{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Ph\u00e2n \u0111\u1ea1m ur\u00ea c\u00f3 c\u00f4ng th\u1ee9c h\u00f3a h\u1ecdc l\u00e0 $(N{H}_{2})_{2}CO$. Th\u00e0nh ph\u1ea7n $\\%$ theo kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a nguy\u00ean t\u1ed1 $N$ l\u00e0","select":["A. $46,67\\%$ ","B. $23,33\\%$ ","C. $33,33\\%$ ","D. $45,67\\%$ "],"hint":"","explain":"<span class='basic_left'>Kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a $(N{{H}_{2}})_{2}CO$ l\u00e0 $M=14.2+1.4+12+16=60$ (g\/mol)<br\/>Trong 1 mol $(N{{H}_{2}})_{2}CO$ c\u00f3 2 mol nguy\u00ean t\u1eed $N$<br\/>$\\Rightarrow \\%{{m}_{N}}=\\dfrac{2.14}{60}.100=46,67\\%$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2100},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"M\u1ed9t lo\u1ea1i oxit \u0111\u1ed3ng m\u00e0u \u0111\u1ecf g\u1ea1ch c\u00f3 $M=144$ g\/mol. Oxit n\u00e0y c\u00f3 th\u00e0nh ph\u1ea7n theo kh\u1ed1i l\u01b0\u1ee3ng l\u00e0 $88,89\\%Cu$ v\u00e0 $11,11\\%O$. C\u00f4ng th\u1ee9c h\u00f3a h\u1ecdc c\u1ee7a oxit \u0111\u1ed3ng l\u00e0 ","select":["A. $CuO$ ","B. $C{{u}_{2}}O$ ","C. $C{{u}_{3}}O$ ","D. $C{{u}_{2}}{{O}_{3}}$ "],"hint":"","explain":"<span class='basic_left'>Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a m\u1ed7i nguy\u00ean t\u1ed1 c\u00f3 trong 1 mol oxit \u0111\u1ed3ng:<br\/>${{m}_{Cu}}=\\dfrac{88,89.144}{100}=128$ (g)<br\/>${{m}_{O}}=\\dfrac{11,11.144}{100}=16$ (g)<br\/>S\u1ed1 mol nguy\u00ean t\u1eed c\u1ee7a m\u1ed7i nguy\u00ean t\u1ed1 c\u00f3 trong 1 mol oxit \u0111\u1ed3ng :<br\/>${{n}_{Cu}}=\\dfrac{128}{64}=2(mol)$<br\/>${{n}_{O}}=\\dfrac{16}{16}=1(mol)$<br\/>C\u00f4ng th\u1ee9c h\u00f3a h\u1ecdc c\u1ee7a oxit \u0111\u1ed3ng l\u00e0 $C{{u}_{2}}O$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2101},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"C\u00f4ng th\u1ee9c h\u00f3a h\u1ecdc c\u1ee7a glucoz\u01a1 l\u00e0 ${{C}_{6}}{{H}_{12}}{{O}_{6}}$. C\u00f3 bao nhi\u00eau mol nguy\u00ean t\u1eed cacbon trong $0,05$ mol glucoz\u01a1","select":["A. $0,5$ mol ","B. $0,04$ mol ","C. $0,03$ mol ","D. $0,3$ mol "],"hint":"","explain":"<span class='basic_left'>C\u1ee9 1 mol ${{C}_{6}}{{H}_{12}}{{O}_{6}}$ c\u00f3 6 mol $C$<br\/>V\u1eady 0,05 mol ${{C}_{6}}{{H}_{12}}{{O}_{6}}$ c\u00f3 n mol $C$<br\/>$\\Rightarrow $ $n=0,05.6=0,3$ mol<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2102},{"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\u1ee3p ch\u1ea5t kh\u00ed $X$ c\u00f3 th\u00e0nh ph\u1ea7n $\\%$ theo kh\u1ed1i l\u01b0\u1ee3ng l\u00e0 $50\\%S$ v\u00e0 $50\\%O$. Bi\u1ebft ${{d}_{X\/{{O}_{2}}}}=2$. C\u00f4ng th\u1ee9c h\u00f3a h\u1ecdc c\u1ee7a $X$ l\u00e0 ","select":["A. $S{{O}_{3}}$ ","B. $SO$ ","C. $S{{O}_{2}}$ ","D. ${{S}_{2}}{{O}_{3}}$ "],"hint":"","explain":"<span class='basic_left'>Kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a $X$ l\u00e0 ${{M}_{X}}=32.{{d}_{X\/{{O}_{2}}}}=32.2=64$<br\/>Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a m\u1ed7i nguy\u00ean t\u1ed1 c\u00f3 trong 1 mol h\u1ee3p ch\u1ea5t kh\u00ed:<br\/>${{m}_{S}}=\\dfrac{50.64}{100}=32$ (g)<br\/>${{m}_{O}}=\\dfrac{50.64}{100}=32$ (g)<br\/>S\u1ed1 mol nguy\u00ean t\u1eed c\u1ee7a m\u1ed7i nguy\u00ean t\u1ed1 c\u00f3 trong 1 mol h\u1ee3p ch\u1ea5t kh\u00ed :<br\/>${{n}_{S}}=\\dfrac{32}{32}=1(mol)$<br\/>${{n}_{O}}=\\dfrac{32}{16}=2(mol)$<br\/>V\u1eady c\u00f4ng th\u1ee9c h\u00f3a h\u1ecdc c\u1ee7a h\u1ee3p ch\u1ea5t kh\u00ed l\u00e0 $S{{O}_{2}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2103},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"\u0110\u1ed1t ch\u00e1t ho\u00e0n to\u00e0n $m$ g h\u1ee3p ch\u1ea5t h\u1eefu c\u01a1 $X$ thu \u0111\u01b0\u1ee3c $2,86$ g kh\u00ed $Y$ c\u00f3 th\u00e0nh ph\u1ea7n $\\%$ theo kh\u1ed1i l\u01b0\u1ee3ng l\u00e0 $27,27\\%C$ v\u00e0 $72,73\\%O$. Bi\u1ebft r\u1eb1ng ${{d}_{Y\/{{H}_{2}}}}=22$. C\u00f3 bao nhi\u00eau mol nguy\u00ean t\u1eed oxi c\u00f3 trong $2,86$ g kh\u00ed $Y$ ","select":["A. $0,065$ mol ","B. $0,13$ mol ","C. $0,3$ mol ","D. $0,23$ mol "],"hint":"","explain":"<span class='basic_left'>Kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a kh\u00ed $Y$ l\u00e0 ${{M}_{Y}}=2.{{d}_{Y\/{{H}_{2}}}}=2.22=44$<br\/>Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a m\u1ed7i nguy\u00ean t\u1ed1 c\u00f3 trong 1 mol h\u1ee3p ch\u1ea5t kh\u00ed:<br\/>${{m}_{C}}=\\dfrac{27,27.44}{100}=12$ (g)<br\/>${{m}_{O}}=\\dfrac{72,73.44}{100}=32$ (g)<br\/>S\u1ed1 mol nguy\u00ean t\u1eed c\u1ee7a m\u1ed7i nguy\u00ean t\u1ed1 c\u00f3 trong 1 mol h\u1ee3p ch\u1ea5t kh\u00ed :<br\/>${{n}_{C}}=\\dfrac{12}{12}=1(mol)$<br\/>${{n}_{O}}=\\dfrac{32}{16}=2(mol)$<br\/>V\u1eady c\u00f4ng th\u1ee9c h\u00f3a h\u1ecdc c\u1ee7a h\u1ee3p ch\u1ea5t kh\u00ed l\u00e0 $C{{O}_{2}}$<br\/>Ta c\u00f3: $n=\\dfrac{2,86}{44}=0,065(mol)$<br\/>C\u1ee9 1 mol $C{{O}_{2}}$ c\u00f3 2 mol $O$<br\/>V\u1eady 0,065 mol $C{{O}_{2}}$ c\u00f3 n mol $O$<br\/>$\\Rightarrow $ $n=0,065.2=0,13$ mol<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2104},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"M\u1ed9t h\u1ee3p ch\u1ea5t h\u1eefu c\u00f3 c\u00f3 th\u00e0nh ph\u1ea7n c\u00e1c nguy\u00ean t\u1ed1 l\u00e0 $40\\%Cu$ ; $20\\%S$ v\u00e0 $40\\%O$. C\u00f4ng th\u1ee9c h\u00f3a h\u1ecdc c\u1ee7a h\u1ee3p ch\u1ea5t l\u00e0 ( bi\u1ebft kh\u1ed1i l\u01b0\u1ee3ng mol l\u00e0 160 ) ","select":["A. $C{{u}_{2}}S{{O}_{4}}$ ","B. $C{{u}_{2}}{{(S{{O}_{4}})}_{3}}$ ","C. $Cu{{(S{{O}_{4}})}_{2}}$ ","D. $CuS{{O}_{4}}$ "],"hint":"","explain":"<span class='basic_left'>Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a m\u1ed7i nguy\u00ean t\u1ed1 c\u00f3 trong 1 mol h\u1ee3p ch\u1ea5t:<br\/>${{m}_{Cu}}=\\dfrac{40.160}{100}=64$ (g)<br\/>${{m}_{S}}=\\dfrac{20.160}{100}=32$ (g)<br\/>${{m}_{O}}=\\dfrac{40.160}{100}=64$ (g)<br\/>S\u1ed1 mol nguy\u00ean t\u1eed c\u1ee7a m\u1ed7i nguy\u00ean t\u1ed1 c\u00f3 trong 1 mol h\u1ee3p ch\u1ea5t :<br\/>${{n}_{Cu}}=\\dfrac{64}{64}=1(mol)$<br\/>${{n}_{S}}=\\dfrac{32}{32}=1(mol)$<br\/>${{n}_{S}}=\\dfrac{64}{16}=4(mol)$<br\/>V\u1eady c\u00f4ng th\u1ee9c h\u00f3a h\u1ecdc c\u1ee7a h\u1ee3p ch\u1ea5t l\u00e0 $CuS{{O}_{4}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2105},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"H\u1ee3p ch\u1ea5t $A$ c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng mol l\u00e0 $94$ g\/mol, c\u00f3 th\u00e0nh ph\u1ea7n c\u00e1c nguy\u00ean t\u1ed1 l\u00e0 $82,98\\%K$, c\u00f2n l\u1ea1i l\u00e0 oxi. C\u00f4ng th\u1ee9c h\u00f3a h\u1ecdc c\u1ee7a $A$ l\u00e0 ","select":["A. ${{K}_{2}}O$ ","B. $K{{O}_{2}}$ ","C. $K{{O}_{3}}$ ","D. ${{K}_{2}}{{O}_{3}}$ "],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3 : $\\%O=100\\%-82,98\\%=17,02\\%$<br\/>Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a m\u1ed7i nguy\u00ean t\u1ed1 c\u00f3 trong 1 mol h\u1ee3p ch\u1ea5t:<br\/>${{m}_{K}}=\\dfrac{82,98.94}{100}=78$ (g)<br\/>${{m}_{O}}=\\dfrac{17,02.94}{100}=16$ (g)<br\/>S\u1ed1 mol nguy\u00ean t\u1eed c\u1ee7a m\u1ed7i nguy\u00ean t\u1ed1 c\u00f3 trong 1 mol h\u1ee3p ch\u1ea5t :<br\/>${{n}_{K}}=\\dfrac{78}{39}=2(mol)$<br\/>${{n}_{O}}=\\dfrac{16}{16}=1(mol)$<br\/>C\u00f4ng th\u1ee9c h\u00f3a h\u1ecdc c\u1ee7a h\u1ee3p ch\u1ea5t l\u00e0 ${{K}_{2}}O$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2106},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"Th\u00e0nh ph\u1ea7n $\\%$ theo kh\u1ed1i l\u01b0\u1ee3ng m\u1ed7i nguy\u00ean t\u1ed1 h\u00f3a h\u1ecdc c\u00f3 trong h\u1ee3p ch\u1ea5t $A{{l}_{2}}{{O}_{3}}$ l\u00e0","select":["A. $80\\%Al$ v\u00e0 $20\\%O$ ","B. $62,94\\%Al$ v\u00e0 $37,06\\%O$ ","C. $52,94\\%Al$ v\u00e0 $47,06\\%O$ ","D. $70\\%Al$ v\u00e0 $30\\%O$ "],"hint":"","explain":"<span class='basic_left'>Kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a $A{{l}_{2}}{{O}_{3}}$ l\u00e0 $M=27.2+16.3=102$<br\/>Trong 1 mol $A{{l}_{2}}{{O}_{3}}$ c\u00f3 2 mol nguy\u00ean t\u1eed $Al$<br\/>Trong 1 mol $A{{l}_{2}}{{O}_{3}}$ c\u00f3 3 mol nguy\u00ean t\u1eed $O$<br\/>$\\Rightarrow \\%{{m}_{Al}}=\\dfrac{2.27}{102}.100=52,94\\%$<br\/>$\\Rightarrow \\%{{m}_{O}}=\\dfrac{3.16}{102}.100=47,06\\%$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2107},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"M\u1ed9t h\u1ee3p ch\u1ea5t kh\u00ed $A$ c\u00f3 th\u00e0nh ph\u1ea7n $\\%$ theo kh\u1ed1i l\u01b0\u1ee3ng l\u00e0 $82,35\\%N$ v\u00e0 $17,65\\%H$, bi\u1ebft r\u1eb1ng ${{d}_{A\/{{H}_{2}}}}=8,5$. S\u1ed1 nguy\u00ean t\u1eed c\u1ee7a m\u1ed7i nguy\u00ean t\u1ed1 c\u00f3 trong $1,12$ l\u00edt kh\u00ed $A$ \u1edf \u0111ktc l\u00e0 ","select":["A. ${{5.10}^{22}}$ nguy\u00ean t\u1eed $N$ v\u00e0 ${{11.10}^{22}}$ nguy\u00ean t\u1eed $H$ ","B. ${{4.10}^{22}}$ nguy\u00ean t\u1eed $N$ v\u00e0 ${{8.10}^{22}}$ nguy\u00ean t\u1eed $H$ ","C. ${{2.10}^{22}}$ nguy\u00ean t\u1eed $N$ v\u00e0 ${{7.10}^{22}}$ nguy\u00ean t\u1eed $H$ ","D. ${{3.10}^{22}}$ nguy\u00ean t\u1eed $N$ v\u00e0 ${{9.10}^{22}}$ nguy\u00ean t\u1eed $H$ "],"hint":"","explain":"<span class='basic_left'>Kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a $A$ l\u00e0 ${{M}_{A}}=2.{{d}_{A\/{{H}_{2}}}}=2.8,5=17$<br\/>Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a m\u1ed7i nguy\u00ean t\u1ed1 c\u00f3 trong 1 mol h\u1ee3p ch\u1ea5t kh\u00ed:<br\/>${{m}_{N}}=\\dfrac{82,35.17}{100}=14$ (g)<br\/>${{m}_{H}}=\\dfrac{17,65.17}{100}=3$ (g)<br\/>S\u1ed1 mol nguy\u00ean t\u1eed c\u1ee7a m\u1ed7i nguy\u00ean t\u1ed1 c\u00f3 trong 1 mol h\u1ee3p ch\u1ea5t kh\u00ed :<br\/>${{n}_{N}}=\\dfrac{14}{14}=1(mol)$<br\/>${{n}_{H}}=\\dfrac{3}{1}=3(mol)$<br\/>V\u1eady c\u00f4ng th\u1ee9c h\u00f3a h\u1ecdc c\u1ee7a h\u1ee3p ch\u1ea5t kh\u00ed l\u00e0 $N{{H}_{3}}$<br\/>Ta c\u00f3: ${{n}_{A}}=\\dfrac{V}{22,4}=\\dfrac{1,12}{22,4}=0,05(mol)$<br\/>C\u1ee9 1 mol $N{{H}_{3}}$ c\u00f3 1 mol $N$<br\/>V\u1eady 0,05 mol $N{{H}_{3}}$ c\u00f3 n mol $N$<br\/>$\\Rightarrow$ $n=0,05.1=0,05$ mol<br\/>$\\Rightarrow$ S\u1ed1 nguy\u00ean t\u1eed $N$ trong 0,05 mol $N{{H}_{3}}$ l\u00e0 ${{0,05.6.10}^{23}}={{3.10}^{22}}$<br\/>T\u01b0\u01a1ng t\u1ef1 v\u1edbi nguy\u00ean t\u1eed $H$<br\/>$\\Rightarrow $ s\u1ed1 mol nguy\u00ean t\u1eed $H$ trong 0,05 mol $N{{H}_{3}}$ l\u00e0 0,15 mol<br\/>$\\Rightarrow$ S\u1ed1 nguy\u00ean t\u1eed $H$ trong 0,05 mol $N{{H}_{3}}$ l\u00e0 ${{0,15.6.10}^{23}}={{9.10}^{22}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":2108},{"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 $30,6$ g $A{{l}_{2}}{{O}_{3}}$. Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a m\u1ed7i nguy\u00ean t\u1ed1 c\u00f3 trong $30,6$ g $A{{l}_{2}}{{O}_{3}}$ l\u00e0 ","select":["A. ${{m}_{Al}}=16,2$ gam v\u00e0 ${{m}_{O}}=14,4$ gam","B. ${{m}_{Al}}=19,6$ gam v\u00e0 ${{m}_{O}}=13,5$ gam ","C. ${{m}_{Al}}=17,2$ gam v\u00e0 ${{m}_{O}}=16,4$ gam ","D. ${{m}_{Al}}=19,67$ gam v\u00e0 ${{m}_{O}}=15,84$ gam "],"hint":"","explain":"<span class='basic_left'>$30,6$ g $A{{l}_{2}}{{O}_{3}}$ c\u00f3 s\u1ed1 mol ph\u00e2n t\u1eed l\u00e0 <br\/>$\\dfrac{30,6}{27.2+16.3}=0,3(mol)$<br\/>C\u1ee9 1 mol $A{{l}_{2}}{{O}_{3}}$ c\u00f3 2 mol $Al$<br\/>V\u1eady 0,3 mol $A{{l}_{2}}{{O}_{3}}$ c\u00f3 n mol $Al$<br\/>$\\Rightarrow$ $n=0,3.2=0,6$ mol<br\/>$\\Rightarrow$ Kh\u1ed1i l\u01b0\u1ee3ng nguy\u00ean t\u1eed $Al$ l\u00e0 $m=n.M=0,6.27=16,2$ g<br\/>T\u01b0\u01a1ng t\u1ef1 v\u1edbi nguy\u00ean t\u1ed1 $O$ ta c\u00f3 s\u1ed1 mol oxi l\u00e0 $0,9$ mol<br\/>$\\Rightarrow$ Kh\u1ed1i l\u01b0\u1ee3ng nguy\u00ean t\u1eed $O$ l\u00e0 $m=n.M=0,9.16=14,4$ g<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2109}],"lesson":{"save":0,"level":2}}