hint":"","explain":"<span class='basic_left'>$CO_2$ kh\u00f4ng duy tr\u00ec s\u1ef1 ch\u00e1y n\u00ean d\u00f9ng \u0111\u1ec3 d\u1eadp t\u1eaft \u0111\u00e1m ch\u00e1y<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":2698},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Tr\u1eadt t\u1ef1 t\u0103ng d\u1ea7n m\u1ee9c \u0111\u1ed9 ho\u1ea1t \u0111\u1ed9ng h\u00f3a h\u1ecdc c\u1ee7a c\u00e1c kim lo\u1ea1i l\u00e0:","select":["A. Na, Fe, Cu","B. Zn, Al, Ag","C. Cu, Fe, Na","D. Ag, Al, Zn"],"hint":"","explain":"<span class='basic_left'>D\u00e3y ho\u1ea1t \u0111\u1ed9ng h\u00f3a h\u1ecdc c\u1ee7a kim lo\u1ea1i: (m\u1ee9c \u0111\u1ed9 ho\u1ea1t \u0111\u1ed9ng h\u00f3a h\u1ecdc gi\u1ea3m d\u1ea7n) <br\/> K, Na, Mg, Al, Zn, Fe, Pb, H, Cu, Ag, Au<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2699},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" H\u1ee3p ch\u1ea5t n\u00e0o sau \u0111\u00e2y ph\u1ea3n \u1ee9ng \u0111\u01b0\u1ee3c v\u1edbi Clo:","select":["A. $NaCl$ ","B. $NaOH$ ","C. $CaCO_3$ ","D. $HCl$ "],"hint":"","explain":"<span class='basic_left'>$2NaOH+C{{l}_{2}}\\to NaCl+NaClO+{{H}_{2}}O$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2700},{"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 v\u00e0 axit axetic \u0111\u1ec1u t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi:","select":["A. $Na$ ","B. $NaOH$ ","C. $NaCl$ ","D. $Na_2CO_3$"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & 2{{C}_{2}}{{H}_{5}}OH+2Na\\to 2{{C}_{2}}{{H}_{5}}ONa+{{H}_{2}}\\uparrow \\\\ & 2C{{H}_{3}}\\text{COO}H+2Na\\to 2C{{H}_{3}}\\text{COO}Na+{{H}_{2}}\\uparrow \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2701},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" N\u1ebfu l\u1ea5y 8,96 gam etilen th\u00ec ph\u1ea3n \u1ee9ng t\u1ed1i \u0111a v\u1edbi bao nhi\u00eau gam brom trong dung d\u1ecbch ? ","select":["A. 51,2 gam. ","B. 49,2 gam. ","C. 34 gam. ","D. 60,2 gam."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{C}_{2}}{{H}_{4}}+B{{r}_{2}}\\to {{C}_{2}}{{H}_{4}}B{{r}_{2}} \\\\ & {{n}_{B{{r}_{2}}}}={{n}_{{{C}_{2}}{{H}_{4}}}}=0,32\\,mol \\\\ & \\Rightarrow {{m}_{B{{r}_{2}}}}=51,2\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2702},{"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 nh\u1eadn bi\u1ebft $H_2SO_4, NaOH, NaCl, NaNO_3$. Ta d\u00f9ng ","select":["A. Phenolphtalein v\u00e0 dung d\u1ecbch $CuSO_4$.","B. Qu\u1ef3 t\u00edm v\u00e0 dung d\u1ecbch $AgNO_3$.","C. Qu\u1ef3 t\u00edm v\u00e0 dung d\u1ecbch $BaCl_2$. ","D. Dung d\u1ecbch $CuSO_4$ v\u00e0 dung dich $BaCl_2$."],"hint":"","explain":"<span class='basic_left'>Nh\u00fang qu\u1ef3 t\u00edm v\u00e0o c\u00e1c dung d\u1ecbch. Dung d\u1ecbch n\u00e0o l\u00e0m qu\u1ef3 chuy\u1ec3n \u0111\u1ecf l\u00e0 dung d\u1ecbch $H_2SO_4$. Dung d\u1ecbch n\u00e0o l\u00e0m qu\u1ef3 chuy\u1ec3n xanh l\u00e0 dung d\u1ecbch $NaOH$, 2 dung d\u1ecbch c\u00f2n l\u1ea1i kh\u00f4ng l\u00e0m qu\u1ef3 chuy\u1ec3n m\u00e0u. <br\/> Cho 2 dung d\u1ecbch c\u00f2n l\u1ea1i t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch $AgNO_3$. Dung d\u1ecbch n\u00e0o c\u00f3 k\u1ebft t\u1ee7a tr\u1eafng t\u1ea1o th\u00e0nh l\u00e0 dung d\u1ecbch $NaCl$, dung d\u1ecbch c\u00f2n l\u1ea1i kh\u00f4ng c\u00f3 hi\u1ec7n t\u01b0\u1ee3ng g\u00ec l\u00e0 $NaNO_3$ <br\/>$HCl+AgN{{O}_{3}}\\to AgCl\\downarrow +HN{{O}_{3}}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":2703},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Th\u1ec3 t\u00edch dung d\u1ecbch HCl 0,4M c\u1ea7n \u0111\u1ec3 trung h\u00f2a 200ml dung d\u1ecbch NaOH 0,3M l\u00e0 ","select":["A. 450 ml ","B. 150 ml.","C. 300 ml. ","D. 267 ml."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & HCl+NaOH\\to NaCl+{{H}_{2}}O \\\\ & {{n}_{HCl}}={{n}_{NaOH}}=0,06\\,mol \\\\ & {{V}_{HCl}}=\\dfrac{0,06}{0,4}=0,15\\,L=150\\,ml \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2704},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Cho 60 gam dung d\u1ecbch $HCl$ t\u00e1c d\u1ee5ng v\u1edbi $Na_2CO_3$ v\u1eeba \u0111\u1ee7, thu \u0111\u01b0\u1ee3c 3,36 l\u00edt kh\u00ed (\u0111ktc). N\u1ed3ng \u0111\u1ed9 ph\u1ea7n tr\u0103m c\u1ee7a dung d\u1ecbch $HCl$ l\u00e0 ","select":["A. 1,825%.","B. 9,13%.","C. 5%. ","D. 18,25%."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & N{{a}_{2}}C{{O}_{3}}+2HCl\\to 2NaCl+C{{O}_{2}}\\uparrow +{{H}_{2}}O \\\\ & {{n}_{HCl}}=2{{n}_{\\uparrow }}=2.0,15=0,3\\,mol \\\\ & \\Rightarrow {{m}_{HCl}}=10,95\\,gam\\Rightarrow C{{\\%}_{\\text{dd}\\,\\text{HCl}}}=\\dfrac{10,95}{60}.100\\%=18,25\\% \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2705},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Trong \u0111i\u1ec1u ki\u1ec7n c\u00f3 x\u00fac t\u00e1c, V l\u00edt etilen (\u0111ktc) h\u1ee3p n\u01b0\u1edbc th\u00e0nh r\u01b0\u1ee3u etylic, l\u01b0\u1ee3ng r\u01b0\u1ee3u thu \u0111\u01b0\u1ee3c t\u00e1c d\u1ee5ng h\u1ebft v\u1edbi Na t\u1ea1o th\u00e0nh 11,2 l\u00edt $H_2$ (\u0111ktc). Gi\u00e1 tr\u1ecb c\u1ee7a V l\u00e0 ","select":["A. 11,2. ","B. 22,4. ","C. 33,6. ","D. 4,48."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{C}_{2}}{{H}_{4}}+{{H}_{2}}O\\xrightarrow[{{t}^{0}}]{xt}{{C}_{2}}{{H}_{5}}OH \\\\ & 2{{C}_{2}}{{H}_{5}}OH+2Na\\to 2{{C}_{2}}{{H}_{5}}ONa+{{H}_{2}}\\uparrow \\\\ & {{n}_{{{C}_{2}}{{H}_{4}}}}={{n}_{{{C}_{2}}H_ 5OH}}=2{{n}_{{{H}_{2}}}}=1\\,mol\\Rightarrow {{V}_{{{C}_{2}}{{H}_{4}}}}=22,4\\,L \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2706},{"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 trung h\u00f2a 10ml dung d\u1ecbch $CH_3COOH$ c\u1ea7n 15,2 ml dung d\u1ecbch $NaOH$ 0,2M. V\u1eady n\u1ed3ng \u0111\u1ed9 c\u1ee7a dung d\u1ecbch $CH_3COOH$ l\u00e0 ","select":["A. 0,05 M","B. 0,10 M.","C. 0,304 M.","D. 0,215 M."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & C{{H}_{3}}COOH+NaOH\\to C{{H}_{3}}COONa+{{H}_{2}}O \\\\ & {{n}_{C{{H}_{3}}COOH}}={{n}_{NaOH}}=0,2.\\dfrac{15,2}{1000}=3,{{04.10}^{-3}}\\,mol \\\\ & \\Rightarrow {{C}_{M\\,\\text{dd}\\,\\,C{{H}_{3}}COOH}}=\\dfrac{3,{{04.10}^{-3}}}{{{10.10}^{-3}}}=0,304 M \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2707},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" A l\u00e0 h\u1ee3p ch\u1ea5t c\u00f3 c\u00f4ng th\u1ee9c $XO_2$, trong \u0111\u00f3 %X (theo kh\u1ed1i l\u01b0\u1ee3ng) l\u00e0 27,27%. A l\u00e0:","select":["A. $CO_2$ ","B. $SO_2$ ","C. $SiO_2$ ","D. $NO_2$ "],"hint":"","explain":"<span class='basic_left'>$\\%X=\\dfrac{X}{X+32}.100\\%=27,27\\%\\Rightarrow X\\approx 12\\,(C)$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2708},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" H\u00f2a tan 2,84 gam h\u1ed7n h\u1ee3p g\u1ed3m $CaCO_3$ v\u00e0 $MgCO_3$ c\u1ea7n 30 ml $H_2SO_4$ 1M. V\u1eady th\u00e0nh ph\u1ea7n % v\u1ec1 kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a h\u1ed7n h\u1ee3p mu\u1ed1i ban \u0111\u1ea7u l\u00e0 ","select":["A. 70,42% v\u00e0 29,58% .","B. 71% v\u00e0 29%.","C. 72,5% v\u00e0 27,5%.","D. 75% v\u00e0 25% ."],"hint":"","explain":"<span class='basic_left'>G\u1ecdi s\u1ed1 mol c\u1ee7a $CaCO_3$ v\u00e0 $MgCO_3$ trong h\u1ed7n h\u1ee3p l\u1ea7n l\u01b0\u1ee3t l\u00e0 x, y <br\/> Kh\u1ed1i l\u01b0\u1ee3ng h\u1ed7n h\u1ee3p l\u00e0 2,84 gam n\u00ean ta c\u00f3: 100x + 84y = 2,84 (1) <br\/> $\\begin{aligned} & CaC{{O}_{3}}+{{H}_{2}}S{{O}_{4}}\\to CaS{{O}_{4}}+C{{O}_{2}}\\uparrow +{{H}_{2}}O \\\\ & \\,\\,\\,\\,\\,\\,x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x \\\\ & MgC{{O}_{3}}+{{H}_{2}}S{{O}_{4}}\\to MgS{{O}_{4}}+C{{O}_{2}}\\uparrow +{{H}_{2}}O \\\\ & \\,\\,\\,\\,\\,\\,\\,y\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,y \\\\ & {{n}_{{{H}_{2}}S{{O}_{4}}}}=0,03\\,\\,mol\\Rightarrow x+y=0,03\\,\\,\\,\\,\\,\\,(2) \\\\ \\end{aligned}$ <br\/> Gi\u1ea3i (1) v\u00e0 (2) ta c\u00f3 x = 0,02 mol; y=0,01 mol <br\/> $\\begin{aligned} & {{m}_{CaC{{O}_{3}}}}=0,02.100=2\\,gam\\Rightarrow \\%{{m}_{CaC{{O}_{3}}}}=\\dfrac{2}{2,84}.100\\%\\approx 70,42\\% \\\\ & \\Rightarrow \\%{{m}_{MgC{{O}_{3}}}}=29,58\\% \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2709},{"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 h\u1ebft x gam $C_2H_5OH$ thu \u0111\u01b0\u1ee3c 0,25 mol $CO_2$. \u0110\u1ed1t ch\u00e1y h\u1ebft y gam $CH_3COOH$ thu \u0111\u01b0\u1ee3c 0,25 mol $CO_2$. Cho x gam $C_2H_5OH$ t\u00e1c d\u1ee5ng v\u1edbi y gam $CH_3COOH$ (gi\u1ea3 s\u1eed hi\u1ec7u su\u1ea5t ph\u1ea3n \u1ee9ng l\u00e0 100%). Kh\u1ed1i l\u01b0\u1ee3ng este thu \u0111\u01b0\u1ee3c l\u00e0 ","select":["A. 9 gam. ","B. 10 gam.","C. 11 gam. ","D. 12 gam."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{C}_{2}}{{H}_{5}}OH+3{{O}_{2}}\\xrightarrow{{{t}^{0}}}2C{{O}_{2}}+3{{H}_{2}}O \\\\ & {{n}_{{{C}_{2}}{{H}_{5}}OH}}=\\dfrac{1}{2}{{n}_{C{{O}_{2}}}}=0,125\\,mol \\\\ & C{{H}_{3}}COOH+2{{O}_{2}}\\xrightarrow{{{t}^{0}}}2C{{O}_{2}}+2{{H}_{2}}O \\\\ & {{n}_{C{{H}_{3}}COOH}}=\\dfrac{1}{2}{{n}_{C{{O}_{2}}}}=0,125\\,mol \\\\ & C{{H}_{3}}COOH+{{C}_{2}}{{H}_{5}}OHC{{H}_{3}}COO{{C}_{2}}{{H}_{5}}+{{H}_{2}}O \\\\ \\end{aligned}$ <br\/> T\u1ef7 l\u1ec7 : $\\dfrac{{{n}_{C{{H}_{3}}COOH}}}{1}=\\dfrac{{{n}_{{{C}_{2}}{{H}_{5}}OH}}}{1}$ => C\u1ea3 2 ch\u1ea5t \u0111\u1ec1u ph\u1ea3n \u1ee9ng h\u1ebft.<br\/> $\\begin{aligned} & {{n}_{C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}}}={{n}_{C{{H}_{3}}COOH}}={{n}_{{{C}_{2}}{{H}_{5}}OH}}=0,125\\,mol \\\\ & \\Rightarrow {{m}_{C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}}}=11\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2710}],"lesson":{"save":0,"level":2}}