{"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":"Kh\u00ed nh\u1eb9 nh\u1ea5t trong c\u00e1c ch\u1ea5t kh\u00ed l\u00e0 ","select":["A. Hi\u0111ro ","B. Oxi ","C. Heri ","D. Nit\u01a1 "],"hint":"","explain":"<span class='basic_left'>Hi\u0111ro l\u00e0 kh\u00ed nh\u1eb9 nh\u1ea5t trong c\u00e1c ch\u1ea5t kh\u00ed<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2240},{"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\u00ednh ch\u1ea5t v\u1eady l\u00fd n\u00e0o d\u01b0\u1edbi \u0111\u00e2y kh\u00f4ng ph\u1ea3i c\u1ee7a Hi\u0111ro? ","select":["A. L\u00e0 ch\u1ea5t kh\u00ed kh\u00f4ng m\u00e0u, kh\u00f4ng m\u00f9i, kh\u00f4ng v\u1ecb ","B. Tan r\u1ea5t \u00edt trong n\u01b0\u1edbc ","C. Tan nhi\u1ec1u trong n\u01b0\u1edbc ","D. Nh\u1eb9 h\u01a1n kh\u00f4ng kh\u00ed "],"hint":"","explain":"<span class='basic_left'>Kh\u00ed hi\u0111ro l\u00e0 ch\u1ea5t kh\u00ed kh\u00f4ng m\u00e0u, kh\u00f4ng m\u00f9i, kh\u00f4ng v\u1ecb, nh\u1eb9 nh\u1ea5t trong c\u00e1c kh\u00ed, tan r\u1ea5t \u00edt trong n\u01b0\u1edbc<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2241},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Kh\u00ed hi\u0111ro k\u1ebft h\u1ee3p v\u1edbi kh\u00ed n\u00e0o sau \u0111\u00e2y theo t\u1ef7 l\u1ec7 th\u1ec3 t\u00edch $\\dfrac{{{V}_{{{H}_{2}}}}}{{{V}_{{{X}_{2}}}}}=2$ t\u1ea1o th\u00e0nh h\u1ed7n h\u1ee3p n\u1ed5 ? ","select":["A. ${{N}_{2}}$ ","B. ${{O}_{2}}$ ","C. $C{{O}_{2}}$ ","D. ${{Cl}_{2}}$ "],"hint":"","explain":"<span class='basic_left'>H\u1ed7n h\u1ee3p kh\u00ed hi\u0111ro v\u00e0 kh\u00ed oxi l\u00e0 h\u1ed7n h\u1ee3p n\u1ed5<br\/>PTHH : $2{{H}_{2}}+{{O}_{2}}\\xrightarrow{{{t}^{o}}}2{{H}_{2}}O$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2242},{"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\u1ea3n \u1ee9ng : ${{H}_{2}}+CuO\\xrightarrow{{{t}^{o}}}{{H}_{2}}O+Cu$. Hi\u0111ro \u0111\u00f3ng vai tr\u00f2 l\u00e0 ","select":["A. Ch\u1ea5t oxi h\u00f3a ","B. Ch\u1ea5t x\u00fac t\u00e1c ","C. Ch\u1ea5t t\u1ea1o m\u00f4i tr\u01b0\u1eddng ","D. Ch\u1ea5t kh\u1eed "],"hint":"","explain":"<span class='basic_left'>Kh\u00ed hi\u0111ro \u0111\u00e3 chi\u1ebfm nguy\u00ean t\u1ed1 oxi trong h\u1ee3p ch\u1ea5t CuO. Hi\u0111ro c\u00f3 t\u00ednh kh\u1eed hay l\u00e0 ch\u1ea5t kh\u1eed<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2243},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"\u1ede \u0111i\u1ec1u ki\u1ec7n th\u01b0\u1eddng, Hi\u0111ro l\u00e0 ch\u1ea5t \u1edf tr\u1ea1ng th\u00e1i n\u00e0o? ","select":["A. R\u1eafn ","B. L\u1ecfng ","C. Kh\u00ed ","D. D\u1ea1ng h\u1ee3p ch\u1ea5t v\u1edbi c\u00e1c ch\u1ea5t kh\u00e1c "],"hint":"","explain":"<span class='basic_left'>\u1ede \u0111i\u1ec1u ki\u1ec7n th\u01b0\u1eddng, c\u00e1c nguy\u00ean t\u1eed hydro k\u1ebft h\u1ee3p v\u1edbi nhau t\u1ea1o th\u00e0nh nh\u1eefng ph\u00e2n t\u1eed g\u1ed3m hai nguy\u00ean t\u1eed ${{H}_{2}}$<br\/>Hay \u1edf \u0111i\u1ec1u ki\u1ec7n ti\u00eau chu\u1ea9n, hidro l\u00e0 m\u1ed9t ch\u1ea5t kh\u00ed l\u01b0\u1ee1ng nguy\u00ean t\u1eed kh\u00f4ng m\u00e0u, kh\u00f4ng m\u00f9i, kh\u00f4ng v\u1ecb v\u00e0 l\u00e0 m\u1ed9t phi kim<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2244},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Tr\u01b0\u1eddng h\u1ee3p m\u1ed9t qu\u1ea3 b\u00f3ng ch\u1ee9a 1 (g) ${{H}_{2}}$ th\u00ec c\u00f3 \u00edt nh\u1ea5t l\u00e0 ","select":["A. ${{3.10}^{23}}$ ph\u00e2n t\u1eed ${{H}_{2}}O$ ","B. ${{6.10}^{23}}$ ph\u00e2n t\u1eed ${{H}_{2}}$ ","C. ${{3.10}^{23}}$ ph\u00e2n t\u1eed $C{{H}_{4}}$ ","D. ${{3.10}^{23}}$ ph\u00e2n t\u1eed ${{H}_{2}}$ "],"hint":"","explain":"<span class='basic_left'>${{n}_{{{H}_{2}}}}=\\dfrac{1}{2}=0,5(mol)$<br\/>C\u1ee9 1 mol ${{H}_{2}}$ ch\u1ee9a ${{6.10}^{23}}$ ph\u00e2n t\u1eed ${{H}_{2}}$<br\/> V\u1eady 0,5 mol ${{H}_{2}}$ ch\u1ee9a x ph\u00e2n t\u1eed ${{H}_{2}}$<br\/>$\\to$ x = ${{0,5.6.10}^{23}}$ = ${{3.10}^{23}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2245},{"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\u1ec3 tr\u00e1nh hi\u1ec7n t\u01b0\u1ee3ng n\u1ed5 m\u1ea1nh, tr\u01b0\u1edbc khi \u0111\u1ed1t hidro ta ph\u1ea3i th\u1ef1c hi\u1ec7n thao t\u00e1c n\u00e0o trong c\u00e1c thao t\u00e1c d\u01b0\u1edbi \u0111\u00e2y ? ","select":["A. Th\u1eed \u0111\u1ed9 tinh khi\u1ebft c\u1ee7a Hi\u0111ro tr\u01b0\u1edbc khi \u0111\u1ed1t ","B. \u0110\u1ed1t kh\u00ed Hi\u0111ro khi v\u1eeba \u0111i\u1ec1u ch\u1ebf ","C. Ch\u1edd m\u1ed9t th\u1eddi gian m\u1edbi \u0111\u1ed1t ","D. T\u1ea5t c\u1ea3 \u0111\u1ec1u \u0111\u00fang "],"hint":"","explain":"<span class='basic_left'>\u0110\u1ec3 tr\u00e1nh hi\u1ec7n t\u01b0\u1ee3ng n\u1ed5 m\u1ea1nh, tr\u01b0\u1edbc khi \u0111\u1ed1t hidro ta ph\u1ea3i th\u1eed xem kh\u00ed hidro c\u00f3 l\u1eabn kh\u00ed oxi kh\u00f4ng b\u1eb1ng c\u00e1ch thu kh\u00ed ${{H}_{2}}$ \u0111\u00f3 v\u00e0o \u1ed1ng nghi\u1ec7m r\u1ed3i \u0111\u1ed1t \u1edf mi\u1ec7ng \u1ed1ng nghi\u1ec7m. N\u1ebfu ${{H}_{2}}$ l\u00e0 tinh khi\u1ebft th\u00ec ch\u1ec9 nghe th\u1ea5y ti\u1ebfng n\u1ed5 nh\u1ecf, n\u1ebfu ${{H}_{2}}$ c\u00f3 l\u1eabn ${{O}_{2}}$ ( ho\u1eb7c kh\u00f4ng kh\u00ed ) ti\u1ebfng n\u1ed5 m\u1ea1nh<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2246},{"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 nh\u1eefng oxit sau : $CuO,MgO,A{{l}_{2}}{{O}_{3}},N{{a}_{2}}O,PbO,F{{e}_{2}}{{O}_{3}},ZnO$. Hi\u0111ro kh\u1eed \u0111\u01b0\u1ee3c bao nhi\u00eau oxit \u1edf nhi\u1ec7t \u0111\u1ed9 cao? ","select":["A. $5$ ","B. $6$ ","C. $4$ ","D. $3$ "],"hint":"","explain":"<span class='basic_left'>${{H}_{2}}$ ch\u1ec9 c\u00f3 kh\u1ea3 n\u0103ng kh\u1eed oxit c\u1ee7a kim lo\u1ea1i trung b\u00ecnh v\u00e0 y\u1ebfu ( t\u1eeb Zn tr\u1edf xu\u1ed1ng )<br\/>${{H}_{2}}$ kh\u00f4ng kh\u1eed \u0111\u01b0\u1ee3c oxit c\u1ee7a kim lo\u1ea1i m\u1ea1nh nh\u01b0 $MgO,A{{l}_{2}}{{O}_{3}},N{{a}_{2}}O$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2247},{"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 8 gam $CuO$ t\u00e1c d\u1ee5ng v\u1edbi ${{H}_{2}}$ \u1edf nhi\u1ec7t \u0111\u1ed9 cao. Sau ph\u1ea3n \u1ee9ng th\u1ea5y c\u00f3 m (g) ch\u1ea5t r\u1eafn. T\u00ednh m, ch\u1ea5t r\u1eafn \u0111\u00f3 l\u00e0 ch\u1ea5t n\u00e0o? ","select":["A. $Cu$ v\u00e0 $m=0,64(g)$ ","B. $Cu$ v\u00e0 $m=6,4(g)$ ","C. $CuO$ v\u00e0 $m=4(g)$ ","D. Kh\u00f4ng x\u00e1c \u0111\u1ecbnh \u0111\u01b0\u1ee3c "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{CuO}}=\\dfrac{8}{80}=0,1(mol) \\\\ & CuO+{{H}_{2}}\\xrightarrow{{{t}^{o}}}{{H}_{2}}O+Cu \\\\ & \\begin{matrix} 0,1mol & {} & \\to & {} & {} & 0,1mol & {} & {} & {} & {} \\\\\\end{matrix} \\\\ & \\to {{m}_{Cu}}=0,1.64=6,4(g) \\\\ \\end{aligned}$<br\/>V\u1eady ch\u1ea5t r\u1eafn l\u00e0 Cu <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2248},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"S\u1ea3n ph\u1ea9m thu \u0111\u01b0\u1ee3c sau khi nung s\u1eaft (III) oxit trong hi\u0111ro l\u00e0 ","select":["A. $F{{e}_{2}}{{O}_{3}}$ ","B. $Fe$ ","C. ${{H}_{2}}$ ","D. Kh\u00f4ng ph\u1ea3n \u1ee9ng "],"hint":"","explain":"<span class='basic_left'>PTHH : $F{{e}_{2}}{{O}_{3}}+{{3H}_{2}}\\xrightarrow{{{t}^{o}}}{{3H}_{2}}O+2Fe$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2249},{"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 \u1ee9ng d\u1ee5ng sau \u0111\u00e2y, \u1ee9ng d\u1ee5ng n\u00e0o l\u00e0 c\u1ee7a hi\u0111ro ? ","select":["A. H\u00e0n c\u1eaft kim lo\u1ea1i ","B. L\u00e0m nguy\u00ean li\u1ec7u s\u1ea3n xu\u00e2t $N{{H}_{3}}$, $HCl$, ch\u1ea5t h\u1eefu c\u01a1 ","C. Kh\u1eed oxi c\u1ee7a m\u1ed9t s\u1ed1 oxit kim lo\u1ea1i ","D. T\u1ea5t c\u1ea3 c\u00e1c \u0111\u00e1p \u00e1n tr\u00ean \u0111\u1ec1u \u0111\u00fang "],"hint":"","explain":"<span class='basic_left'>\u1ee8ng d\u1ee5ng c\u1ee7a hi\u0111ro l\u00e0 h\u00e0n c\u1eaft kim lo\u1ea1i, n\u1ea1p v\u00e0o kh\u00ed c\u1ea7u, kh\u1eed oxi c\u1ee7a m\u1ed9t s\u1ed1 oxit kim lo\u1ea1i, s\u1ea3n xu\u1ea5t amoniac, ph\u00e2n \u0111\u1ea1m, s\u1ea3n xu\u1ea5t nhi\u00ean li\u1ec7u, s\u1ea3n xu\u1ea5t axit clohidric.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":2250},{"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\u00f9ng hi\u0111ro \u0111\u1ec3 kh\u1eed a gam $CuO$ thu \u0111\u01b0\u1ee3c b gam Cu. Cho l\u01b0\u1ee3ng \u0111\u1ed3ng n\u00e0y t\u00e1c d\u1ee5ng v\u1edbi clo ${{Cl}_{2}}$ thu \u0111\u01b0\u1ee3c 33,75 gam $Cu{{Cl}_{2}}$. Gi\u00e1 tr\u1ecb a v\u00e0 b l\u00e0 ","select":["A. $a=30(g)$ v\u00e0 $b=26(g)$ ","B. $a=20(g)$ v\u00e0 $b=16(g)$ ","C. $a=40(g)$ v\u00e0 $b=36(g)$ ","D. $a=50(g)$ v\u00e0 $b=46(g)$ "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{CuC{{l}_{2}}}}=\\dfrac{33,75}{64+35,5.2}=0,25(mol) \\\\ & Cu+C{{l}_{2}}\\xrightarrow{{{t}^{o}}}CuC{{l}_{2}} \\\\ & \\begin{matrix} 0,25mol & \\leftarrow & 0,25mol & {} & {} & {} & {} \\\\\\end{matrix} \\\\ & CuO+{{H}_{2}}\\xrightarrow{{{t}^{o}}}{{H}_{2}}O+Cu \\\\ & \\begin{matrix} 0,25mol & \\leftarrow & {} & {} & 0,25mol & {} & {} \\\\\\end{matrix} \\\\ & \\to \\left\\{ \\begin{aligned} & a=0,25.80=20(g) \\\\ & b=0,25.64=16(g) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2251},{"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 m\u1ed9t h\u1ed7n h\u1ee3p g\u1ed3m $40\\%F{{e}_{2}}{{O}_{3}}$ v\u00e0 $60\\%CuO$ v\u1ec1 kh\u1ed1i l\u01b0\u1ee3ng. Ng\u01b0\u1eddi ta d\u00f9ng ${{H}_{2}}$ d\u01b0 \u0111\u1ec3 kh\u1eed 30 gam h\u1ed7n h\u1ee3p \u0111\u00f3. S\u1ed1 mol ${{H}_{2}}$ \u0111\u00e3 tham gia ph\u1ea3n \u1ee9ng l\u00e0 ","select":["A. $0,45(mol)$ ","B. $0,225(mol)$ ","C. $0,55(mol)$ ","D. $0,65(mol)$ "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{m}_{F{{e}_{2}}{{O}_{3}}}}=\\dfrac{40.30}{100}=12(g)\\to {{n}_{F{{e}_{2}}{{O}_{3}}}}=\\dfrac{12}{56.2+16.3}=0,075(mol) \\\\ & {{m}_{CuO}}=\\dfrac{60.30}{100}=18(g)\\to {{n}_{CuO}}=\\dfrac{18}{64+16}=0,225(mol) \\\\ & F{{e}_{2}}{{O}_{3}}+3{{H}_{2}}\\xrightarrow{{{t}^{o}}}2Fe+3{{H}_{2}}O(1) \\\\ & \\begin{matrix} 0,075\\to 0,225 & {} & {} & {} & {} & {} & {} & mol & {} & {} \\\\\\end{matrix} \\\\ & CuO+{{H}_{2}}\\xrightarrow{{{t}^{o}}}Cu+{{H}_{2}}O(2) \\\\ & \\begin{matrix} 0,225\\to 0,225 & {} & {} & {} & {} & {} & mol & {} & {} & {} \\\\\\end{matrix} \\\\ & \\to {{n}_{{{H}_{2}}}}={{n}_{{{H}_{2(1)}}}}+{{n}_{{{H}_{2(2)}}}}=0,225+0,225=0,45(mol) \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2252},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Trong v\u1ecf tr\u00e1i \u0111\u1ea5t, hidro chi\u1ebfm $1\\%$ v\u1ec1 kh\u1ed1i l\u01b0\u1ee3ng v\u00e0 silic chi\u1ebfm $26\\%$ v\u1ec1 kh\u1ed1i l\u01b0\u1ee3ng. Nh\u1eadn x\u00e9t n\u00e0o sau \u0111\u00e2y \u0111\u00fang ? ","select":["A. S\u1ed1 nguy\u00ean t\u1eed c\u1ee7a silic nhi\u1ec1u h\u01a1n s\u1ed1 nguy\u00ean t\u1eed c\u1ee7a hi\u0111ro ","B. S\u1ed1 nguy\u00ean t\u1eed c\u1ee7a hi\u0111ro b\u1eb1ng s\u1ed1 nguy\u00ean t\u1eed c\u1ee7a silic ","C. S\u1ed1 nguy\u00ean t\u1eed c\u1ee7a hi\u0111ro nhi\u1ec1u h\u01a1n s\u1ed1 nguy\u00ean t\u1eed c\u1ee7a silic ","D. S\u1ed1 nguy\u00ean t\u1eed c\u1ee7a hi\u0111ro b\u1eb1ng s\u1ed1 nguy\u00ean t\u1eed c\u1ee7a silic v\u00e0 b\u1eb1ng ${{6.10}^{23}}$ "],"hint":"","explain":"<span class='basic_left'>\u0110\u1eb7t kh\u1ed1i l\u01b0\u1ee3ng v\u1edf tr\u00e1i \u0111\u1ea5t l\u00e0 x ( gam ) :<br\/>$\\begin{aligned} & {{m}_{H}}=\\dfrac{x}{100}\\to {{n}_{H}}=\\dfrac{x}{100.1}(mol) \\\\ & {{m}_{Si}}=\\dfrac{26.x}{100}\\to {{n}_{Si}}=\\dfrac{26.x}{100.28}(mol) \\\\ & \\to \\dfrac{{{n}_{H}}}{{{n}_{Si}}}=\\dfrac{x}{100.1}.\\dfrac{100.28}{26.x}=\\dfrac{28}{26}=\\dfrac{14}{13} \\\\ & \\to {{n}_{H}}=\\dfrac{14}{13}{{n}_{Si}}\\to {{n}_{H}}>{{n}_{Si}} \\\\ \\end{aligned}$<br\/>V\u1eady s\u1ed1 nguy\u00ean t\u1eed c\u1ee7a hi\u0111ro nhi\u1ec1u h\u01a1n s\u1ed1 nguy\u00ean t\u1eed c\u1ee7a silic<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2253},{"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\u1ed1 gam n\u01b0\u1edbc thu \u0111\u01b0\u1ee3c khi cho 5,04 l\u00edt kh\u00ed hi\u0111ro t\u00e1c d\u1ee5ng v\u1edbi 10,64 l\u00edt kh\u00ed oxi ( c\u00e1c th\u1ec3 t\u00edch kh\u00ed \u0111o \u1edf \u0111ktc ) l\u00e0","select":["A. $4,05(g)$ ","B. $8,9(g)$ ","C. $4,9(g)$ ","D. $5,06(g)$ "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & 2{{H}_{2}}+{{O}_{2}}\\xrightarrow{{{t}^{o}}}2{{H}_{2}}O \\\\ & \\left\\{ \\begin{aligned} & {{n}_{{{H}_{2}}}}=\\dfrac{5,04}{22,4}=0,225(mol) \\\\ & {{n}_{{{O}_{2}}}}=\\dfrac{10,64}{22,4}=0,475(mol) \\\\ \\end{aligned} \\right.\\to \\dfrac{0,225}{2}<\\dfrac{0,475}{1}(1) \\\\ \\end{aligned}$<br\/>T\u1eeb t\u1ec9 s\u1ed1 \u1edf (1) th\u00ec l\u01b0\u1ee3ng oxi c\u00f2n d\u01b0 v\u00e0 hi\u0111ro ph\u1ea3n \u1ee9ng h\u1ebft<br\/>S\u1ed1 mol c\u1ee7a n\u01b0\u1edbc thu \u0111\u01b0\u1ee3c s\u1ebd t\u00ednh theo l\u01b0\u1ee3ng h\u1ebft<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh :<br\/>$\\begin{aligned} & \\to {{n}_{{{H}_{2}}O}}={{n}_{{{H}_{2}}}}=0,225(mol) \\\\ & \\to {{m}_{{{H}_{2}}O}}=0,225.18=4,05(g) \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2254},{"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\u1ec3 ph\u00e2n bi\u1ec7t hai kh\u00ed ${{O}_{2}}$ v\u00e0 $C{{O}_{2}}$ ri\u00eang bi\u1ec7t . Ng\u01b0\u1eddi ta s\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p n\u00e0o trong c\u00e1c ph\u01b0\u01a1ng ph\u00e1p sau \u0111\u00e2y ?","select":["A. S\u1ee5c kh\u00ed qua h\u1ed3 tinh b\u1ed9t ","B. S\u1ee5c kh\u00ed qua n\u01b0\u1edbc v\u00f4i trong $Ca{{OH}_{2}}$","C. S\u1ee5c kh\u00ed qua n\u01b0\u1edbc nguy\u00ean ch\u1ea5t ","D. S\u1ee5c kh\u00ed qua dung d\u1ecbch $NaOH$ "],"hint":"","explain":"<span class='basic_left'>S\u1ee5c l\u1ea7n l\u01b0\u1ee3t c\u00e1c kh\u00ed qua dung d\u1ecbch n\u01b0\u1edbc v\u00f4i trong $Ca{{OH}_{2}}$ d\u01b0, kh\u00ed n\u00f2a l\u00e0m \u0111\u1ee5c n\u01b0\u1edbc v\u00f4i trong d\u01b0 l\u00e0 $C{{O}_{2}}$ c\u00f2n l\u1ea1i l\u00e0 ${{O}_{2}}$<br\/>PTHH : $C{{O}_{2}}+Ca{{(OH)}_{2}}\\to CaC{{O}_{3}}\\downarrow +{{H}_{2}}O$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2255},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Ng\u01b0\u1eddi ta d\u00f9ng hi\u0111ro \u0111\u1ec3 kh\u1eed k\u1ebdm (II) oxit th\u00e0nh k\u1ebdm. \u0110\u1ec3 \u0111i\u1ec1u ch\u1ebf 55,25 gam k\u1ebdm th\u00ec th\u1ec3 t\u00edch kh\u00ed hi\u0111ro c\u1ea7n d\u00f9ng \u1edf \u0111ktc l\u00e0 ","select":["A. $19,04$ l\u00edt ","B. $19$ l\u00edt ","C. $20,04$ l\u00edt ","D. $24,04$ l\u00edt "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{Zn}}=\\dfrac{55,25}{65}=0,85(mol) \\\\ & ZnO+{{H}_{2}}\\xrightarrow{{{t}^{o}}}Zn+{{H}_{2}}O \\\\ & \\begin{matrix} {} & {} & 0,85\\leftarrow & 0,85 & {} & {} & mol \\\\\\end{matrix} \\\\ & \\to {{V}_{{{H}_{2}}}}=0,85.22,4=19,04(l) \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2256},{"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\u00ed $X$ c\u00f3 t\u1ef7 kh\u1ed1i h\u01a1i so v\u1edbi ${{H}_{2}}$ l\u00e0 $8,5$. $X$ l\u00e0 kh\u00ed n\u00e0o cho d\u01b0\u1edbi \u0111\u00e2y ","select":["A. $S{{O}_{2}}$ ","B. ${{O}_{2}}$ ","C. ${{Cl}_{2}}$ ","D. $N{{H}_{3}}$ "],"hint":"","explain":"<span class='basic_left'>${{d}_{X\/{{H}_{2}}}}=8,5\\to {{M}_{X}}=8,5.2=17$<br\/>$S{{O}_{2}}$ c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng mol b\u1eb1ng 64 (g\/mol)<br\/>${{O}_{2}}$ c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng mol b\u1eb1ng 32 (g\/mol)<br\/>${{Cl}_{2}}$ c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng mol b\u1eb1ng 71 (g\/mol)<br\/>$N{{H}_{3}}$ c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng mol b\u1eb1ng 17 (g\/mol)<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2257},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Kh\u1eed $5,64$ h\u1ed7n h\u1ee3p g\u1ed3m c\u00f3 $CuO$ v\u00e0 $ZnO$ b\u1eb1ng kh\u00ed ${{H}_{2}}$, thu \u0111\u01b0\u1ee3c $1,26$ gam ${{H}_{2}}O$. Th\u00e0nh ph\u1ea7n ph\u1ea7n tr\u0103m theo kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a c\u00e1c oxit c\u00f3 trong h\u1ed7n h\u1ee3p ban \u0111\u1ea7u l\u00e0","select":["A. $\\%{{m}_{CuO}}=40\\%,\\%{{m}_{ZnO}}=60\\%$ ","B. $\\%{{m}_{CuO}}=44,55\\%,\\%{{m}_{ZnO}}=55,45\\%$ ","C. $\\%{{m}_{CuO}}=42,55\\%,\\%{{m}_{ZnO}}=57,45\\%$ ","D. $\\%{{m}_{CuO}}=50\\%,\\%{{m}_{ZnO}}=50\\%$ "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{{{H}_{2}}O}}=\\dfrac{1,26}{18}=0,07(mol) \\\\ & CuO+{{H}_{2}}\\xrightarrow{{{t}^{o}}}Cu+{{H}_{2}}O \\\\ & \\begin{matrix} x(mol) & {} & \\leftarrow & {} & x(mol) & {} & {} \\\\\\end{matrix} \\\\ & ZnO+{{H}_{2}}\\xrightarrow{{{t}^{o}}}Zn+{{H}_{2}}O \\\\ & (0,07-\\begin{matrix} x)mol & \\leftarrow & (0,07-x)mol & {} & {} & {} & {} \\\\\\end{matrix} \\\\ & \\to {{m}_{CuO}}+{{m}_{ZnO}}=5,64 \\\\ & \\to 80.x+81.(0,07-x)=5,64 \\\\ & \\to x=0,03(mol) \\\\ \\end{aligned}$ <br\/>V\u1eady th\u00e0nh ph\u1ea7n ph\u1ea7n tr\u0103m theo kh\u1ed7i l\u01b0\u1ee3ng c\u1ee7a c\u00e1c oxit trong h\u1ed7n h\u1ee3p \u0111\u01b0\u1ee3c t\u00ednh nh\u01b0 sau : <br\/>$\\begin{aligned} & {{m}_{CuO}}=0,03.80=2,4(g) \\\\ & \\to \\%{{m}_{CuO}}=\\dfrac{2,4}{5,64}.100=42,55\\% \\\\ & \\to \\%{{m}_{ZnO}}=100\\%-42,55\\%=57,45\\% \\\\ \\end{aligned}$ <br\/><br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2258},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"H\u1ed7n h\u1ee3p kh\u00ed ${{H}_{2}}$ v\u00e0 ${{O}_{2}}$ khi ch\u00e1y l\u1ea1i g\u00e2y ti\u1ebfng n\u1ed5 v\u00ec ","select":["A. Hi\u0111ro ch\u00e1y m\u00e3nh li\u1ec7t trong oxi ","B. Ph\u1ea3n \u1ee9ng t\u1ecfa nhi\u1ec1u nhi\u1ec7t ","C. Ch\u1ea5t tham gia ph\u1ea3n \u1ee9ng \u0111\u1ec1u \u1edf th\u1ec3 kh\u00ed, n\u00ean khi ch\u00e1y g\u00e2y ra ti\u1ebfng n\u1ed5 ","D. Th\u1ec3 t\u00edch n\u01b0\u1edbc m\u1edbi t\u1ea1o th\u00e0nh b\u1ecb d\u00e3n n\u1edf \u0111\u1ed9t ng\u1ed9t, g\u00e2y ra s\u1ef1 ch\u1ea5n \u0111\u1ed9ng kh\u00f4ng kh\u00ed, \u0111\u00f3 l\u00e0 ti\u1ebfng n\u1ed5 m\u00e0 ta nghe \u0111\u01b0\u1ee3c "],"hint":"","explain":"<span class='basic_left'>H\u1ed7n h\u1ee3p kh\u00ed ${{H}_{2}}$ v\u00e0 ${{O}_{2}}$ l\u00e0 h\u1ed7n h\u1ee3p n\u1ed5 khi ch\u00e1y v\u00ec khi ch\u00e1y h\u1ed7n h\u1ee3p kh\u00ed n\u00e0y ch\u00e1y r\u1ea5t nhanh v\u00e0 t\u1ecfa nhi\u1ec1u nhi\u1ec7t, nhi\u1ec7t n\u00e0y l\u00e0m cho th\u1ec3 t\u00edch h\u01a1i n\u01b0\u1edbc t\u1ea1o th\u00e0nh sau ph\u1ea3n \u1ee9ng t\u0103ng l\u00ean \u0111\u1ed9t ng\u1ed9t nhi\u1ec1u l\u1ea7n, do \u0111\u00f3 l\u00e0m ch\u1ea5n \u0111\u1ed9ng m\u1ea1nh kh\u00f4ng kh\u00ed, g\u00e2y ra ti\u1ebfng n\u1ed5<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":2259}],"lesson":{"save":0,"level":2}}