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{"segment":[{"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 KOH ph\u1ea3n \u1ee9ng v\u1edbi d\u00e3y oxit:","select":["A. $CO_2; SO_2; P_2O_5; Fe_2O_3$","B. $Fe_2O_3; SO_2; SO_3; MgO$","C. $P_2O_5; CO_2; Al_2O_3; SO_3$ ","D. $P_2O_5; CO_2; CuO; SO_3$"],"hint":"","explain":"<span class='basic_left'>Dung d\u1ecbch ki\u1ec1m ph\u1ea3n \u1ee9ng \u0111\u01b0\u1ee3c v\u1edbi c\u00e1c oxit axit v\u00e0 oxit l\u01b0\u1ee1ng t\u00ednh. 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":1}]}],"id_ques":1851},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" D\u00e3y c\u00e1c baz\u01a1 b\u1ecb nhi\u1ec7t ph\u00e2n hu\u1ef7 t\u1ea1o th\u00e0nh oxit baz\u01a1 t\u01b0\u01a1ng \u1ee9ng v\u00e0 n\u01b0\u1edbc:","select":["A. $Cu(OH)_2; Zn(OH)_2; Al(OH)_3; Mg(OH)_2$ ","B. $Cu(OH)_2; Zn(OH)_2; Al(OH)_3; NaOH$","C. $Fe(OH)_3; Cu(OH)_2; KOH; Mg(OH)_2$ ","D. $Fe(OH)_3; Cu(OH)_2; Ba(OH)_2; Mg(OH)_2$"],"hint":"","explain":"<span class='basic_left'>C\u00e1c baz\u01a1 kh\u00f4ng tan \u0111\u1ec1u b\u1ecb nhi\u1ec7t ph\u00e2n h\u1ee7y. 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":1}]}],"id_ques":1852},{"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 baz\u01a1 l\u00e0m phenolphtalein ho\u00e1 h\u1ed3ng:","select":["A. $NaOH; Ca(OH)_2; Zn(OH)_2; Mg(OH)_2$ ","B. $NaOH; Ca(OH)_2; KOH; LiOH $","C. $LiOH; Ba(OH)_2; KOH; Al(OH)_3$ ","D. $LiOH; Ba(OH)_2; Ca(OH)_2; Fe(OH)_3$"],"hint":"","explain":"<span class='basic_left'>C\u00e1c baz\u01a1 c\u1ee7a kim lo\u1ea1i ki\u1ec1m v\u00e0 ki\u1ec1m th\u1ed5 \u0111\u1ec1u l\u00e0m phenolphtalein ho\u00e1 h\u1ed3ng. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n B th\u1ecfa m\u00e3n<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":1853},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Dung d\u1ecbch KOH KH\u00d4NG C\u00d3 t\u00ednh ch\u1ea5t ho\u00e1 h\u1ecdc n\u00e0o sau \u0111\u00e2y?","select":["A. L\u00e0m qu\u1ef3 t\u00edm ho\u00e1 xanh ","B. T\u00e1c d\u1ee5ng v\u1edbi oxit axit t\u1ea1o th\u00e0nh mu\u1ed1i v\u00e0 n\u01b0\u1edbc","C. T\u00e1c d\u1ee5ng v\u1edbi axit t\u1ea1o th\u00e0nh mu\u1ed1i v\u00e0 n\u01b0\u1edbc ","D. B\u1ecb nhi\u1ec7t ph\u00e2n hu\u1ef7 t\u1ea1o ra oxit baz\u01a1 v\u00e0 n\u01b0\u1edbc"],"hint":"","explain":"<span class='basic_left'>Dung d\u1ecbch ki\u1ec1m kh\u00f4ng b\u1ecb ph\u00e2n h\u1ee7y b\u1edfi nhi\u1ec7t<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":1854},{"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 c\u00e1c dung d\u1ecbch c\u00f3 pH > 7 l\u00e0","select":["A. $HCl, HNO_3$ ","B. $NaCl, KNO_3$","C. $NaOH, Ba(OH)_2 $","D. $H_2O, NaCl$ "],"hint":"","explain":"<span class='basic_left'>C\u00e1c dung d\u1ecbch baz\u01a1 \u0111\u1ec1u c\u00f3 pH > 7. 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":1855},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Cho c\u00e1c baz\u01a1 sau: $Fe(OH)_3, Al(OH)_3, Cu(OH)_2, Zn(OH)_2$. Khi nung n\u00f3ng c\u00e1c baz\u01a1 tr\u00ean t\u1ea1o ra d\u00e3y oxit baz\u01a1 t\u01b0\u01a1ng \u1ee9ng l\u00e0:","select":["A. $FeO, Al_2O_3, CuO, ZnO$ ","B. $Fe_2O_3, Al_2O_3, CuO, ZnO$","C. $Fe_3O_4, Al_2O_3, CuO, ZnO$ ","D. $Fe_2O_3, Al_2O_3, Cu2O, ZnO$"],"hint":"","explain":"<span class='basic_left'>C\u00e1c baz\u01a1 kh\u00f4ng tan b\u1ecb nhi\u1ec7t ph\u00e2n h\u1ee7y t\u1ea1o oxit t\u01b0\u01a1ng \u1ee9ng t\u01b0\u01a1ng \u1ee9ng v\u00e0 n\u01b0\u1edbc. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n B th\u1ecfa m\u00e3n<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":1856},{"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 baz\u01a1 v\u1eeba t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi dung d\u1ecbch HCl, v\u1eeba t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi dung d\u1ecbch KOH","select":["A. $Ba(OH)_2$v\u00e0 $NaOH$ ","B. $NaOH$ v\u00e0 $Cu(OH)_2$","C. $Al(OH)_3$ v\u00e0 $Zn(OH)_2$ ","D. $Zn(OH)_2$ v\u00e0 $Mg(OH)_2$"],"hint":"","explain":"<span class='basic_left'>C\u00e1c baz\u01a1 l\u01b0\u1ee1ng t\u00ednh nh\u01b0 $Al(OH)_3, Zn(OH)_2$,.... v\u1eeba t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi axit, v\u1eeba t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi baz\u01a1<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1857},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Cho dung d\u1ecbch ch\u1ee9a $0,9$ mol $NaOH$ v\u00e0o dung d\u1ecbch c\u00f3 ch\u1ee9a $ a $ mol $H_3PO_4$. Sau ph\u1ea3n \u1ee9ng ch\u1ec9 thu \u0111\u01b0\u1ee3c mu\u1ed1i $Na_3PO_4$ v\u00e0 $H_2O$. Gi\u00e1 tr\u1ecb c\u1ee7a $a$ l\u00e0","select":["A. 0,3 mol ","B. 0,4 mol ","C. 0,6 mol ","D. 0,9 mol "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& 3NaOH+{{H}_{3}}P{{O}_{4}}\\to N{{a}_{3}}P{{O}_{4}}+3{{H}_{2}}O \\\\ & {{n}_{{{H}_{3}}P{{O}_{4}}}}=\\dfrac{1}{3}{{n}_{NaOH}}=0,3\\,mol \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1858},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Baz\u01a1 tan v\u00e0 kh\u00f4ng tan c\u00f3 t\u00ednh ch\u1ea5t ho\u00e1 h\u1ecdc chung l\u00e0:","select":["A. L\u00e0m qu\u1ef3 t\u00edm ho\u00e1 xanh ","B. T\u00e1c d\u1ee5ng v\u1edbi oxit axit t\u1ea1o th\u00e0nh mu\u1ed1i v\u00e0 n\u01b0\u1edbc","C. T\u00e1c d\u1ee5ng v\u1edbi axit t\u1ea1o th\u00e0nh mu\u1ed1i v\u00e0 n\u01b0\u1edbc","D. B\u1ecb nhi\u1ec7t ph\u00e2n hu\u1ef7 t\u1ea1o ra oxit baz\u01a1 v\u00e0 n\u01b0\u1edbc"],"hint":"","explain":"<span class='basic_left'>C\u1ea3 baz\u01a1 tan v\u00e0 kh\u00f4ng tan \u0111\u1ec1u t\u00e1c d\u1ee5ng v\u1edbi axit t\u1ea1o th\u00e0nh mu\u1ed1i v\u00e0 n\u01b0\u1edbc<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":1859},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" $NaOH$ c\u00f3 th\u1ec3 l\u00e0m kh\u00f4 ch\u1ea5t kh\u00ed \u1ea9m sau","select":["A. $CO_2$ ","B. $SO_2$","C. $N_2$ ","D. $HCl$"],"hint":"","explain":"<span class='basic_left'>$NaOH$ l\u00e0m kh\u00f4 ch\u1ea5t kh\u00ed m\u00e0 ch\u1ea5t \u0111\u00f3 kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi $NaOH$. <br\/> 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":1860},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" C\u1eb7p ch\u1ea5t n\u00e0o sau \u0111\u00e2y t\u1ed3n t\u1ea1i trong m\u1ed9t dung d\u1ecbch (kh\u00f4ng c\u00f3 x\u1ea3y ra ph\u1ea3n \u1ee9ng v\u1edbi nhau)?","select":["A. $NaOH$ v\u00e0 $Mg(OH)_2$ ","B. $KOH$ v\u00e0 $Na_2CO_3$","C. $Ba(OH)_2$ v\u00e0 $Na_2SO_4$ ","D. $Na_3PO_4$ v\u00e0 $Ca(OH)_2$"],"hint":"","explain":"<span class='basic_left'>C\u00e1c ch\u1ea5t t\u1ed3n t\u1ea1i \u0111\u01b0\u1ee3c trong $1$ dung d\u1ecbch ph\u1ea3i l\u00e0 c\u00e1c ch\u1ea5t tan trong n\u01b0\u1edbc v\u00e0 kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi nhau. <br\/> Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n B th\u1ecfa m\u00e3n<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1861},{"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 trung ho\u00e0 $200ml$ h\u1ed7n h\u1ee3p ch\u1ee9a $HCl$ $0,3M$ v\u00e0 $H_2SO_4$ $0,1M$ c\u1ea7n d\u00f9ng $V $ (ml) dung d\u1ecbch $Ba(OH)_2$ $0,2M$. Gi\u00e1 tr\u1ecb c\u1ee7a $V$ l\u00e0:","select":["A. 400 ml","B. 350 ml","C. 300 ml ","D. 250 ml "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{HCl}}=0,06\\,mol \\\\ & {{n}_{{{H}_{2}}S{{O}_{4}}}}=0,02\\,mol \\\\ & 2HCl+Ba{{(OH)}_{2}}\\to BaC{{l}_{2}}+2{{H}_{2}}O \\\\ & {{H}_{2}}S{{O}_{4}}+Ba{{(OH)}_{2}}\\to BaS{{O}_{4}}\\downarrow +{{H}_{2}}O \\\\ & {{n}_{Ba{{(OH)}_{2}}}}={{n}_{{{H}_{2}}S{{O}_{4}}}}+\\dfrac{1}{2}{{n}_{HCl}}=0,02+0,03=0,05\\,mol \\\\ & {{V}_{Ba{{(OH)}_{2}}}}=\\dfrac{0,05}{0,2}=0,25\\,l=250\\,ml \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":1862},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" S\u1ee5c $2,24$ l\u00edt kh\u00ed $CO_2$ v\u00e0o dung d\u1ecbch ch\u1ee9a $0,2$ mol $NaOH$. Dung d\u1ecbch thu \u0111\u01b0\u1ee3c sau ph\u1ea3n \u1ee9ng ch\u1ee9a:","select":["A. $NaHCO_3$ ","B. $Na_2CO_3$","C. $Na_2CO_3$ v\u00e0 $NaOH$ ","D. $NaHCO_3$ v\u00e0 $NaOH$"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{C{{O}_{2}}}}=0,1\\,mol \\\\ & C{{O}_{2}}+2NaOH\\to N{{a}_{2}}C{{O}_{3}}+{{H}_{2}}O \\\\ & {{n}_{NaOH}}=2{{n}_{C{{O}_{2}}}} \\\\ \\end{aligned}$ <br\/>C\u1ea3 $CO_2$ v\u00e0 $NaOH$ \u0111\u1ec1u h\u1ebft. Dung d\u1ecbch thu \u0111\u01b0\u1ee3c ch\u1ee9a $Na_2CO_3$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1863},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Cho $200ml$ dung d\u1ecbch $KOH$ $1M$ t\u00e1c d\u1ee5ng v\u1edbi $200ml$ dung d\u1ecbch $H_2SO_4$ $1M$, sau ph\u1ea3n \u1ee9ng cho th\u00eam m\u1ed9t m\u1ea3nh $Mg$ d\u01b0 v\u00e0o s\u1ea3n ph\u1ea9m th\u1ea5y tho\u00e1t ra m\u1ed9t th\u1ec3 t\u00edch kh\u00ed $H_2$ (\u0111ktc) l\u00e0:","select":["A. $2,24$ l\u00edt ","B. $4,48$ l\u00edt","C. $3,36$ l\u00edt ","D. $6,72$ l\u00edt"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{KOH}}=0,2\\,mol \\\\ & {{n}_{{{H}_{2}}S{{O}_{4}}}}=0,2\\,mol \\\\ & 2KOH+{{H}_{2}}S{{O}_{4}}\\to {{K}_{2}}S{{O}_{4}}+2{{H}_{2}}O \\\\ & \\dfrac{{{n}_{{{H}_{2}}S{{O}_{4}}}}}{1}>\\dfrac{{{n}_{KOH}}}{2} \\\\ & {{H}_{2}}S{{O}_{4}}\\text{ d} \\\\ & {{\\text{n}}_{{{H}_{2}}S{{O}_{4}}(\\text{d})}}=0,2-\\dfrac{0,2}{2}=0,1\\,mol \\\\ & Mg+{{H}_{2}}S{{O}_{4}}\\,(d)\\to MgS{{O}_{4}}+{{H}_{2}} \\\\ & {{n}_{{{H}_{2}}}}={{n}_{{{H}_{2}}S{{O}_{4\\,\\text{d}}}}}=0,1\\,mol \\\\ & {{V}_{{{H}_{2}}}}=2,24\\,l \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1864},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" D\u1eabn $1,68$ l\u00edt kh\u00ed $CO_2$ (\u0111ktc) v\u00e0o $x$ g dung d\u1ecbch $KOH$ $5,6\\%$. \u0110\u1ec3 thu \u0111\u01b0\u1ee3c mu\u1ed1i $KHCO_3$ duy nh\u1ea5t th\u00ec $x$ c\u00f3 gi\u00e1 tr\u1ecb l\u00e0:","select":["A. 75 g ","B. 150 g ","C. 225 g ","D. 300 g"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{C{{O}_{2}}}}=0,075\\,mol \\\\ & C{{O}_{2}}+KOH\\to KHC{{O}_{3}} \\\\ & {{n}_{KOH}}={{n}_{C{{O}_{2}}}}=0,075\\,mol \\\\ & {{m}_{KOH}}=4,2\\,gam \\\\ & {{m}_{dd\\,KOH}}=\\dfrac{4,2.100\\%}{5,6\\%}=75\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1865},{"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 nh\u1eadn bi\u1ebft dd $KOH$ v\u00e0 dd $Ba(OH)_2$ ta d\u00f9ng thu\u1ed1c th\u1eed l\u00e0","select":["A. Phenolphtalein","B. Qu\u1ef3 t\u00edm","C. dd $H_2SO_4$ ","D. dd $HCl$"],"hint":"","explain":"<span class='basic_left'>\u0110\u1ec3 ph\u00e2n bi\u1ec7t dung d\u1ecbch c\u00f3 ch\u1ee9a $Ba$ th\u01b0\u1eddng d\u00f9ng dung d\u1ecbch c\u00f3 ch\u1ee9a $SO_4$ v\u00ec t\u1ea1o th\u00e0nh k\u1ebft t\u1ee7a tr\u1eafng $BaSO_4$.<br\/> 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":1866},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho $200ml$ dung d\u1ecbch $Ba(OH)_2$ $0,4M$ v\u00e0o $250ml$ dung d\u1ecbch $H_2SO_4$ $0,3M$. Kh\u1ed1i l\u01b0\u1ee3ng k\u1ebft t\u1ee7a thu \u0111\u01b0\u1ee3c l\u00e0:","select":["A. 17,645 g ","B. 16,475 g","C. 17,475 g ","D. 18,645 g"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{Ba(O{{H}_{2}}}}=0,08\\,mol \\\\ & {{n}_{{{H}_{2}}S{{O}_{4}}}}=0,075\\,mol \\\\ & Ba{{(OH)}_{2}}+{{H}_{2}}S{{O}_{4}}\\to BaS{{O}_{4}}\\downarrow +{{H}_{2}}O \\\\ & \\dfrac{{{n}_{Ba{{(OH)}_{2}}}}}{1}>\\dfrac{{{n}_{{{H}_{2}}S{{O}_{4}}}}}{1} \\\\ \\end{aligned}$<br\/> $H_2SO_4$ ph\u1ea3n \u1ee9ng h\u1ebft. T\u00ednh theo $H_2SO_4$ <br\/>$\\begin{aligned}& {{n}_{BaS{{O}_{4}}}}={{n}_{{{H}_{2}}S{{O}_{4}}}}=0,075\\,mol \\\\ & {{m}_{\\downarrow }}=17,475\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1867},{"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 $19,6g$ $Cu(OH)_2$ thu \u0111\u01b0\u1ee3c m\u1ed9t ch\u1ea5t r\u1eafn m\u00e0u \u0111en, d\u00f9ng kh\u00ed $H_2$ d\u01b0 kh\u1eed ch\u1ea5t r\u1eafn m\u00e0u \u0111en \u0111\u00f3 thu \u0111\u01b0\u1ee3c m\u1ed9t ch\u1ea5t r\u1eafn m\u00e0u \u0111\u1ecf c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng l\u00e0:","select":["A. 8,4 g ","B. 9,6 g ","C. 12,8 g ","D. 16 g"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{Cu{{(OH)}_{2}}}}=0,2\\,mol \\\\ & Cu{{(OH)}_{2}}\\xrightarrow{{{t}^{0}}}CuO+{{H}_{2}}O \\\\ & CuO+{{H}_{2}}\\xrightarrow{{{t}^{0}}}Cu+{{H}_{2}}O \\\\ & {{n}_{Cu}}={{n}_{Cuo}}={{n}_{Cu{{(OH)}_{2}}}}=0,2\\,mol \\\\ & {{m}_{Cu}}=12,8\\,gam \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1868},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho $4,48$ l\u00edt $CO_2$ t\u00e1c d\u1ee5ng h\u1ebft v\u1edbi $300$ ml dung d\u1ecbch $NaOH$ $1M$. Kh\u1ed1i l\u01b0\u1ee3ng mu\u1ed1i khan thu \u0111\u01b0\u1ee3c l\u00e0","select":["A. 6,4 g ","B. 10,6 g ","C. 19 g ","D. 23 g"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{C{{O}_{2}}}}=0,2\\,mol \\\\ & {{n}_{NaOH}}=0,3\\,mol \\\\ & C{{O}_{2}}+2NaOH\\to N{{a}_{2}}C{{O}_{3}}+{{H}_{2}}O \\\\ & \\dfrac{{{n}_{C{{O}_{2}}}}}{1}>\\dfrac{{{n}_{NaOH}}}{2} \\\\ & C{{O}_{2}}\\,\\text{d }\\text{.}\\,{{\\text{n}}_{N{{a}_{2}}C{{O}_{3}}}}=0,15\\,mol \\\\ & {{\\text{n}}_{C{{O}_{2}}\\text{ d}}}=0,2-\\dfrac{0,3}{2}=0,05\\,mol \\\\ & C{{O}_{2}}\\,d\\,+N{{a}_{2}}C{{O}_{3}}+{{H}_{2}}O\\to 2NaHC{{O}_{3}} \\\\ & \\dfrac{{{n}_{C{{O}_{2}}\\,\\text{d}}}}{1}<\\dfrac{{{n}_{N{{a}_{2}}C{{O}_{3}}}}}{1} \\\\ & N{{a}_{2}}C{{O}_{3}}\\,\\text{d}\\text{.}\\, \\\\ & {{\\text{n}}_{N{{a}_{2}}C{{O}_{3}}\\,(\\text{d)}}}=\\,0,15-0,05=0,1\\,mol \\\\ & {{n}_{NaHC{{O}_{3}}}}=2{{n}_{C{{O}_{2}}\\,(\\text{d)}}}=0,1\\,mol \\\\ & {{m}_{\\text{mu }\\!\\!\\grave{\\mathrm{e}}\\!\\!\\text{ i}}}=0,1.106+0,1.84=19\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1869},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Cho $14,6$ gam h\u1ed7n h\u1ee3p $Zn$ v\u00e0 $ZnO$ t\u00e1c d\u1ee5ng v\u1eeba \u0111\u1ee7 v\u1edbi $200$ ml dung d\u1ecbch $NaOH$ $2M$. Th\u1ec3 t\u00edch kh\u00ed thu \u0111\u01b0\u1ee3c \u1edf \u0111ktc l\u00e0","select":["A. 2,24 l\u00edt ","B. 4,48 l\u00edt","C. 3,36 l\u00edt ","D. 6,72 l\u00edt "],"hint":"","explain":"<span class='basic_left'>G\u1ecdi $x, y$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 s\u1ed1 mol c\u1ee7a $Zn$ v\u00e0 $ZnO$ trong h\u1ed7n h\u1ee3p <br\/>Kh\u1ed1i l\u01b0\u1ee3ng h\u1ed7n h\u1ee3p b\u1eb1ng $14,6$ gam n\u00ean ta c\u00f3: <br\/> $65x + 81y = 14,6$ (1) <br\/>$\\begin{aligned}& Zn+2NaOH\\to N{{a}_{2}}Zn{{O}_{2}}+{{H}_{2}} \\\\ & ZnO+2NaOH\\to N{{a}_{2}}Zn{{O}_{2}}+{{H}_{2}}O \\\\ & {{n}_{NaOH}}=0,4\\,mol \\\\ & =>\\,2x+2y=0,4\\,\\,\\,\\,\\,\\,\\,(2) \\\\ \\end{aligned}$ <br\/>T\u1eeb (1) v\u00e0 (2) gi\u1ea3i ra x = y = 0,1 mol <br\/>$\\begin{aligned}& {{n}_{{{H}_{2}}}}={{n}_{Zn}}=0,1\\,mol \\\\ & {{V}_{{{H}_{2}}}}=2,24\\,l \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1870}],"lesson":{"save":0,"level":2}}

Điểm của bạn.

Câu hỏi này theo dạng chọn đáp án đúng, sau khi đọc xong câu hỏi, bạn bấm vào một trong số các đáp án mà chương trình đưa ra bên dưới, sau đó bấm vào nút gửi để kiểm tra đáp án và sẵn sàng chuyển sang câu hỏi kế tiếp

Trả lời đúng trong khoảng thời gian quy định bạn sẽ được + số điểm như sau:
Trong khoảng 1 phút đầu tiên + 5 điểm
Trong khoảng 1 phút -> 2 phút + 4 điểm
Trong khoảng 2 phút -> 3 phút + 3 điểm
Trong khoảng 3 phút -> 4 phút + 2 điểm
Trong khoảng 4 phút -> 5 phút + 1 điểm

Quá 5 phút: không được cộng điểm

Tổng thời gian làm mỗi câu (không giới hạn)

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