{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"Kh\u00ed butan ${{C}_{4}}{{H}_{10}}$ c\u00f3 trong th\u00e0nh ph\u1ea7n kh\u00ed m\u1ecf d\u1ea7u. Th\u1ec3 t\u00edch kh\u00ed ${{O}_{2}}$ (\u0111ktc) c\u1ea7n \u0111\u1ec3 \u0111\u1ed1t ch\u00e1y $2,9$ g ${{C}_{4}}{{H}_{10}}$ l\u00e0( Bi\u1ebft s\u1ea3n ph\u1ea9m c\u1ee7a ph\u1ea3n \u1ee9ng \u0111\u1ed1t ch\u00e1y l\u00e0 $C{{O}_{2}}$ v\u00e0 ${{H}_{2}}O$ ) ","select":["A. $9,28$ l\u00edt ","B. $7,28$ l\u00edt ","C. $14$ l\u00edt ","D. $28$ l\u00edt "],"hint":"","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc c\u1ee7a ph\u1ea3n \u1ee9ng l\u00e0 $2{{C}_{4}}{{H}_{10}}+13{{O}_{2}}\\to 8C{{O}_{2}}+10{{H}_{2}}O$<br\/>S\u1ed1 mol c\u1ee7a ${{C}_{4}}{{H}_{10}}$ l\u00e0<br\/>$n=\\dfrac{m}{M}=\\dfrac{2,9}{58}=0,05(mol)$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc ta c\u00f3:<br\/>C\u1ee9 2 mol ${{C}_{4}}{{H}_{10}}$ tham gia ph\u1ea3n \u1ee9ng, s\u1ebd thu \u0111\u01b0\u1ee3c 13 mol ${{O}_{2}}$<br\/>V\u1eady 0,05 mol ${{C}_{4}}{{H}_{10}}$ tham gia ph\u1ea3n \u1ee9ng s\u1ebd thu \u0111\u01b0\u1ee3c x mol ${{O}_{2}}$<br\/>$\\Rightarrow x=\\dfrac{0,05.13}{2}=0,325(mol)$<br\/>V\u1eady th\u1ec3 t\u00edch ${{O}_{2}}$ c\u1ea7n \u0111\u1ec3 \u0111\u1ed1t ${{C}_{4}}{{H}_{10}}$ l\u00e0 $V=n.22,4=0,325.22,4=7,28$ l\u00edt<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2120},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":" X\u00e1c \u0111\u1ecbnh c\u00f4ng th\u1ee9c c\u1ee7a h\u1ee3p ch\u1ea5t 2 nguy\u00ean t\u1ed1 g\u1ed3m $M$ v\u00e0 $O$. Trong \u0111\u00f3 nguy\u00ean t\u1ed1 $M$ c\u00f3 h\u00f3a tr\u1ecb $VII$ v\u00e0 h\u1ee3p ch\u1ea5t c\u00f3 $M=272$. H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n m\u00e0 em cho l\u00e0 \u0111\u00fang","select":["A. ${{Br}_{2}}{{O}_{7}}$ ","B. ${{Cl}_{2}}{{O}_{7}}$ ","C. ${{I}_{2}}{{O}_{7}}$ ","D. ${{F}_{2}}{{O}_{7}}$ "],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3: M h\u00f3a tr\u1ecb VII v\u00e0 O h\u00f3a tr\u1ecb II<br\/>$\\Rightarrow$ theo quy t\u1eafc h\u00f3a tr\u1ecb ta c\u00f3 c\u00f4ng th\u1ee9c c\u1ee7a h\u1ee3p ch\u1ea5t d\u1ea1ng ${{M}_{2}}{{O}_{7}}$<br\/>$\\begin{aligned} & \\Rightarrow M=2{{M}_{R}}+7{{M}_{O}}=272 \\\\ & \\Rightarrow 2{{M}_{R}}=272-7.16=160 \\\\ & \\Rightarrow {{M}_{R}}=80 \\\\ \\end{aligned}$<br\/>$\\Rightarrow$ M l\u00e0 Brom (Br)<br\/>V\u1eady c\u00f4ng th\u1ee9c c\u1ee7a h\u1ee3p ch\u1ea5t l\u00e0 ${{Br}_{2}}{{O}_{7}}$<br\/><br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2121},{"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\u1ec3 \u0111i\u1ec1u ch\u1ebf kh\u00ed oxi trong ph\u00f2ng th\u00ed nghi\u1ec7m, ng\u01b0\u1eddi ta nung thu\u1ed1c t\u00edm $(KMn{{O}_{4}})$. Sau ph\u1ea3n \u1ee9ng ngo\u00e0i kh\u00ed ${{O}_{2}}$ c\u00f2n thu \u0111\u01b0\u1ee3c 2 ch\u1ea5t r\u1eafn c\u00f3 c\u00f4ng th\u1ee9c l\u00e0 ${{K}_{2}}Mn{{O}_{4}}$ v\u00e0 $Mn{{O}_{2}}$. Kh\u1ed1i l\u01b0\u1ee3ng $KMn{{O}_{4}}$ c\u1ea7n \u0111\u1ec3 \u0111i\u1ec1u ch\u1ebf $2,8$ l\u00edt ${{O}_{2}}$ (\u0111ktc) v\u00e0 kh\u1ed1i l\u01b0\u1ee3ng hai ch\u1ea5t r\u1eafn l\u00e0 ","select":["A. ${{m}_{KMn{{O}_{4}}}}=35,5(g);{{m}_{({{K}_{2}}Mn{{O}_{4}}+Mn{{O}_{2}})}}=39,5(g)$ ","B. ${{m}_{KMn{{O}_{4}}}}=40(g);{{m}_{({{K}_{2}}Mn{{O}_{4}}+Mn{{O}_{2}})}}=30(g)$ ","C. ${{m}_{KMn{{O}_{4}}}}=38,5(g);{{m}_{({{K}_{2}}Mn{{O}_{4}}+Mn{{O}_{2}})}}=34,5(g)$ ","D. ${{m}_{KMn{{O}_{4}}}}=39,5(g);{{m}_{({{K}_{2}}Mn{{O}_{4}}+Mn{{O}_{2}})}}=35,5(g)$ "],"hint":"","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc c\u1ee7a ph\u1ea3n \u1ee9ng l\u00e0 $2KMn{{O}_{4}}\\xrightarrow{{{t}^{o}}}{{K}_{2}}Mn{{O}_{4}}+Mn{{O}_{2}}+{{O}_{2}}$<br\/>S\u1ed1 mol c\u1ee7a ${{O}_{2}}$ l\u00e0 $n=\\dfrac{2,8}{22,4}=0,125(mol)$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc ta c\u00f3 ${{n}_{KMn{{O}_{4}}}}=2.0,125=0,25(mol)$<br\/>$\\Rightarrow {{m}_{KMn{{O}_{4}}}}=0,25.(39+55+16.4)=39,5(g)$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh lu\u1eadt b\u1ea3o to\u00e0n kh\u1ed1i l\u01b0\u1ee3ng<br\/>$\\Rightarrow {{m}_{{{K}_{2}}Mn{{O}_{4}}}}+{{m}_{Mn{{O}_{2}}}}={{m}_{KMn{{O}_{4}}}}-{{m}_{{{O}_{2}}}}(g)=39,5-0,125.32=35,5(g)$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":2122},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"Cho $4,8$ gam kim lo\u1ea1i $A$ ph\u1ea3n \u1ee9ng v\u1edbi $2,24$ l\u00edt kh\u00ed ${{O}_{2}}$ (\u0111ktc), sau ph\u1ea3n \u1ee9ng t\u1ea1o th\u00e0nh $AO$. Kim lo\u1ea1i $A$ v\u00e0 S\u1ed1 ph\u00e2n t\u1eed $AO$ t\u1ea1o th\u00e0nh l\u00e0","select":["A. $Ca$ v\u00e0 ${{1,2.10}^{23}}$ ","B. $Mg$ v\u00e0 ${{12.10}^{23}}$ ","C. $Mg$ v\u00e0 ${{1,2.10}^{23}}$ ","D. $Ba$ v\u00e0 ${{1,2.10}^{22}}$ "],"hint":"","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc c\u1ee7a ph\u1ea3n \u1ee9ng l\u00e0 $2A+{{O}_{2}}\\to 2AO$<br\/>S\u1ed1 mol c\u1ee7a ${{O}_{2}}$ l\u00e0 $n=\\dfrac{2,24}{22,4}=0,1(mol)$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc ta c\u00f3 ${{n}_{AO}}=2.0,1=0,2(mol)$ v\u00e0 ${{n}_{A}}=2.0,1=0,2(mol)$<br\/>Kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a $A$ l\u00e0 >$M=\\dfrac{m}{n}=\\dfrac{4,8}{0,2}=24$<br\/>$\\Rightarrow$ $A$ l\u00e0 magie $(Mg)$<br\/>$\\Rightarrow$ S\u1ed1 ph\u00e2n t\u1eed $MgO$ c\u00f3 trong 0,2 mol $MgO$ l\u00e0 ${{0,2.6.10}^{23}}={{1,2.10}^{23}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2123},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"\u1ede \u0111i\u1ec1u ki\u1ec7n $({{t}^{o}}={{0}^{o}}C;p=1atm)$ ta c\u00f3 $1$ gam ${{H}_{2}}$ v\u00e0 $16$ gam ${{O}_{2}}$. M\u1ec7nh \u0111\u1ec1 n\u00e0o d\u01b0\u1edbi \u0111\u00e2y l\u00e0 \u0111\u00fang? ","select":["A. ${{V}_{{{H}_{2}}}}={{V}_{{{O}_{2}}}}$ ","B. ${{V}_{{{H}_{2}}}}={{V}_{{{O}_{2}}}}=22,4(l)$ ","C. ${{V}_{{{H}_{2}}}}\\ne {{V}_{{{O}_{2}}}}$ ","D. ${{V}_{{{H}_{2}}}}=1,2(l);{{V}_{{{O}_{2}}}}=22,4(l)$ "],"hint":"","explain":"<span class='basic_left'>\u1ede \u0111i\u1ec1u ki\u1ec7n $({{t}^{o}}={{0}^{o}}C;p=1atm)$ \u0111\u01b0\u1ee3c g\u1ecdi l\u00e0 \u0111i\u1ec1u ki\u1ec7n ti\u00eau chu\u1ea9n (\u0111ktc)<br\/>Ta c\u00f3: S\u1ed1 mol ${H}_{2}$ l\u00e0: $n = \\dfrac{m}{M} = \\dfrac{1}{2} = 0,5 (mol)$ <br\/>Th\u1ec3 t\u00edch kh\u00ed ${H}_{2}$ l\u00e0: $V = n . 22,4 = 0, 5 . 22,4 = 11,2 (l)$ <br\/>S\u1ed1 mol ${O}_{2}$ l\u00e0: $n = \\dfrac{m}{M} = \\dfrac{16}{32} = 0,5 (mol)$<br\/> Th\u1ec3 t\u00edch kh\u00ed ${O}_{2}$ l\u00e0: $V = n. 22,4 = 0,5 . 22,4 = 11,2 (l)$<br\/>V\u1eady ${{V}_{{{O}_{2}}}}={{V}_{{{H}_{2}}}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2124},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"Kh\u00ed $X$ c\u00f3 t\u1ef7 kh\u1ed1i ${{d}_{X\/kk}}>1$ l\u00e0 ","select":["A. ${{H}_{2}}$ ","B. $C{{H}_{4}}$ ","C. ${{C}_{2}}{{H}_{2}}$ ","D. $C{{O}_{2}}$ "],"hint":"","explain":"<span class='basic_left'>So s\u00e1nh kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a c\u00e1c kh\u00ed v\u1edbi kh\u00f4ng kh\u00ed (kk) ta c\u00f3<br\/>${{M}_{C{{O}_{2}}}}>{{M}_{kk}}>{{M}_{{{C}_{2}}{{H}_{2}}}}>{{M}_{C{{H}_{4}}}}>{{M}_{{{H}_{2}}}}$<br\/>V\u1eady kh\u00ed $C{{O}_{2}}$ l\u00e0 kh\u00ed n\u1eb7ng h\u01a1n kh\u00f4ng kh\u00ed<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2125},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"Cho $2,8$ gam oxit kim lo\u1ea1i $R$ h\u00f3a tr\u1ecb $II$ ph\u1ea3n \u1ee9ng v\u1edbi dung d\u1ecbch axit $HCl$, sau ph\u1ea3n \u1ee9ng thu \u0111\u01b0\u1ee3c mu\u1ed1i $R{{Cl}_{2}}$ v\u00e0 n\u01b0\u1edbc. T\u00ean c\u1ee7a kim lo\u1ea1i $R$ l\u00e0 ( bi\u1ebft sau ph\u1ea3n \u1ee9ng thu \u0111\u01b0\u1ee3c $0,9$ gam n\u01b0\u1edbc )","select":["A. Crom ","B. \u0110\u1ed3ng ","C. Canxi ","D. Ch\u00ec "],"hint":"","explain":"<span class='basic_left'>R h\u00f3a tr\u1ecb II v\u00e0 O c\u00f3 h\u00f3a tr\u1ecb II<br\/> Theo quy t\u1eafc h\u00f3a tr\u1ecb ta c\u00f3 CTHH c\u1ee7a oxit R l\u00e0 $RO$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc c\u1ee7a ph\u1ea3n \u1ee9ng l\u00e0 $RO+2HCl\\to RC{{l}_{2}}+{{H}_{2}}O$<br\/>Theo \u0111\u1ea7u b\u00e0i ta c\u00f3: ${{n}_{{{H}_{2}}O}}=\\dfrac{0,9}{18}=0,05(mol)$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc, s\u1ed1 mol oxit kim lo\u1ea1i l\u00e0 ${{n}_{RO}}={{n}_{{{H}_{2}}O}}=0,05(mol)$<br\/>Kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a $RO$ l\u00e0 $M=\\dfrac{m}{n}=\\dfrac{2,8}{0,05}=56$<br\/> Kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a O l\u00e0 16. <br\/> V\u1eady kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a $R$ = 56 - 16 = 40<br\/> V\u1eady R l\u00e0 Canxi <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2126},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"\u0110\u1ed1t ch\u00e1y ho\u00e0n to\u00e0n $1,98$ gam h\u1ed7n h\u1ee3p $C{{H}_{4}};{{C}_{2}}{{H}_{6}}$ thu \u0111\u01b0\u1ee3c $2,912$ l\u00edt kh\u00ed $C{{O}_{2}}$ (\u0111ktc) v\u00e0 n\u01b0\u1edbc. $\\%$ kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a t\u1eebng ph\u00e2n t\u1eed trong h\u1ed1n h\u1ee3p l\u00e0( Bi\u1ebft r\u1eb1ng \u0111\u1ed1t ch\u00e1y h\u00f4n h\u1ee3p l\u00e0 qu\u00e1 tr\u00ecnh x\u1ea3y ra ph\u1ea3n \u1ee9ng v\u1edbi oxi ) ","select":["A. $\\%{{m}_{C{{H}_{4}}}}=24,24\\%$ v\u00e0 $\\%{{m}_{{{C}_{2}}{{H}_{6}}}}=75,76\\%$ ","B. $\\%{{m}_{C{{H}_{4}}}}=75,76\\%$ v\u00e0 $\\%{{m}_{{{C}_{2}}{{H}_{6}}}}=24,24\\%$ ","C. $\\%{{m}_{C{{H}_{4}}}}=70\\%$ v\u00e0 $\\%{{m}_{{{C}_{2}}{{H}_{6}}}}=30\\%$ ","D. $\\%{{m}_{C{{H}_{4}}}}=30\\%$ v\u00e0 $\\%{{m}_{{{C}_{2}}{{H}_{6}}}}=70\\%$ "],"hint":"","explain":"<span class='basic_left'>C\u00e1c ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc c\u1ee7a ph\u1ea3n \u1ee9ng l\u00e0<br\/>$\\begin{aligned} & 2C{{H}_{4}}+4{{O}_{2}}\\to 2C{{O}_{2}}+4{{H}_{2}}O(1) \\\\ & 2{{C}_{2}}{{H}_{6}}+5{{O}_{2}}\\to 2C{{O}_{2}}+6{{H}_{2}}O(2) \\\\ \\end{aligned}$<br\/>S\u1ed1 mol c\u1ee7a $C{{O}_{2}}$ l\u00e0 $n=\\dfrac{2,912}{22,4}=0,13(mol)$<br\/>G\u1ecdi s\u1ed1 mol c\u1ee7a $C{{O}_{2}}$ thu \u0111\u01b0\u1ee3c \u1edf ph\u1ea3n \u1ee9ng (1) l\u00e0 x (mol)<br\/>$\\Rightarrow$ s\u1ed1 mol $C{{O}_{2}}$ thu \u0111\u01b0\u1ee3c \u1edf ph\u1ea3n \u1ee9ng (2) l\u00e0 $(0,13-x)$ mol<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh ph\u1ea3n \u1ee9ng ta c\u00f3<br\/>$\\left\\{ \\begin{aligned} & {{n}_{C{{H}_{4}}(1)}}={{n}_{C{{O}_{2}}(1)}}=x \\\\ & {{n}_{{{C}_{2}}{{H}_{6}}(2)}}=\\dfrac{1}{2}{{n}_{C{{O}_{2}}(2)}}=\\dfrac{1}{2}(0,13-x) \\\\ \\end{aligned} \\right.$<br\/>M\u1eb7t kh\u00e1c :<br\/> $\\begin{aligned} & {{m}_{hh}}={{m}_{C{{H}_{4}}}}+{{m}_{{{C}_{2}}{{H}_{6}}}}=1,98(g) \\\\ & \\Rightarrow 16.x+30.\\dfrac{1}{2}.(0,13-x)=1,98(*) \\\\ \\end{aligned}$<br\/>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh (*) ta \u0111\u01b0\u1ee3c $x=0,03$ mol<br\/>$\\begin{aligned} & \\Rightarrow \\left\\{ \\begin{aligned} & {{n}_{C{{H}_{4}}}}=0,03(mol) \\\\ & {{n}_{{{C}_{2}}{{H}_{6}}}}=0,05(mol) \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & {{m}_{C{{H}_{4}}}}=0,03.16=0,46(g) \\\\ & {{m}_{{{C}_{2}}{{H}_{6}}}}=0,05.30=1,5(g) \\\\ \\end{aligned} \\right. \\\\ & \\Rightarrow \\left\\{ \\begin{aligned} & \\%{{m}_{C{{H}_{4}}}}=\\dfrac{{{m}_{C{{H}_{4}}}}}{{{m}_{hh}}}.100=\\dfrac{0,46}{1,98}.100=24,24\\% \\\\ & \\%{{m}_{{{C}_{2}}{{H}_{6}}}}=100\\%-24,24\\%=75,76\\% \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2127},{"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 $3,2$ g $Cu$ v\u00e0o b\u00ecnh k\u00edn ch\u1ee9a \u0111\u1ea7y kh\u00ed ${{O}_{2}}$ c\u00f3 dung t\u00edch $784$ ml (\u0111ktc). Nung n\u00f3ng b\u00ecnh \u0111\u1ec3 ph\u1ea3n \u1ee9ng x\u1ea3y ra ho\u00e0n to\u00e0n, l\u1ea5y ch\u1ea5t r\u1eafn trong b\u00ecnh ra c\u00e2n \u0111\u01b0\u1ee3c $a$ gam. Bi\u1ebft r\u1eb1ng qu\u00e1 tr\u00ecnh nung x\u1ea3y ra ph\u1ea3n \u1ee9ng h\u00f3a h\u1ecdc sau: $2Cu+{{O}_{2}}\\to 2CuO$). Gi\u00e1 tr\u1ecb $a$ l\u00e0 ","select":["A. $4,32$ gam ","B. $4$ gam ","C. $4,23$ gam ","D. $4,6$ gam "],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3 : <br\/>$\\left\\{ \\begin{aligned} & {{n}_{{{O}_{2}}}}=\\dfrac{784}{1000.22,4}=0,035(mol) \\\\ & {{n}_{Cu}}=\\dfrac{3,2}{64}=0,05 \\\\ \\end{aligned} \\right.$<br\/>Theo t\u1ec9 l\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh ${{n}_{Cu}}=0,05:2=0,025<{{n}_{{{O}_{2}}}}=0,035$<br\/>V\u1eady sau ph\u1ea3n \u1ee9ng $Cu$ ph\u1ea3n \u1ee9ng h\u1ebft v\u00e0 oxi c\u00f2n d\u01b0<br\/>V\u1eady ch\u1ea5t r\u1eafn c\u00f2n l\u1ea1i sau ph\u1ea3n \u1ee9ng ch\u1ec9 c\u00f3 $CuO$<br\/>Theo ph\u01b0\u01a1ng tr\u00ecnh ta c\u00f3 : ${{n}_{CuO}}={{n}_{Cu}}=0,05(mol)$<br\/>$\\to {{m}_{CuO}}=a=0,05.80=4(g)$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2128},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"H\u1ee3p ch\u1ea5t $A$ c\u00f3 ${{M}_{A}}=56(g\/mol)$ v\u00e0 th\u00e0nh ph\u1ea7n c\u00e1c nguy\u00ean t\u1ed1 theo kh\u1ed1i l\u01b0\u1ee3ng l\u00e0 $71,43\\%Ca$ ; $28,57\\%O$. CTHH c\u1ee7a $A$ l\u00e0. ","select":["A. ${{Ca}_{2}}{{O}_{3}}$ ","B. ${{Ca}_{2}}O$ ","C. $Ca{{O}_{2}}$ ","D. $CaO$ "],"hint":"","explain":"<span class='basic_left'>Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a m\u1ed7i nguy\u00ean t\u1ed1 c\u00f3 trong 1 mol $A$:<br\/>${{m}_{Ca}}=\\dfrac{71,43.56}{100}=40$ (g)<br\/>${{m}_{O}}=\\dfrac{28,57.56}{100}=16$ (g)<br\/>S\u1ed1 mol nguy\u00ean t\u1eed c\u1ee7a m\u1ed7i nguy\u00ean t\u1ed1 c\u00f3 trong 1 mol $A$ :<br\/>${{n}_{Ca}}=\\dfrac{40}{40}=1(mol)$<br\/>${{n}_{O}}=\\dfrac{16}{16}=1(mol)$<br\/>V\u1eady CTHH l\u00e0 $CaO$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2129}],"lesson":{"save":0,"level":2}}