{"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":"Ph\u1ea3n \u1ee9ng h\u00f3a h\u1ecdc trong \u0111\u00f3 ch\u1ec9 c\u00f3 m\u1ed9t ch\u1ea5t m\u1edbi \u0111\u01b0\u1ee3c t\u1ea1o th\u00e0nh t\u1eeb hai hay nhi\u1ec1u ch\u1ea5t ban \u0111\u1ea7u g\u1ecdi l\u00e0 ","select":["A. Ph\u1ea3n \u1ee9ng h\u00f3a h\u1ee3p ","B. Ph\u1ea3n \u1ee9ng ph\u00e2n h\u1ee7y ","C. Ph\u1ea3n \u1ee9ng th\u1ebf ","D. Ph\u1ea3n \u1ee9ng trao \u0111\u1ed5i "],"hint":"","explain":"<span class='basic_left'>Ph\u1ea3n \u1ee9ng h\u00f3a h\u1ee3p l\u00e0 ph\u1ea3n \u1ee9ng h\u00f3a h\u1ecdc trong \u0111\u00f3 ch\u1ec9 c\u00f3 m\u1ed9t ch\u1ea5t m\u1edbi \u0111\u01b0\u1ee3c t\u1ea1o th\u00e0nh t\u1eeb hai hay nhi\u1ec1u ch\u1ea5t ban \u0111\u1ea7u<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2140},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Ph\u1ea3n \u1ee9ng h\u00f3a h\u1ecdc n\u00e0o sau \u0111\u00e2y l\u00e0 ph\u1ea3n \u1ee9ng h\u00f3a h\u1ee3p ","select":["A. $2KCl{{O}_{3}}\\to 2KCl+3{{O}_{2}}$ ","B. $Cu+2AgN{{O}_{3}}\\to Cu{{(N{{O}_{3}})}_{2}}+2Ag$ ","C. $CaO+C{{O}_{2}}\\to CaC{{O}_{3}}$ ","D. $AgN{{O}_{3}}+NaCl\\to AgCl+NaN{{O}_{3}}$ "],"hint":"","explain":"<span class='basic_left'>Ph\u1ea3n \u1ee9ng h\u00f3a h\u1ee3p l\u00e0 ph\u1ea3n \u1ee9ng h\u00f3a h\u1ecdc trong \u0111\u00f3 ch\u1ec9 c\u00f3 m\u1ed9t ch\u1ea5t m\u1edbi \u0111\u01b0\u1ee3c t\u1ea1o th\u00e0nh t\u1eeb hai hay nhi\u1ec1u ch\u1ea5t ban \u0111\u1ea7u<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2141},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Oxit c\u1ee7a m\u1ed9t nguy\u00ean t\u1ed1 $R$ c\u00f3 h\u00f3a tr\u1ecb $II$ ch\u1ee9a $20\\%O$ ( v\u1ec1 kh\u1ed1i l\u01b0\u1ee3ng ). C\u00f4ng th\u1ee9c h\u00f3a h\u1ecdc c\u1ee7a oxit \u0111\u00f3 l\u00e0","select":["A. $FeO$ ","B. $CuO$ ","C. $MgO$ ","D. $PbO$ "],"hint":"","explain":"<span class='basic_left'>R c\u00f3 h\u00f3a tr\u1ecb II v\u00e0 oxi c\u00f3 h\u00f3a tr\u1ecb II<br\/>Theo quy t\u1eafc h\u00f3a tr\u1ecb ta c\u00f3 c\u00f4ng th\u1ee9c oxit d\u1ea1ng: $RO$<br\/>Ta c\u00f3 :<br\/> $\\begin{aligned} & \\%O=\\dfrac{{{M}_{O}}}{{{M}_{RO}}}.100\\%=\\dfrac{{{M}_{O}}}{{{M}_{R}}+{{M}_{O}}}.100\\%=\\dfrac{16}{{{M}_{R}}+16}.100\\%=20\\% \\\\ & \\Rightarrow {{M}_{R}}=\\dfrac{16.100}{20}-16=64 \\\\ \\end{aligned}$<br\/>V\u1eady R l\u00e0 kim lo\u1ea1i \u0111\u1ed3ng (Cu)<br\/>$\\to $C\u00f4ng th\u1ee9c h\u00f3a h\u1ecdc c\u1ee7a oxit l\u00e0 $CuO$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2142},{"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 c\u00e1c c\u00f4ng th\u1ee9c h\u00f3a h\u1ecdc sau : $S{{O}_{2}},CuO,F{{e}_{2}}{{O}_{3}},NaOH,{{P}_{2}}{{O}_{5}},{{N}_{2}},Al,AgN{{O}_{3}}$. S\u1ed1 c\u00f4ng th\u1ee9c l\u00e0 c\u00f4ng th\u1ee9c c\u1ee7a oxit l\u00e0 ","select":["A. $6$ ","B. $3$ ","C. $5$ ","D. $4$ "],"hint":"","explain":"<span class='basic_left'>Oxit l\u00e0 h\u1ee3p ch\u1ea5t c\u1ee7a hai nguy\u00ean t\u1ed1, trong \u0111\u00f3 c\u00f3 m\u1ed9t nguy\u00ean t\u1ed1 l\u00e0 oxi<br\/> C\u00e1c c\u00f4ng th\u1ee9c c\u1ee7a oxit l\u00e0 : $S{{O}_{2}},CuO,F{{e}_{2}}{{O}_{3}},{{P}_{2}}{{O}_{5}}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2143},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Ng\u01b0\u1eddi l\u00ednh c\u1ee9u h\u1ecfa d\u00f9ng b\u00ecnh \u0111\u1eb7c bi\u1ec7t ch\u1ee9a kh\u00ed oxi \u0111\u1ec3 : ","select":["A. D\u1eadp t\u1eaft \u0111\u00e1m ch\u00e1y ","B. Tr\u00e1nh b\u1ecb b\u1ecfng ","C. H\u00f4 h\u1ea5p ","D. Li\u00ean l\u1ea1c v\u1edbi ng\u01b0\u1eddi b\u00ean ngo\u00e0i "],"hint":"","explain":"<span class='basic_left'>Qu\u00e1 tr\u00ecnh ch\u00e1y l\u00e0 nh\u1eefng qu\u00e1 tr\u00ecnh l\u00fd, h\u00f3a h\u1ecdc ph\u1ee9c t\u1ea1p c\u1ee7a ph\u1ea3n \u1ee9ng ch\u00e1y gi\u1eefa c\u00e1c ch\u1ea5t \u00f4xy h\u00f3a t\u1ea1o th\u00e0nh s\u1ea3n ph\u1ea9m ch\u00e1y. C\u00e1c s\u1ea3n ph\u1ea9m ch\u00e1y t\u00f9y theo \u0111i\u1ec1u ki\u1ec7n ch\u00e1y m\u00e0 ta c\u00f3 c\u00e1c s\u1ea3n ph\u1ea9m ch\u00e1y ho\u00e0n to\u00e0n nh\u01b0 c\u00e1c ch\u1ea5t: $S{{O}_{2}},C{{O}_{2}},{{H}_{2}}O,HCl,{{N}_{2}}$ \u2026Hay c\u00e1c s\u1ea3n ph\u1ea9m ch\u00e1y kh\u00f4ng ho\u00e0n to\u00e0n nh\u01b0 c\u00e1c ch\u1ea5t: $CO,C{{H}_{4}},{{H}_{2}}S$ \u2026<br\/>T\u1ea5t c\u1ea3 c\u00e1c kh\u00ed sinh ra trong qu\u00e1 tr\u00ecnh ch\u00e1y \u0111\u1ec1u \u1ea3nh h\u01b0\u1edfng \u0111\u1ebfn s\u1ef1 h\u00f4 h\u1ea5p, \u0111\u1ed9c h\u1ea1i v\u1edbi c\u01a1 th\u1ec3 con ng\u01b0\u1eddi v\u00e0 khi ch\u00e1y l\u01b0\u1ee3ng oxi s\u1ebd b\u1ecb thi\u1ebfu v\u00ec oxi trong kh\u00f4ng kh\u00ed \u0111\u00e3 cung c\u1ea5p cho qu\u00e1 tr\u00ecnh ch\u00e1y v\u00ec v\u1eady c\u00e1c l\u00ednh c\u1ee9u h\u00f3a c\u1ea7n \u0111eo b\u00ecnh ch\u1ee9a oxi \u0111\u1ec3 ph\u1ee5c v\u1ee5 cho qu\u00e1 tr\u00ecnh h\u00f4 h\u1ea5p<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2144},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc n\u00e0o sau \u0111\u00e2y bi\u1ec3u di\u1ec5n qu\u00e1 tr\u00ecnh oxi h\u00f3a : ","select":["A. $CaO+{{H}_{2}}O\\to Ca{{(OH)}_{2}}$ ","B. ${{C}_{12}}{{H}_{22}}{{O}_{11}}+12{{O}_{2}}\\xrightarrow{{{t}^{o}}}12C{{O}_{2}}+11{{H}_{2}}O$ ","C. $BaC{{l}_{2}}+N{{a}_{2}}S{{O}_{4}}\\to BaS{{O}_{4}}+2NaCl$ ","D. $2KCl{{O}_{3}}\\xrightarrow{{{t}^{o}}}2KCl+3{{O}_{2}}$ "],"hint":"","explain":"<span class='basic_left'>Qu\u00e1 tr\u00ecnh oxi h\u00f3a l\u00e0 s\u1ef1 t\u00e1c d\u1ee5ng c\u1ee7a oxi v\u1edbi m\u1ed9t ch\u1ea5t<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2145},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"\u1ee8ng d\u1ee5ng quan tr\u1ecdng nh\u1ea5t c\u1ee7a kh\u00ed oxi l\u00e0 ","select":["A. S\u1ef1 h\u00f4 h\u1ea5p ","B. D\u1eadp t\u1eaft c\u00e1c \u0111\u00e1m ch\u00e1y ","C. S\u1ef1 \u0111\u1ed1t nhi\u00ean li\u1ec7u ","D. C\u1ea3 $A$ v\u00e0 $C$ "],"hint":"","explain":"<span class='basic_left'><br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D .<\/span><\/span> ","column":2}]}],"id_ques":2146},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"S\u1ef1 oxi h\u00f3a l\u00e0 ","select":["A. S\u1ef1 t\u00e1c d\u1ee5ng c\u1ee7a m\u1ed9t ch\u1ea5t v\u1edbi oxi ","B. S\u1ef1 t\u00e1c d\u1ee5ng c\u1ee7a \u0111\u01a1n ch\u1ea5t v\u1edbi oxi ","C. S\u1ef1 t\u00e1c d\u1ee5ng c\u1ee7a h\u1ee3p ch\u1ea5t v\u1edbi oxi ","D. S\u1ef1 t\u00e1c d\u1ee5ng c\u1ee7a nhi\u1ec1u ch\u1ea5t v\u1edbi nhau "],"hint":"","explain":"<span class='basic_left'>S\u1ef1 t\u00e1c d\u1ee5ng c\u1ee7a oxi v\u1edbi m\u1ed9t ch\u1ea5t l\u00e0 s\u1ef1 oxi h\u00f3a<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2147},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Khi cho c\u00e2y n\u1ebfn \u0111ang ch\u00e1y v\u00e0o m\u1ed9t l\u1ecd th\u1ee7y tinh ch\u1ee9a oxi nguy\u00ean ch\u1ea5t. Hi\u1ec7n t\u01b0\u1ee3ng x\u1ea3y ra ti\u1ebfp theo l\u00e0 ","select":["A. C\u00e2y n\u1ebfn b\u1ecb t\u1eaft ngay ","B. C\u00e2y n\u1ebfn ch\u00e1y b\u00ecnh th\u01b0\u1eddng ","C. C\u00e2y n\u00ean ch\u00e1y s\u00e1ng ch\u00f3i ","D. C\u00e2y n\u1ebfn ch\u00e1y m\u1ed9t l\u00fac r\u1ed3i t\u1eaft "],"hint":"","explain":"<span class='basic_left'>Khi c\u00e2y n\u1ebfn ch\u00e1y trong kh\u00f4ng kh\u00ed v\u1edbi l\u01b0\u1ee3ng oxi th\u1ea5p th\u00ec n\u00ean ch\u00e1y b\u00ecnh th\u01b0\u1eddng nh\u01b0ng khi cho v\u00e0o b\u00ecnh th\u1ee7y tinh c\u00f3 ch\u1ee9a oxi nguy\u00ean ch\u1ea5t th\u00ec c\u00e2y n\u1ebfn s\u1ebd ch\u00e1y v\u1edbi ng\u1ecdn l\u1eeda s\u00e1ng ch\u00f3i v\u00ec khi \u0111\u00f3 n\u1ebfn \u0111\u01b0\u1ee3c cung c\u1ea5p l\u01b0\u1ee3ng oxi nhi\u1ec1u h\u01a1n cho s\u1ef1 ch\u00e1y<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2148},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"\u0110\u1ed1t ch\u00e1y $3,1$ gam photpho trong b\u00ecnh ch\u1ee9a $2,912$ l\u00edt kh\u00ed oxi (\u0111ktc). Sau ph\u1ea3n \u1ee9ng ch\u1ea5t n\u00e0o c\u00f2n d\u01b0 ? H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n m\u00e0 em cho l\u00e0 \u0111\u00fang.","select":["A. Photpho ","B. C\u1ea3 hai ch\u1ea5t \u0111\u1ec1u d\u01b0 ","C. Oxi ","D. Kh\u00f4ng x\u00e1c \u0111\u1ecbnh \u0111\u01b0\u1ee3c "],"hint":"","explain":"<span class='basic_left'>S\u1ed1 mol photpho tham gia ph\u1ea3n \u1ee9ng l\u00e0 <br\/>${{n}_{P}}=\\dfrac{m}{M}=\\dfrac{3,1}{31}=0,1(mol)$<br\/>S\u1ed1 mol oxi tham gia ph\u1ea3n \u1ee9ng l\u00e0 <br\/>${{n}_{{{O}_{2}}}}=\\dfrac{V}{22,4}=\\dfrac{2,912}{22,4}=0,13(mol)$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh ph\u1ea3n \u1ee9ng : $4P+5{{O}_{2}}\\xrightarrow{{{t}^{o}}} 2{{P}_{2}}{{O}_{5}}$<br\/>X\u00e9t t\u1ec9 l\u1ec7 ph\u1ea3n \u1ee9ng s\u1ed1 mol chia cho h\u1ec7 s\u1ed1 ph\u1ea3n \u1ee9ng ta c\u00f3 : <br\/>$\\dfrac{0,1}{4}<\\dfrac{0,13}{5}\\Rightarrow $ Oxi d\u01b0, P ph\u1ea3n \u1ee9ng v\u1eeba \u0111\u1ee7 <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2149},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"\u0110\u1ed1t ch\u00e1y ho\u00e0n to\u00e0n m\u1ed9t d\u00e2y s\u1eaft c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng b\u1eb1ng $0,84$ gam trong kh\u00f4ng kh\u00ed d\u01b0. Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a s\u1ea3n ph\u1ea9m thu \u0111\u01b0\u1ee3c sau ph\u1ea3n \u1ee9ng l\u00e0 ","select":["A. $1,2(g)$ ","B. $1,08(g)$ ","C. $2,4(g)$ ","D. $1,16(g)$ "],"hint":"","explain":"<span class='basic_left'>Ph\u1ea3n \u1ee9ng x\u1ea3y ra ho\u00e0n to\u00e0n trong kh\u00f4ng kh\u00ed d\u01b0 n\u00ean s\u1ea3n ph\u1ea9m cu\u1ed1i c\u00f9ng c\u1ee7a qu\u00e1 tr\u00ecnh \u0111\u1ed1t ch\u00e1y l\u00e0 $F{{e}_{3}}{{O}_{4}}$<br\/>PTHH : $3Fe+2{{O}_{2}}\\xrightarrow{{{t}^{o}}} F{{e}_{3}}{{O}_{4}}$<br\/>${{n}_{Fe}}=\\dfrac{m}{M}=\\dfrac{0,84}{56}=0,015(mol)$<br\/>Theo ph\u01b0\u01a1ng trinh h\u00f3a h\u1ecdc : ${{n}_{F{{e}_{3}}{{O}_{4}}}}=\\dfrac{1}{3}{{n}_{Fe}}=0,005(mol)$<br\/>$\\to$ ${{m}_{F{{e}_{3}}{{O}_{4}}}}=n.M=0,005.232=1,16(g)$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2150},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Khi \u0111\u1ed1t ch\u00e1y l\u01b0u hu\u1ef3nh trong kh\u00f4ng kh\u00ed th\u00ec thu \u0111\u01b0\u1ee3c s\u1ea3n ph\u1ea9m l\u00e0 kh\u00ed l\u01b0u hu\u1ef3nh \u0111ioxit $(S{{O}_{2}})$. Th\u1ec3 t\u00edch kh\u00ed $S{{O}_{2}}$ (\u0111ktc) thu \u0111\u01b0\u1ee3c khi \u0111\u1ed1t ch\u00e1y ho\u00e0n to\u00e0n $5,12$ gam b\u1ed9t l\u01b0u hu\u1ef3nh l\u00e0 ","select":["A. $3,136$ l\u00edt ","B. $2,912$ l\u00edt ","C. $3,808$ l\u00edt ","D. $3,584$ l\u00edt "],"hint":"","explain":"<span class='basic_left'>PTHH : $S+{{O}_{2}}\\xrightarrow{{{t}^{o}}} S{{O}_{2}}$<br\/>${{n}_{S}}=\\dfrac{m}{M}=\\dfrac{5,12}{32}=0,16(mol)$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc : ${{n}_{S{{O}_{2}}}}={{n}_{S}}=0,16(mol)$<br\/>$\\to {{V}_{S{{O}_{2}}}}=n.22,4=0,16.22,4=3,584$ l\u00edt<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2151},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"\u0110\u1ed1t ch\u00e1y m\u1ed9t kim lo\u1ea1i $M$ h\u00f3a tr\u1ecb $I$ c\u1ea7n d\u00f9ng $2,016$ l\u00edt kh\u00ed oxi (\u0111ktc) thu \u0111\u01b0\u1ee3c $16,92$ gam oxit. T\u00ean c\u1ee7a kim lo\u1ea1i $M$ l\u00e0 ","select":["A. Natri ","B. B\u1ea1c ","C. Kali ","D. Canxi "],"hint":"","explain":"<span class='basic_left'>Kim lo\u1ea1i $M$ c\u00f3 h\u00f3a tr\u1ecb I v\u00e0 oxi c\u00f3 h\u00f3a tr\u1ecb II<br\/>Theo quy t\u1eafc h\u00f3a tr\u1ecb ta c\u00f3 oxit d\u1ea1ng ${{M}_{2}}O$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc l\u00e0 $4M+{{O}_{2}}\\xrightarrow{{{t}^{o}}} 2{{M}_{2}}O$<br\/>${{n}_{{{O}_{2}}}}=\\dfrac{V}{22,4}=\\dfrac{2,016}{22,4}=0,09(mol)$<br\/>Theo t\u1ef7 l\u1ec7 h\u1ec7 s\u1ed1 ph\u1ea3n \u1ee9ng ta c\u00f3 : ${{n}_{M}}=4{{n}_{{{O}_{2}}}}=4.0,09=0,36(mol)$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh lu\u1eadt b\u1ea3o to\u00e0n kh\u1ed1i l\u01b0\u1ee3ng:<br\/>${{m}_{M}}+{{m}_{{{O}_{2}}}}={{m}_{{{M}_{2}}O}}\\Rightarrow {{m}_{M}}=16,92-0,09.32=14,04(g)$<br\/>Kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a M l\u00e0 $M=\\dfrac{m}{n}=\\dfrac{14,04}{0,36}=39$<br\/>V\u1eady M l\u00e0 kali (K)<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2152},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Th\u1ec3 t\u00edch kh\u00ed oxi c\u1ea7n thi\u1ebft \u0111\u1ec3 \u0111\u1ed1t ch\u00e1y ho\u00e0n to\u00e0n kh\u00ed metan $C{{H}_{4}}$ c\u00f3 trong $1,5{{m}^{3}}$ kh\u00ed ch\u1ee9a $2\\%$ t\u1ea1p ch\u1ea5t kh\u00f4ng ch\u00e1y l\u00e0 ( c\u00e1c th\u1ec3 t\u00edch kh\u00ed \u0111\u1ec1u \u0111o \u1edf \u0111ktc ) ","select":["A. $2940(l)$ ","B. $2840(l)$ ","C. $2900(l)$ ","D. $2740(l)$ "],"hint":"","explain":"<span class='basic_left'>PTHH : $C{{H}_{4}}+2{{O}_{2}}\\xrightarrow{{{t}^{o}}} C{{O}_{2}}+2{{H}_{2}}O$<br\/>\u0110\u1ed5i $1,5{{m}^{3}}=1,5.1000d{{m}^{3}}=1500d{{m}^{3}}=1500(l)$<br\/>Th\u1ec3 t\u00edch $C{{H}_{4}}$ nguy\u00ean ch\u1ea5t l\u00e0 ${{V}_{C{{H}_{4}}}}=1500.\\dfrac{100-2}{100}=1470(l)$<br\/>Ta c\u00f3 t\u1ef7 l\u1ec7 v\u1ec1 s\u1ed1 mol c\u0169ng ch\u00ednh l\u00e0 t\u1ef7 l\u1ec7 v\u1ec1 th\u1ec3 t\u00edch<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh ta c\u00f3 : ${{V}_{{{O}_{2}}}}=2.{{V}_{C{{H}_{4}}}}=2.1470=2940(l)$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2153},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Qu\u00e1 tr\u00ecnh n\u00e0o sau \u0111\u00e2y l\u00e0m t\u0103ng l\u01b0\u1ee3ng oxi trong kh\u00f4ng kh\u00ed? ","select":["A. S\u1ef1 ch\u00e1y c\u1ee7a than, c\u1ee7i, b\u1ebfp ga ","B. S\u1ef1 quang h\u1ee3p c\u1ee7a c\u00e2y xanh ","C. S\u1ef1 g\u1ec9 c\u1ee7a c\u00e1c v\u1eadt d\u1ee5ng b\u1eb1ng s\u1eaft ","D. S\u1ef1 h\u00f4 h\u1ea5p c\u1ee7a \u0111\u1ed9ng v\u1eadt "],"hint":"","explain":"<span class='basic_left'>Quang h\u1ee3p \u1edf c\u00e2y xanh l\u00e0 qu\u00e1 tr\u00ecnh do n\u0103ng l\u01b0\u1ee3ng \u00e1nh s\u00e1ng m\u1eb7t tr\u1eddi \u0111\u01b0\u1ee3c di\u1ec7p l\u1ee5c h\u1ea5p th\u1ee5 \u0111\u1ec3 t\u1ed5ng h\u1ee3p cacbohi\u0111rat v\u00e0 gi\u1ea3i ph\u00f3ng oxi t\u1eeb kh\u00ed cacbonic v\u00e0 n\u01b0\u1edbc<br\/>Kh\u00ed cacbon \u0111ioxit + n\u01b0\u1edbc $\\to $ glucoz\u01a1 + kh\u00ed oxi<br\/>Nh\u1edd c\u00f3 ph\u1ea3n \u1ee9ng n\u00e0y m\u00e0 kh\u00f4ng kh\u00ed \u0111\u01b0\u1ee3c trong l\u00e0nh, do ch\u1ea5t c\u00f3 h\u1ea1i l\u00e0 kh\u00ed cacbon \u0111ioxit gi\u1ea3m \u0111i, ch\u1ea5t c\u1ea7n thi\u1ebft cho s\u1ef1 h\u00f4 h\u1ea5p l\u00e0 kh\u00ed oxi t\u0103ng l\u00ean<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2154},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"\u0110\u1ed1t ch\u00e1y ho\u00e0n to\u00e0n $6,2$ gam photpho trong b\u00ecnh ch\u1ee9a $8,96$ l\u00edt kh\u00ed oxi \u1edf (\u0111ktc), s\u1ea3n ph\u1ea9m thu \u0111\u01b0\u1ee3c l\u00e0 ch\u1ea5t r\u1eafn m\u00e0u tr\u1eafng. N\u1ebfu hi\u1ec7u su\u1ea5t ph\u1ea3n \u1ee9ng l\u00e0 $85\\%$ th\u00ec kh\u1ed1i l\u01b0\u1ee3ng ch\u1ea5t r\u1eafn thu \u0111\u01b0\u1ee3c l\u00e0 ","select":["A. $12,07(g)$ ","B. $11,36(g)$ ","C. $13,44(g)$ ","D. $15,7(g)$ "],"hint":"","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh ph\u1ea3n \u1ee9ng ch\u00e1y l\u00e0 $4P+5{{O}_{2}}\\xrightarrow{{{t}^{o}}} 2{{P}_{2}}{{O}_{5}}$<br\/>Ta c\u00f3 : <br\/>$\\left\\{ \\begin{aligned} & {{n}_{P}}=\\dfrac{m}{M}=\\dfrac{6,2}{31}=0,2(mol) \\\\ & {{n}_{{{O}_{2}}}}=\\dfrac{V}{22,4}=\\dfrac{8,96}{22,4}=0,4(mol) \\\\ \\end{aligned} \\right.$<br\/>So s\u00e1nh t\u1ef7 l\u1ec7 s\u1ed1 mol theo ph\u01b0\u01a1ng tr\u00ecnh : $\\dfrac{0,4}{5}>\\dfrac{0,2}{4}$<br\/>$\\to$ L\u01b0\u1ee3ng oxi d\u01b0, P ph\u1ea3n \u1ee9ng h\u1ebft.<br\/>Kh\u1ed1i l\u01b0\u1ee3ng ch\u1ea5t r\u1eafn \u0111\u01b0\u1ee3c t\u00ednh theo photpho : ${{n}_{{{P}_{2}}{{O}_{5}}}}=\\dfrac{2}{4}.0,2=0,1(mol)$<br\/>Kh\u1ed1i l\u01b0\u1ee3ng th\u1ef1c th\u1ebf ${{P}_{2}}{{O}_{5}}$ thu \u0111\u01b0\u1ee3c l\u00e0 ${{m}_{{{P}_{2}}{{O}_{5}}}}=0,1.142.\\dfrac{85}{100}=12,07(g)$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2155},{"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 $1,344$ l\u00edt oxi (\u0111ktc) ph\u1ea3n \u1ee9ng ho\u00e0n to\u00e0n v\u1edbi m\u1ed9t kim lo\u1ea1i $Y$ h\u00f3a tr\u1ecb $III$, thu \u0111\u01b0\u1ee3c $6,08$ gam oxit. C\u00f4ng th\u1ee9c h\u00f3a h\u1ecdc c\u1ee7a oxit l\u00e0 ","select":["A. $F{{e}_{2}}{{O}_{3}}$ ","B. $A{{l}_{2}}{{O}_{3}}$ ","C. $C{{r}_{2}}{{O}_{3}}$ ","D. $Z{{n}_{2}}{{O}_{3}}$ "],"hint":"","explain":"<span class='basic_left'>Kim lo\u1ea1i Y c\u00f3 h\u00f3a tr\u1ecb III v\u00e0 oxi c\u00f3 h\u00f3a tr\u1ecb II<br\/>Theo quy t\u1eafc h\u00f3a tr\u1ecb ta c\u00f3 oxit d\u1ea1ng ${{Y}_{2}}{{O}_{3}}$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc l\u00e0 $4Y+3{{O}_{2}}\\xrightarrow{{{t}^{o}}} 2{{Y}_{2}}{{O}_{3}}$<br\/>${{n}_{{{O}_{2}}}}=\\dfrac{V}{22,4}=\\dfrac{1,344}{22,4}=0,06(mol)$<br\/>Theo t\u1ef7 l\u1ec7 h\u1ec7 s\u1ed1 ph\u1ea3n \u1ee9ng ta c\u00f3 : ${{n}_{Y}}=\\dfrac{4}{3}{{n}_{{{O}_{2}}}}=\\dfrac{4}{3}.0,06=0,08(mol)$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh lu\u1eadt b\u1ea3o to\u00e0n kh\u1ed1i l\u01b0\u1ee3ng:<br\/>${{m}_{Y}}+{{m}_{{{O}_{2}}}}={{m}_{{{Y}_{2}}{{O}_{3}}}}\\Rightarrow {{m}_{Y}}=6,08-0,06.32=4,16(g)$<br\/>Kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a Y l\u00e0 $M=\\dfrac{m}{n}=\\dfrac{4,16}{0,08}=52$<br\/>V\u1eady Y l\u00e0 Crom (Cr)<br\/>CTHH c\u1ee7a oxit l\u00e0 $C{{r}_{2}}{{O}_{3}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2156},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"\u0110\u1ed1t ch\u00e1y ho\u00e0n to\u00e0n $V$ l\u00edt kh\u00ed hi\u0111rocacbon ${{C}_{x}}{{H}_{y}}$, thu \u0111\u01b0\u1ee3c $10,08$ l\u00edt kh\u00ed $C{{O}_{2}}$ v\u00e0 $10,8$ gam ${{H}_{2}}O$. C\u00f4ng th\u1ee9c ph\u00e2n t\u1eed hi\u0111rocacbon l\u00e0 , bi\u1ebft hi\u0111rocacbon n\u00e0y c\u00f3 t\u1ef7 kh\u1ed1i so v\u1edbi heli b\u1eb1ng 11, c\u00e1c kh\u00ed \u0111\u01b0\u1ee3c \u0111o \u1edf \u0111ktc","select":["A. ${{C}_{4}}{{H}_{8}}$ ","B. ${{C}_{3}}{{H}_{8}}$ ","C. ${{C}_{3}}{{H}_{6}}$ ","D. ${{C}_{4}}{{H}_{6}}$ "],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3 : <br\/>$\\left\\{ \\begin{aligned} & {{n}_{C{{O}_{2}}}}=\\dfrac{10,08}{22,4}=0,45(mol) \\\\ & {{n}_{{{H}_{2}}O}}=\\dfrac{10,8}{18}=0,6(mol) \\\\ & {{M}_{{{C}_{x}}{{H}_{y}}}}=11.4=44(g\/mol) \\\\ \\end{aligned} \\right.$<br\/>PTHH : ${{C}_{x}}{{H}_{y}}+(x+\\dfrac{y}{4}){{O}_{2}}\\to xC{{O}_{2}}+\\dfrac{y}{2}{{H}_{2}}O$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh : $\\dfrac{x}{0,5y}=\\dfrac{{{n}_{C{{O}_{2}}}}}{{{n}_{{{H}_{2}}O}}}=\\dfrac{0,45}{0,6}=\\dfrac{3}{4}\\Rightarrow \\dfrac{x}{y}=\\dfrac{3}{8}$<br\/>C\u00f4ng th\u1ee9c ph\u00e2n t\u1eed c\u1ee7a hi\u0111rocacbon c\u00f3 d\u1ea1ng ${{({{C}_{3}}{{H}_{8}})}_{n}}$ v\u00e0 c\u00f3 M=44(g\/mol)<br\/>V\u1eady 44n=44 $\\to$ n=1<br\/>C\u00f4ng th\u1ee9c ph\u00e2n t\u1eed c\u1ee7a hi\u0111rocacbon l\u00e0 ${{C}_{3}}{{H}_{8}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2157},{"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 $5$ gam l\u01b0u hu\u1ef3nh v\u00e0o b\u00ecnh c\u00f3 dung t\u00edch $2,8$ l\u00edt ch\u1ee9a kh\u00f4ng kh\u00ed ( \u1edf \u0111ktc ), r\u1ed3i \u0111\u1ed1t ch\u00e1y ho\u00e0n to\u00e0n l\u01b0\u1ee3ng l\u01b0u hu\u1ef3nh tr\u00ean thu \u0111\u01b0\u1ee3c $m$ gam l\u01b0u hu\u1ef3nh \u0111ioxit. Cho bi\u1ebft oxi chi\u1ebfm $20\\%$ th\u1ec3 t\u00edch kh\u00f4ng kh\u00ed, hi\u1ec7u su\u1ea5t ph\u1ea3n \u1ee9ng \u0111\u1ea1t $75\\%$. Gi\u00e1 tr\u1ecb $m$ l\u00e0 ","select":["A. $1,2(g)$ ","B. $1,6(g)$ ","C. $2,2(g)$ ","D. $3,2(g)$ "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{S}}=\\dfrac{5}{32}=0,15625(mol) \\\\ & {{n}_{{{O}_{2}}}}=\\dfrac{2,8.20}{22,4.100}=0,025(mol) \\\\ \\end{aligned}$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc c\u1ee7a ph\u1ea3n \u1ee9ng : <br\/>$S+{{O}_{2}}\\xrightarrow{{{t}^{o}}} S{{O}_{2}}$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh ta c\u00f3 t\u1ef7 l\u1ec7 v\u1ec1 s\u1ed1 mol $\\dfrac{0,15625}{1}>\\dfrac{0,025}{1}\\Rightarrow $: <br\/>$\\to$ S d\u01b0 n\u00ean t\u00ednh kh\u1ed1i l\u01b0\u1ee3ng $S{{O}_{2}}$ theo ${{O}_{2}}$<br\/>${{n}_{S{{O}_{2}}}}={{n}_{{{O}_{2}}}}=0,025(mol)$<br\/>Kh\u1ed1i l\u01b0\u1ee3ng $S{{O}_{2}}$ th\u1ef1c t\u1ebf thu \u0111\u01b0\u1ee3c l\u00e0 $m=64.0,025.\\dfrac{75}{100}=1,2(g)$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2158},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Trong ph\u00f2ng th\u00ed nghi\u1ec7m ng\u01b0\u1eddi ta th\u01b0\u1eddng \u0111i\u1ec1u ch\u1ebf kh\u00ed oxi t\u1eeb c\u00e1c mu\u1ed1i $KCl{{O}_{3}},KMn{{O}_{4}}$. Nung $a$ gam $KCl{{O}_{3}}$ v\u00e0 $b$ gam $KMn{{O}_{4}}$ Thu \u0111\u01b0\u1ee3c c\u00f9ng m\u1ed9t l\u01b0\u1ee3ng ${{O}_{2}}$. T\u1ef7 l\u1ec7 $a\/b$ l\u00e0 ","select":["A. $\\dfrac{a}{b}=\\dfrac{3}{4}$ ","B. $\\dfrac{a}{b}=\\dfrac{5}{8}$ ","C. $\\dfrac{a}{b}=\\dfrac{948}{245}$ ","D. $\\dfrac{a}{b}=\\dfrac{245}{948}$ "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & 2KCl{{O}_{3}}\\xrightarrow{{{t}^{o}}}2KCl+3{{O}_{2}} \\\\ & \\begin{matrix} 2mol & {} & {} & {} & {} & 3mol \\\\\\end{matrix} \\\\ & \\begin{matrix} \\dfrac{a}{122,5}mol & \\to & {} & \\dfrac{3a}{2.122,5}mol & {} & {} \\\\\\end{matrix} \\\\ \\end{aligned}$<br\/>$\\begin{aligned} & 2KMn{{O}_{4}}\\xrightarrow{{{t}^{o}}}{{K}_{2}}Mn{{O}_{4}}+{{O}_{2}}+Mn{{O}_{2}} \\\\ & \\begin{matrix} 2mol & {} & {} & {} & {} & {} & 1mol & {} \\\\\\end{matrix} \\\\ & \\begin{matrix} \\dfrac{b}{158}mol & {} & {} & \\to & {} & \\dfrac{b}{2.158}mol & {} & {} \\\\\\end{matrix} \\\\ \\end{aligned}$<br\/>Mu\u1ed1n \u0111\u01b0\u1ee3c c\u00f9ng m\u1ed9t l\u01b0\u1ee3ng oxi : $\\dfrac{3a}{2.122,5}=\\dfrac{b}{2.158}\\Rightarrow \\dfrac{a}{b}=\\dfrac{245}{948}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2159}],"lesson":{"save":0,"level":2}}