{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"C\u00e1c kh\u00ed \u1ea9m n\u00e0o d\u01b0\u1edbi \u0111\u00e2y \u0111\u01b0\u1ee3c l\u00e0m kh\u00f4 b\u1eb1ng $CaO$ ","select":["A. $H_2; O_2 ; N_2$ . ","B. $H_2; CO_2; N_2$.","C. $H_2; O_2; SO_2$","D. $CO_2 ; SO_2; HCl$."],"hint":"","explain":"<span class='basic_left'>C\u00e1c kh\u00ed \u0111\u01b0\u1ee3c l\u00e0m kh\u00f4 b\u1eb1ng $CaO$ th\u00ec kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi $CaO$. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n A th\u1ecfa m\u00e3n<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2151},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Ch\u1ea5t n\u00e0o sau \u0111\u00e2y t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi $HCl$ v\u00e0 $CO_2$ ","select":["A. S\u1eaft ","B. Nh\u00f4m","C. K\u1ebdm ","D. Dung d\u1ecbch NaOH."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& NaOH+HCl\\to NaCl+{{H}_{2}}O \\\\ & C{{O}_{2}}+2NaOH\\to N{{a}_{2}}C{{O}_{3}}+{H}_{2}O \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2152},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Ph\u01b0\u01a1ng ph\u00e1p n\u00e0o sau \u0111\u00e2y \u0111\u01b0\u1ee3c d\u00f9ng \u0111\u1ec3 \u0111i\u1ec1u ch\u1ebf canxioxit trong c\u00f4ng nghi\u1ec7p.","select":["A. Nung \u0111\u00e1 v\u00f4i \u1edf nhi\u1ec7t \u0111\u1ed9 cao l\u00e0 trong c\u00f4ng nghi\u1ec7p ho\u1eb7c l\u00f2 th\u1ee7 c\u00f4ng ","B. Nung $CaSO_4$ trong l\u00f2 c\u00f4ng nghi\u1ec7p ","C. Nung \u0111\u00e1 v\u00f4i tr\u00ean ng\u1ecdn l\u1eeda \u0111\u00e8n c\u1ed3n. ","D. Cho canxi t\u00e1c d\u1ee5ng tr\u1ef1c ti\u1ebfp v\u1edbi oxi. "],"hint":"","explain":"<span class='basic_left'>$CaO $ \u0111\u01b0\u1ee3c \u0111i\u1ec1u ch\u1ebf trong c\u00f4ng nghi\u1ec7p b\u1eb1ng c\u00e1ch nung \u0111\u00e1 v\u00f4i trong l\u00f2 theo ph\u1ea3n \u1ee9ng: <br\/>$CaC{{O}_{3}}\\xrightarrow{{{t}^{0}}}CaO+C{{O}_{2}}\\uparrow $ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":1}]}],"id_ques":2153},{"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 n\u00e0o sau \u0111\u00e2y khi tan trong n\u01b0\u1edbc cho dung d\u1ecbch, l\u00e0m qu\u1ef3 t\u00edm h\u00f3a \u0111\u1ecf ","select":["A. $KOH$ ","B. $KNO_3$","C. $SO_3$ ","D. $CaO$"],"hint":"","explain":"<span class='basic_left'>$SO_3$ tan trong n\u01b0\u1edbc t\u1ea1o th\u00e0nh dung d\u1ecbch axit sunfuric l\u00e0m qu\u1ef3 t\u00edm h\u00f3a \u0111\u1ecf.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2154},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" M\u1ed9t h\u1ed7n h\u1ee3p r\u1eafn g\u1ed3m $Fe_2O_3$ v\u00e0 $CaO$, \u0111\u1ec3 h\u00f2a tan ho\u00e0n to\u00e0n h\u1ed7n h\u1ee3p n\u00e0y ng\u01b0\u1eddi ta ph\u1ea3i d\u00f9ng d\u01b0:","select":["A. N\u01b0\u1edbc. ","B. Dung d\u1ecbch NaOH.","C. Dung d\u1ecbch HCl.","D. Dung d\u1ecbch NaCl."],"hint":"","explain":"<span class='basic_left'>Dung d\u1ecbch $HCl $ v\u1eeba t\u00e1c d\u1ee5ng v\u1edbi $Fe_2O_3$ v\u1eeba t\u00e1c d\u1ee5ng v\u1edbi $CaO$, n\u00ean c\u00f3 th\u1ec3 d\u00f9ng dung d\u1ecbch $HCl$ \u0111\u1ec3 h\u00f2a tan h\u1ed7n h\u1ee3p<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2155},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" T\u1ea1i sao \u1edf \u0111k th\u01b0\u1eddng, c\u00e1c v\u1eadt d\u1ee5ng l\u00e0m b\u1eb1ng nh\u00f4m kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi n\u01b0\u1edbc","select":["A. Do $Al$ th\u1ee5 \u0111\u1ed9ng trong n\u01b0\u1edbc","B. Do l\u1edbp m\u00e0ng oxit b\u1ea3o v\u1ec7.","C. Do b\u1ecdt kh\u00ed $H_2$ t\u1ea1o th\u00e0nh c\u00e1c l\u1edbp m\u00e0ng.","D. Do khi $Al$ ph\u1ea3n \u1ee9ng v\u1edbi $H_2O$ t\u1ea1o ra l\u1edbp m\u00e0ng hi\u0111roxit b\u1ea3o v\u1ec7."],"hint":"","explain":"<span class='basic_left'>Nh\u00f4m r\u1ea5t b\u1ec1n \u1edf \u0111i\u1ec1u ki\u1ec7n th\u01b0\u1eddng do c\u00f3 l\u1edbp m\u00e0ng oxit r\u1ea5t b\u1ec1n b\u1ea3o v\u1ec7.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":2156},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Cho $5,4g$ $Al$ v\u00e0o $500ml$ dd $H_2SO_4$. T\u00ednh n\u1ed3ng \u0111\u1ed9 mol\/l c\u1ee7a ch\u1ea5t thu \u0111c sau ph\u1ea3n \u1ee9ng (coi th\u1ec3 t\u00edch c\u1ee7a dung d\u1ecbch kh\u00f4ng thay \u0111\u1ed5i):","select":["A. $0,2M$","B. $0,6M$","C. $1M$","D. $0,4M$"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& 2Al+3{{H}_{2}}S{{O}_{4}}\\to A{{l}_{2}}{{(S{{O}_{4}})}_{3}}+3{{H}_{2}} \\\\ & {{n}_{A{{l}_{2}}{{(S{{O}_{4}})}_{3}}}}=\\dfrac{1}{2}{{n}_{Al}}=\\dfrac{1}{2}.0,2=0,1\\,mol \\\\ & {{C}_{M\\,\\,A{{l}_{2}}{{(S{{O}_{4}})}_{3}}}}=\\dfrac{0,1}{0,5}=0,2\\,M \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2157},{"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 c\u1ee7a $NaOH$ c\u00f3 trong $200ml$ dd $NaOH \\,\\,2M$ l\u00e0:","select":["A. $16g$ ","B. $23g$.","C. $12g$. ","D. $1,6g$."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{NaOH}}=C_M.V=0,4\\,mol \\\\ & {{m}_{NaOH}}=0,4.40=16\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2158},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Trong c\u00e1c kim lo\u1ea1i sau kim lo\u1ea1i c\u00f3 t\u00ednh kh\u1eed m\u1ea1nh nh\u1ea5t l\u00e0:","select":["A. $Fe$","B. $Na$","C. $K$","D. $Al$"],"hint":"","explain":"<span class='basic_left'>D\u00e3y ho\u1ea1t \u0111\u1ed9ng h\u00f3a h\u1ecdc x\u1ebfp theo chi\u1ec1u t\u00ednh kh\u1eed gi\u1ea3m d\u1ea7n: $K, Na, Mg, Al, Zn, Fe, Pb, H, Cu, Ag, Au$ <br\/> <span class='basic_pink'>Ch\u1ecdn \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2159},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Nh\u1eefng c\u1eb7p ch\u1ea5t n\u00e0o sau \u0111\u00e2y c\u00f9ng t\u1ed3n t\u1ea1i trong m\u1ed9t dung d\u1ecbch","select":["A. $KCl$ v\u00e0 $NaNO_3$","B. $KOH$ v\u00e0 $HCl$","C. $Na_3PO_4$ v\u00e0 $CaCl_2$","D. $HBr$ v\u00e0 $AgNO_3$."],"hint":"","explain":"<span class='basic_left'>C\u00e1c ch\u1ea5t c\u00f9ng t\u1ed3n t\u1ea1i \u0111\u01b0\u1ee3c trong 1 dung d\u1ecbch khi ch\u00fang kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi nhau. <br\/> Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n A th\u1ecfa m\u00e3n.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2160},{"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 2 l\u1ecd m\u1ea5t nh\u00e3n $H_2SO_4$ v\u00e0 $Na_2SO_4$, ta s\u1eed d\u1ee5ng thu\u1ed1c th\u1eed n\u00e0o sau \u0111\u00e2y","select":["A. $HCl$","B. Gi\u1ea5y qu\u1ef3 t\u00edm","C. $NaOH$","D. $BaCl_2$"],"hint":"","explain":"<span class='basic_left'>Nh\u00fang qu\u1ef3 t\u00edm v\u00e0o 2 dung d\u1ecbch. <br\/> Dung d\u1ecbch n\u00e0o khi\u1ebfn qu\u1ef3 chuy\u1ec3n \u0111\u1ecf th\u00ec \u0111\u00f3 ch\u00ednh l\u00e0 $H_2SO_4$.<br\/> Dung d\u1ecbch c\u00f2n l\u1ea1i kh\u00f4ng l\u00e0m qu\u1ef3 chuy\u1ec3n m\u00e0u l\u00e0 dung d\u1ecbch $Na_2SO_4$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2161},{"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 ho\u00e0n to\u00e0n $29,4$ gam \u0111\u1ed3ng (II) hidroxit b\u1eb1ng dd axit sunfuric. S\u1ed1 gam mu\u1ed1i thu \u0111\u01b0\u1ee3c sau ph\u1ea3n \u1ee9ng l\u00e0:","select":["A. $48$ gam","B. $9,6$ gam","C. $4,8$ gam ","D. $24$ gam"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& Cu{{(OH)}_{2}}+{{H}_{2}}S{{O}_{4}}\\to CuS{{O}_{4}}+2{{H}_{2}}O \\\\ & {{n}_{CuS{{O}_{4}}}}={{n}_{Cu{{(OH)}_{2}}}}=0,3mol \\\\ & {{m}_{CuS{{O}_{4}}}}=48\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2162},{"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 ch\u1ea5t h\u00f3a h\u1ecdc chung c\u1ee7a kim lo\u1ea1i g\u1ed3m:","select":["A. T\u00e1c d\u1ee5ng v\u1edbi phi kim, t\u00e1c d\u1ee5ng v\u1edbi axit","B. T\u00e1c d\u1ee5ng v\u1edbi phi kim, t\u00e1c d\u1ee5ng v\u1edbi baz\u01a1, t\u00e1c d\u1ee5ng v\u1edbi mu\u1ed1i","C. T\u00e1c d\u1ee5ng v\u1edbi phi kim, t\u00e1c d\u1ee5ng v\u1edbi axit, t\u00e1c d\u1ee5ng v\u1edbi mu\u1ed1i","D. T\u00e1c d\u1ee5ng v\u1edbi oxit baz\u01a1, t\u00e1c d\u1ee5ng v\u1edbi axit"],"hint":"","explain":"<span class='basic_left'>H\u1ea7u h\u1ebft c\u00e1c kim lo\u1ea1i c\u00f3 c\u00e1c t\u00ednh ch\u1ea5t h\u00f3a h\u1ecdc chung:<br\/>T\u00e1c d\u1ee5ng v\u1edbi phi kim, t\u00e1c d\u1ee5ng v\u1edbi axit, t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch mu\u1ed1i.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2163},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" $40ml$ dd $H_2SO_4$ $8M$ \u0111\u01b0\u1ee3c pha lo\u00e3ng \u0111\u1ebfn $160ml$. N\u1ed3ng \u0111\u1ed9 mol c\u1ee7a dd $H_2SO_4$ sau khi pha lo\u00e3ng l\u00e0 bao nhi\u00eau: ","select":["A. $2M $","B. $1M $ ","C. $0,1M $","D. $0,2M$"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{{{H}_{2}}S{{O}_{4}}}}=8.0,04 = 0,32\\,mol \\\\ & {{C}_{M\\,\\text{dd}}}=\\dfrac{0,32}{0,16}=2\\,M \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2164},{"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 $5,04 l$ kh\u00ed oxi \u1edf \u0111ktc c\u1ea7n ph\u1ea3i d\u00f9ng bao nhi\u00eau gam $KClO_3$?","select":["A. $18g$","B. $18,4g$","C. $18,375g$ ","D. $20,3g$"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& 2KCl{{O}_{3}}\\xrightarrow{{{t}^{0}}}2KCl+3{{O}_{2}} \\\\ & {{n}_{KCl{{O}_{3}}}}=\\dfrac{2}{3}{{n}_{{{O}_{2}}}}=0,15\\,mol \\\\ & {{m}_{KCl{{O}_{3}}}}=18,375gam \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2165},{"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 $1,3g$ k\u1ebdm v\u00e0o $14,7g$ dung d\u1ecbch $H_2SO_4$ $20\\%$. Khi ph\u1ea3n \u1ee9ng k\u1ebft th\u00fac kh\u1ed1i l\u01b0\u1ee3ng hi\u0111ro thu \u0111\u01b0\u1ee3c l\u00e0","select":["A. $0,03g$","B. $0,04g$","C. $0,05g$ ","D. $0,06g$"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& Zn+{{H}_{2}}S{{O}_{4}}\\to ZnS{{O}_{4}}+{{H}_{2}} \\\\ & {{n}_{Zn}}=0,02\\,mol \\\\ & {{m}_{{{H}_{2}}S{{O}_{4}}}}=\\dfrac{14,7.20\\%}{100\\%}\\,=2,94gam \\\\ & {{n}_{{{H}_{2}}S{{O}_{4}}}}=0,03\\,mol \\\\ & \\dfrac{{{n}_{Zn}}}{1}<\\dfrac{{{n}_{{{H}_{2}}S{{O}_{4}}}}}{1} \\\\ \\end{aligned}$ <br\/> Suy ra, $Zn$ h\u1ebft, t\u00ednh theo $Zn$<br\/>$\\begin{aligned}& {{n}_{{{H}_{2}}}}={{n}_{Zn}}=0,02\\,mol \\\\ & {{m}_{{{H}_{2}}}}=0,04\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2166},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho $x$ gam s\u1eaft tan ho\u00e0n to\u00e0n trong dd c\u00f3 ch\u1ee9a $H_2SO_4$ \u0111\u1eb7c n\u00f3ng thu \u0111\u01b0\u1ee3c $5,04$ l\u00edt kh\u00ed $SO_2$ (s\u1ea3n ph\u1ea9m kh\u1eed duy nh\u1ea5t). Gi\u00e1 tr\u1ecb c\u1ee7a $x$ l\u00e0:","select":["A. $5,6$ gam","B. $7$ gam","C. $8,4$ gam","D. $11,2$ gam"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& 2Fe+6{{H}_{2}}S{{O}_{4\\text{\u0111}}}\\xrightarrow{{{t}^{0}}}F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+3S{{O}_{2}}\\uparrow +6{{H}_{2}}O \\\\ & {{n}_{Fe}}=\\dfrac{2}{3}{{n}_{S{{O}_{2}}}}=0,15\\,mol \\\\ & {{m}_{Fe}}=8,4\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2167},{"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 $11,2$ gam $KOH$ $20\\%$, th\u00ec c\u1ea7n l\u1ea5y bao nhi\u00eau gam dung d\u1ecbch axit $H_2SO_4$ $35\\%$","select":["A. 9 gam","B. 4,6 gam ","C. 5,6 gam ","D. 1,7 gam"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& 2KOH+{{H}_{2}}S{{O}_{4}}\\to {{K}_{2}}S{{O}_{4}}+2{{H}_{2}}O \\\\ & {{m}_{KOH}}=\\dfrac{11,2.20\\%}{100\\%}=2,24\\,gam \\\\ & {{n}_{{{H}_{2}}S{{O}_{4}}}}=\\dfrac{1}{2}{{n}_{KOH}}=0,02\\,mol \\\\ & {{m}_{{{H}_{2}}S{{O}_{4}}}}=1,96gam \\\\ & {{m}_{\\text{dd}\\,{{H}_{2}}S{{O}_{4}}\\,}}=\\dfrac{1,96.100\\%}{35\\%}=5,6\\,gam \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2168},{"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 $1,44g$ kim lo\u1ea1i $M$ h\u00f3a tr\u1ecb $II$ b\u1eb1ng $250ml$ dung d\u1ecbch $H_2SO_4$ $0,3M$. \u0110\u1ec3 trung h\u00f2a l\u01b0\u1ee3ng axit d\u01b0 c\u1ea7n d\u00f9ng $60ml$ dung d\u1ecbch $NaOH$ $0,5M$. \u0110\u00f3 l\u00e0 kim lo\u1ea1i g\u00ec? ","select":["A. Ca ","B. Mg ","C. Zn ","D. Ba"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& M+{{H}_{2}}S{{O}_{4}}\\to {{M}_{2}}S{{O}_{4}}+{{H}_{2}}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(1) \\\\ & {{H}_{2}}S{{O}_{4\\text{d\u01b0}}}+2NaOH\\to N{{a}_{2}}S{{O}_{4}}+2{{H}_{2}}O\\,\\,\\,\\,\\,\\,\\,(2) \\\\ & {{n}_{{{H}_{2}}S{{O}_{4\\,(\\text{\u1edf}\\,\\,2)}}}}=\\dfrac{1}{2}{{n}_{NaOH}}=\\dfrac{1}{2}.0,03=0,015\\,mol \\\\ & {{n}_{{{H}_{2}}S{{O}_{4\\,(\\text{b\u0111}\\,\\text{)}}}}}=0,075\\,mol \\\\ & {{n}_{{{H}_{2}}S{{O}_{4(\\text{\u1edf}\\,\\,1)}}\\,\\,}}=0,06\\,mol \\\\ & {{n}_{M}}={{n}_{{{H}_{2}}S{{O}_{4(\\text{\u1edf}\\,\\,1)}}\\,\\,}}=0,06\\,mol \\\\ & {{M}_{M}}=\\dfrac{1,44}{0,06}=24\\,(Mg) \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2169},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Kh\u1eed ho\u00e0n to\u00e0n $15,15g$ h\u1ed7n h\u1ee3p 2 oxit l\u00e0 $CuO$ v\u00e0 $PbO$ b\u1eb1ng kh\u00ed $CO$ \u1edf nhi\u1ec7t \u0111\u1ed9 cao. Kh\u00ed sinh ra sau ph\u1ea3n \u1ee9ng \u0111\u01b0\u1ee3c d\u1eabn v\u00e0o b\u00ecnh \u0111\u1ef1ng dd $Ca(OH)_2$ d\u01b0 thu \u0111\u01b0\u1ee3c $10g$ k\u1ebft t\u1ee7a. Kh\u1ed1i l\u01b0\u1ee3ng $Cu$ thu \u0111\u01b0\u1ee3c l\u00e0:","select":["A. 2,3g ","B. 2,4g","C. 3,2g","D. 2,5g"],"hint":"","explain":"<span class='basic_left'>G\u1ecdi $x, y$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 s\u1ed1 mol c\u1ee7a $CuO$ v\u00e0 $PbO$ trong h\u1ed7n h\u1ee3p.<br\/>Kh\u1ed1i l\u01b0\u1ee3ng h\u1ed7n h\u1ee3p b\u1eb1ng $4$ gam n\u00ean $80x +223y = 15,15$ (1)<br\/>$\\begin{aligned}& CuO+CO\\xrightarrow{{{t}^{0}}}Cu+C{{O}_{2}} \\\\ & x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x \\\\ & PbO+CO\\xrightarrow{{{t}^{0}}}Pb+C{{O}_{2}} \\\\ & y\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,y \\\\ & C{{O}_{2}}+Ca{{(OH)}_{2}}\\to CaC{{O}_{3}}\\downarrow +{{H}_{2}}O \\\\ & {{n}_{C{{O}_{2}}}}={{n}_{\\downarrow }}=0,1\\,mol \\\\ & =>x+y=0,1\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(2) \\\\ \\end{aligned}$<br\/> Gi\u1ea3i (1) v\u00e0 (2) ta c\u00f3 $x = y = 0,05$ mol<br\/>${{m}_{Cu}}=3,2\\,gam$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2170}],"lesson":{"save":0,"level":2}}