{"segment":[{"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\u1ee3p ch\u1ea5t c\u1ee7a hai nguy\u00ean t\u1ed1, trong \u0111\u00f3 c\u00f3 m\u1ed9t nguy\u00ean t\u1ed1 l\u00e0 oxi \u0111\u01b0\u1ee3c g\u1ecdi l\u00e0 ","select":["A. Axit ","B. Baz\u01a1 ","C. Oxit ","D. Mu\u1ed1i "],"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\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2160},{"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 g\u1ed3m c\u00e1c ch\u1ea5t thu\u1ed9c lo\u1ea1i oxit axit l\u00e0 : ","select":["A. $S{{O}_{2}},C{{O}_{2}},N{{a}_{2}}O,{{P}_{2}}{{O}_{5}},S{{O}_{3}}$ ","B. $S{{O}_{2}},C{{O}_{2}},N{{O}_{2}},{{P}_{2}}{{O}_{5}},S{{O}_{3}}$ ","C. $CaO,C{{O}_{2}},N{{a}_{2}}O,{{P}_{2}}{{O}_{5}},S{{O}_{3}}$ ","D. $C{{O}_{2}},N{{a}_{2}}O,{{P}_{2}}{{O}_{5}},S{{O}_{3}},F{{e}_{2}}{{O}_{3}}$ "],"hint":"","explain":"<span class='basic_left'>Oxit axit th\u01b0\u1eddng l\u00e0 oxit c\u1ee7a phi kim v\u00e0 t\u01b0\u01a1ng \u1ee9ng v\u1edbi m\u1ed9t axit<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2161},{"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 d\u00e3y c\u00e1c oxit sau : $C{{O}_{2}},N{{a}_{2}}O,{{P}_{2}}{{O}_{5}},S{{O}_{3}},F{{e}_{2}}{{O}_{3}},S{{O}_{2}},N{{O}_{2}},NO,CO,CaO,CuO$. S\u1ed1 oxit l\u00e0 oxit baz\u01a1 l\u00e0 ","select":["A. $4$ ","B. $7$ ","C. $5$ ","D. $6$ "],"hint":"","explain":"<span class='basic_left'>Oxit baz\u01a1 l\u00e0 oxit c\u1ee7a kim lo\u1ea1i v\u00e0 t\u01b0\u01a1ng \u1ee9ng v\u1edbi m\u1ed9t baz\u01a1<br\/> Oxit baz\u01a1 l\u00e0: $ N{{a}_{2}}O,F{{e}_{2}}{{O}_{3}},CaO,CuO$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2162},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"T\u00ean g\u1ecdi c\u1ee7a oxit axit ${{P}_{2}}{{O}_{5}}$ l\u00e0 ","select":["A. Photpho oxit ","B. \u0110iphotpho trioxit ","C. \u0110iphotpho tetraoxxit ","D. \u0110iphotpho pentaoxit "],"hint":"","explain":"<span class='basic_left'>T\u00ean g\u1ecdi c\u1ee7a m\u1ed9t oxit axit: T\u00ean phi kim( c\u00f3 ti\u1ec1n t\u1ed1 ch\u1ec9 s\u1ed1 nguy\u00ean t\u1eed phi kim ) + oxit( c\u00f3 ti\u1ec1n t\u1ed1 ch\u1ec9 s\u1ed1 nguy\u00ean t\u1eed oxi )<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2163},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Baz\u01a1 t\u01b0\u01a1ng \u1ee9ng c\u1ee7a oxit baz\u01a1 $F{{e}_{2}}{{O}_{3}}$ l\u00e0 ","select":["A. $Fe{{(OH)}_{2}}$ ","B. $Fe{{(OH)}_{3}}$ ","C. $Fe{{(OH)}_{4}}$ ","D. $FeOH$ "],"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 B.<\/span><\/span> ","column":2}]}],"id_ques":2164},{"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 s\u01a1 \u0111\u1ed3 chuy\u1ec3n h\u00f3a sau: L\u01b0u hu\u1ef3nh $\\xrightarrow{(1)}$ l\u01b0u hu\u1ef3nh \u0111ioxit $\\xrightarrow{(2)}$ axit sunfur\u01a1. Ph\u01b0\u01a1ng tr\u00ecnh bi\u1ec3u di\u1ec5n qu\u00e1 tr\u00ecnh chuy\u1ec3n h\u00f3a tr\u00ean l\u00e0","select":["A. $\\begin{aligned} & (1)2S+3{{O}_{2}}\\xrightarrow{{{t}^{o}}}2S{{O}_{3}} \\\\ & (2)S{{O}_{3}}+{{H}_{2}}O\\to {{H}_{2}}S{{O}_{3}} \\\\ \\end{aligned}$ ","B. $\\begin{aligned} & (1)S+{{O}_{2}}\\xrightarrow{{{t}^{o}}}S{{O}_{2}} \\\\ & (2)S{{O}_{2}}+{{H}_{2}}O\\to {{H}_{2}}S{{O}_{4}} \\\\ \\end{aligned}$ ","C. $\\begin{aligned} & (1)2S+3{{O}_{2}}\\xrightarrow{{{t}^{o}}}2S{{O}_{3}} \\\\ & (2)S{{O}_{3}}+{{H}_{2}}O\\to {{H}_{2}}S{{O}_{4}} \\\\ \\end{aligned}$ ","D. $\\begin{aligned} & (1)S+{{O}_{2}}\\xrightarrow{{{t}^{o}}}S{{O}_{2}} \\\\ & (2)S{{O}_{2}}+{{H}_{2}}O\\to {{H}_{2}}S{{O}_{3}} \\\\ \\end{aligned}$ "],"hint":"","explain":"<span class='basic_left'>L\u01b0u hu\u1ef3nh ( S)<br\/>L\u01b0u hu\u1ef3nh \u0111ioxit ($S{{O}_{2}}$)<br\/>Axit sunfur\u01a1 (${{H}_{2}}S{{O}_{3}}$)<br\/>Ph\u1ea3n \u1ee9ng (1) l\u00e0 ph\u1ea3n \u1ee9ng \u0111\u1ed1t ch\u00e1y l\u01b0u hu\u1ef3nh trong kh\u00ed oxi<br\/>Ph\u1ea3n \u1ee9ng (2) l\u00e0 ph\u1ea3n \u1ee9ng oxit axit t\u00e1c d\u1ee5ng v\u1edbi n\u01b0\u1edbc thu \u0111\u01b0\u1ee3c axit t\u01b0\u01a1ng \u1ee9ng<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2165},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"T\u1ef7 l\u1ec7 v\u1ec1 kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a nit\u01a1 v\u00e0 oxi trong m\u1ed9t oxit c\u1ee7a nit\u01a1 l\u00e0 $7$ : $4$. C\u00f4ng th\u1ee9c c\u1ee7a oxit l\u00e0","select":["A. ${{N}_{2}}{{O}_{5}}$ ","B. ${{N}_{2}}{{O}_{3}}$ ","C. ${{N}_{2}}O$ ","D. $N{{O}_{2}}$ "],"hint":"","explain":"<span class='basic_left'>G\u1ecdi c\u00f4ng th\u1ee9c t\u1ed5ng qu\u00e1t c\u1ee7a oxit l\u00e0 ${{N}_{x}}{{O}_{y}}$<br\/>$\\dfrac{{{m}_{N}}}{{{m}_{O}}}=\\dfrac{14.x}{16.y}=\\dfrac{7}{4}\\Rightarrow \\dfrac{x}{y}=\\dfrac{2}{1}$<br\/>V\u1eady CTHH c\u1ee7a oxit l\u00e0 ${{N}_{2}}O$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2166},{"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 $4,5$ gam ch\u1ea5t h\u1eefu c\u01a1 $X$ thu \u0111\u01b0\u1ee3c $6,6$ gam kh\u00ed $C{{O}_{2}}$ v\u00e0 $2,7$ gam ${{H}_{2}}O$. Th\u1ec3 t\u00edch oxi c\u1ea7n \u0111\u1ec3 \u0111\u1ed1t ch\u00e1y ho\u00e0n to\u00e0n h\u1ee3p ch\u1ea5t h\u1eefu c\u01a1 l\u00e0 ","select":["A. $3,36(l)$ ","B. $1,12(l)$ ","C. $2,24(l)$ ","D. $4,48(l)$ "],"hint":"","explain":"<span class='basic_left'>PTHH : ${{C}_{x}}{{H}_{y}}+(x+\\dfrac{y}{4}){{O}_{2}}\\to xC{{O}_{2}}+\\dfrac{y}{2}{{H}_{2}}O$<br\/>Theo \u0111\u1ecbnh lu\u1eadt b\u1ea3o to\u00e0n kh\u1ed1i l\u01b0\u1ee3ng :<br\/>${{m}_{_{X}}}+{{m}_{{{O}_{2}}}}={{m}_{C{{O}_{2}}}}+{{m}_{{{H}_{2}}O}}\\Rightarrow {{m}_{{{O}_{2}}}}=6,6+2,7-4,5=4,8(g)$<br\/>$\\to {{n}_{{{O}_{2}}}}=\\dfrac{m}{M}=\\dfrac{4,8}{32}=0,15(mol)$<br\/>$\\to {{V}_{{{O}_{2}}}}=n.22,4=0,15.22,4=3,36(l)$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2167},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"M\u1ed9t s\u1ed1 c\u00f4ng th\u1ee9c h\u00f3a h\u1ecdc \u0111\u01b0\u1ee3c vi\u1ebft nh\u01b0 sau : $NaO,A{{l}_{2}}{{O}_{3}},FeO,CaO,Z{{n}_{2}}O,MgO,M{{g}_{2}}O,{{N}_{2}}O,PO,SO,{{S}_{2}}O$. S\u1ed1 c\u00f4ng th\u1ee9c oxit vi\u1ebft sai l\u00e0 ","select":["A. $5$ ","B. $6$ ","C. $7$ ","D. $4$ "],"hint":"","explain":"<span class='basic_left'>Na h\u00f3a tr\u1ecb I<br\/>Al h\u00f3a tr\u1ecb III<br\/>Fe h\u00f3a tr\u1ecb II<br\/>Ca h\u00f3a tr\u1ecb II<br\/>Zn h\u00f3a tr\u1ecb II<br\/>Mg h\u00f3a tr\u1ecb II<br\/>N h\u00f3a tr\u1ecb I, II, III, V<br\/>P h\u00f3a tr\u1ecb III, V<br\/>S h\u00f3a tr\u1ecb II, IV, VI<br\/>O h\u00f3a tr\u1ecb II<br\/>Theo quy t\u1eafc h\u00f3a tr\u1ecb th\u00ec nh\u1eefng oxit b\u1ecb sai c\u00f4ng th\u1ee9c l\u00e0 $NaO,,Z{{n}_{2}}O,M{{g}_{2}}O,PO,SO,{{S}_{2}}O$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2168},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Axit t\u01b0\u01a1ng \u1ee9ng c\u1ee7a oxit axit $S{{O}_{3}}$ l\u00e0 ","select":["A. ${{H}_{2}}S{{O}_{4}}$ ","B. ${{H}_{2}}S{{O}_{3}}$ ","C. ${{H}_{2}}S{{O}_{2}}$ ","D. $HS{{O}_{4}}$ "],"hint":"","explain":"<span class='basic_left'>Ph\u00e2n t\u1eed axit g\u1ed3m c\u00f3 m\u1ed9t hay nhi\u1ec1u nguy\u00ean t\u1eed hi\u0111ro li\u00ean k\u1ebft v\u1edbi g\u1ed1c axit, c\u00e1c nguy\u00ean t\u1eed hi\u0111ro 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":2169},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Cho $28,4$ gam \u0111iphotpho pentaoxit ${{P}_{2}}{{O}_{5}}$ v\u00e0o c\u1ed1c ch\u1ee9a $90$ gam ${{H}_{2}}O$ \u0111\u1ec3 t\u1ecda th\u00e0nh axit photphoric ${{H}_{3}}P{{O}_{4}}$. Kh\u1ed1i l\u01b0\u1ee3ng axit ${{H}_{3}}P{{O}_{4}}$ t\u1ea1o th\u00e0nh l\u00e0","select":["A. $58,8(g)$ ","B. $39,2(g)$ ","C. $19,6(g)$ ","D. $30(g)$ "],"hint":"","explain":"<span class='basic_left'>PTHH : ${{P}_{2}}{{O}_{5}}+3{{H}_{2}}O\\to 2{{H}_{3}}P{{O}_{4}}$<br\/>$\\begin{aligned} & {{n}_{{{P}_{2}}{{O}_{5}}}}=\\dfrac{28,4}{142}=0,2(mol) \\\\ & {{n}_{{{H}_{2}}O}}=\\dfrac{90}{18}=5(mol) \\\\ \\end{aligned}$<br\/>L\u1eadp t\u1ef7 s\u1ed1 theo ph\u01b0\u01a1ng tr\u00ecnh : $\\dfrac{5}{3}>\\dfrac{0,2}{1}$<br\/>$\\to$ ${{H}_{2}}O$ d\u01b0, ${{P}_{2}}{{O}_{5}}$ ph\u1ea3n \u1ee9ng h\u1ebft<br\/>V\u1eady kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a ${{H}_{3}}P{{O}_{4}}$ t\u00ednh theo l\u01b0\u1ee3ng ${{P}_{2}}{{O}_{5}}$<br\/>${{n}_{{{H}_{3}}P{{O}_{4}}}}=2{{n}_{{{P}_{2}}{{O}_{5}}}}=0,4(mol)$<br\/>$\\to {{m}_{{{H}_{3}}P{{O}_{4}}}}=0,4.98=39,2(g)$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2170},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"T\u00ean g\u1ecdi c\u1ee7a c\u00f4ng th\u1ee9c $C{{r}_{2}}{{O}_{3}}$ l\u00e0 ","select":["A. Crom \u0111ioxit ","B. Crom oxit ","C. Crom(II) oxit ","D. Crom(III) oxit "],"hint":"","explain":"<span class='basic_left'>T\u00ean g\u1ecdi c\u1ee7a oxit baz\u01a1 : T\u00ean kim lo\u1ea1i(k\u00e8m theo h\u00f3a tr\u1ecb) + oxit<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2171},{"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\u1ee5c $1,792$ l\u00edt h\u1ed7n h\u1ee3p kh\u00ed $C{{O}_{2}}$ v\u00e0 $S{{O}_{2}}$ (\u0111ktc) v\u00e0o n\u01b0\u1edbc thu \u0111\u01b0\u1ee3c $1,86$ gam axit ${{H}_{2}}C{{O}_{3}}$ v\u00e0 $m$ gam ${{H}_{2}}S{{O}_{3}}$. $\\%$ th\u1ec3 t\u00edch c\u1ee7a $S{{O}_{2}}$ trong h\u1ed7n h\u1ee3p kh\u00ed l\u00e0","select":["A. $62,5\\%$ ","B. $37,5\\%$ ","C. $62\\%$ ","D. $64,5\\%$ "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{hh}}=\\dfrac{1,792}{22,4}=0,08(mol) \\\\ & {{n}_{{{H}_{2}}C{{O}_{3}}}}=\\dfrac{1,86}{62}=0,03(mol) \\\\ \\end{aligned}$<br\/>$\\begin{aligned} & C{{O}_{2}}+{{H}_{2}}O\\to {{H}_{2}}C{{O}_{3}} \\\\ & \\begin{matrix} 0,03mol & \\leftarrow & 0,03mol & {} & {} \\\\\\end{matrix} \\\\ & S{{O}_{2}}+{{H}_{2}}O\\to {{H}_{2}}S{{O}_{3}} \\\\ \\end{aligned}$<br\/>$\\to {{n}_{S{{O}_{2}}}}=0,08-0,03=0,05(mol)$<br\/>Ph\u1ea7n tr\u0103m th\u1ec3 t\u00edch c\u0169ng ch\u00ednh l\u00e0 ph\u1ea7n tr\u0103m s\u1ed1 mol<br\/>$\\to \\%{{V}_{S{{O}_{2}}}}=\\dfrac{0,05}{0,08}.100\\%=62,5\\%$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2172},{"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 $a$ gam h\u1ed7n h\u1ee3p hai kim lo\u1ea1i $Fe$ v\u00e0 $Cu$ c\u00f3 s\u1ed1 mol b\u1eb1ng nhau, ph\u1ea3n \u1ee9ng ho\u00e0n to\u00e0n v\u1edbi l\u01b0\u1ee3ng oxi d\u01b0. K\u1ebft th\u00fac ph\u1ea3n \u1ee9ng thu \u0111\u01b0\u1ee3c h\u1ed7n h\u1ee3p ch\u1ea5t r\u1eafn c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng t\u0103ng so v\u1edbi h\u1ed7n h\u1ee3p ban \u0111\u1ea7u l\u00e0 $4$ gam. Gi\u00e1 tr\u1ecb $a$ l\u00e0 ","select":["A. $12(g)$ ","B. $13,857(g)$ ","C. $12,857(g)$ ","D. $14(g)$ "],"hint":"","explain":"<span class='basic_left'>G\u1ecdi s\u1ed1 mol c\u1ee7a Fe v\u00e0 Cu l\u00e0 x mol<br\/>Ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc c\u1ee7a ph\u1ea3n \u1ee9ng :<br\/>$\\begin{aligned} & 3Fe+2{{O}_{2}}\\xrightarrow{{{t}^{o}}}F{{e}_{3}}{{O}_{4}} \\\\ & \\begin{matrix} x\\to & \\dfrac{2}{3}x & {} & {} & {} \\\\\\end{matrix} \\\\ & 2Cu+{{O}_{2}}\\xrightarrow{{{t}^{o}}}2CuO \\\\ & \\begin{matrix} x\\to & \\dfrac{1}{2}x & {} & {} & {} \\\\\\end{matrix} \\\\ \\end{aligned}$<br\/>Kh\u1ed1i l\u01b0\u1ee3ng ch\u1ea5t r\u1eafn t\u0103ng = kh\u1ed1i l\u01b0\u1ee3ng oxi tham gia ph\u1ea3n \u1ee9ng = 4 g<br\/>$\\dfrac{2}{3}x+\\dfrac{1}{2}x=\\dfrac{4}{32}\\Rightarrow x=\\dfrac{3}{28}(mol)$<br\/>$\\to a=\\dfrac{3}{28}.(56+64)=12,857(g)$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2173},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"M\u1ed9t thanh s\u1eaft c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng $168$ gam, sau khi \u0111\u1ec3 ngo\u00e0i kh\u00f4ng kh\u00ed m\u1ed9t th\u1eddi gian, kh\u1ed1i l\u01b0\u1ee3ng thanh s\u1eaft t\u0103ng so v\u1edbi ban \u0111\u1ea7u l\u00e0 $6$ gam. S\u1ed1 ph\u00e2n t\u1eed kh\u00ed ${{O}_{2}}$ l\u00e0 ","select":["A. ${{1,125.10}^{21}}$ ","B. ${{1,125.10}^{24}}$ ","C. ${{1,125.10}^{22}}$ ","D. ${{1,125.10}^{23}}$ "],"hint":"","explain":"<span class='basic_left'>Khi thanh s\u1eaft \u0111\u1ec3 ngo\u00e0i kh\u00f4ng kh\u00ed s\u1ebd b\u1ecb oxi h\u00f3a t\u1ea1o th\u00e0nh oxit s\u1eaft.<br\/>Qu\u00e1 tr\u00ecnh oxi h\u00f3a ch\u00ednh l\u00e0 s\u1eaft t\u00e1c d\u1ee5ng v\u1edbi oxi trong kh\u00f4ng kh\u00ed<br\/>$\\to$ Kh\u1ed1i l\u01b0\u1ee3ng thanh s\u1eaft t\u0103ng = kh\u1ed1i l\u01b0\u1ee3ng oxi tham gia ph\u1ea3n \u1ee9ng = 6 g<br\/>$\\to {{n}_{{{O}_{2}}}}=\\dfrac{6}{32}=0,1875(mol)$<br\/>$\\to$ S\u1ed1 ph\u00e2n t\u1eed kh\u00ed oxi l\u00e0 ${{0,1875.6.10}^{23}}={{1,125.10}^{23}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2174},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Oxit c\u1ee7a nguy\u00ean t\u1ed1 $R$ c\u00f3 h\u00f3a tr\u1ecb $III$ ch\u1ee9a $70\\%$ v\u1ec1 kh\u1ed1i l\u01b0\u1ee3ng nguy\u00ean t\u1ed1 $R$. Oxit tr\u00ean thu\u1ed9c lo\u1ea1i oxit n\u00e0o? ","select":["A. Oxit baz\u01a1 ","B. Oxit axit ","C. Oxit l\u01b0\u1ee1ng t\u00ednh ","D. oxit trung t\u00ednh "],"hint":"","explain":"<span class='basic_left'>R c\u00f3 h\u00f3a tr\u1ecb III v\u00e0 O 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 : ${{R}_{2}}{{O}_{3}}$<br\/>$\\begin{aligned} & \\to \\%R=\\dfrac{2{{M}_{R}}}{2{{M}_{R}}+3.16}.100=70 \\\\ & \\to {{M}_{R}}=56 \\\\ \\end{aligned}$<br\/>v\u1eady R l\u00e0 s\u1eaft (Fe)<br\/>$\\to$ CTHH c\u1ee7a oxit l\u00e0 $F{{e}_{2}}{{O}_{3}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2175},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"M\u1ed9t h\u1ee3p ch\u1ea5t oxit c\u00f3 ch\u1ee9a $50\\%$ v\u1ec1 kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a $S$. C\u00f4ng th\u1ee9c h\u00f3a h\u1ecdc c\u1ee7a oxit l\u00e0 ","select":["A. $S{{O}_{3}}$ ","B. $SO$ ","C. $S{{O}_{2}}$ ","D. ${{S}_{2}}{{O}_{3}}$ "],"hint":"","explain":"<span class='basic_left'>G\u1ecdi CTHH c\u1ee7a oxit d\u1ea1ng ${{S}_{x}}{{O}_{y}}$<br\/>$\\begin{aligned} & \\%S=\\dfrac{x.{{M}_{S}}}{{{M}_{{{S}_{x}}{{O}_{y}}}}}.100=50 \\\\ & \\%O=\\dfrac{y.{{M}_{O}}}{{{M}_{{{S}_{x}}{{O}_{y}}}}}.100=50 \\\\ & \\to \\dfrac{x}{y}=\\dfrac{1}{2} \\\\ \\end{aligned}$<br\/>$\\to$ CTHH l\u00e0 $S{{O}_{2}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2176},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"M\u1ed9t oxit c\u1ee7a photpho c\u00f3 th\u00e0nh ph\u1ea7n ph\u1ea7n tr\u0103m kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a $P$ b\u1eb1ng $56,36\\%$. Bi\u1ebft ph\u00e2n t\u1eed kh\u1ed1i c\u1ee7a oxit b\u1eb1ng 110 \u0111vC. C\u00f4ng th\u1ee9c h\u00f3a h\u1ecdc c\u1ee7a oxit l\u00e0 ","select":["A. ${{P}_{2}}{{O}_{5}}$ ","B. ${{P}_{2}}{{O}_{3}}$ ","C. ${{P}_{2}}{{O}_{4}}$ ","D. $P{{O}_{2}}$ "],"hint":"","explain":"<span class='basic_left'>Ph\u1ea7n tr\u0103m kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a oxi l\u00e0 $100\\%$ - $56,36\\%$ = $43,64\\%$<br\/>$\\begin{aligned} & {{m}_{P}}=\\dfrac{110.56,36}{100}=62(g) \\\\ & {{m}_{O}}=\\dfrac{110.43.64}{100}=48(g) \\\\ \\end{aligned}$<br\/>S\u1ed1 mol m\u1ed7i nguy\u00ean t\u1eed c\u1ee7a m\u1ed7i nguy\u00ean t\u1ed1 c\u00f3 trong 1 mol h\u1ee3p ch\u1ea5t<br\/>$\\begin{aligned} & {{n}_{P}}=\\dfrac{62}{31}=2(mol) \\\\ & {{n}_{O}}=\\dfrac{48}{16}=3(mol) \\\\ \\end{aligned}$<br\/>CTHH c\u1ee7a h\u1ee3p ch\u1ea5t l\u00e0 ${{P}_{2}}{{O}_{3}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2177},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"B\u00ecnh \u0111\u1ef1ng ga d\u00f9ng \u0111\u1ec3 \u0111un n\u1ea5u trong gia \u0111\u00ecnh c\u00f3 ch\u1ee9a $12,5$ kg butan ${{C}_{4}}{{H}_{10}}$ \u1edf tr\u1ea1ng th\u00e1i l\u1ecfng, do \u0111\u01b0\u1ee3c n\u00e9n d\u01b0\u1edbi \u00e1p su\u1ea5t cao. Th\u1ec3 t\u00edch kh\u00f4ng kh\u00ed c\u1ea7n d\u00f9ng \u1edf \u0111ktc \u0111\u1ec3 \u0111\u1ed1t ch\u00e1y h\u1ebft l\u01b0\u1ee3ng nhi\u00eau li\u1ec7u c\u00f3 trong b\u00ecnh ( Bi\u1ebft oxi chi\u1ebfm $20\\%$ th\u1ec3 t\u00edch kh\u00f4ng kh\u00ed, ph\u1ea3n \u1ee9ng ch\u00e1y c\u1ee7a butan cho $C{{O}_{2}}$ v\u00e0 ${{H}_{2}}O$) ","select":["A. $146884(l)$ ","B. $136884(l)$ ","C. $126884(l)$ ","D. $156884(l)$ "],"hint":"","explain":"<span class='basic_left'>${{n}_{{{C}_{4}}{{H}_{10}}}}=\\dfrac{12500}{58}=215,5(mol)$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc : <br\/>$\\begin{aligned} & 2{{C}_{4}}{{H}_{10}}+13{{O}_{2}}\\to 8C{{O}_{2}}+10{{H}_{2}}O \\\\ & \\begin{matrix} 215,5\\to & \\dfrac{13}{2}.215,5 & {} & \\begin{matrix} {} & {} & (mol) \\\\\\end{matrix} \\\\\\end{matrix} \\\\ \\end{aligned}$<br\/>Th\u1ec3 t\u00edch kh\u00f4ng kh\u00ed c\u1ea7n d\u00f9ng l\u00e0 <br\/>${{V}_{kk}}=\\dfrac{13}{2}.215,5.22,4.\\dfrac{100}{20}=156884(l)$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2178},{"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 c\u00e1c oxi baz\u01a1 sau : $CuO,FeO,N{{a}_{2}}O,MgO$. C\u00f4ng th\u1ee9c c\u1ee7a c\u00e1c baz\u01a1 t\u01b0\u01a1ng \u1ee9ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 ","select":["A. $CuOH,Fe{{(OH)}_{2}},NaOH,Mg{{(OH)}_{2}}$ ","B. $Cu{{(OH)}_{2}},FeOH,NaOH,Mg{{(OH)}_{2}}$ ","C. $Cu{{(OH)}_{2}},Fe{{(OH)}_{2}},NaOH,Mg{{(OH)}_{2}}$ ","D. $Cu{{(OH)}_{2}},Fe{{(OH)}_{2}},NaOH,MgOH$ "],"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\/>Cu h\u00f3a tr\u1ecb II<br\/>Fe h\u00f3a tr\u1ecb II<br\/>Na h\u00f3a tr\u1ecb I<br\/>Mg h\u00f3a tr\u1ecb II<br\/>Nh\u00f3m OH h\u00f3a tr\u1ecb I<br\/>Theo quy t\u1eafc h\u00f3a tr\u1ecb ta c\u00f3 : $Cu{{(OH)}_{2}},Fe{{(OH)}_{2}},NaOH,Mg{{(OH)}_{2}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2179}],"lesson":{"save":0,"level":2}}