{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"D\u00e3y dung d\u1ecbch n\u00e0o d\u01b0\u1edbi \u0111\u00e2y, d\u00e3y h\u1ee3p ch\u1ea5t n\u00e0o l\u00e0 axit ? ","select":["A. ${{H}_{2}}S{{O}_{4}},{{H}_{3}}P{{O}_{4}},HCl,HN{{O}_{3}}$ ","B. ${{H}_{3}}P{{O}_{4}},NaOH,HBr,HN{{O}_{3}}$ ","C. $Ca{{(OH)}_{2}},NaOH,NaCl,HN{{O}_{3}}$ ","D. $CaC{{l}_{2}},KOH,FeO,AgN{{O}_{3}}$ "],"hint":"","explain":"<span class='basic_left'>Axit l\u00e0m qu\u1ef3 t\u00edm chuy\u1ec3n th\u00e0nh \u0111\u1ecf<br\/>Ph\u00e2n t\u1eed axit g\u1ed3m c\u00f3 m\u1ed9t hay nhi\u1ec1u nguy\u00ean t\u1eed hidro li\u00ean k\u1ebft v\u1edbi g\u1ed1c axit, c\u00e1c nguy\u00ean t\u1eed hidro n\u00e0y c\u00f3 th\u1ec3 thay th\u1ebf b\u1eb1ng c\u00e1c nguy\u00ean t\u1eed kim lo\u1ea1i.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2340},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"G\u1ed1c axit v\u00e0 h\u00f3a tr\u1ecb c\u1ee7a g\u1ed1c axit trong axit ${{H}_{2}}Si{{O}_{3}}$ l\u00e0 ","select":["A. G\u1ed1c axit l\u00e0 $Si{{O}_{3}}$ c\u00f3 h\u00f3a tr\u1ecb III ","B. G\u1ed1c axit l\u00e0 $Si{{O}_{3}}$ c\u00f3 h\u00f3a tr\u1ecb II ","C. G\u1ed1c axit l\u00e0 $Si{{O}_{3}}$ c\u00f3 h\u00f3a tr\u1ecb I ","D. G\u1ed1c axit l\u00e0 $Si{{O}_{3}}$ c\u00f3 h\u00f3a tr\u1ecb IV "],"hint":"","explain":"<span class='basic_left'>Ng\u01b0\u1eddi ta quy \u01b0\u1edbc g\u00e1n cho H h\u00f3a tr\u1ecb I trong m\u1ecdi h\u1ee3p ch\u1ea5t<br\/>G\u1ecdi h\u00f3a tr\u1ecb c\u1ee7a $Si{{O}_{3}}$ l\u00e0 b <br\/>Theo quy t\u1eafc h\u00f3a tr\u1ecb ta c\u00f3 : <br\/>$\\begin{aligned} & 2\\times I=1\\times b \\\\ & \\to b=II \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2341},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Th\u00e0nh ph\u1ea7n ph\u00e2n t\u1eed c\u1ee7a baz\u01a1 g\u1ed3m ","select":["A. M\u1ed9t nguy\u00ean t\u1eed kim lo\u1ea1i v\u00e0 nhi\u1ec1u nh\u00f3m hi\u0111roxit (-OH) ","B. M\u1ed9t hay nhi\u1ec1u nguy\u00ean t\u1eed kim lo\u1ea1i v\u00e0 m\u1ed9t hay nhi\u1ec1u nh\u00f3m hi\u0111roxit (-OH) ","C. M\u1ed9t hay nhi\u1ec1u nguy\u00ean t\u1eed kim lo\u1ea1i v\u00e0 nhi\u1ec1u nh\u00f3m hi\u0111roxit (-OH) ","D. M\u1ed9t nguy\u00ean t\u1eed kim lo\u1ea1i v\u00e0 m\u1ed9t hay nhi\u1ec1u nh\u00f3m hi\u0111roxit (-OH) "],"hint":"","explain":"<span class='basic_left'>Ph\u00e2n t\u1eed baz\u01a1 g\u1ed3m c\u00f3 m\u1ed9t nguy\u00ean t\u1eed kim lo\u1ea1i li\u00ean k\u1ebft v\u1edbi m\u1ed9t hay nhi\u1ec1u nh\u00f3m hi\u0111roxit (-OH)<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":2342},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Cho d\u00e3y c\u00e1c h\u1ee3p ch\u1ea5t sau : $HCl,NaOH,HN{{O}_{3}},KOH,{{H}_{2}}S{{O}_{4}},Ca{{(OH)}_{2}},Ba{{(OH)}_{2}},NaCl$. C\u00f3 bao nhi\u00eau h\u1ee3p ch\u1ea5t l\u00e0m \u0111\u1ed5i m\u00e0u qu\u1ef3 t\u00edm th\u00e0nh xanh ? ","select":["A. 5 ","B. 3 ","C. 4 ","D. 6 "],"hint":"","explain":"<span class='basic_left'>Baz\u01a1 l\u00e0m \u0111\u1ed5i m\u00e0u qu\u1ef3 t\u00edm th\u00e0nh xanh<br\/>Ph\u00e2n t\u1eed baz\u01a1 g\u1ed3m c\u00f3 m\u1ed9t nguy\u00ean t\u1eed kim lo\u1ea1i li\u00ean k\u1ebft v\u1edbi m\u1ed9t hay nhi\u1ec1u nh\u00f3m hi\u0111roxit (-OH)<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2343},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"C\u00f3 3 l\u1ecd kh\u00f4ng nh\u00e3n \u0111\u1ef1ng c\u00e1c dung d\u1ecbch sau : axit, mu\u1ed1i \u0103n, ki\u1ec1m ( baz\u01a1 ). Ph\u01b0\u01a1ng ph\u00e1p n\u00e0o sau \u0111\u00e2y c\u00f3 th\u1ec3 ph\u00e2n bi\u1ec7t \u0111\u01b0\u1ee3c 3 l\u1ecd \u0111\u1ef1ng dung d\u1ecbch tr\u00ean ? ","select":["A. Qu\u1ef3 t\u00edm ","B. H\u1ed3 tinh b\u1ed9t ","C. Phenolphatlein ","D. C\u1ea3 A v\u00e0 C "],"hint":"","explain":"<span class='basic_left'>D\u00f9ng qu\u1ef3 t\u00edm ta nh\u1eadn bi\u1ebft \u0111\u01b0\u1ee3c l\u1ecd d\u1ef1ng axit l\u00e0 l\u1ecd l\u00e0m qu\u1ef3 t\u00edm chuy\u1ec3n \u0111\u1ecf, c\u00f2n l\u1ecd \u0111\u1ef1ng ki\u1ec1m l\u00e0m qu\u1ef3 t\u00edm chuy\u1ec3n xanh, l\u1ecd \u0111\u1ef1ng mu\u1ed1i kh\u00f4ng l\u00e0m \u0111\u1ed5i m\u00e0u qu\u1ef3 t\u00edm<br\/>\u1ede \u0111\u00e2y kh\u00f4ng d\u00f9ng phenolphatlein v\u00ec c\u1ea3 axit v\u00e0 mu\u1ed1i \u0103n \u0111\u1ec1u kh\u00f4ng l\u00e0m \u0111\u1ed5i m\u00e0u phenolphatlein, ch\u1ec9 c\u00f3 ki\u1ec1m l\u00e0m phenolphatlein chuy\u1ec3n m\u00e0u h\u1ed3ng<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2344},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"D\u00e3y c\u00e1c h\u1ee3p ch\u1ea5t n\u00e0o sau \u0111\u00e2y l\u00e0 mu\u1ed1i axit ? ","select":["A. $NaHS{{O}_{4}},CaC{{O}_{3}},AgCl,Ba{{(HC{{O}_{3}})}_{2}}$ ","B. $NaHS{{O}_{4}},KHC{{O}_{3}},CaHP{{O}_{4}},Ba{{(HC{{O}_{3}})}_{2}}$ ","C. $N{{a}_{2}}S{{O}_{4}},CaC{{O}_{3}},AgCl,KBr$ ","D. $N{{a}_{2}}S{{O}_{4}},CaC{{O}_{3}},Ba{{(OH)}_{2}},HCl$ "],"hint":"","explain":"<span class='basic_left'>Mu\u1ed1i axit l\u00e0 mu\u1ed1i m\u00e0 trong \u0111\u00f3 g\u1ed1c axit c\u00f2n nguy\u00ean t\u1eed H ch\u01b0a \u0111\u01b0\u1ee3c thay th\u1ebf b\u1eb1ng nguy\u00ean t\u1eed kim lo\u1ea1i.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":2345},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"C\u00f3 3 ch\u1ea5t r\u1eafn l\u00e0 Ag, Zn, CuO \u0111\u1ef1ng ri\u00eang bi\u1ec7t trong 3 l\u1ecd b\u1ecb m\u1ea5t nh\u00e3n. \u0110\u1ec3 nh\u1eadn bi\u1ebft 3 ch\u1ea5t r\u1eafn tr\u00ean, ta d\u00f9ng thu\u1ed1c th\u1eed l\u00e0 ","select":["A. Dung d\u1ecbch HCl ","B. Dung d\u1ecbch NaOH ","C. Dung d\u1ecbch NaCl ","D. Kh\u00ed hi\u0111ro "],"hint":"","explain":"<span class='basic_left'>Khi cho t\u1eebng ch\u1ea5t t\u00e1c d\u1ee5ng v\u1edbi HCl :<br\/>Ch\u1ea5t kh\u00f4ng t\u00e1c d\u1ee5ng ( kh\u00f4ng tan ) l\u00e0 Ag<br\/>Ch\u1ea5t tan, cho kh\u00ed bay ra l\u00e0 Zn : $Zn+2HCl\\to ZnC{{l}_{2}}+{{H}_{2}}$<br\/>Ch\u1ea5t tan nh\u01b0ng kh\u00f4ng c\u00f3 kh\u00ed tho\u00e1t ra v\u00e0 t\u1ea1o th\u00e0nh dung d\u1ecbch c\u00f3 m\u00e0u xanh l\u00e0 CuO : $CuO+2HCl\\to CuC{{l}_{2}}+{{H}_{2}}O$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2346},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Cho d\u00e3y chuy\u1ec3n h\u00f3a sau : $S\\xrightarrow{(1)}S{{O}_{2}}\\xrightarrow{(2)}{{H}_{2}}S{{O}_{3}}$. Ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc bi\u1ec3u di\u1ec5n chuy\u1ec3n h\u00f3a l\u00e0 ","select":["A. $\\begin{aligned} & (1)2S+{{O}_{2}}\\xrightarrow{{{t}^{o}}}S{{O}_{3}} \\\\ & (2)S{{O}_{3}}+{{H}_{2}}O\\to {{H}_{2}}S{{O}_{3}} \\\\ \\end{aligned}$ ","B. $\\begin{aligned} & (1)2S+{{O}_{2}}\\xrightarrow{{{t}^{o}}}2S{{O}_{2}} \\\\ & (2)S{{O}_{2}}+{{H}_{2}}O\\to {{H}_{2}}S{{O}_{4}} \\\\ \\end{aligned}$ ","C. $\\begin{aligned} & (1)S+{{O}_{2}}\\xrightarrow{{{t}^{o}}}S{{O}_{2}} \\\\ & (2)S{{O}_{2}}+{{H}_{2}}O\\to {{H}_{2}}S{{O}_{3}} \\\\ \\end{aligned}$ ","D. $\\begin{aligned} & (1)2S+3{{O}_{2}}\\xrightarrow{{{t}^{o}}}2S{{O}_{3}} \\\\ & (2)S{{O}_{3}}+{{H}_{2}}O\\to {{H}_{2}}S{{O}_{4}} \\\\ \\end{aligned}$ "],"hint":"","explain":"<span class='basic_left'>(1) ph\u1ea3n \u1ee9ng \u0111i\u1ec1u ch\u1ebf oxit axit : $S+{{O}_{2}}\\xrightarrow{{{t}^{o}}}S{{O}_{2}}$<br\/>(2) ph\u1ea3n \u1ee9ng \u0111i\u1ec1u ch\u1ebf axit b\u1eb1ng c\u00e1ch cho oxit axit t\u00e1c d\u1ee5ng v\u1edbi n\u01b0\u1edbc <br\/>$S{{O}_{2}}+{{H}_{2}}O\\to {{H}_{2}}S{{O}_{3}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2347},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Trong s\u1ed1 nh\u1eefng ch\u1ea5t d\u01b0\u1edbi \u0111\u00e2y, ch\u1ea5t n\u00e0o l\u00e0m qu\u1ef3 t\u00edm h\u00f3a \u0111\u1ecf ? ","select":["A. \u0110\u01b0\u1eddng ","B. Gi\u1ea5m \u0103n ","C. N\u01b0\u1edbc v\u00f4i trong ","D. Mu\u1ed1i \u0103n "],"hint":"","explain":"<span class='basic_left'>Axit axetic c\u00f3 trong gi\u1ea5m \u0103n $\\to$ gi\u1ea5m \u0103n l\u00e0m qu\u1ef3 t\u00edm h\u00f3a \u0111\u1ecf<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2348},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"C\u00f3 th\u1ec3 \u0111i\u1ec1u ch\u1ebf \u0111\u01b0\u1ee3c bao nhi\u00eau mol axit sunfuric khi cho 320 gam l\u01b0u hu\u1ef3nh trioxit $S{{O}_{3}}$ t\u00e1c d\u1ee5ng v\u1edbi n\u01b0\u1edbc ? ","select":["A. 3 mol ","B. 5 mol ","C. 4 mol ","D. 4,5 mol "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{S{{O}_{3}}}}=\\dfrac{320}{80}=4(mol) \\\\ & S{{O}_{3}}+{{H}_{2}}O\\to {{H}_{2}}S{{O}_{4}} \\\\ & \\begin{matrix} 4mol & \\to & 4mol & {} \\\\\\end{matrix} \\\\ & \\to {{n}_{{{H}_{2}}S{{O}_{4}}}}=4(mol) \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2349},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Th\u1ec3 t\u00edch kh\u00ed hidro tho\u00e1t ra (\u0111ktc) khi cho 13 gam k\u1ebdm t\u00e1c d\u1ee5ng h\u1ebft v\u1edbi axit sunfuric l\u00e0 ","select":["A. 3,36 l\u00edt ","B. 2,24 l\u00edt ","C. 5,6 l\u00edt ","D. 4,48 l\u00edt "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{Zn}}=\\dfrac{13}{65}=0,2(mol) \\\\ & Zn+{{H}_{2}}S{{O}_{4}}\\to ZnS{{O}_{4}}+{{H}_{2}} \\\\ & \\begin{matrix} 0,2mol & {} & \\to & {} & {} & 0,2mol & {} \\\\\\end{matrix} \\\\ & \\to {{V}_{{{H}_{2}}}}=0,2.22,4=4,48(l) \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2350},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Ng\u00e2m 2,7 gam b\u1ed9t nh\u00f4m trong dung d\u1ecbch ch\u1ee9a 39,2 gam axit sunfuric. L\u01b0\u1ee3ng kh\u00ed hi\u0111ro thu \u0111\u01b0\u1ee3c \u1edf ph\u1ea3n \u1ee9ng tr\u00ean c\u00f3 th\u1ec3 d\u00f9ng \u0111\u1ec3 kh\u1eed t\u1ed1i \u0111a bao nhi\u00eau gam b\u1ed9t ch\u00ec(II) oxit ? ","select":["A. 45,33 gam ","B. 33,45 gam ","C. 44,67 gam ","D. 34,56 gam "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{Al}}=\\dfrac{2,7}{27}=0,1(mol) \\\\ & {{n}_{{{H}_{2}}S{{O}_{4}}}}=\\dfrac{39,2}{98}=0,4(mol) \\\\ & 2Al+3{{H}_{2}}S{{O}_{4}}\\to A{{l}_{2}}{{(S{{O}_{4}})}_{3}}+3{{H}_{2}} \\\\ \\end{aligned}$<br\/>L\u1eadp t\u1ef7 s\u1ed1 : $\\dfrac{{{n}_{Al}}}{2}<\\dfrac{{{n}_{{{H}_{2}}S{{O}_{4}}}}}{3}$<br\/>$\\to$ Al ph\u1ea3n \u1ee9ng h\u1ebft, ${{H}_{2}}S{{O}_{4}}$ d\u01b0, l\u01b0\u1ee3ng ${{H}_{2}}$ sinh ra \u0111\u01b0\u1ee3c t\u00ednh theo l\u01b0\u1ee3ng h\u1ebft<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh : ${{n}_{{{H}_{2}}}}=\\dfrac{3}{2}{{n}_{Al}}=0,15(mol)$<br\/>PTHH : $PbO+{{H}_{2}}\\xrightarrow{{{t}^{o}}}Pb+{{H}_{2}}O$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh kh\u1eed b\u1ed9t ch\u00ec(II) oxit : <br\/>$\\begin{aligned} & {{n}_{PbO}}={{n}_{{{H}_{2}}}}=0,15(mol) \\\\ & \\to {{m}_{PbO}}=0,15.223=33,45(g) \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2351},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Cho h\u00f2a tan ho\u00e0n to\u00e0n 20 gam h\u1ed7n h\u1ee3p kim lo\u1ea1i g\u1ed3m Al v\u00e0 Cu trong dung d\u1ecbch HCl d\u01b0 thu \u0111\u01b0\u1ee3c 6,72 l\u00edt kh\u00ed hi\u0111ro (\u0111ktc). Bi\u1ebft Cu kh\u00f4ng tan trong HCl. Th\u00e0nh ph\u1ea7n ph\u1ea7n tr\u0103m kh\u1ed1i l\u01b0\u1ee3ng kim lo\u1ea1i trong h\u1ed7n h\u1ee3p l\u00e0 ","select":["A. $\\%{{m}_{Al}}=27\\%,\\%{{m}_{Cu}}=73\\%$ ","B. $\\%{{m}_{Al}}=73\\%,\\%{{m}_{Cu}}=27\\%$ ","C. $\\%{{m}_{Al}}=70\\%,\\%{{m}_{Cu}}=30\\%$ ","D. $\\%{{m}_{Al}}=20\\%,\\%{{m}_{Cu}}=80\\%$ "],"hint":"","explain":"<span class='basic_left'>Theo b\u00e0i ra Cu kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi HCl, ch\u1ec9 c\u00f3 Al t\u00e1c d\u1ee5ng v\u1edbi HCl<br\/>$\\begin{aligned} & {{n}_{{{H}_{2}}}}=\\dfrac{6,72}{22,4}=0,3(mol) \\\\ & 2Al+6HCl\\to 2AlC{{l}_{3}}+3{{H}_{2}} \\\\ \\end{aligned}$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh : ${{n}_{Al}}=\\dfrac{2}{3}{{n}_{{{H}_{2}}}}=0,2(mol)$<br\/>$\\begin{aligned} & \\to {{m}_{Al}}=0,2.27=5,4(g) \\\\ & \\to \\%{{m}_{Al}}=\\dfrac{5,4}{20}.100=27\\% \\\\ & \\to \\%{{m}_{Cu}}=100\\%-27\\%=73\\% \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2352},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Cho 11 gam h\u1ed7n h\u1ee3p g\u1ed3m Al v\u00e0 Fe t\u00e1c d\u1ee5ng v\u1eeba \u0111\u1ee7 v\u1edbi dung d\u1ecbch HCl t\u1ea1o th\u00e0nh 8,96 l\u00edt kh\u00ed hi\u0111ro tho\u00e1t ra \u1edf \u0111ktc. D\u1eabn to\u00e0n b\u1ed9 l\u01b0\u1ee3ng kh\u00ed tr\u00ean qua 16 gam b\u1ed9t CuO \u0111un n\u00f3ng \u0111\u1ebfn ph\u1ea3n \u1ee9ng k\u1ebft th\u00fac. Kh\u1ed1i l\u01b0\u1ee3ng Cu thu \u0111\u01b0\u1ee3c l\u00e0 ","select":["A. 13 gam ","B. 14 gam ","C. 12,8 gam ","D. 32 gam "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{{{H}_{2}}}}=\\dfrac{8,96}{22,4}=0,4(mol) \\\\ & {{n}_{CuO}}=\\dfrac{16}{80}=0,2(mol) \\\\ & CuO+{{H}_{2}}\\xrightarrow{{{t}^{o}}}Cu+{{H}_{2}}O \\\\ \\end{aligned}$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh c\u1ee9 1 mol CuO t\u00e1c d\u1ee5ng v\u1edbi 1 mol kh\u00ed hi\u0111ro v\u1eady sau khi ph\u1ea3n \u1ee9ng k\u1ebft th\u00fac CuO h\u1ebft, ${{H}_{2}}$ d\u01b0<br\/>$\\to$ l\u01b0\u1ee3ng Cu sinh ra \u0111\u01b0\u1ee3c t\u00ednh theo l\u01b0\u1ee3ng h\u1ebft<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh : <br\/>$\\begin{aligned} & {{n}_{Cu}}={{n}_{CuO}}=0,2(mol) \\\\ & \\to {{m}_{Cu}}=0,2.64=12,8(g) \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2353},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Kh\u1eed ho\u00e0n to\u00e0n 19,7 gam h\u1ed7n h\u1ee3p g\u1ed3m $F{{e}_{3}}{{O}_{4}},ZnO$ c\u1ea7n d\u00f9ng v\u1eeba \u0111\u1ee7 6,72 l\u00edt ${{H}_{2}}$ thu \u0111\u01b0\u1ee3c h\u1ed7n h\u1ee3p kim lo\u1ea1i. \u0110\u1ec3 c\u00f3 \u0111\u01b0\u1ee3c l\u01b0\u1ee3ng tr\u00ean ph\u1ea3i d\u00f9ng bao nhi\u00eau gam kim lo\u1ea1i Mg v\u00e0 axit sunfuric? Bi\u1ebft axit d\u00f9ng d\u01b0 $10\\%$ ","select":["A. ${{m}_{Mg}}=8,4(g),{{m}_{{{H}_{2}}S{{O}_{4}}}}=32,67(g)$ ","B. ${{m}_{Mg}}=7,2(g),{{m}_{{{H}_{2}}S{{O}_{4}}}}=29,4(g)$ ","C. ${{m}_{Mg}}=6,8(g),{{m}_{{{H}_{2}}S{{O}_{4}}}}=33,45(g)$ ","D. ${{m}_{Mg}}=7,2(g),{{m}_{{{H}_{2}}S{{O}_{4}}}}=32,67(g)$ "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{{{H}_{2}}}}=\\dfrac{6,72}{22,4}=0,3(mol) \\\\ & Mg+{{H}_{2}}S{{O}_{4}}\\to MgS{{O}_{4}}+{{H}_{2}} \\\\ \\end{aligned}$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh ph\u1ea3n \u1ee9ng : ${{n}_{Mg}}={{n}_{{{H}_{2}}S{{O}_{4}}}}={{n}_{{{H}_{2}}}}=0,3(mol)$<br\/>Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a Mg v\u00e0 axit sunfuric tham gia ph\u1ea3n \u1ee9ng l\u00e0 <br\/>$\\begin{align} & {{m}_{Mg}}=0,3.24=7,2(g) \\\\ & {{m}_{{{H}_{2}}S{{O}_{4}}}}=0,3.98=29,4(g) \\\\ \\end{align}$<br\/>Bi\u1ebft axit d\u00f9ng d\u01b0 $10\\%$ n\u00ean kh\u1ed1i l\u01b0\u1ee3ng axit sunfuric th\u1ef1c t\u1ebf ban \u0111\u1ea7u l\u00e0 $\\dfrac{29,4.100}{90}=32,67(g)$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2354},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"C\u00f4ng th\u1ee9c h\u00f3a h\u1ecdc c\u1ee7a mu\u1ed1i c\u00f3 t\u00ean l\u00e0 nh\u00f4m (III) nitrat ","select":["A. $A{{l}_{2}}{{(N{{O}_{3}})}_{3}}$ ","B. $A{{l}_{3}}{{(N{{O}_{3}})}_{4}}$ ","C. $AlN{{O}_{3}}$ ","D. $Al{{(N{{O}_{3}})}_{3}}$ "],"hint":"","explain":"<span class='basic_left'>Nh\u00f4m(III) nitrat : nh\u00f4m h\u00f3a tr\u1ecb III c\u00f2n nh\u00f3m $N{{O}_{3}}$ h\u00f3a tr\u1ecb I<br\/>G\u1ecdi c\u00f4ng th\u1ee9c mu\u1ed1i c\u00f3 d\u1ea1ng $A{{l}_{x}}{{(N{{O}_{3}})}_{y}}$<br\/>Theo quy t\u1eafc h\u00f3a tr\u1ecb :<br\/>$III\\times x=I\\times y\\to \\dfrac{x}{y}=\\dfrac{I}{III}=\\dfrac{1}{3}$<br\/>V\u1eady c\u00f4ng th\u1ee9c h\u00f3a h\u1ecdc c\u1ee7a mu\u1ed1i l\u00e0 $Al{{(N{{O}_{3}})}_{3}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2355},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"H\u00f2a tan 3,6 gam m\u1ed9t kim lo\u1ea1i A h\u00f3a tr\u1ecb II b\u1eb1ng m\u1ed9t l\u01b0\u1ee3ng d\u01b0 axit HCl thu \u0111\u01b0\u1ee3c 3,36 l\u00edt kh\u00ed ${{H}_{2}}$ (\u0111ktc). T\u00ean kim lo\u1ea1i A l\u00e0 ","select":["A. \u0110\u1ed3ng ","B. Magie ","C. S\u1eaft ","D. Nh\u00f4m "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{{{H}_{2}}}}=\\dfrac{3,36}{22,4}=0,15(mol) \\\\ & A+2HCl\\to AC{{l}_{2}}+{{H}_{2}} \\\\ & \\begin{matrix} 0,15mol & \\leftarrow & {} & 0,15mol & {} & {} & {} \\\\\\end{matrix} \\\\ & \\to {{M}_{A}}=\\dfrac{3,6}{0,15}=24 \\\\ \\end{aligned}$<br\/>V\u1eady A l\u00e0 Magie (Mg)<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2356},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Cho 11,3 gam h\u1ed7n h\u1ee3p g\u1ed3m Zn v\u00e0 Mg t\u00e1c d\u1ee5ng v\u1eeba \u0111\u1ee7 v\u1edbi dung d\u1ecbch HCl t\u1ecda th\u00e0nh 6,72 l\u00edt kh\u00ed ${{H}_{2}}$ tho\u00e1t ra \u1edf \u0111ktc. Kh\u1ed1i l\u01b0\u1ee3ng m\u1ed7i kim lo\u1ea1i trong h\u1ed7n h\u1ee3p l\u00e0 ","select":["A. ${{m}_{Zn}}=6,5(g),{{m}_{Mg}}=4,8(g)$ ","B. ${{m}_{Zn}}=4,8(g),{{m}_{Mg}}=6,5(g)$ ","C. ${{m}_{Zn}}=5,4(g),{{m}_{Mg}}=5,9(g)$ ","D. ${{m}_{Zn}}=5,9(g),{{m}_{Mg}}=5,4(g)$ "],"hint":"","explain":"<span class='basic_left'>G\u1ecdi s\u1ed1 mol c\u1ee7a ${{H}_{2}}$ sinh ra \u1edf ph\u1ea3n \u1ee9ng Zn t\u00e1c d\u1ee5ng v\u1edbi HCl l\u00e0 x (mol)<br\/>$\\begin{aligned} & {{n}_{{{H}_{2}}}}=\\dfrac{6,72}{22,4}=0,3(mol) \\\\ & Zn+2HCl\\to ZnC{{l}_{2}}+{{H}_{2}} \\\\ & \\begin{matrix} x & {} & \\leftarrow & {} & {} & {} & x \\\\\\end{matrix} \\\\ & Mg+2HCl\\to MgC{{l}_{2}}+{{H}_{2}} \\\\ & \\begin{matrix} 0,3-x & \\leftarrow & {} & {} & {} & 0,3-x & {} \\\\\\end{matrix} \\\\ & {{m}_{hh}}={{m}_{Zn}}+{{m}_{Mg}}=11,3 \\\\ & \\to 65x+24.(0,3-x)=11,3 \\\\ & \\to x=0,1(mol) \\\\ \\end{aligned}$<br\/>Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a m\u1ed7i kim lo\u1ea1i l\u00e0 <br\/>$\\begin{aligned} & {{m}_{Zn}}=0,1.65=6,5(g) \\\\ & \\to {{m}_{Mg}}=11,3-6,5=4,8(g) \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2357},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"H\u00f2a tan 8 gam h\u1ed7n h\u1ee3p oxit $CaO,F{{e}_{2}}{{O}_{3}}$ v\u00e0o n\u01b0\u1edbc th\u00ec thu \u0111\u01b0\u1ee3c 5,92 gam ki\u1ec1m. Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a $F{{e}_{2}}{{O}_{3}}$ l\u00e0 ","select":["A. 4,48 gam ","B. 5,6 gam ","C. 3,52 gam ","D. 4,56 gam "],"hint":"","explain":"<span class='basic_left'>$F{{e}_{2}}{{O}_{3}}$ kh\u00f4ng ta trong n\u01b0\u1edbc ch\u1ec9 c\u00f3 $CaO$ tan<br\/>$\\to$ Ki\u1ec1m thu \u0111\u01b0\u1ee3c ch\u1ec9 c\u00f3 $Ca{{OH}_{2}}$<br\/>$\\begin{aligned} & {{n}_{Ca{{(OH)}_{2}}}}=\\dfrac{5,92}{40+17.2}=0,08(mol) \\\\ & CaO+{{H}_{2}}O\\to Ca{{(OH)}_{2}} \\\\ & \\begin{matrix} 0,08 & \\leftarrow & {} & 0,08 \\\\\\end{matrix} \\\\ & \\to {{m}_{CaO}}=0,08.(40+16)=4,48(g) \\\\ & \\to {{m}_{F{{e}_{2}}{{O}_{3}}}}=8-4,48=3,52(g) \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2358},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Cho 1,35 gam nh\u00f4m t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch ch\u1ee9a 7,3 gam HCl. Kh\u1ed1i l\u01b0\u1ee3ng mu\u1ed1i t\u1ea1o th\u00e0nh l\u00e0 ","select":["A. 5,689 gam ","B. 6,984 gam ","C. 6,756 gam ","D. 6,675 gam "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{Al}}=\\dfrac{1,35}{27}=0,05(mol) \\\\ & {{n}_{HCl}}=\\dfrac{7,3}{36,5}=0,2(mol) \\\\ & 2Al+6HCl\\to 2AlC{{l}_{3}}+3{{H}_{2}} \\\\ & \\to \\dfrac{0,05}{2}<\\dfrac{0,2}{6}(*) \\\\ \\end{aligned}$<br\/>Theo b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh (*) sau ph\u1ea3n \u1ee9ng Al h\u1ebft, HCl d\u01b0<br\/>L\u01b0\u1ee3ng mu\u1ed1i sinh ra \u0111\u01b0\u1ee3c t\u00ednh theo l\u01b0\u1ee3ng h\u1ebft<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh ph\u1ea3n \u1ee9ng :<br\/>$\\begin{aligned} & {{n}_{AlC{{l}_{3}}}}={{n}_{Al}}=0,05(mol) \\\\ & \\to {{m}_{AlC{{l}_{3}}}}=0,05.(27+35,5.3)=6,675(g) \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2359}],"lesson":{"save":0,"level":2}}