{"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":"C\u00f4ng th\u1ee9c chuy\u1ec3n \u0111\u1ed5i gi\u1eefa l\u01b0\u1ee3ng ch\u1ea5t $(n)$ v\u00e0 kh\u1ed1i l\u01b0\u1ee3ng ch\u1ea5t $(m)$ ","select":["A. $n=\\frac{M}{m}(mol)$ ","B. $n=\\dfrac{m}{M}(mol)$ ","C. $m=\\dfrac{n}{M}(mol)$ ","D. $M=n.m(mol)$ "],"hint":"","explain":"<span class='basic_left'>C\u00f4ng th\u1ee9c chuy\u1ec3n \u0111\u1ed5i gi\u1eefa l\u01b0\u1ee3ng ch\u1ea5t $(n)$ v\u00e0 kh\u1ed1i l\u01b0\u1ee3ng ch\u1ea5t $(m)$<br\/>$n=\\dfrac{m}{M}(mol)$<br\/>Trong \u0111\u00f3:<br\/>+ $M$ l\u00e0 kh\u1ed1i l\u01b0\u1ee3ng mol (g\/mol)<br\/>+ $n$ l\u00e0 s\u1ed1 ch\u1ea5t ch\u1ea5t (mol)<br\/>+ $m$ l\u00e0 kh\u1ed1i l\u01b0\u1ee3ng ch\u1ea5t (gam)<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2080},{"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\u1ed1i l\u01b0\u1ee3ng c\u1ee7a $0,75$ mol $CuS{{O}_{4}}$ l\u00e0 ","select":["A. $130$ gam ","B. $110$ gam ","C. $120$ gam ","D. $140$ gam "],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{aligned} & \\left\\{ \\begin{aligned} & n=0,75 \\\\ & M=64+32+16.4=160 \\\\ \\end{aligned} \\right. \\\\ & \\to m=n.M=0,75.160=120g \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2081},{"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 s\u1ed1 mol $(n)$ c\u1ee7a $32$ g $Cu$? Bi\u1ebft ${{M}_{Cu}}=64$. H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang. ","select":["A. $0,5$ mol ","B. $0,25$ mol ","C. $0,2$ mol ","D. $2$ mol "],"hint":"","explain":"<span class='basic_left'>\u00c1p d\u1ee5ng c\u00f4ng th\u1ee9c chuy\u1ec3n \u0111\u1ed5i gi\u1eefa l\u01b0\u1ee3ng ch\u1ea5t $n$ v\u00e0 kh\u1ed1i l\u01b0\u1ee3ng $m$<br\/>$\\to n=\\dfrac{m}{M}=\\dfrac{32}{64}=0,5(mol)$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2082},{"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 kh\u00ed (\u0111ktc) c\u1ee7a $0,725$ mol $C{{O}_{2}}$ l\u00e0 ","select":["A. $16,75$ l\u00edt ","B. $3,92$ l\u00edt ","C. $11,2$ l\u00edt ","D. $16,24$ l\u00edt "],"hint":"","explain":"<span class='basic_left'>\u00c1p d\u1ee5ng c\u00f4ng th\u1ee9c chuy\u1ec3n \u0111\u1ed5i gi\u1eefa l\u01b0\u1ee3ng ch\u1ea5t $n$ v\u00e0 th\u1ec3 t\u00edch c\u1ee7a ch\u1ea5t kh\u00ed $V$ \u1edf \u0111ktc<br\/>$n=\\dfrac{V}{22,4}\\to V=n.22,4=0,725.22,4=16,24(lit)$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2083},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"M\u1ed9t ph\u00e2n t\u1eed kh\u00ed ${{X}_{2}}$ c\u00f3 th\u1ec3 t\u00edch l\u00e0 $6,72$ l\u00edt (\u0111ktc) v\u00e0 kh\u1ed1i l\u01b0\u1ee3ng $21,3$ gam. C\u00f4ng th\u1ee9c ph\u00e2n t\u1eed c\u1ee7a ${{X}_{2}}$ l\u00e0","select":["A. ${{Cl}_{2}}$ ","B. ${{Br}_{2}}$ ","C. ${{O}_{2}}$ ","D. ${{H}_{2}}$ "],"hint":"","explain":"<span class='basic_left'>\u00c1p d\u1ee5ng c\u00f4ng th\u1ee9c chuy\u1ec3n \u0111\u1ed5i gi\u1eefa l\u01b0\u1ee3ng ch\u1ea5t $n$ v\u00e0 th\u1ec3 t\u00edch c\u1ee7a ch\u1ea5t kh\u00ed $V$ \u1edf \u0111ktc<br\/>$n=\\dfrac{V}{22,4}=\\dfrac{6,72}{22,4}=0,3(mol)$<br\/>\u00c1p d\u1ee5ng c\u00f4ng th\u1ee9c chuy\u1ec3n \u0111\u1ed5i gi\u1eefa l\u01b0\u1ee3ng ch\u1ea5t $n$ v\u00e0 kh\u1ed1i l\u01b0\u1ee3ng $m$<br\/>$n=\\dfrac{m}{M}\\to {{M}_{{{X}_{2}}}}=\\dfrac{21,3}{0,3}=71$<br\/>Kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a nguy\u00ean t\u1eed X l\u00e0<br\/>${{M}_{X}}=\\dfrac{71}{2}=35,5$<br\/> V\u1eady X l\u00e0 Clo $(Cl)$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2084},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"H\u1ed7n h\u1ee3p kh\u00ed A g\u1ed3m 2 kh\u00ed $S{{O}_{2}}$ v\u00e0 $S{{O}_{3}}$ c\u00f3 th\u1ec3 t\u00edch $11,2$ l\u00edt v\u00e0 kh\u1ed1i l\u01b0\u1ee3ng $36,8$ gam. Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a m\u1ed7i kh\u00ed l\u00e0 ","select":["A. ${{m}_{S{{O}_{2}}}}=11,3$ gam v\u00e0 ${{m}_{S{{O}_{3}}}}=25,5$ gam ","B. ${{m}_{S{{O}_{2}}}}=14,9$ gam v\u00e0 ${{m}_{S{{O}_{3}}}}=21,9$ gam ","C. ${{m}_{S{{O}_{2}}}}=12,8$ gam v\u00e0 ${{m}_{S{{O}_{3}}}}=24$ gam","D. ${{m}_{S{{O}_{2}}}}=21,5$ gam v\u00e0 ${{m}_{S{{O}_{3}}}}=15,3$ gam "],"hint":"","explain":"<span class='basic_left'>S\u1ed1 mol c\u1ee7a h\u1ed7n h\u1ee3p l\u00e0 ${{n}_{hh}}=\\dfrac{V}{22,4}=\\dfrac{11,2}{22,4}=0,5$ mol<br\/>G\u1ecdi s\u1ed1 mol c\u1ee7a $S{{O}_{2}}$ l\u00e0 x mol<br\/>$\\to $ S\u1ed1 mol c\u1ee7a $S{{O}_{3}}$ l\u00e0 $0,5-x$ mol<br\/>$\\to $Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a h\u1ed7n h\u1ee3p l\u00e0 <br\/>$\\begin{aligned} & {{m}_{S{{O}_{2}}}}+{{m}_{S{{O}_{3}}}}=36,8 \\\\ & 64x+80(0,5-x)=36,8 \\\\ \\end{aligned}$<br\/>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ta \u0111\u01b0\u1ee3c $x=0,2$<br\/>$\\to $ Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a $S{{O}_{2}}$ l\u00e0 ${{m}_{S{{O}_{2}}}}=0,2.64=12,8$ gam <br\/>$\\to{{m}_{S{{O}_{3}}}}=36,8-12,8=24$ gam<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2085},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"S\u1ed1 nguy\u00ean t\u1eed c\u1ee7a 28 gam s\u1eaft $(Fe)$ l\u00e0 ","select":["A. ${{4.10}^{23}}$ ","B. ${{3.10}^{23}}$ ","C. ${{5.10}^{23}}$ ","D. ${{6.10}^{23}}$ "],"hint":"","explain":"<span class='basic_left'>S\u1ed1 mol c\u1ee7a $Fe$ l\u00e0 $n=\\dfrac{m}{M}=\\dfrac{28}{56}=0,5(mol)$<br\/>C\u1ee9 1 mol $Fe$ ch\u1ee9a ${{6.10}^{23}}$ ph\u00e2n t\u1eed $Fe$<br\/>v\u1eady 0,5 mol $Fe$ ch\u1ee9a x ph\u00e2n t\u1eed $Fe$<br\/>$\\to x=\\dfrac{{{0,5.6.10}^{23}}}{1}={{3.10}^{23}}$<br\/>V\u1eady s\u1ed1 ph\u00e2n t\u1eed $Fe$ l\u00e0 ${{3.10}^{23}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2086},{"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 kh\u00ed \u1edf \u0111ktc c\u1ee7a ${{9.10}^{23}}$ ph\u00e2n t\u1eed ${{H}_{2}}$ ","select":["A. $67,2$ l\u00edt ","B. $11,2$ l\u00edt ","C. $44,8$ l\u00edt ","D. $33,6$ l\u00edt "],"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 $n$ mol ${{H}_{2}}$ ch\u1ee9a ${{9.10}^{23}}$ ph\u00e2n t\u1eed ${{H}_{2}}$<br\/>$\\to n=\\dfrac{{{1.9.10}^{23}}}{{{6.10}^{23}}}=1,5$ mol<br\/>Th\u1ec3 t\u00edch c\u1ee7a ${{H}_{2}}$ \u1edf \u0111ktc l\u00e0 $V=n.22,4=1,5.22,4=33,6$ l\u00edt<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2087},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"K\u1ebft lu\u1eadn n\u00e0o \u0111\u00fang, n\u1ebfu hai ch\u1ea5t kh\u00ed kh\u00e1c nhau m\u00e0 c\u00f3 th\u1ec3 t\u00edch b\u1eb1ng nhau( \u0111o c\u00f9ng \u0111i\u1ec1u ki\u1ec7n nhi\u1ec7t \u0111\u1ed9, \u00e1p su\u1ea5t ) th\u00ec","select":["A. Ch\u00fang c\u00f3 c\u00f9ng s\u1ed1 mol ch\u1ea5t ","B. Ch\u00fang c\u00f3 c\u00f9ng kh\u1ed1i l\u01b0\u1ee3ng ","C. Ch\u00fang c\u00f3 c\u00f9ng nguy\u00ean t\u1eed kh\u1ed1i ","D. Kh\u00f4ng th\u1ec3 k\u1ebft lu\u1eadn \u0111\u01b0\u1ee3c \u0111i\u1ec1u g\u00ec "],"hint":"","explain":"<span class='basic_left'>Theo ph\u01b0\u01a1ng tr\u00ecnh tr\u1ea1ng th\u00e1i c\u1ee7a kh\u00ed l\u00fd t\u01b0\u1edfng: $PV=nRT$<br\/>trong \u0111\u00f3<br\/>P l\u00e0 \u00e1p su\u1ea5t kh\u00ed $(atm)$<br\/>V l\u00e0 th\u1ec3 t\u00edch kh\u00ed $(l\u00edt)$<br\/>n l\u00e0 s\u1ed1 mol kh\u00ed $(mol)$<br\/>R l\u00e0 h\u1eb1ng s\u1ed1 kh\u00ed v\u00e0 R = 0,082 (l\u00edt.atm)\/(mol.K) <br\/>T l\u00e0 nhi\u1ec7t \u0111\u1ed9 tuy\u1ec7t \u0111\u1ed1i( T = ${{t}^{o}}$ + 273 $(K)$<br\/>Nh\u1eadn x\u00e9t: t\u1eeb ph\u01b0\u01a1ng tr\u00ecnh ta th\u1ea5y khi \u0111o \u1edf c\u00f9ng \u0111i\u1ec1u ki\u1ec7n nhi\u1ec7t \u0111\u1ed9 v\u00e0 \u00e1p su\u1ea5t v\u00e0 th\u1ec3 t\u00edch c\u1ee7a hai kh\u00ed b\u1eb1ng nhau th\u00ec ch\u00fang c\u00f3 c\u00f9ng s\u1ed1 mol ch\u1ea5t $(n=\\dfrac{PV}{RT})$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2088},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"T\u00ednh kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a ch\u1ea5t $A$, bi\u1ebft r\u1eb1ng $0,5$ mol ch\u1ea5t n\u00e0y c\u00f3 c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng $32$ g. Ch\u1ea5t n\u00e0y l\u00e0 ch\u1ea5t g\u00ec? H\u00e3y khoanh tr\u00f2n v\u00e0o \u0111\u00e1p \u00e1n m\u00e0 em cho l\u00e0 \u0111\u00fang.","select":["A. $M=56$ v\u00e0 $Fe$ ","B. $M=52$ v\u00e0 $Cr$ ","C. $M=64$ v\u00e0 $Cu$ ","D. $M=65$ v\u00e0 $Zn$ "],"hint":"","explain":"<span class='basic_left'>\u00c1p d\u1ee5ng c\u00f4ng th\u1ee9c chuy\u1ec3n \u0111\u1ed5i gi\u1eefa l\u01b0\u1ee3ng ch\u1ea5t $n$ v\u00e0 kh\u1ed1i l\u01b0\u1ee3ng ch\u1ea5t $m$:<br\/>$n=\\dfrac{m}{M}\\to M=\\dfrac{32}{0,5}=64$<br\/>V\u1eady $A$ l\u00e0 \u0111\u1ed3ng $(Cu)$<br\/><br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2089}],"lesson":{"save":0,"level":2}}