{"segment":[{"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 g\u1ed3m c\u00e1c kim lo\u1ea1i t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi dung d\u1ecbch $H_2SO_4$ lo\u00e3ng l\u00e0:","select":["A. Fe, Cu, Mg.","B. Zn, Fe, Cu.","C. Zn, Fe, Al.","D. Fe, Zn, Ag"],"hint":"","explain":"<span class='basic_left'>C\u00e1c kim lo\u1ea1i t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi dung d\u1ecbch $H_2SO_4$ lo\u00e3ng l\u00e0 c\u00e1c kim lo\u1ea1i m\u1ea1nh : K, Na, Mg, Al, Zn, Fe,.....<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1791},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Nh\u00f3m ch\u1ea5t t\u00e1c d\u1ee5ng v\u1edbi n\u01b0\u1edbc v\u00e0 v\u1edbi dung d\u1ecbch $HCl$ l\u00e0:","select":["A. $Na_2O, SO_3, CO_2$ ","B. $K_2O, P_2O_5, CaO$ ","C. $BaO, SO_3, P_2O_5$ ","D. $CaO, BaO, Na_2O$"],"hint":"","explain":"<span class='basic_left'>C\u00e1c oxit t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi n\u01b0\u1edbc v\u00e0 dung d\u1ecbch axit l\u00e0 c\u00e1c oxit c\u1ee7a kim lo\u1ea1i ki\u1ec1m, ki\u1ec1m th\u1ed5. 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":1792},{"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 oxit t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch HCl t\u1ea1o th\u00e0nh mu\u1ed1i v\u00e0 n\u01b0\u1edbc l\u00e0:","select":["A. $CO_2, SO_2, CuO$ ","B. $Na_2O, SO_2, CaO$ ","C. $CuO, Na_2O, CaO$ ","D. $CaO, CuO, SO_2$"],"hint":"","explain":"<span class='basic_left'>C\u00e1c oxit baz\u01a1 t\u00e1c d\u1ee5ng v\u1edbi axit HCl t\u1ea1o mu\u1ed1i v\u00e0 n\u01b0\u1edbc. 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":1793},{"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 kh\u00f4ng t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi dung d\u1ecbch $H_2SO_4$ lo\u00e3ng l\u00e0:","select":["A. $Zn, ZnO, Zn(OH)_2$ ","B. $Cu, CO_2, CuCl_2$","C. $Na_2O, NaOH, Na_2CO_3$","D. $MgO, MgCO_3, Mg(OH)_2$"],"hint":"","explain":"<span class='basic_left'>Kim lo\u1ea1i y\u1ebfu nh\u01b0 Cu, Ag kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi axit $H_2SO_4$ lo\u00e3ng. Ch\u1ecdn B<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1794},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" $CuO$ t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch $H_2SO_4$ t\u1ea1o th\u00e0nh: ","select":["A. Dung d\u1ecbch kh\u00f4ng m\u00e0u ","B Dung d\u1ecbch c\u00f3 m\u00e0u l\u1ee5c nh\u1ea1t","C. Dung d\u1ecbch c\u00f3 m\u00e0u xanh lam ","D. Dung d\u1ecbch c\u00f3 m\u00e0u v\u00e0ng n\u00e2u"],"hint":"","explain":"<span class='basic_left'> $CuO$ t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch $H_2SO_4$ theo ph\u1ea3n \u1ee9ng: <br\/> $CuO+{{H}_{2}}S{{O}_{4}}\\to CuS{{O}_{4}}+{{H}_{2}}O$ <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":1}]}],"id_ques":1795},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" K\u1ebdm t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch axit clohi\u0111ric sinh ra:","select":["A. Dung d\u1ecbch c\u00f3 m\u00e0u xanh lam v\u00e0 ch\u1ea5t kh\u00ed m\u00e0u n\u00e2u ","B. Dung d\u1ecbch kh\u00f4ng m\u00e0u v\u00e0 ch\u1ea5t kh\u00ed c\u00f3 m\u00f9i h\u1eafc.","C. Dung d\u1ecbch c\u00f3 m\u00e0u v\u00e0ng n\u00e2u v\u00e0 ch\u1ea5t kh\u00ed kh\u00f4ng m\u00e0u ","D. Dung d\u1ecbch kh\u00f4ng m\u00e0u v\u00e0 ch\u1ea5t kh\u00ed ch\u00e1y \u0111\u01b0\u1ee3c trong kh\u00f4ng kh\u00ed."],"hint":"","explain":"<span class='basic_left'>Zn l\u00e0 kim lo\u1ea1i m\u1ea1nh t\u00e1c d\u1ee5ng v\u1edbi HCl theo ph\u1ea3n \u1ee9ng: <br\/> $Zn+2HCl\\to Zn{Cl}_{2}+{{H}_{2}}$ <br\/>Dung d\u1ecbch $ZnCl_2$ kh\u00f4ng m\u00e0u. Kh\u00ed $H_2$ l\u00e0 kh\u00ed kh\u00f4ng m\u00e0u, ch\u00e1y \u0111\u01b0\u1ee3c trong kh\u00f4ng kh\u00ed. <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":1796},{"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 ph\u1ea3n \u1ee9ng \u0111\u01b0\u1ee3c v\u1edbi dung d\u1ecbch HCl t\u1ea1o ra m\u1ed9t ch\u1ea5t kh\u00ed c\u00f3 m\u00f9i h\u1eafc, n\u1eb7ng h\u01a1n kh\u00f4ng kh\u00ed v\u00e0 l\u00e0m \u0111\u1ee5c n\u01b0\u1edbc v\u00f4i trong:","select":["A. $Zn$ ","B. $Na_2SO_3$","C. $FeS$","D. $Na_2CO_3$"],"hint":"","explain":"<span class='basic_left'>Kh\u00ed c\u00f3 m\u00f9i h\u1eafc l\u00e0 kh\u00ed $SO_2$. Ch\u1ea5t t\u00e1c d\u1ee5ng v\u1edbi $HCl$ sinh ra $SO_2$ l\u00e0 $Na_2SO_3$ <br\/> $N{{a}_{2}}S{{O}_{3}}+2HCl\\to 2NaCl+S{{O}_{2}}+{{H}_{2}}O$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1797},{"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 ph\u00e2n bi\u1ec7t 2 dung d\u1ecbch $HCl$ v\u00e0 $H_2SO_4$ lo\u00e3ng. Ta d\u00f9ng m\u1ed9t kim lo\u1ea1i:","select":["A. Mg ","B. Cu","C. Ba ","D. Zn"],"hint":"","explain":"<span class='basic_left'>Ba t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch $HCl$ c\u00f3 k\u1ebft t\u1ee7a $BaSO_4$ v\u00e0 sinh ra kh\u00ed kh\u00f4ng m\u00e0u, c\u00f2n $HCl$ t\u00e1c d\u1ee5ng v\u1edbi $Ba$ c\u00f3 kh\u00ed kh\u00f4ng m\u00e0u <br\/>$\\begin{aligned}& Ba+2HCl\\to BaC{{l}_{2}}+{{H}_{2}}\\uparrow \\\\ & Ba+{{H}_{2}}S{{O}_{4}}\\to BaS{{O}_{4}}\\downarrow +{{H}_{2}}\\uparrow \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1798},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Trung h\u00f2a 200 ml dung d\u1ecbch $H_2SO_4$ 1M b\u1eb1ng dung d\u1ecbch $NaOH$ 20%. Kh\u1ed1i l\u01b0\u1ee3ng dung d\u1ecbch $NaOH$ c\u1ea7n d\u00f9ng l\u00e0:","select":["A. 100 g","B. 80 g","C. 90 g","D. 150 g"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{{{H}_{2}}S{{O}_{4}}}}=0,2.1=0,2\\,mol \\\\ & {{H}_{2}}S{{O}_{4}}+2NaOH\\to N{{a}_{2}}S{{O}_{4}}+2{{H}_{2}}O \\\\ & {{n}_{NaOH}}=2\\,{{n}_{{{H}_{2}}S{{O}_{4}}}}=2.0,2=0,4\\,mol \\\\ & {{m}_{NaOH}}=0,4.40=16\\,gam \\\\ & {{m}_{\\text{dd}\\,NaOH}}=\\dfrac{16.100\\%}{20\\%}=80\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1799},{"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 axit sau: $HCl, H_2SO_4, H_2S, HNO_3$. Axit y\u1ebfu l\u00e0:","select":["A. $HCl$ ","B. $H_2SO_4$","C. $H_2S$","D. $HNO_3$"],"hint":"","explain":"<span class='basic_left'>C\u00e1c axit y\u1ebfu: $HF, H_2SO_3, H_2CO_3, H_2S,$...<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1800},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Cho 0,1 mol kim lo\u1ea1i k\u1ebdm v\u00e0o dung d\u1ecbch HCl d\u01b0. Kh\u1ed1i l\u01b0\u1ee3ng mu\u1ed1i thu \u0111\u01b0\u1ee3c l\u00e0:","select":["A. 13,6 g","B. 1,36 g","C. 20,4 g","D. 27,2 g"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& Zn+2HCl\\to ZnC{{l}_{2}}+{{H}_{2}} \\\\ & {{n}_{ZnC{{l}_{2}}}}={{n}_{Zn}}=0,1\\,mol \\\\ & {{m}_{ZnC{{l}_{2}}}}=0,1.136=13,6\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1801},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Dung d\u1ecbch A c\u00f3 pH < 7 v\u00e0 t\u1ea1o ra k\u1ebft t\u1ee7a khi t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch Bari nitrat $Ba(NO_3)_2$ . Ch\u1ea5t A l\u00e0","select":["A. $HCl$ ","B. $H_2SO_4$","C. $Na_2SO_4$","D. $Ca(OH)_2$"],"hint":"","explain":"<span class='basic_left'>Dung d\u1ecbch c\u00f3 pH < 7 l\u00e0 dung d\u1ecbch axit, t\u00e1c d\u1ee5ng v\u1edbi $Ba(NO_3)_2$ t\u1ea1o k\u1ebft t\u1ee7a l\u00e0 axit $H_2SO_4$: <br\/>${{H}_{2}}S{{O}_{4}}+Ba{{(N{{O}_{3}})}_{2}}\\to 2HN{{O}_{3}}+BaS{{O}_{4}}\\downarrow $ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1802},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" D\u00f9ng qu\u00ec t\u00edm \u0111\u1ec3 ph\u00e2n bi\u1ec7t \u0111\u01b0\u1ee3c c\u1eb7p ch\u1ea5t n\u00e0o sau \u0111\u00e2y:","select":["A. Dung d\u1ecbch $HCl$ v\u00e0 dung d\u1ecbch $KOH$ ","B. Dung d\u1ecbch $HCl$ v\u00e0 dung d\u1ecbch $H_2SO_4$","C. Dung d\u1ecbch $Na_2SO_4$ v\u00e0 dung d\u1ecbch $NaCl$ ","D. Dung d\u1ecbch $NaOH$ v\u00e0 dung d\u1ecbch $KOH$."],"hint":"","explain":"<span class='basic_left'>Q\u00f9y t\u00edm h\u00f3a \u0111\u1ecf khi g\u1eb7p axit, h\u00f3a xanh khi g\u1eb7p dung d\u1ecbch ki\u1ec1m, kh\u00f4ng \u0111\u1ed5i m\u00e0u khi g\u1eb7p dung d\u1ecbch mu\u1ed1i. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n A th\u1ecfa m\u00e3n. Qu\u1ef3 t\u00edm g\u1eb7p dung d\u1ecbch HCl h\u00f3a \u0111\u1ecf, g\u1eb7p dung d\u1ecbch NaOH h\u00f3a xanh.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":1}]}],"id_ques":1803},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Kim lo\u1ea1i X t\u00e1c d\u1ee5ng v\u1edbi HCl sinh ra kh\u00ed hi\u0111ro. D\u1eabn kh\u00ed hi\u0111ro qua oxit c\u1ee7a kim lo\u1ea1i Y \u0111un n\u00f3ng th\u00ec thu \u0111\u01b0\u1ee3c kim lo\u1ea1i Y. Hai kim lo\u1ea1i X v\u00e0 Y l\u1ea7n l\u01b0\u1ee3t l\u00e0:","select":["A. Cu, Ca","B. Pb, Cu","C. Pb, Ca ","D. Ag, Cu"],"hint":"","explain":"<span class='basic_left'>C\u00e1c kim lo\u1ea1i y\u1ebfu nh\u01b0 $Cu, Ag$ kh\u00f4ng t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi dung d\u1ecbch $HCl$ sinh ra kh\u00ed $H_2$. C\u00e1c oxit c\u1ee7a kim lo\u1ea1i y\u1ebfu ($CuO, Ag_2O,$...) m\u1edbi ph\u1ea3n \u1ee9ng v\u1edbi $H_2$ sinh ra kim lo\u1ea1i. <br\/>$\\begin{aligned}& Pb+2HCl\\to PbC{{l}_{2}}+{{H}_{2}} \\\\ & {{H}_{2}}+CuO\\to Cu+{{H}_{2}}O \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1804},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho 300ml dung d\u1ecbch HCl 1M v\u00e0o 300ml dung d\u1ecbch NaOH 0,5M. N\u1ebfu cho qu\u00ec t\u00edm v\u00e0o dung d\u1ecbch sau ph\u1ea3n \u1ee9ng th\u00ec qu\u00ec t\u00edm chuy\u1ec3n sang","select":["A. M\u00e0u xanh. ","B. Kh\u00f4ng \u0111\u1ed5i m\u00e0u","C. M\u00e0u \u0111\u1ecf ","D. M\u00e0u v\u00e0ng nh\u1ea1t"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{HCl}}=0,3\\,mol \\\\ & {{n}_{NaOH}}=0,3.0,5=0,15\\,mol \\\\ & HCl+NaOH\\to NaCl+{{H}_{2}}O \\\\ & \\dfrac{{{n}_{HCl}}}{1}>\\dfrac{{{n}_{NaOH}}}{1} \\\\ \\end{aligned}$ <br\/> => HCl d\u01b0, dung d\u1ecbch l\u00e0m qu\u1ef3 chuy\u1ec3n \u0111\u1ecf <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1805},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Cho 4,8 gam kim lo\u1ea1i magie t\u00e1c d\u1ee5ng v\u1eeba \u0111\u1ee7 v\u1edbi dung d\u1ecbch axit sunfuric. Th\u1ec3 t\u00edch kh\u00ed Hi\u0111ro thu \u0111\u01b0\u1ee3c \u1edf \u0111ktc l\u00e0:","select":["A. 44,8 l\u00edt ","B. 4,48 l\u00edt","C. 2,24 l\u00edt ","D. 22,4 l\u00edt"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{Mg}}=\\dfrac{4,8}{24}=0,2\\,mol \\\\ & Mg+{{H}_{2}}S{{O}_{4}}\\to MgS{{O}_{4}}+{{H}_{2}} \\\\ & {{n}_{{{H}_{2}}}}={{n}_{Mg}}=0,2\\,mol \\\\ & {{V}_{{{H}_{2}}}}=0,2,22,4=4,48\\,l \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1806},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Cho 21 gam MgCO3 t\u00e1c d\u1ee5ng v\u1edbi m\u1ed9t l\u01b0\u1ee3ng v\u1eeba \u0111\u1ee7 dung d\u1ecbch HCl 2M.Th\u1ec3 t\u00edch dung d\u1ecbch HCl \u0111\u00e3 d\u00f9ng l\u00e0:","select":["A. 2,5 l\u00edt ","B. 0,25 l\u00edt","C. 3,5 l\u00edt ","D. 1,5 l\u00edt"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{MgC{{O}_{3}}}}=\\dfrac{21}{84}=0,25\\,mol \\\\ & MgC{{O}_{3}}+2HCl\\to MgC{{l}_{2}}+C{{O}_{2}}+{{H}_{2}}O \\\\ & {{n}_{HCl}}=2{{n}_{MgC{{O}_{3}}}}=0,25.2=0,5\\,mol \\\\ & {{V}_{HCl}}=\\dfrac{0,5}{2}=0,25\\,l \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1807},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Cho 0,2 mol Canxi oxit t\u00e1c d\u1ee5ng v\u1edbi 500ml dung d\u1ecbch HCl 1M. Kh\u1ed1i l\u01b0\u1ee3ng mu\u1ed1i thu \u0111\u01b0\u1ee3c l\u00e0:","select":["A. 2,22 g","B. 22,2 g","C. 23,2 g","D. 22,3 g"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{HCl}}=0,5\\,mol \\\\ & CaO+2HCl\\to CaC{{l}_{2}}+{{H}_{2}}O \\\\ & \\dfrac{{{n}_{CaO}}}{1}<\\dfrac{{{n}_{HCl}}}{2} \\\\ \\end{aligned}$ <br\/> => CaO h\u1ebft, t\u00ednh theo CaO <br\/>$\\begin{aligned}& {{n}_{CaC{{l}_{2}}}}={{n}_{CaO}}=0,2\\,mol \\\\ & {{m}_{CaC{{l}_{2}}}}=0,2.111=22,2\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1808},{"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 v\u1eeba h\u1ebft 24 gam h\u1ed7n h\u1ee3p $CuO$ v\u00e0 $Fe_2O_3$ trong 200 ml dung d\u1ecbch $HCl$ 4M . Kh\u1ed1i l\u01b0\u1ee3ng m\u1ed7i oxit trong h\u1ed7n h\u1ee3p l\u00e0:","select":["A. 4 g v\u00e0 16 g","B. 10 g v\u00e0 10 g","C. 8 g v\u00e0 16 g ","D. 14 g v\u00e0 6 g"],"hint":"","explain":"<span class='basic_left'>G\u1ecdi x, y l\u1ea7n l\u01b0\u1ee3t l\u00e0 s\u1ed1 mol c\u1ee7a CuO, Fe2O3<br\/>Kh\u1ed1i l\u01b0\u1ee3ng h\u1ed7n h\u1ee3p b\u1eb1ng 24 gam n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>80x + 160y = 24 (1)<br\/>$\\begin{aligned}& {{n}_{HCl}}=0,2.4=0,8mol \\\\ & CuO+2HCl\\to CuC{{l}_{2}}+{{H}_{2}}O \\\\ & x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2x \\\\ & F{{e}_{2}}{{O}_{3}}+6HCl\\to 2FeC{{l}_{3}}+3{{H}_{2}}O \\\\ & y\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,6y \\\\ & {{n}_{HCl}}=0,8\\,mol \\\\ & =>\\,2x+6y=0,8(2) \\\\ \\end{aligned}$ <br\/>T\u1eeb (1) v\u00e0 (2) gi\u1ea3i ra x = y = 0,1 mol<br\/>$\\begin{aligned}& {{m}_{CuO}}=0,1.80=8\\,gam \\\\ & {{m}_{F{{e}_{2}}{{O}_{3}}}}=24-8=16\\,gam \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1809},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Ho\u00e0 tan 12,1 g h\u1ed7n h\u1ee3p b\u1ed9t kim lo\u1ea1i Zn v\u00e0 Fe c\u1ea7n 400ml dung d\u1ecbch HCl 1M. Kh\u1ed1i l\u01b0\u1ee3ng h\u1ed7n h\u1ee3p mu\u1ed1i thu \u0111\u01b0\u1ee3c sau ph\u1ea3n \u1ee9ng l\u00e0","select":["A. 26,3 g","B. 40,5 g","C. 19,2 g","D. 22,8 g"],"hint":"","explain":"<span class='basic_left'> $\\begin{aligned} & {{n}_{HCl}}=0,4\\,mol \\\\ & Zn+2HCl\\to ZnC{{l}_{2}}+{{H}_{2}} \\\\ & Fe+2HCl\\to FeC{{l}_{2}}+{{H}_{2}} \\\\ & \\text{T\u1ed5ng }{{\\text{n}}_{{{H}_{2}}}}=\\dfrac{1}{2}{{n}_{HCl}}=0,2\\,mol \\\\ & {{m}_{\\text{mu\u1ed1i}}}={{m}_{\\text{h\u1ed7n h\u1ee3p kim lo\u1ea1i}}}+{{m}_{HCl}}-{{m}_{{{H}_{2}}}} \\\\ & {{m}_{\\text{mu\u1ed1i}}}=12,1+0,4.36,5-0,2.2=26,3\\,gam \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1810}],"lesson":{"save":0,"level":2}}