{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Dung d\u1ecbch axit clohi\u0111ric t\u00e1c d\u1ee5ng v\u1edbi s\u1eaft t\u1ea1o th\u00e0nh:","select":["A. S\u1eaft (II) clorua v\u00e0 kh\u00ed hi\u0111r\u00f4. ","B. S\u1eaft (III) clorua v\u00e0 kh\u00ed hi\u0111r\u00f4.","C. S\u1eaft (II) Sunfua v\u00e0 kh\u00ed hi\u0111r\u00f4.","D. S\u1eaft (II) clorua v\u00e0 n\u01b0\u1edbc."],"hint":"","explain":"<span class='basic_left'>dung d\u1ecbch HCl t\u00e1c d\u1ee5ng v\u1edbi Fe theo ph\u1ea3n \u1ee9ng: <br\/> $Fe+2HCl\\to FeC{{l}_{2}}+{{H}_{2}}\\uparrow $ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":1}]}],"id_ques":1811},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Dung d\u1ecbch axit clohi\u0111ric t\u00e1c d\u1ee5ng v\u1edbi \u0111\u1ed3ng (II) hi\u0111r\u00f4xit t\u1ea1o th\u00e0nh dung d\u1ecbch m\u00e0u:","select":["A. V\u00e0ng \u0111\u1eadm. ","B. \u0110\u1ecf.","C. Xanh lam. ","D. Da cam."],"hint":"","explain":"<span class='basic_left'>$2HCl+Cu{{(OH)}_{2}}\\to CuC{{l}_{2}}+{{2H}_{2}}O$ <br\/>C\u00e1c mu\u1ed1i c\u1ee7a Cu c\u00f3 m\u00e0u xanh lam<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1812},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Mu\u1ed1n pha lo\u00e3ng axit sunfuric \u0111\u1eb7c ta ph\u1ea3i:","select":["A. R\u00f3t n\u01b0\u1edbc v\u00e0o axit \u0111\u1eb7c. ","B. R\u00f3t t\u1eeb t\u1eeb n\u01b0\u1edbc v\u00e0o axit \u0111\u1eb7c.","C. R\u00f3t nhanh axit \u0111\u1eb7c v\u00e0o n\u01b0\u1edbc. ","D. R\u00f3t t\u1eeb t\u1eeb axit \u0111\u1eb7c v\u00e0o n\u01b0\u1edbc"],"hint":"","explain":"<span class='basic_left'>Khi pha lo\u00e3ng dung d\u1ecbch $H_2SO_4$ \u0111\u1eb7c ph\u1ea3i r\u00f3t t\u1eeb t\u1eeb axit v\u00e0o n\u01b0\u1edbc v\u00e0 khu\u1ea5y \u0111\u1ec1u. Tuy\u1ec7t \u0111\u1ed1i kh\u00f4ng l\u00e0m NG\u01af\u1ee2C L\u1ea0I.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":1813},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Khi nh\u1ecf t\u1eeb t\u1eeb $H_2SO_4$ \u0111\u1eadm \u0111\u1eb7c v\u00e0o \u0111\u01b0\u1eddng ch\u1ee9a trong c\u1ed1c hi\u1ec7n t\u01b0\u1ee3ng quan s\u00e1t \u0111\u01b0\u1ee3c l\u00e0:","select":["A. S\u1ee7i b\u1ecdt kh\u00ed, \u0111\u01b0\u1eddng kh\u00f4ng tan. ","B. M\u00e0u tr\u1eafng c\u1ee7a \u0111\u01b0\u1eddng m\u1ea5t d\u1ea7n, kh\u00f4ng s\u1ee7i b\u1ecdt.","C. M\u00e0u \u0111en xu\u1ea5t hi\u1ec7n v\u00e0 c\u00f3 b\u1ecdt kh\u00ed sinh ra. ","D. M\u00e0u \u0111en xu\u1ea5t hi\u1ec7n, kh\u00f4ng c\u00f3 b\u1ecdt kh\u00ed sinh ra."],"hint":"","explain":"<span class='basic_left'>Axit $H_2SO_4$ \u0111\u1eb7c c\u00f3 t\u00ednh h\u00e1o n\u01b0\u1edbc, s\u1ebd l\u1ea5y n\u01b0\u1edbc c\u1ee7a \u0111\u01b0\u1eddng t\u1ea1o th\u00e0nh C (m\u00e0u \u0111en). Sau \u0111\u00f3 1 ph\u1ea7n $C$ t\u00e1c d\u1ee5ng v\u1edbi $H_2SO_4$ \u0111\u1eb7c sinh ra kh\u00ed $CO_2$ v\u00e0 $SO_2$ s\u1ee7i b\u1ecdt kh\u00ed l\u00e0m $C$ d\u00e2ng l\u00ean: <br\/>$\\begin{aligned}& {{C}_{12}}{{H}_{22}}{{O}_{11}}\\xrightarrow{{{H}_{2}}S{{O}_{4}}}12C+11{{H}_{2}}O \\\\ & C+2{{H}_{2}}S{{O}_{4}}\\to C{{O}_{2}}+2S{{O}_{2}}+2{{H}_{2}}O \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":1814},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Nh\u1ecf t\u1eeb t\u1eeb dung d\u1ecbch axit clohi\u0111ric v\u00e0o c\u1ed1c \u0111\u1ef1ng m\u1ed9t m\u1ea9u \u0111\u00e1 v\u00f4i cho \u0111\u1ebfn d\u01b0 axit. Hi\u1ec7n t\u01b0\u1ee3ng n\u00e0o sau \u0111\u00e2y x\u1ea3y ra ?","select":["A. S\u1ee7i b\u1ecdt kh\u00ed, \u0111\u00e1 v\u00f4i kh\u00f4ng tan. ","B. \u0110\u00e1 v\u00f4i tan d\u1ea7n, kh\u00f4ng s\u1ee7i b\u1ecdt kh\u00ed. ","C. Kh\u00f4ng s\u1ee7i b\u1ecdt kh\u00ed, \u0111\u00e1 v\u00f4i kh\u00f4ng tan. ","D. S\u1ee7i b\u1ecdt kh\u00ed, \u0111\u00e1 v\u00f4i tan d\u1ea7n."],"hint":"","explain":"<span class='basic_left'>$HCl$ t\u00e1c d\u1ee5ng v\u1edbi \u0111\u00e1 v\u00f4i $(CaCO_3)$ sinh ra kh\u00ed $CO_2$ theo ph\u1ea3n \u1ee9ng: <br\/> $2HCl+CaC{{O}_{3}}\\to CaC{{l}_{2}}+{{H}_{2}}O+C{{O}_{2}}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":1815},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" D\u00e3y c\u00e1c ch\u1ea5t thu\u1ed9c lo\u1ea1i axit l\u00e0:","select":["A. $HCl, H_2SO_4, Na_2S, H_2S.$","B. $Na_2SO_4, H_2SO_4, HNO_3, H_2S$","C. $HCl, H_2SO_4, HNO_3, Na_2S.$ ","D. $HCl, H_2SO_4, HNO_3, H_2S.$"],"hint":"","explain":"<span class='basic_left'>Axit l\u00e0 h\u1ee3p ch\u1ea5t khi H li\u00ean k\u1ebft v\u1edbi c\u00e1c g\u1ed1c axit. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n D th\u1ecfa m\u00e3n<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1816},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" \u0110\u1ec3 nh\u1eadn bi\u1ebft g\u1ed1c sunfat $(= SO_4)$ ng\u01b0\u1eddi ta d\u00f9ng mu\u1ed1i n\u00e0o sau \u0111\u00e2y ?","select":["A. $BaCl_2.$","B. $NaCl.$","C. $CaCl_2$. ","D. $MgCl_2$"],"hint":"","explain":"<span class='basic_left'>C\u00e1c mu\u1ed1i c\u00f3 g\u1ed1c sunfat \u0111\u01b0\u1ee3c nh\u1eadn bi\u1ebft b\u1eb1ng c\u00e1c mu\u1ed1i c\u1ee7a $Ba$ sinh ra $BaSO_4$ k\u1ebft t\u1ee7a tr\u1eafng<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1817},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Cho 5,6 g s\u1eaft t\u00e1c d\u1ee5ng v\u1edbi axit clohi\u0111ric d\u01b0, sau ph\u1ea3n \u1ee9ng th\u1ec3 t\u00edch kh\u00ed $H_2$ thu \u0111\u01b0\u1ee3c (\u1edf \u0111ktc):","select":["A. 1,12 l\u00edt ","B. 2,24 l\u00edt.","C. 11,2 l\u00edt. ","D. 22,4 l\u00edt."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{{{H}_{2}}}}=\\dfrac{5,6}{56}=0,1\\,mol \\\\ & Fe+2HCl\\to FeC{{l}_{2}}+{{H}_{2}} \\\\ & {{n}_{{{H}_{2}}}}={{n}_{Fe}}=0,1\\,mol \\\\ & {{V}_{{{H}_{2}}}}=0,1.22,4=2,24\\,l \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1818},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Khi cho 1 m\u1ea9u $Cu$ v\u00e0o \u1ed1ng ngi\u1ec7m ch\u1ee9a dung d\u1ecbch $H_2SO_4$ lo\u00e3ng hi\u1ec7n t\u01b0\u1ee3ng quan s\u00e1t \u0111\u01b0\u1ee3c l\u00e0:","select":["A. C\u00f3 kh\u00ed m\u00f9i h\u1eafc tho\u00e1t ra, dung d\u1ecbch m\u00e0u xanh","B. C\u00f3 kh\u00ed m\u00f9i h\u1eafc tho\u00e1t ra, dung d\u1ecbch kh\u00f4ng m\u00e0u","C. Kh\u00f4ng c\u00f3 hi\u1ec7n t\u01b0\u1ee3ng g\u00ec ","D. C\u00f3 kh\u00ed kh\u00f4ng m\u00e0u tho\u00e1t ra, dung d\u1ecbch m\u00e0u xanh."],"hint":"","explain":"<span class='basic_left'>$Cu$ kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi $H_2SO_4$ lo\u00e3ng n\u00ean kh\u00f4ng c\u00f3 hi\u1ec7n t\u01b0\u1ee3ng g\u00ec <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":1819},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" \u0110\u1ec3 l\u00e0m s\u1ea1ch dung d\u1ecbch $FeCl_2$ c\u00f3 l\u1eabn t\u1ea1p ch\u1ea5t $CuCl_2$ ta d\u00f9ng:","select":["A. $H_2SO_4$ .","B. $HCl$.","C. $Al$. ","D. $Fe$."],"hint":"","explain":"<span class='basic_left'>\u0110\u1ec3 s\u1ea1ch dung d\u1ecbch $FeCl_2$ c\u00f3 l\u1eabn t\u1ea1p ch\u1ea5t $CuCl_2$ ta cho h\u1ed7n h\u1ee3p t\u00e1c d\u1ee5ng v\u1edbi ch\u1ea5t m\u00e0 $CuCl_2$ t\u00e1c d\u1ee5ng c\u00f2n $FeCl_2$ kh\u00f4ng t\u00e1c d\u1ee5ng. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n D th\u1ecfa m\u00e3n<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":1820},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" D\u00e3y c\u00e1c oxit t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi dung d\u1ecbch HCl:","select":["A. $CO, CaO, CuO, FeO .$","B. $NO, Na_2O, CuO, Fe_2O_3$","C. $SO_2, CaO, CuO, FeO$","D. $CuO, CaO, Na_2O, FeO$"],"hint":"","explain":"<span class='basic_left'>C\u00e1c oxit baz\u01a1 s\u1ebd t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi dung d\u1ecbch $HCl$. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n D th\u1ecfa m\u00e3n<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":1821},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Trung ho\u00e0 200g dung d\u1ecbch $HCl$ 3,65% b\u1eb1ng dung d\u1ecbch $KOH$ 1M . Th\u1ec3 t\u00edch dung d\u1ecbch $KOH$ 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'>$\\begin{aligned}& {{m}_{HCl}}=\\dfrac{200.3,65%}{100%}=7,3\\,gam \\\\ & {{n}_{HCl}}=\\dfrac{7,3}{36,5}=0,2\\,mol \\\\ & HCl+KOH\\to KCl+{{H}_{2}}O \\\\ & {{n}_{KOH}}={{n}_{HCl}}=0,2\\,mol \\\\ & {{V}_{KOH}}=\\dfrac{0,2}{1}=0,2\\,l=200\\,ml \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1822},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Pha dung d\u1ecbch ch\u1ee9a 1 g NaOH v\u1edbi dung d\u1ecbch ch\u1ee9a 1 g HCl sau ph\u1ea3n \u1ee9ng thu \u0111\u01b0\u1ee3c dung d\u1ecbch c\u00f3 m\u00f4i tr\u01b0\u1eddng:","select":["A. Ax\u00edt ","B. Trung t\u00ednh","C. Baz\u01a1 ","D. Kh\u00f4ng x\u00e1c \u0111\u1ecbnh."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{HCl}}=\\dfrac{1}{40}=0,025\\,mol \\\\ & {{n}_{HCl}}=\\dfrac{1}{36,5}\\approx 0,03\\,mol \\\\ & NaOH+HCl\\to NaCl+{{H}_{2}}O \\\\ & \\dfrac{{{n}_{NaOH}}}{1}<\\dfrac{{{n}_{HCl}}}{1} \\\\ \\end{aligned}$ <br\/>V\u1eady HCl d\u01b0, dung d\u1ecbch c\u00f3 m\u00f4i tr\u01b0\u1eddng axit<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1823},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Cho 5,6 gam s\u1eaft t\u00e1c d\u1ee5ng v\u1edbi 5,6 l\u00edt kh\u00ed $Cl_2$ (\u0111ktc). Sau ph\u1ea3n \u1ee9ng thu \u0111\u01b0\u1ee3c m\u1ed9t l\u01b0\u1ee3ng mu\u1ed1i clorua l\u00e0:","select":["A. 16,25 g ","B. 15,25 g","C. 17,25 g. ","D. 16,20 g."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{Fe}}=\\dfrac{5,6}{56}=0,1\\,mol \\\\ & {{n}_{C{{l}_{2}}}}=\\dfrac{5,6}{22,4}=0,25\\,mol \\\\ & 2Fe+3C{{l}_{2}}\\to 2FeC{{l}_{3}} \\\\ & \\dfrac{{{n}_{Fe}}}{2}<\\dfrac{{{n}_{C{{l}_{2}}}}}{3} \\\\ \\end{aligned}$ <br\/>Fe h\u1ebft, t\u00ednh theo Fe <br\/>$\\begin{aligned}& {{n}_{FeC{{l}_{3}}}}={{n}_{Fe}}=0,1\\,mol \\\\ & {{m}_{FeC{{l}_{3}}}}=0,1.162,5=16,25\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1824},{"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 l\u00e0m kh\u00f4 m\u1ed9t m\u1eabu kh\u00ed $SO_2$ \u1ea9m c\u00f3 (l\u1eabn h\u01a1i n\u01b0\u1edbc) ta d\u1eabn m\u1eabu kh\u00ed n\u00e0y qua:","select":["A. $NaOH$ \u0111\u1eb7c ","B. N\u01b0\u1edbc v\u00f4i trong d\u01b0","C. $H_2SO_4$ \u0111\u1eb7c. ","D. Dung d\u1ecbch $HCl$"],"hint":"","explain":"<span class='basic_left'>\u0110\u1ec3 l\u00e0m kh\u00f4 m\u1ed9t ch\u1ea5t kh\u00ed ta cho kh\u00ed \u0111\u00f3 qua ch\u1ea5t h\u00e1o n\u01b0\u1edbc v\u00e0 ch\u1ea5t h\u00e1o n\u01b0\u1edbc \u0111\u00f3 kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi kh\u00ed l\u00e0m kh\u00f4. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n C th\u1ecfa m\u00e3n<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1825},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho c\u00f9ng m\u1ed9t l\u01b0\u1ee3ng s\u1eaft v\u00e0 k\u1ebdm t\u00e1c d\u1ee5ng h\u1ebft v\u1edbi axit clohi\u0111ric:","select":["A. L\u01b0\u1ee3ng $H_2$ tho\u00e1t ra t\u1eeb s\u1eaft nhi\u1ec1u h\u01a1n k\u1ebdm . ","B. L\u01b0\u1ee3ng $H_2$ tho\u00e1t ra t\u1eeb k\u1ebdm nhi\u1ec1u h\u01a1n s\u1eaft.","C. L\u01b0\u1ee3ng $H_2$ thu \u0111\u01b0\u1ee3c t\u1eeb s\u1eaft v\u00e0 k\u1ebdm nh\u01b0 nhau. ","D. L\u01b0\u1ee3ng $H_2$ tho\u00e1t ra t\u1eeb s\u1eaft g\u1ea5p 2 l\u1ea7n l\u01b0\u1ee3ng $H_2$ tho\u00e1t ra t\u1eeb k\u1ebdm"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& Fe+2HCl\\to FeC{{l}_{2}}+{{H}_{2}} \\\\ & Zn+2HCl\\to ZnC{{l}_{2}}+{{H}_{2}} \\\\ \\end{aligned}$ <br\/>S\u1ed1 mol $H_2$ sinh ra \u0111\u1ec1u b\u1eb1ng s\u1ed1 mol c\u1ee7a $Fe, Zn$ n\u00ean l\u01b0\u1ee3ng $H_2$ sinh ra l\u00e0 nh\u01b0 nhau.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":1826},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho magi\u00ea t\u00e1c d\u1ee5ng v\u1edbi axit sunfuric \u0111\u1eb7c n\u00f3ng sinh ra kh\u00ed $SO_2$. T\u1ed5ng h\u1ec7 s\u1ed1 trong ph\u01b0\u01a1ng tr\u00ecnh ho\u00e1 h\u1ecdc l\u00e0:","select":["A. 5 ","B. 6","C. 7 ","D. 8"],"hint":"","explain":"<span class='basic_left'>$Mg+2{{H}_{2}}S{{O}_{4}}\\to MgS{{O}_{4}}+S{{O}_{2}}+2{{H}_{2}}O$ <br\/>T\u1ed5ng h\u1ec7 s\u1ed1 = 1 +2 + 1 + 1 +2 = 7<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1827},{"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 kh\u00f4ng th\u1ec3 \u0111\u1ed3ng th\u1eddi t\u1ed3n t\u1ea1i trong m\u1ed9t dung d\u1ecbch:","select":["A. $NaOH, K_2SO_4$ ","B. $HCl, Na_2SO_4$.","C. $H_2SO_4, KNO_3.$","D. $HCl, AgNO_3.$"],"hint":"","explain":"<span class='basic_left'>C\u00e1c ch\u1ea5t tan ph\u1ea3n \u1ee9ng v\u1edbi nhua s\u1ebd kh\u00f4ng t\u1ed3n t\u1ea1i \u0111\u01b0\u1ee3c trong 1 dung d\u1ecbch. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n D th\u1ecfa m\u00e3n <br\/>$HCl+AgN{{O}_{3}}\\to AgCl\\downarrow +HN{{O}_{3}}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":1828},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Ho\u00e0 tan h\u1ebft 3,6 g m\u1ed9t kim lo\u1ea1i M ho\u00e1 tr\u1ecb II b\u1eb1ng dung d\u1ecbch $H_2SO_4$ lo\u00e3ng \u0111\u01b0\u1ee3c 3,36 l\u00edt $H_2$ (\u0111ktc). Kim lo\u1ea1i l\u00e0:","select":["A. Zn","B. Mg","C. Fe ","D. Ca"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& M+{{H}_{2}}S{{O}_{4}}\\to MS{{O}_{4}}+{{H}_{2}} \\\\ & {{n}_{{{H}_{2}}}}=\\dfrac{3,36}{22,4}=0,15\\,mol \\\\ & {{n}_{M}}={{n}_{{{H}_{2}}}}=0,15\\,mol \\\\ & {{M}_{M}}=\\dfrac{3,6}{0,15}=24\\,(Mg) \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1829},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Cho 8 g h\u1ed7n h\u1ee3p $Fe$ v\u00e0 $Mg$ t\u00e1c d\u1ee5ng ho\u00e0n to\u00e0n v\u1edbi dung d\u1ecbch $H_2SO_4$ \u0111\u1eb7c n\u00f3ng d\u01b0 sinh ra 5,6 l\u00edt kh\u00ed $SO_2$ (\u0111ktc). Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a Mg trong h\u1ed7n h\u1ee3p l\u00e0:","select":["A. 1,2 g ","B. 2,4 g","C. 3,6 g ","D. 4,8 g"],"hint":"","explain":"<span class='basic_left'>G\u1ecdi x, y l\u1ea7n l\u01b0\u1ee3t l\u00e0 s\u1ed1 mol c\u1ee7a Fe, Mg trong h\u1ed7n h\u1ee3p. <br\/>Kh\u1ed1i l\u01b0\u1ee3ng h\u1ed7n h\u1ee3p b\u1eb1ng 8 gam n\u00ean ta c\u00f3: <br\/>56x + 24y = 8 (1) <br\/>$\\begin{aligned}& {{n}_{S{{O}_{2}}}}=\\dfrac{5,6}{22,4}=0,25\\,mol \\\\ & Mg+2{{H}_{2}}S{{O}_{4}}\\to MgS{{O}_{4}}+S{{O}_{2}}+2{{H}_{2}}O \\\\ & 2Fe+6{{H}_{2}}S{{O}_{4}}\\to F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+3S{{O}_{2}}+6{{H}_{2}}O \\\\ & {{n}_{S{{O}_{2}}}}=0,25\\,\\, \\\\ & =>\\,\\,\\,\\,1,5x+y=0,25(2) \\\\ \\end{aligned}$ <br\/> T\u1eeb (1) v\u00e0 (2) gi\u1ea3i ra x = y = 0,1 mol <br\/>${{m}_{Mg}}=0,1.24=2,4\\,gam$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1830}],"lesson":{"save":0,"level":2}}