{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Ch\u1ea5t kh\u00ed n\u00e0o sau \u0111\u00e2y c\u00f3 th\u1ec3 g\u00e2y ch\u1ebft ng\u01b0\u1eddi v\u00ec ng\u0103n c\u1ea3n s\u1ef1 v\u1eadn chuy\u1ec3n oxi trong m\u00e1u ","select":["A. $CO$ ","B. $CO_2$","C. $SO_2$ ","D. $NO$"],"hint":"","explain":"<span class='basic_left'>CO l\u00e0 kh\u00ed c\u1ef1c \u0111\u1ed9c ng\u0103n c\u1ea3n s\u1ef1 v\u1eadn chuy\u1ec3n oxi trong m\u00e1u. N\u1ebfu h\u00edt ph\u1ea3i kh\u00ed CO v\u1edbi n\u1ed3ng \u0111\u1ed9 cao c\u00f3 th\u1ec3 g\u00e2y t\u1eed vong<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2311},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Ph\u1ea3n \u1ee9ng gi\u1eefa $Cl_2$ v\u00e0 dung d\u1ecbch $NaOH$ d\u00f9ng \u0111\u1ec3 \u0111i\u1ec1u ch\u1ebf ","select":["A. thu\u1ed1c t\u00edm","B. n\u01b0\u1edbc javen","C. clorua v\u00f4i","D. kali clorat"],"hint":"","explain":"<span class='basic_left'>$C{{l}_{2}}+2NaOH\\to NaCl+NaClO+{{H}_{2}}O$ <br\/H\u1ed7n h\u1ee3p thu \u0111\u01b0\u1ee3c ch\u00ednh l\u00e0 n\u01b0\u1edbc javen<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2312},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Trong th\u1ef1c t\u1ebf, ng\u01b0\u1eddi ta c\u00f3 th\u1ec3 d\u00f9ng cacbon \u0111\u1ec3 kh\u1eed oxit kim lo\u1ea1i n\u00e0o trong s\u1ed1 c\u00e1c oxit kim lo\u1ea1i d\u01b0\u1edbi \u0111\u00e2y \u0111\u1ec3 s\u1ea3n xu\u1ea5t kim lo\u1ea1i ","select":["A. $Al_2O_3$ ","B. $Na_2O$ ","C. $MgO$ ","D. $Fe_3O_4$"],"hint":"","explain":"<span class='basic_left'>CO c\u00f3 th\u1ec3 kh\u1eed c\u00e1c oxit kim lo\u1ea1i sau Al trong d\u00e3y ho\u1ea1t \u0111\u1ed9ng h\u00f3a h\u1ecdc<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2313},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Trong c\u00e1c ch\u1ea5t sau ch\u1ea5t n\u00e0o c\u00f3 th\u1ec3 tham gia ph\u1ea3n \u1ee9ng v\u1edbi Clo ","select":["A. Oxi ","B. Dung d\u1ecbch NaOH.","C. CuO","D. NaCl"],"hint":"","explain":"<span class='basic_left'>$C{{l}_{2}}+2NaOH\\to NaCl+NaClO+{{H}_{2}}O$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2314},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Nhi\u1ec7t ph\u00e2n ho\u00e0n to\u00e0n mu\u1ed1i $KHCO_3$ s\u1ea3n ph\u1ea9m thu \u0111\u01b0\u1ee3c g\u1ed3m:","select":["A. $K_2CO_3, H_2O$ ","B. $K_2O, CO_2, H_2O$","C. $K_2CO_3, CO_2, H_2O$","D. $K_2O, CO, H_2$"],"hint":"","explain":"<span class='basic_left'>$2KHC{{O}_{3}}\\xrightarrow{{{t}^{0}}}{{K}_{2}}C{{O}_{3}}+C{{O}_{2}}+{{H}_{2}}O$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2315},{"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 ho\u00e0n to\u00e0n dung d\u1ecbch ch\u1ee9a 16,8 gam $MgCO_3$ trong dung d\u1ecbch axit sunfuric lo\u00e3ng thu \u0111\u01b0\u1ee3c V l\u00edt kh\u00ed. Gi\u00e1 tr\u1ecb c\u1ee7a V l\u00e0","select":["A. 2,24 l\u00edt","B. 4,48 l\u00edt","C. 5,6 l\u00edt","D. 7,92 l\u00edt"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & MgC{{O}_{3}}+{H}_{2}{SO}_{4}\\to Mg{{SO}_{4}}+C{{O}_{2}}+{{H}_{2}}O \\\\ & {{n}_{C{{O}_{2}}}}={{n}_{MgC{{O}_{3}}}}=0,2\\,mol \\\\ & {{V}_{C{{O}_{2}}}}=4,48\\,l \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2316},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" L\u01b0u hu\u1ef3nh l\u00e0 ch\u1ea5t r\u1eafn m\u00e0u:","select":["A. Tr\u1eafng","B. \u0110\u1ecf","C. V\u00e0ng ","D. Kh\u00f4ng m\u00e0u"],"hint":"","explain":"<span class='basic_left'>L\u01b0u hu\u1ef3nh l\u00e0 ch\u1ea5t r\u1eafn m\u00e0u v\u00e0ng<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2317},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Phi kim ho\u1ea1t \u0111\u1ed9ng h\u00f3a h\u1ecdc m\u1ea1nh nh\u1ea5t l\u00e0","select":["A. Clo","B. Nit\u01a1","C. Flo","D. Oxi"],"hint":"","explain":"<span class='basic_left'>Flo l\u00e0 phi kim ho\u1ea1t \u0111\u1ed9ng h\u00f3a h\u1ecdc m\u1ea1nh nh\u1ea5t<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2318},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Kh\u1ed1i l\u01b0\u1ee3ng kh\u00ed oxi c\u1ea7n \u0111\u1ec3 \u0111\u1ed1t ch\u00e1y ho\u00e0n to\u00e0n 3,36 l\u00edt kh\u00ed hi\u0111ro l\u00e0:","select":["A. 2,4 gam","B. 6,4 gam ","C. 8 gam ","D. 9,6 gam"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & 2{{H}_{2}}+{{O}_{2}}\\xrightarrow{{{t}^{0}}}2{{H}_{2}}O \\\\ & {{n}_{{{O}_{2}}}}=\\dfrac{1}{2}{{n}_{{{H}_{2}}}}=\\dfrac{1}{2}.0,15=0,075\\,mol \\\\ & {{m}_{{{O}_{2}}}}=2,4\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2319},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" C\u00f3 2 b\u00ecnh kh\u00ed ch\u01b0a kh\u00ed $CO$ v\u00e0 $CO_2$. Thu\u1ed1c th\u1eed n\u00e0o sau \u0111\u00e2y \u0111\u1ec3 ph\u00e2n bi\u1ebft \u0111\u01b0\u1ee3c 2 b\u00ecnh kh\u00ed.","select":["A. $NaOH$","B. $CuCl_2$","C. $Ba(OH)_2$","D. $NaCl$"],"hint":"","explain":"<span class='basic_left'>D\u1eabn 2 kh\u00ed qua dung d\u1ecbch $Ba(OH)_2$. B\u00ecnh kh\u00ed t\u1ea1o k\u1ebft t\u1ee7a tr\u1eafng l\u00e0 b\u00ecnh ch\u1ee9a $CO_2$. <br\/>$C{{O}_{2}}+Ba{{(OH)}_{2}}\\to BaC{{O}_{3}}\\downarrow +{{H}_{2}}O$ <br\/>B\u00ecnh c\u00f2n l\u1ea1i kh\u00f4ng c\u00f3 hi\u1ec7n t\u01b0\u1ee3ng g\u00ec l\u00e0 $CO$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2320},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Nh\u00f3m VIA g\u1ed3m c\u00e1c nguy\u00ean t\u1ed1 ","select":["A. Kim lo\u1ea1i m\u1ea1nh ","B. Kim lo\u1ea1i m\u1ea1nh ","C. Phi kim m\u1ea1nh ","D. Kh\u00ed hi\u1ebfm"],"hint":"","explain":"<span class='basic_left'>C\u00e1c nguy\u00ean t\u1ed1 nh\u00f3m VIA l\u00e0 c\u00e1c nguy\u00ean t\u1ed1 nh\u00f3m oxi, l\u00e0 c\u00e1c phi kim m\u1ea1nh<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2321},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Thu\u1ed1c d\u00f9ng \u0111\u1ec3 ch\u1eefa \u0111au d\u1ea1 d\u00e0y c\u00f3 th\u00e0nh ph\u1ea7n ch\u00ednh l\u00e0","select":["A. $Na_2CO_3$ ","B. $NaHCO_3$","C. $NaCl$","D. $K_2CO_3$"],"hint":"","explain":"<span class='basic_left'>Thu\u1ed1c ch\u1eefa \u0111au d\u1ea1 d\u00e0y c\u00f3 ch\u1ee9a $NaHCO_3$l\u00e0m gi\u1ea3m n\u1ed3ng \u0111\u1ed9 axit trong d\u1ea1 d\u00e0y <br\/>$NaHC{{O}_{3}}+HCl\\to NaCl+C{{O}_{2}}\\uparrow +{{H}_{2}}O$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2322},{"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 ph\u00e2n 100 gam $CaCO_3$ \u0111\u01b0\u1ee3c 33 gam $CO_2$. Hi\u1ec7u su\u1ea5t c\u1ee7a ph\u1ea3n \u1ee9ng l\u00e0 ","select":["A. 75%.","B. 33%.","C. 67%.","D. 42%."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & CaC{{O}_{3}}\\xrightarrow{{{t}^{0}}}CaO+C{{O}_{2}} \\\\ & 100\\,g\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,44\\,g \\\\ & H=\\dfrac{{{m}_{C{{O}_{2\\,TT}}}}}{{{m}_{C{{O}_{2}}\\,LT}}}=\\dfrac{33}{44}.100\\%=75\\% \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2323},{"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 6 gam $C$ th\u00e0nh $CO_2$. Cho to\u00e0n b\u1ed9 s\u1ea3n ph\u1ea9m h\u1ea5p th\u1ee5 v\u00e0o dung d\u1ecbch n\u01b0\u1edbc v\u00f4i trong d\u01b0. Kh\u1ed1i l\u01b0\u1ee3ng k\u1ebft t\u1ee7a t\u1ea1o th\u00e0nh l\u00e0 ","select":["A. 15 gam. ","B. 25 gam.","C. 50 gam ","D. 40 gam."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & C+{{O}_{2}}\\xrightarrow{{{t}^{0}}}C{{O}_{2}} \\\\ & C{{O}_{2}}+Ca{{(OH)}_{2}}\\to CaC{{O}_{3}}\\downarrow +{{H}_{2}}O \\\\ & {{n}_{\\downarrow }}={{n}_{C{{O}_{2}}}}={{n}_{C}}=0,5\\,mol \\\\ & {{m}_{\\downarrow }}=50\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2324},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho 69,6 gam $MnO_2$ t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch $HCl$ \u0111\u1eb7c d\u01b0 thu \u0111\u01b0\u1ee3c bao nhi\u00eau l\u00edt kh\u00ed $Cl_2$ (\u0111ktc) ","select":["A. 4,48 l\u00edt","B. 6,72 l\u00edt","C. 17,92 l\u00edt. ","D. 13,44 l\u00edt."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& Mn{{O}_{2}}+4HC{{l}_{\\text{\u0111}}}\\xrightarrow{{{t}^{0}}}MnC{{l}_{2}}+C{{l}_{2}}+2{{H}_{2}}O \\\\ & {{n}_{C{{l}_{2}}}}={{n}_{Mn{{O}_{2}}}}=0,8\\,\\,mol \\\\ & {{V}_{C{{l}_{2}}}}=17,92\\,l \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2325},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Th\u1ed5i t\u1eeb t\u1eeb \u0111\u1ebfn d\u01b0 kh\u00ed cacbonic v\u00e0o dung d\u1ecbch n\u01b0\u1edbc v\u00f4i trong. Hi\u1ec7n t\u01b0\u1ee3ng quan s\u00e1t \u0111\u01b0\u1ee3c l\u00e0","select":["A. Dung d\u1ecbch n\u01b0\u1edbc v\u00f4i trong v\u1ea9n \u0111\u1ee5c.","B. Kh\u00f4ng c\u00f3 hi\u1ec7n t\u01b0\u1ee3ng g\u00ec","C. Dung d\u1ecbch v\u1ea9n \u0111\u1ee5c sau \u0111\u00f3 tr\u1edf n\u00ean trong su\u1ed1t ","D. Dung d\u1ecbch trong su\u1ed1t sau \u0111\u00f3 v\u1ea9n \u0111\u1ee5c."],"hint":"","explain":"<span class='basic_left'>Ban \u0111\u1ea7u:$C{{O}_{2}}+Ca{{(OH)}_{2}}\\to CaC{{O}_{3}}\\downarrow +{{H}_{2}}O$ khi\u1ebfn dung d\u1ecbch v\u1ea9n \u0111\u1ee5c<br\/>Sau \u0111\u00f3 $CO_2$ d\u01b0 th\u00ec: $2C{{O}_{2}}+Ca{{(OH)}_{2}}\\to Ca{{(HC{{O}_{3}})}_{2}}$ dung d\u1ecbch tr\u1edf n\u00ean trong su\u1ed1t.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2326},{"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 224 ml kh\u00ed HCl (\u0111ktc) v\u00e0o 200 ml n\u01b0\u1edbc. Bi\u1ebft r\u1eb1ng th\u1ec3 t\u00edch c\u1ee7a dung d\u1ecbch thay \u0111\u1ed5i kh\u00f4ng \u0111\u00e1ng k\u1ec3. Dung d\u1ecbch HCl thu \u0111\u01b0\u1ee3c c\u00f3 n\u1ed3ng \u0111\u1ed9 mol l\u00e0","select":["A. 0,5 ","B. 0,05","C. 0,1 ","D. 0,01"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{HCl}}=0,01\\,mol \\\\ & {{C}_{M\\,\\,\\text{dd}}}=\\dfrac{0,01}{0,2}=0,05\\,M \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2327},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" H\u00e0ng n\u0103m th\u1ebf gi\u1edbi c\u1ea7n ti\u00eau th\u1ee5 kho\u1ea3ng 45 tri\u1ec7u t\u1ea5n clo. N\u1ebfu l\u01b0\u1ee3ng clo ch\u1ec9 \u0111\u01b0\u1ee3c \u0111i\u1ec1u ch\u1ebf t\u1eeb mu\u1ed1i \u0103n NaCl theo ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc: $2NaCl\\xrightarrow{\\text{\u0111}pnc}2Na+C{{l}_{2}}$.V\u1edbi hi\u1ec7u su\u1ea5t l\u00e0 95 % th\u00ec l\u01b0\u1ee3ng mu\u1ed1i \u0103n t\u1ed1i thi\u1ec3u c\u1ea7n l\u00e0","select":["A. 72,01 tri\u1ec7u t\u1ea5n","B. 78,06 tri\u1ec7u t\u1ea5n ","C. 75,05 tri\u1ec7u t\u1ea5n ","D. 80,50 tri\u1ec7u t\u1ea5n"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & 2NaCl\\xrightarrow{\\text{\u0111}pnc}2Na+C{{l}_{2}} \\\\ & 117\\text{t.t\u1ea5n}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,71\\text{t.t\u1ea5n} \\\\ & x\\text{t.t\u1ea5n}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,45 \\text{t.t\u1ea5n} \\\\ & x=\\dfrac{45.117}{71}\\approx 74,15\\text{tri\u1ec7u t\u1ea5n} \\\\ & H=95\\%\\Rightarrow {{m}_{NaCl}}=\\dfrac{74,15.100\\%}{95\\%}\\approx 78,06 \\text{ tri\u1ec7u t\u1ea5n} \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2328},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" C\u00f9ng d\u00f9ng m gam ch\u1ea5t X t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch axit clohi\u0111ric \u0111\u1ec3 t\u1ea1o ra kh\u00ed clo. \u0110\u1ec3 thu \u0111\u01b0\u1ee3c l\u01b0\u1ee3ng clo l\u1edbn nh\u1ea5t th\u00ec ch\u1ea5t X l\u00e0:","select":["A. $KMnO_4$","B. $MnO_2$","C. $KClO_3$","D. C\u00e1c ch\u1ea5t cho l\u01b0\u1ee3ng kh\u00ed nh\u01b0 nhau "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & 2KMn{{O}_{4}}+16HC{{l}_{\\text{\u0111}}}\\xrightarrow{{{t}^{0}}}2KCl+2MnC{{l}_{2}}+5C{{l}_{2}}\\uparrow +8{{H}_{2}}O \\\\ & \\dfrac{m}{158}\\,mol\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\dfrac{5}{2}.\\dfrac{m}{158}\\,mol\\, \\\\ & KCl{{O}_{3}}+6HC{{l}_{\\text{\u0111}}}\\xrightarrow{{{t}^{0}}}KCl+3C{{l}_{2}}\\uparrow +3{{H}_{2}}O \\\\ & \\dfrac{m}{122,5}\\,mol\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3.\\dfrac{m}{122,5}\\,mol\\,\\,\\, \\\\ & Mn{{O}_{2}}+4HC{{l}_{\\text{\u0111}}}\\xrightarrow{{{t}^{0}}}2MnC{{l}_{2}}+C{{l}_{2}}\\uparrow +2{{H}_{2}}O \\\\ & \\dfrac{m}{87}\\,mol\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\dfrac{m}{87}\\,mol\\,\\,\\, \\\\ \\end{aligned}$<br\/>V\u1eady d\u00f9ng $KClO_3$ s\u1ebd cho l\u01b0\u1ee3ng kh\u00ed clo l\u1edbn nh\u1ea5t.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2329},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Th\u1ed5i 4,48 l\u00edt kh\u00ed $CO_2$ v\u00e0o 150 ml dung d\u1ecbch n\u01b0\u1edbc v\u00f4i trong 1M thu \u0111\u01b0\u1ee3c kh\u1ed1i l\u01b0\u1ee3ng mu\u1ed1i l\u00e0:","select":["A. 10 gam","B. 8,1 gam","C. 18,1 gam","D. 21,2 gam"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{C{{O}_{2}}}}=0,2\\,mol \\\\ & {{n}_{Ca{{(OH)}_{2}}}}=0,15\\,mol \\\\ & 1<\\dfrac{{{n}_{C{{O}_{2}}}}}{{{n}_{Ca{{(OH)}_{2}}}}}=\\dfrac{4}{3}<2 \\\\ \\end{aligned}$ <br\/>Ph\u1ea3n \u1ee9ng t\u1ea1o 2 mu\u1ed1i. <br\/>G\u1ecdi x, y l\u1ea7n l\u01b0\u1ee3t l\u00e0 s\u00f4 mol c\u1ee7a mu\u1ed1i trung h\u00f2a v\u00e0 mu\u1ed1i axit thu \u0111\u01b0\u1ee3c.<br\/>$\\begin{aligned} & C{{O}_{2}}+Ca{{(OH)}_{2}}\\to CaC{{O}_{3}}\\downarrow +{{H}_{2}}O \\\\ & x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x \\\\ & 2C{{O}_{2}}+Ca{{(OH)}_{2}}\\to Ca{{(HC{{O}_{3}})}_{2}} \\\\ & 2y\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,y\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,y \\\\ & {{n}_{C{{O}_{2}}}}=0,2\\Rightarrow x+2y=0,2\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(1) \\\\ & {{n}_{Ca{{(OH)}_{2}}}}=0,15\\Rightarrow x+y=0,15\\,\\,\\,\\,\\,\\,(2) \\\\ \\end{aligned}$ <br\/>Gi\u1ea3i (1) v\u00e0 (2) ta c\u00f3 x = 0,1; y= 0,05 <br\/>${{m}_{\\text{?}}}=0,1.100+0,05.162=18,1\\,gam$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2330}],"lesson":{"save":0,"level":2}}