{"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":"H\u00e3y cho bi\u1ebft kh\u00ed ${{Cl}_{2}}$ n\u1eb7ng h\u01a1n kh\u00ed ${{H}_{2}}$ bao nhi\u00eau l\u1ea7n? H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n m\u00e0 em cho l\u00e0 \u0111\u00fang? ","select":["A. $22$ l\u1ea7n ","B. $71$ l\u1ea7n ","C. $36,5$ l\u1ea7n ","D. $35,5$ l\u1ea7n "],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>${{d}_{C{{l}_{2}}\/{{H}_{2}}}}=\\dfrac{{{M}_{C{{l}_{2}}}}}{{{M}_{{{H}_{2}}}}}=\\dfrac{2.35,5}{2}=35,5$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2090},{"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 $A$ c\u00f3 c\u00f4ng th\u1ee9c d\u1ea1ng l\u00e0: $R{{O}_{2}}$. Bi\u1ebft ${{d}_{A\/kk}}=1,5862$. C\u00f4ng th\u1ee9c c\u1ee7a kh\u00ed $A$ ","select":["A. $S{{O}_{2}}$ ","B. $N{{O}_{2}}$ ","C. $C{{O}_{2}}$ ","D. $P{{O}_{2}}$ "],"hint":"","explain":"<span class='basic_left'>Theo b\u00e0i ra ta c\u00f3:<br\/>$\\begin{aligned} & \\left\\{ \\begin{aligned} & {{M}_{A}}=29.{{d}_{A\/kk}} \\\\ & {{M}_{A}}={{M}_{R}}+2.16 \\\\ \\end{aligned} \\right.\\to \\left\\{ \\begin{aligned} & {{M}_{A}}=29.1,5862\\approx 46 \\\\ & {{M}_{R}}={{M}_{A}}-32 \\\\ \\end{aligned} \\right. \\\\ & \\to {{M}_{R}}=46-32=14 \\\\ \\end{aligned}$<br\/>$\\to $ R l\u00e0 nit\u01a1 $(N)$<br\/>$\\to $C\u00f4ng th\u1ee9c h\u00f3a h\u1ecdc c\u1ee7a $A$ l\u00e0 $N{{O}_{2}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2091},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"Cho kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a kh\u00ed $A$ l\u00e0 ${{M}_{A}}$ v\u00e0 kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a kh\u00ed B l\u00e0 ${{M}_{B}}$. T\u1ef7 kh\u1ed1i c\u1ee7a kh\u00ed $A$ \u0111\u1ed1i v\u1edbi kh\u00ed $B$ l\u00e0 ${{d}_{A\/B}}=\\dfrac{{{M}_{A}}}{{{M}_{B}}}$. K\u1ebft lu\u1eadn n\u00e0o sau \u0111\u00e2y \u0111\u00fang? ","select":["A. ${{d}_{A\/B}}<1$ th\u00ec kh\u00ed $A$ nh\u1eb9 h\u01a1n kh\u00ed $B$ ","B. ${{d}_{A\/B}}>1$ th\u00ec kh\u00ed $A$ n\u1eb7ng h\u01a1n kh\u00ed $B$","C. ${{d}_{A\/B}}=1$ th\u00ec kh\u00ed $A$ n\u1eb7ng b\u1eb1ng kh\u00ed $B$ ","D. T\u1ea5t c\u1ea3 c\u00e1c \u0111\u00e1p \u00e1n tr\u00ean \u0111\u1ec1u \u0111\u00fang "],"hint":"","explain":"<span class='basic_left'><br\/> T\u1ec9 kh\u1ed1i c\u1ee7a kh\u00ed A so v\u1edbi kh\u00ed B k\u00fd hi\u1ec7u l\u00e0 ${{d}_{A\/B}}=\\dfrac{{{M}_{A}}}{{{M}_{B}}}$ <br\/> Tr\u01b0\u1eddng h\u1ee3p 1: ${{d}_{A\/B}}<1$ th\u00ec MA < MB n\u00ean kh\u00ed A nh\u1eb9 h\u01a1n kh\u00ed B. N\u00ean A \u0111\u00fang.<br\/> Tr\u01b0\u1eddng h\u1ee3p 2: ${{d}_{A\/B}}>1$ th\u00ec MA > MB n\u00ean kh\u00ed A n\u1eb7ng h\u01a1n kh\u00ed B. N\u00ean B \u0111\u00fang<br\/>Tr\u01b0\u1eddng h\u1ee3p 3: ${{d}_{A\/B}}=1$ th\u00ec MA = MB n\u00ean kh\u00ed A n\u1eb7ng b\u1eb1ng kh\u00ed B. N\u00ean C \u0111\u00fang<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":2092},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"T\u00ednh kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a kh\u00ed $B$, bi\u1ebft r\u1eb1ng $B$ n\u1eb7ng h\u01a1n kh\u00f4ng kh\u00ed 1,52 l\u1ea7n. H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n m\u00e0 em cho l\u00e0 \u0111\u00fang? ","select":["A. $44,08$ ","B. $32$ ","C. $64$ ","D. $71$ "],"hint":"","explain":"<span class='basic_left'>\u00c1p d\u1ee5ng c\u00f4ng th\u1ee9c t\u00ednh t\u1ec9 kh\u1ed1i<br\/>$\\Rightarrow {{d}_{B\/kk}}=\\dfrac{{{M}_{B}}}{29}=1,52\\Rightarrow {{M}_{B}}=1,52.29=44,08$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2093},{"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 $X$ c\u00f3 th\u1ec3 t\u00edch \u1edf \u0111ktc l\u00e0 $3,36$ l\u00edt v\u00e0 kh\u1ed1i l\u01b0\u1ee3ng l\u00e0 $10,65$ gam. T\u1ec9 kh\u1ed1i ${{d}_{X\/{{O}_{2}}}}$ l\u00e0 ","select":["A. $4,56793$ ","B. $2,21875$ ","C. $3,56924$ ","D. $1,76534$ "],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{aligned} & \\left\\{ \\begin{aligned} & V=n.22,4=3,36 \\\\ & m=n.M=10,65 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & n=\\dfrac{3,36}{22,4}=0,15 \\\\ & M=\\dfrac{10,65}{n} \\\\ \\end{aligned} \\right. \\\\ & \\Rightarrow M=\\dfrac{10,65}{0,15}=71 \\\\ \\end{aligned}$<br\/>$\\Rightarrow {{d}_{X\/{{O}_{2}}}}=2,21875$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2094},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"Kh\u1ea3o s\u00e1t m\u1ed9t qu\u1ea3 b\u00f3ng bay, ng\u01b0\u1eddi ta \u0111\u00f3 \u0111\u01b0\u1ee3c trong \u0111\u00f3 c\u00f3 ${{9.10}^{21}}$ ph\u00e2n t\u1eed kh\u00ed $A$. Bi\u1ebft ${{d}_{A\/{{H}_{2}}}}=2$. Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a $A$ l\u00e0 ","select":["A. $0,08$ gam ","B. $0,07$ gam ","C. $0,06$ gam ","D. $0,04$ gam "],"hint":"","explain":"<span class='basic_left'>C\u1ee9 1 mol $A$ ch\u1ee9a ${{6.10}^{23}}$ ph\u00e2n t\u1eed $A$<br\/>V\u1eady $n$ mol $A$ ch\u1ee9a ${{9.10}^{21}}$ ph\u00e2n t\u1eed $A$<br\/>$\\to n=\\dfrac{{{1.9.10}^{21}}}{{{6.10}^{23}}}=0,015$ mol<br\/>$\\left\\{ \\begin{aligned} & {{d}_{A\/{{H}_{2}}}}=\\dfrac{{{M}_{A}}}{{{M}_{{{H}_{2}}}}}=2 \\\\ & {{m}_{A}}=n.{{M}_{A}}=0,015.{{M}_{A}} \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & {{M}_{A}}=2.2=4 \\\\ & {{m}_{A}}=0,015.4=0,06(g) \\\\ \\end{aligned} \\right.$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2095},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Kh\u00ed $Y$ c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng $1,1$ gam. Bi\u1ebft ${{d}_{Y\/{{H}_{2}}}}=22$. S\u1ed1 mol c\u1ee7a $Y$ l\u00e0","select":["A. $0,025$ mol ","B. $0,05$ mol ","C. $0,015$ mol ","D. $0,09$ mol "],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\left\\{ \\begin{aligned} & {{d}_{Y\/{{H}_{2}}}}=\\dfrac{{{M}_{Y}}}{{{M}_{{{H}_{2}}}}}=22 \\\\ & {{n}_{Y}}=\\dfrac{{{m}_{Y}}}{{{M}_{Y}}}=\\dfrac{1,1}{{{M}_{Y}}} \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & {{M}_{Y}}=22.2=44 \\\\ & {{n}_{Y}}=\\dfrac{1,1}{44}=0,025(mol) \\\\ \\end{aligned} \\right.$<br\/>V\u1eady s\u1ed1 mol c\u1ee7a $Y$ l\u00e0 0,025 mol<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2096},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"Cho $2,52$ g kh\u00ed $Z$, bi\u1ebft ${{d}_{Z\/{{N}_{2}}}}=1$. Th\u1ec3 t\u00edch c\u1ee7a Z \u1edf \u0111ktc l\u00e0 ","select":["A. $0,672$ l\u00edt ","B. $0,896$ l\u00edt ","C. $1,334$ l\u00edt ","D. $2,016$ l\u00edt "],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\left\\{ \\begin{aligned} & {{d}_{Z\/{{N}_{2}}}}=\\dfrac{{{M}_{Z}}}{{{M}_{{{N}_{2}}}}}=1 \\\\ & {{n}_{Z}}=\\dfrac{{{m}_{Z}}}{{{M}_{Z}}}=\\dfrac{2,52}{{{M}_{Z}}} \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & {{M}_{Z}}=1.{{M}_{{{N}_{2}}}}=28 \\\\ & {{n}_{Z}}=\\dfrac{2,52}{28}=0,09(mol) \\\\ \\end{aligned} \\right.$<br\/>$\\Rightarrow V=n.22,4=0,09.22,4=2,016(lit)$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2097},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"Trong ph\u00f2ng th\u00ed nghi\u1ec7m ng\u01b0\u1eddi ta th\u01b0\u1eddng \u0111i\u1ec1u ch\u1ebf kh\u00ed ${{Cl}_{2}}$ b\u1eb1ng c\u00e1ch cho axit clohi\u0111ric \u0111\u1eb7c t\u00e1c d\u1ee5ng v\u1edbi ch\u1ea5t oxi h\u00f3a m\u1ea1nh nh\u01b0 mangan \u0111ioxit r\u1eafn $(Mn{{O}_{2}})$ ho\u1eb7c kali penmanganat $(KMn{{O}_{4}})$.\u0110\u1ec3 thu \u0111\u01b0\u1ee3c kh\u00ed ${{Cl}_{2}}$ v\u00e0o b\u00ecnh ng\u01b0\u1eddi ta s\u1eed d\u1ee5ng c\u00e1ch n\u00e0o sau \u0111\u00e2y. ","select":["A. \u0110\u1eb7t ng\u01b0\u1ee3c b\u00ecnh ","B. \u0110\u1eb7t \u0111\u1ee9ng b\u00ecnh ","C. \u0110\u1eb7t ngang b\u00ecnh ","D. \u0110\u1ea9y n\u01b0\u1edbc "],"hint":"","explain":"<span class='basic_left'>\u00c1p d\u1ee5ng c\u00f4ng th\u1ee9c v\u1ec1 t\u1ec9 kh\u1ed1i:<br\/>${{d}_{C{{l}_{2}}\/kk}}=\\dfrac{71}{29}\\approx 2,448>1$<br\/>Kh\u00ed ${{Cl}_{2}}$ n\u1eb7ng h\u01a1n kh\u00f4ng kh\u00ed n\u00ean s\u1eed d\u1ee5ng c\u00e1ch \u0111\u1eb7t \u0111\u1ee9ng b\u00ecnh s\u1ebd thu \u0111\u01b0\u1ee3c kh\u00ed ${{Cl}_{2}}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2098},{"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 $m$ g m\u1ed9t h\u1ee3p ch\u1ea5t h\u1eefu c\u01a1 $X$ ng\u01b0\u1eddi ta thu \u0111\u01b0\u1ee3c $1,98$ g kh\u00ed $Y$. Bi\u1ebft ${{d}_{Y\/{{O}_{2}}}}=1,375$. S\u1ed1 ph\u00e2n t\u1eed c\u1ee7a kh\u00ed $Y$ l\u00e0 ","select":["A. ${{2,7.10}^{24}}$ ","B. ${{27.10}^{22}}$ ","C. ${{2,7.10}^{23}}$ ","D. ${{2,7.10}^{22}}$ "],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\left\\{ \\begin{aligned} & {{d}_{Y\/{{O}_{2}}}}=\\dfrac{{{M}_{Y}}}{{{M}_{{{O}_{2}}}}}=1,375 \\\\ & {{n}_{Y}}=\\dfrac{{{m}_{Y}}}{{{M}_{Y}}}=\\dfrac{1,98}{{{M}_{Y}}} \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & {{M}_{Y}}=1,375.{{M}_{{{O}_{2}}}}=44 \\\\ & {{n}_{Y}}=\\dfrac{1,98}{44}=0,045(mol) \\\\ \\end{aligned} \\right.$<br\/>C\u1ee9 1 mol $Y$ ch\u1ee9a ${{6.10}^{23}}$ ph\u00e2n t\u1eed $Y$<br\/>V\u1eady 0,045 mol $Y$ ch\u1ee9a x ph\u00e2n t\u1eed $Y$<br\/>$\\to x=\\dfrac{{{0,045.6.10}^{23}}}{1}={{2,7.10}^{22}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2099}],"lesson":{"save":0,"level":2}}