{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"\u0110\u01a1n v\u1ecb c\u1ee7a kh\u1ed1i l\u01b0\u1ee3ng mol l\u00e0 ","select":["A. $gam$ ","B. $mol$ ","C. $g\/mol$ ","D. $kg\/mol$ "],"hint":"","explain":"<span class='basic_left'><br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2070},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Con s\u1ed1 $N$ = ${{6.10}^{23}}$ \u0111\u01b0\u1ee3c g\u1ecdi l\u00e0 ","select":["A. S\u1ed1 Avoga\u0111ro ","B. S\u1ed1 hi\u1ec7u nguy\u00ean t\u1eed ","C. H\u1eb1ng s\u1ed1 kh\u00ed ","D. \u0110\u1ed9 ph\u00e2n t\u00e1n "],"hint":"","explain":"<span class='basic_left'><br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2071},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"Nguy\u00ean t\u1eed $X$ c\u00f3 nguy\u00ean t\u1eed kh\u1ed1i g\u1ea5p 2 l\u1ea7n ph\u00e2n t\u1eed kh\u1ed1i c\u1ee7a ${{O}_{2}}$. Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a 0,075 mol $X$ l\u00e0 ","select":["A. 5,6 gam ","B. 4,8 gam ","C. 6,4 gam ","D. 3,2 gam "],"hint":"","explain":"<span class='basic_left'>Theo b\u00e0i ra ta c\u00f3 nguy\u00ean t\u1eed kh\u1ed1i c\u1ee7a X l\u00e0<br\/>${{M}_{X}}=2{{M}_{{{O}_{2}}}}=2.32=64$<br\/>M\u1eb7t kh\u00e1c kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a m\u1ed9t ch\u1ea5t c\u00f3 tr\u1ecb s\u1ed1 b\u1eb1ng nguy\u00ean t\u1eed kh\u1ed1i ho\u1eb7c ph\u00e2n t\u1eed kh\u1ed1i <br\/>$\\to$ Kh\u1ed1i l\u01b0\u1ee3ng 1 mol nguy\u00ean t\u1eed X l\u00e0 ${{M}_{X}}=64$ gam<br\/>$\\to $Kh\u1ed1i l\u01b0\u1ee3ng 0,075 mol nguy\u00ean t\u1eed $X$ l\u00e0 ${{M}_{X}}=0,075.64=4,8$ gam <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2072},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"Th\u1ec3 t\u00edch \u1edf \u0111ktc c\u1ee7a 0,15 N ph\u00e2n t\u1eed $S{{O}_{2}}$ l\u00e0 ","select":["A. 5,6 l\u00edt ","B. 2,24 l\u00edt ","C. 4,48 l\u00edt ","D. 3,36 l\u00edt "],"hint":"","explain":"<span class='basic_left'>C\u1ee9 1 mol $S{{O}_{2}}$ ch\u1ee9a ${{6.10}^{23}}$ ph\u00e2n t\u1eed $S{{O}_{2}}$<br\/>V\u1eady $n$ mol $S{{O}_{2}}$ ch\u1ee9a ${{0,15.6.10}^{23}}$ ph\u00e2n t\u1eed $S{{O}_{2}}$<br\/>$\\to n=\\dfrac{{{1.0,15.6.10}^{23}}}{{{6.10}^{23}}}=0,15$ mol<br\/>\u1ede \u0111ktc 1 mol kh\u00ed $S{{O}_{2}}$ c\u00f3 th\u1ec3 t\u00edch l\u00e0 22,4 l\u00edt<br\/>V\u1eady \u1edf \u0111ktc 0,15 mol kh\u00ed $S{{O}_{2}}$ c\u00f3 th\u1ec3 t\u00edch l\u00e0 V l\u00edt<br\/>$\\to V=0,15.22,4=3,36$ l\u00edt<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2073},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"M\u1ed9t h\u1ed7n h\u1ee3p kh\u00ed g\u1ed3m ${{N}_{2}},{{H}_{2}}$ c\u00f3 t\u1ed5ng s\u1ed1 nguy\u00ean t\u1eed ${{4,48.10}^{23}}$. S\u1ed1 ph\u00e2n t\u1eed c\u1ee7a ${{N}_{2}}$ l\u00e0 ( bi\u1ebft r\u1eb1ng s\u1ed1 mol c\u1ee7a ${{H}_{2}}$ trong h\u1ed7n h\u1ee3p l\u00e0 0,24 mol ) ","select":["A. ${{4.10}^{23}}$ ","B. ${{2,48.10}^{23}}$ ","C. ${{3,04.10}^{23}}$ ","D. ${{1,48.10}^{23}}$ "],"hint":"","explain":"<span class='basic_left'>C\u1ee9 1 mol ${{H}_{2}}$ ch\u1ee9a ${{6.10}^{23}}$ ph\u00e2n t\u1eed ${{H}_{2}}$<br\/>v\u1eady 0,24 mol ${{H}_{2}}$ ch\u1ee9a x ph\u00e2n t\u1eed ${{H}_{2}}$<br\/>$\\to x=\\dfrac{{{0,24.6.10}^{23}}}{1}={{1,44.10}^{23}}$<br\/>V\u1eady s\u1ed1 ph\u00e2n t\u1eed ${{N}_{2}}$ l\u00e0 ${{3,04.10}^{23}}$<br\/><br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2074},{"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 8,6 g m\u1ed9t h\u1ed7n h\u1ee3p oxit s\u1eaft $(FeO,F{{e}_{2}}{{O}_{3}})$, bi\u1ebft s\u1ed1 mol c\u1ee7a $FeO$ l\u00e0 0,05 mol. S\u1ed1 mol c\u1ee7a $F{{e}_{2}}{{O}_{3}}$ l\u00e0 ","select":["A. 0,03125 mol ","B. 0,01 mol ","C. 0,02 mol ","D. 0,15 mol "],"hint":"","explain":"<span class='basic_left'>C\u1ee9 1 mol $FeO$ c\u00f3 56 + 16 = 72 gam<br\/>v\u1eady 0,05 mol $FeO$ c\u00f3 m gam<br\/>$\\to m = 0,05.72 = 3,6$ gam<br\/>$\\to {{m}_{F{{e}_{2}}{{O}_{3}}}}=8,6-3,6=5$ gam<br\/>C\u1ee9 1 mol $F{{e}_{2}}{{O}_{3}}$ c\u00f3 56.2 + 16.3 = 160 gam<br\/>V\u1eady n mol $F{{e}_{2}}{{O}_{3}}$ c\u00f3 5 gam<br\/>$\\to n=\\dfrac{5.1}{160}=0,03125$ mol<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2075},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"H\u1ed7n h\u1ee3p kh\u00ed g\u1ed3m c\u00f3: 0,3 mol $C{{O}_{2}}$ v\u00e0 0,15 mol ${{O}_{2}}$. T\u1ef7 l\u1ec7 kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a $\\dfrac{{{m}_{C{{O}_{2}}}}}{{{m}_{{{O}_{2}}}}}$ l\u00e0","select":["A. 4,8 ","B. 2,75 ","C. 5,6 ","D. 3 "],"hint":"","explain":"<span class='basic_left'>Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a 1 mol ph\u00e2n t\u1eed $C{{O}_{2}}$ l\u00e0 44 gam<br\/>$\\to $Kh\u1ed1i l\u01b0\u1ee3ng 0,3 mol ph\u00e2n t\u1eed $C{{O}_{2}}$ l\u00e0 ${{m}_{C{{O}_{2}}}}=0,3.44=13,2$ gam<br\/>T\u01b0\u01a1ng t\u1ef1 v\u1edbi ${{O}_{2}}$ ta c\u00f3: ${{m}_{{{O}_{2}}}}=0,15.32=4,8$ gam<br\/>$\\dfrac{{{m}_{C{{O}_{2}}}}}{{{m}_{{{O}_{2}}}}}=\\dfrac{13,2}{4,8}=2,75$<br\/><br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2076},{"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\u1ed1i l\u01b0\u1ee3ng v\u00e0 th\u1ec3 t\u00edch kh\u00ed \u1edf \u0111i\u1ec1u ki\u1ec7n ti\u00eau chu\u1ea9n c\u1ee7a 0,25 N ph\u00e2n t\u1eed $CO$ l\u00e0 ","select":["A. m = 6,7 gam v\u00e0 V = 4,6 l\u00edt ","B. m = 4,9 gam v\u00e0 V = 5,6 l\u00edt ","C. m = 6,8 gam v\u00e0 V = 4,6 l\u00edt ","D. m = 7 gam v\u00e0 V = 5,6 l\u00edt "],"hint":"","explain":"<span class='basic_left'>C\u1ee9 1 mol $CO$ ch\u1ee9a ${{6.10}^{23}}$ ph\u00e2n t\u1eed $CO$<br\/>V\u1eady $n$ mol $CO$ ch\u1ee9a ${{0,25.6.10}^{23}}$ ph\u00e2n t\u1eed $CO$<br\/>$\\to n=\\dfrac{{{1.0,25.6.10}^{23}}}{{{6.10}^{23}}}=0,25$ mol<br\/>\u1ede \u0111ktc 1 mol kh\u00ed $CO$ c\u00f3 th\u1ec3 t\u00edch l\u00e0 22,4 l\u00edt<br\/>V\u1eady \u1edf \u0111ktc 0,25 mol kh\u00ed $CO$ c\u00f3 th\u1ec3 t\u00edch l\u00e0 V l\u00edt<br\/>$\\to V=0,25.22,4=5,6$ l\u00edt<br\/>Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a $CO$ l\u00e0 $m = 0,25.28 = 7$ gam<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2077},{"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 th\u01b0\u1eddng ( ${{20}^{0}}C$ v\u00e0 $1atm$ ), 1 mol ch\u1ea5t kh\u00ed c\u00f3 th\u1ec3 t\u00edch l\u00e0 ","select":["A. 24 l\u00edt ","B. 22,4 l\u00edt ","C. 30 l\u00edt ","D. 29 l\u00edt "],"hint":"","explain":"<span class='basic_left'>\u00c1p d\u1ee5ng ph\u01b0\u01a1ng tr\u00ecnh tr\u1ea1ng th\u00e1i c\u1ee7a kh\u00ed l\u00fd t\u01b0\u1edfng ta c\u00f3: <br\/>$\\begin{aligned} & PV=nRT \\\\ & \\to V=\\dfrac{nRT}{P}=\\dfrac{1.0,082.(20+273)}{1}\\approx 24 \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2078},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"N\u1ebfu ch\u00fang ta c\u00f3 N = ${{6.10}^{23}}$ ( s\u1ed1 Avoga\u0111ro )h\u1ea1t g\u1ea1o, th\u00ec s\u1ebd nu\u00f4i s\u1ed1ng \u0111\u01b0\u1ee3c lo\u00e0i ng\u01b0\u1eddi tr\u00ean tr\u00e1i \u0111\u1ea5t n\u00e0y trong th\u1eddi gian bao l\u00e2u? Bi\u1ebft r\u1eb1ng m\u1ed7i ng\u01b0\u1eddi \u0103n 3 b\u1eefa m\u1ed9t ng\u00e0y v\u00e0 m\u1ed7i b\u1eefa \u0103n 5000 h\u1ea1t g\u1ea1o. ( bi\u1ebft s\u1ed1 d\u00e2n tr\u00ean th\u1ebf gi\u1edbi n\u0103m 2018 l\u00e0 7,61 t\u1ef7 ng\u01b0\u1eddi ) ","select":["A. 20000000 n\u0103m ","B. 10000000 n\u0103m ","C. 14419610 n\u0103m ","D. 9000000 n\u0103m "],"hint":"","explain":"<span class='basic_left'>M\u1ed7i ng\u01b0\u1eddi m\u1ed9t ng\u00e0y \u0103n h\u1ebft: $5000.3 = 15000$ h\u1ea1t g\u1ea1o<br\/>S\u1ed1 d\u00e2n tr\u00ean th\u1ebf gi\u1edbi n\u0103m 2018 l\u00e0 7,61 t\u1ef7 ng\u01b0\u1eddi $({{7,61.10}^{9}})$, m\u1ed9t ng\u00e0y \u0103n h\u1ebft:<br\/>${{7,61.10}^{9}}{{.1,5.10}^{4}}={{11,4.10}^{13}}$ h\u1ea1t g\u1ea1o<br\/>Trong 1 n\u0103m, lo\u00e0i ng\u01b0\u1eddi \u0103n h\u1ebft: <br\/>${{11,4.10}^{13}}.365={{4,161.10}^{16}}$ h\u1ea1t g\u1ea1o<br\/>S\u1ed1 n\u0103m \u0111\u1ec3 lo\u00e0i ng\u01b0\u1eddi tr\u00ean tr\u00e1i \u0111\u1ea5t n\u00e0y \u0103n h\u1ebft N h\u1ea1t g\u1ea1o ( 1 mol h\u1ea1t g\u1ea1o ):<br\/>$\\dfrac{{{6.10}^{23}}}{{{4,161.10}^{16}}}=14419610$ n\u0103m<br\/>v\u1eady c\u00f2n nhi\u1ec1u tri\u1ec7u n\u0103m n\u1eefa lo\u00e0i ng\u01b0\u1eddi m\u1edbi \u0103n h\u1ebft 1 mol h\u1ea1t g\u1ea1o<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2079}],"lesson":{"save":0,"level":2}}