{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"Ph\u01b0\u01a1ng tr\u00ecnh n\u00e0o th\u1ec3 hi\u1ec7n t\u00ednh ch\u1ea5t h\u00f3a h\u1ecdc oxi t\u00e1c d\u1ee5ng v\u1edbi kim lo\u1ea1i : ","select":["A. $4Al+3{{O}_{2}}\\xrightarrow{{{t}^{o}}} 2A{{l}_{2}}{{O}_{3}}$ ","B. $4Fe+3{{O}_{2}}\\xrightarrow{{{t}^{o}}} 2F{{e}_{2}}{{O}_{3}}$ ","C. $2Cu+{{O}_{2}}\\xrightarrow{{{t}^{o}}} 2CuO$ ","D. T\u1ea5t c\u1ea3 c\u00e1c \u0111\u00e1p \u00e1n tr\u00ean "],"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":2130},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"Cho ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc c\u00f2n thi\u1ebfu nh\u01b0 sau : $.......+5{{O}_{2}}\\xrightarrow{{{t}^{o}}} 2{{P}_{2}}{{O}_{5}}$. Ph\u1ea7n c\u00f2n thi\u1ebfu \u1edf (...) l\u00e0: ","select":["A. $PO$ ","B. $4P$ ","C. ${{P}_{2}}O$ ","D. $P$ "],"hint":"","explain":"<span class='basic_left'>\u0110\u00e2y l\u00e0 d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh oxi ph\u1ea3n \u1ee9ng v\u1edbi phi kim v\u00ec nguy\u00ean t\u1ed1 $P$ l\u00e0 phi kim<br\/>$\\Rightarrow$ <br\/>Ch\u1ea5t c\u00f2n thi\u1ebfu \u1edf \u0111\u00e2y l\u00e0 $P$ v\u00e0 h\u1ec7 s\u1ed1 l\u00e0 $4$ <br\/>$4P+5{{O}_{2}}\\xrightarrow{{{t}^{o}}} 2{{P}_{2}}{{O}_{5}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2131},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Cho $1$ gam b\u1ed9t $Fe$ ti\u1ebfp x\u00fac v\u1edbi oxi m\u1ed9t th\u1eddi gian thu \u0111\u01b0\u1ee3c $1,24$ gam h\u1ed7n h\u1ee3p $F{{e}_{2}}{{O}_{3}}$ v\u00e0 $Fe$ d\u01b0. Kh\u1ed1i l\u01b0\u1ee3ng $Fe$ d\u01b0 l\u00e0: ","select":["A. $0,44$ gam ","B. $0,56$ gam ","C. $0,54$ gam ","D. $0,34$ gam "],"hint":"","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh ph\u1ea3n \u1ee9ng l\u00e0 $4Fe+3{{O}_{2}}\\xrightarrow{{{t}^{o}}} 2F{{e}_{2}}{{O}_{3}}$<br\/>Theo \u0111\u1ecbnh lu\u1eadt b\u1ea3o to\u00e0n kh\u1ed1i l\u01b0\u1ee3ng:<br\/>${{m}_{Fe}}+{{m}_{{{O}_{2}}}}={{m}_{hh}}\\Rightarrow {{m}_{{{O}_{2}}}}=1,24-1=0,24(g)$<br\/>$\\to {{n}_{{{O}_{2}}}}=\\dfrac{0,24}{32}=0,0075(mol)$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh s\u1ed1 mol Fe ph\u1ea3n \u1ee9ng l\u00e0 ${{n}_{Fe}}=\\dfrac{4}{3}.0,0075=0,01(mol)$<br\/>$\\to$ Kh\u1ed1i l\u01b0\u1ee3ng $Fe$ tham gia ph\u1ea3n \u1ee9ng l\u00e0 $m = 0,01.56 = 0,56$ gam<br\/>$\\to$ Kh\u1ed1i l\u01b0\u1ee3ng $Fe$ c\u00f2n d\u01b0 l\u00e0 $1 - 0,56 = 0,44$ gam<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2132},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"Ph\u00e1t bi\u1ec3u n\u00e0o sau \u0111\u00e2y \u0111\u00fang ? ","select":["A. Oxi l\u00e0 ch\u1ea5t kh\u00ed tan v\u00f4 h\u1ea1n trong n\u01b0\u1edbc v\u00e0 n\u1eb7ng h\u01a1n kh\u00f4ng kh\u00ed ","B. Oxi l\u00e0 ch\u1ea5t kh\u00ed tan \u00edt trong n\u01b0\u1edbc v\u00e0 nh\u1eb9 h\u01a1n kh\u00f4ng kh\u00ed ","C. \u1ede nhi\u1ec7t \u0111\u1ed9 cao, kh\u00ed oxi d\u1ec5 d\u00e0ng t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi nhi\u1ec1u \u0111\u01a1n ch\u1ea5t (kim lo\u1ea1i, phi kim) v\u00e0 h\u1ee3p ch\u1ea5t ","D. Kh\u00ed oxi d\u1ec5 d\u00e0ng t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi nhi\u1ec1u \u0111\u01a1n ch\u1ea5t (kim lo\u1ea1i, phi kim) v\u00e0 h\u1ee3p ch\u1ea5t "],"hint":"","explain":"<span class='basic_left'>Kh\u00ed oxi l\u00e0 m\u1ed9t \u0111\u01a1n ch\u1ea5t phi kim r\u1ea5t ho\u1ea1t \u0111\u1ed9ng, \u0111\u1eb7c bi\u1ec7t \u1edf nhi\u1ec7t \u0111\u1ed9 cao, d\u1ec5 d\u00e0ng tham gia ph\u1ea3n \u1ee9ng h\u00f3a h\u1ecdc v\u1edbi nhi\u1ec1u phi kim, kim lo\u1ea1i v\u00e0 h\u1ee3p ch\u1ea5t.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2133},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Nung $2,1$ gam b\u1ed9t s\u1eaft trong b\u00ecnh ch\u1ee9a oxi, \u0111\u1ebfn khi ph\u1ea3n \u1ee9ng x\u1ea3y ra ho\u00e0n to\u00e0n thu \u0111\u01b0\u1ee3c $2,9$ gam b\u1ed9t oxit s\u1eaft. C\u00f4ng th\u1ee9c ph\u00e2n t\u1eed c\u1ee7a oxit s\u1eaft l\u00e0 ","select":["A. $F{{e}_{3}}{{O}_{4}}$ ","B. $F{{e}_{2}}{{O}_{3}}$ ","C. $FeO$ ","D. $F{{e}_{2}}O$ "],"hint":"","explain":"<span class='basic_left'>${{n}_{Fe}}=\\dfrac{2,1}{56}=0,0375(mol)$<br\/>Theo \u0111\u1ecbnh lu\u1eadt b\u1ea3o to\u00e0n kh\u1ed1i l\u01b0\u1ee3ng: ${{n}_{O}}=\\dfrac{2,9-2,1}{16}=0,05(mol)$<br\/>G\u1ecdi c\u00f4ng th\u1ee9c t\u1ed5ng qu\u00e1t c\u1ee7a oxit s\u1eaft l\u00e0 $F{{e}_{x}}{{O}_{y}}$<br\/>Ta c\u00f3 t\u1ef7 l\u1ec7: $\\dfrac{x}{y}=\\dfrac{0,0375}{0,05}=\\dfrac{3}{4}$<br\/>V\u1eady oxit c\u1ea7n t\u00ecm l\u00e0 $F{{e}_{3}}{{O}_{4}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2134},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"\u0110\u1ed1t ch\u00e1y $5,6$ l\u00edt ${{C}_{2}}{{H}_{4}}$ \u1edf (\u0111ktc) trong kh\u00f4ng kh\u00ed, sau ph\u1ea3n \u1ee9ng thu \u0111\u01b0\u1ee3c kh\u00ed cacbonic v\u00e0 h\u01a1i n\u01b0\u1edbc. S\u1ed1 ph\u00e2n t\u1eed kh\u00ed cacbonic thu \u0111\u01b0\u1ee3c l\u00e0 ","select":["A. ${{3.10}^{22}}$ ","B. ${{30.10}^{23}}$ ","C. ${{3.10}^{23}}$ ","D. ${{0,3.10}^{23}}$ "],"hint":"","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh ph\u1ea3n \u1ee9ng ch\u00e1y l\u00e0 ${{C}_{2}}{{H}_{4}}+3{{O}_{2}}\\xrightarrow{{{t}^{o}}} 2C{{O}_{2}}+2{{H}_{2}}O$<br\/>${{n}_{{{C}_{2}}{{H}_{4}}}}=\\dfrac{5,6}{22,4}=0,025(mol)$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh ph\u1ea3n \u1ee9ng : ${{n}_{C{{O}_{2}}}}=2.0,25=0,5(mol)$<br\/>V\u1eady s\u1ed1 ph\u00e2n t\u1eed kh\u00ed cacbonic thu \u0111\u01b0\u1ee3c l\u00e0 ${{0,5.6.10}^{23}}={{3.10}^{23}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2135},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"\u0110\u1ed1t ch\u00e1y $6,2$ gam $P$ trong b\u00ecnh ch\u1ee9a $6,72$ l\u00edt kh\u00ed oxi \u1edf (\u0111ktc) t\u1ea1o th\u00e0nh ${{P}_{2}}{{O}_{5}}$. Ch\u1ea5t n\u00e0o c\u00f2n d\u01b0, ch\u1ea5t n\u00e0o c\u00f2n thi\u1ebfu ? ","select":["A. $P$ c\u00f2n d\u01b0, ${{O}_{2}}$ thi\u1ebfu ","B. $P$ c\u00f2n thi\u1ebfu, ${{O}_{2}}$ d\u01b0 ","C. C\u1ea3 hai ch\u1ea5t v\u1eeba \u0111\u1ee7 ","D. T\u1ea5t c\u1ea3 \u0111\u1ec1u sai "],"hint":"","explain":"<span class='basic_left'>S\u1ed1 mol photpho tham gia ph\u1ea3n \u1ee9ng l\u00e0 <br\/>${{n}_{P}}=\\dfrac{m}{M}=\\dfrac{6,2}{31}=0,2(mol)$<br\/>S\u1ed1 mol oxi tham gia ph\u1ea3n \u1ee9ng l\u00e0 <br\/>${{n}_{{{O}_{2}}}}=\\dfrac{V}{22,4}=\\dfrac{6,72}{22,4}=0,3(mol)$<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{0,2}{4}<\\dfrac{0,3}{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 D.<\/span><\/span> ","column":2}]}],"id_ques":2136},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"Ch\u00e1y v\u1edbi ng\u1ecdn l\u1eeda s\u00e1ng ch\u00f3i, t\u1ea1o kh\u00f3i tr\u1eafng d\u00e0y \u0111\u1eb7c b\u00e1m v\u00e0o th\u00e0nh b\u00ecnh. L\u00e0 hi\u1ec7n t\u01b0\u1ee3ng c\u1ee7a ph\u1ea3n \u1ee9ng: ","select":["A. $3Fe+2{{O}_{2}}\\xrightarrow{{{t}^{o}}} F{{e}_{3}}{{O}_{4}}$ ","B. $4P+5{{O}_{2}}\\xrightarrow{{{t}^{o}}} 2{{P}_{2}}{{O}_{5}}$ ","C. $C+{{O}_{2}}\\xrightarrow{{{t}^{o}}} C{{O}_{2}}$ ","D. $C{{H}_{4}}+2{{O}_{2}}\\xrightarrow{{{t}^{o}}} C{{O}_{2}}+2{{H}_{2}}O$ "],"hint":"","explain":"<span class='basic_left'>\u0110\u1ecdc s\u00e1ch gi\u00e1o khoa h\u00f3a 8 ph\u1ea7n t\u00ednh ch\u1ea5t h\u00f3a h\u1ecdc oxi t\u00e1c d\u1ee5ng v\u1edbi ph\u1ed1tpho<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2137},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"C\u1ea7n bao nhi\u00eau gam oxi \u0111\u1ec3 \u0111\u1ed1t ch\u00e1y h\u1ebft $4$ mol l\u01b0u hu\u1ef3nh ? ","select":["A. $120(g)$ ","B. $148(g)$ ","C. $128(g)$ ","D. $160(g)$ "],"hint":"","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh ph\u1ea3n \u1ee9ng ch\u00e1y l\u00e0 <br\/>$S+{{O}_{2}}\\xrightarrow{{{t}^{o}}} S{{O}_{2}}$<br\/>Theo t\u1ec9 l\u1ec7 v\u1ec1 h\u1ec7 s\u1ed1 ph\u01b0\u01a1ng tr\u00ecnh ta c\u00f3 : ${{n}_{{{O}_{2}}}}={{n}_{S}}=4(mol)$<br\/>$\\Rightarrow {{m}_{{{O}_{2}}}}=4.32=128(g)$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2138},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Cho $3,36$ l\u00edt oxi \u1edf (\u0111ktc) ph\u1ea3n \u1ee9ng ho\u00e0n to\u00e0n v\u1edbi m\u1ed9t kim lo\u1ea1i $R$ h\u00f3a tr\u1ecb $III$ thu \u0111\u01b0\u1ee3c $10,2$ gam oxit. K\u00fd hi\u1ec7u h\u00f3a h\u1ecdc c\u1ee7a kim lo\u1ea1i l\u00e0: ","select":["A. $Al$ ","B. $Cr$ ","C. $Fe$ ","D. $Pb$ "],"hint":"","explain":"<span class='basic_left'>Kim lo\u1ea1i $R$ 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 ${{R}_{2}}{{O}_{3}}$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc l\u00e0 $4R+3{{O}_{2}}\\xrightarrow{{{t}^{o}}} 2{{R}_{2}}{{O}_{3}}$<br\/>${{n}_{{{O}_{2}}}}=\\dfrac{V}{22,4}=\\dfrac{3,36}{22,4}=0,15(mol)$<br\/>Theo t\u1ef7 l\u1ec7 h\u1ec7 s\u1ed1 ph\u1ea3n \u1ee9ng ta c\u00f3 : ${{n}_{R}}=\\dfrac{4}{3}{{n}_{{{O}_{2}}}}=\\dfrac{4}{3}.0,15=0,2(mol)$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh lu\u1eadt b\u1ea3o to\u00e0n kh\u1ed1i l\u01b0\u1ee3ng:<br\/>${{m}_{R}}+{{m}_{{{O}_{2}}}}={{m}_{{{P}_{2}}{{O}_{5}}}}\\Rightarrow {{m}_{R}}=10,2-0,15.32=5,4(g)$<br\/>Kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a R l\u00e0 $M=\\dfrac{m}{n}=\\dfrac{5,4}{0,2}=27$<br\/>V\u1eady R l\u00e0 nh\u00f4m (Al)<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2139}],"lesson":{"save":0,"level":2}}