{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Ph\u00e1t bi\u1ec3u n\u00e0o sau \u0111\u00e2y v\u1ec1 oxi l\u00e0 kh\u00f4ng \u0111\u00fang? ","select":["A. Oxi t\u1ea1o oxit baz\u01a1 v\u1edbi t\u1ea5t c\u1ea3 c\u00e1c kim lo\u1ea1i ","B. Oxi l\u00e0 phi kim ho\u1ea1t \u0111\u1ed9ng h\u00f3a h\u1ecdc r\u1ea5t m\u1ea1nh, nh\u1ea5t l\u00e0 \u1edf nhi\u1ec7t \u0111\u1ed9 cao ","C. Oxi l\u00e0 kh\u00ed kh\u00f4ng m\u00e0u, kh\u00f4ng m\u00f9i ","D. Oxi c\u1ea7n thi\u1ebft cho s\u1ef1 s\u1ed1ng "],"hint":"","explain":"<span class='basic_left'>Oxi t\u1ea1o oxit baz\u01a1 v\u1edbi h\u1ea7u h\u1ebft c\u00e1c kim lo\u1ea1i, ch\u1ee9 kh\u00f4ng ph\u1ea3i t\u1ea5t c\u1ea3 c\u1ea3 c\u00e1c kim lo\u1ea1i v\u00ec m\u1ed9t s\u1ed1 kim lo\u1ea1i k\u1ebft h\u1ee3p v\u1edbi oxi t\u1ea1o oxit l\u01b0\u1ee1ng t\u00ednh v\u00ed d\u1ee5 nh\u01b0 : $A{{l}_{2}}{{O}_{3}},ZnO$,oxit axit nh\u01b0 $Cr{{O}_{3}}$<br\/>Oxi l\u00e0 m\u1ed9t \u0111\u01a1n ch\u1ea5t phi kim r\u1ea5t ho\u1eb7t \u0111\u1ed9ng, \u0111\u1eb7c bi\u1ec7t \u1edf nhi\u1ec7t \u0111\u1ed9 cao, d\u1ec5 tham gia ph\u1ea3n \u1ee9ng h\u00f3a h\u1ecdc v\u1edbi nhi\u1ec1u phi kim, nhi\u1ec1u kim lo\u1ea1i v\u00e0 h\u1ee3p ch\u1ea5t<br\/>Oxi l\u00e0 ch\u1ea5t kh\u00ed, kh\u00f4ng m\u00e0u, kh\u00f4ng m\u00f9i, \u00edt tan trong n\u01b0\u1edbc, n\u1eb7ng h\u01a1n kh\u00f4ng kh\u00ed<br\/>Oxi c\u1ea7n thi\u1ebft cho s\u1ef1 h\u00f4 h\u1ea5p v\u00e0 s\u1ef1 \u0111\u1ed1t nhi\u00ean li\u1ec7u<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":1}]}],"id_ques":2220},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Trong ph\u00f2ng th\u00ed nghi\u1ec7m ng\u01b0\u1eddi ta \u0111i\u1ec1u ch\u1ebf oxi b\u1eb1ng c\u00e1ch nhi\u1ec7t ph\u00e2n $KCl{{O}_{3}}$ hay $KMn{{O}_{4}}$ ho\u1eb7c $KN{{O}_{3}}$. V\u00ec l\u00fd do n\u00e0o sau \u0111\u00e2y? ","select":["A. D\u1ec5 ki\u1ebfm v\u00e0 r\u1ebb ti\u1ec1n ","B. Gi\u00e0u oxi v\u00e0 d\u1ec5 b\u1ecb nhi\u1ec7t ph\u00e2n h\u1ee7y ra oxi ","C. Ph\u00f9 h\u1ee3p v\u1edbi thi\u1ebft b\u1ecb hi\u1ec7n \u0111\u1ea1i ","D. Kh\u00f4ng \u0111\u1ed9c h\u1ea1i "],"hint":"","explain":"<span class='basic_left'>Trong ph\u00f2ng th\u00ed nghi\u1ec7m, kh\u00ed oxi \u0111\u01b0\u1ee3c \u0111i\u1ec1u ch\u1ebf b\u1eb1ng c\u00e1ch \u0111un n\u00f3ng nh\u1eefng h\u1ee3p ch\u1ea5t gi\u00e0u oxi v\u00e0 d\u1ec5 b\u1ecb ph\u00e2n h\u1ee7y \u1edf nhi\u1ec7t \u0111\u1ed9 cao nh\u01b0 $KCl{{O}_{3}}$ hay $KMn{{O}_{4}}$ ho\u1eb7c $KN{{O}_{3}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2221},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Ng\u01b0\u1eddi ta thu kh\u00ed oxi b\u1eb1ng c\u00e1ch \u0111\u1ea9y kh\u00f4ng kh\u00ed l\u00e0 nh\u1edd d\u1ef1a v\u00e0o t\u00ednh ch\u1ea5t :","select":["A. Kh\u00ed oxi nh\u1eb9 h\u01a1n kh\u00f4ng kh\u00ed ","B. Kh\u00ed oxi d\u1ec5 tr\u1ed9n l\u1eabn v\u1edbi kh\u00f4ng kh\u00ed ","C. Kh\u00ed oxi \u00edt tan trong n\u01b0\u1edbc ","D. Kh\u00ed oxi n\u1eb7ng h\u01a1n kh\u00f4ng kh\u00ed "],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3 : ${{d}_{{{O}_{2}}\/kk}}=\\dfrac{32}{29}=1,103>1$<br\/>V\u1eady oxi n\u1eb7ng h\u01a1n kh\u00f4ng kh\u00ed ta c\u00f3 th\u1ec3 d\u00f9ng ph\u01b0\u01a1ng ph\u00e1p \u0111\u1ea9y kh\u00f4ng kh\u00ed \u0111\u1ec3 thu oxi<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2222},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"S\u1ef1 ch\u00e1y l\u00e0 ","select":["A. S\u1ef1 oxi h\u00f3a t\u1ecfa nhi\u1ec7t m\u00e0 kh\u00f4ng ph\u00e1t s\u00e1ng ","B. S\u1ef1 t\u1ef1 b\u1ed1c ch\u00e1y ","C. S\u1ef1 oxi h\u00f3a c\u00f3 t\u1ecfa nhi\u1ec7t v\u00e0 ph\u00e1t s\u00e1ng ","D. S\u1ef1 oxi h\u00f3a kh\u00f4ng t\u1ecfa nhi\u1ec7t "],"hint":"","explain":"<span class='basic_left'>S\u1ef1 ch\u00e1y l\u00e0 s\u1ef1 oxi h\u00f3a c\u00f3 t\u1ecfa nhi\u1ec7t v\u00e0 ph\u00e1t s\u00e1ng<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2223},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"H\u00e3y cho bi\u1ebft ${{8,02.10}^{24}}$ ph\u00e2n t\u1eed oxi c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng bao nhi\u00eau gam? ","select":["A. $430,7(g)$ ","B. $440,8(g)$ ","C. $454,9(g)$ ","D. $427,7(g)$ "],"hint":"","explain":"<span class='basic_left'>C\u1ee9 1 mol ph\u00e2n t\u1eed ${{O}_{2}}$ c\u00f3 ${{6.10}^{23}}$ ph\u00e2n t\u1eed ${{O}_{2}}$<br\/>v\u1eady c\u00f3 n mol ph\u00e2n t\u1eed ${{O}_{2}}$ c\u00f3 ${{8,02.10}^{24}}$ ph\u00e2n t\u1eed ${{O}_{2}}$<br\/>$\\to n=\\dfrac{{{1.8,02.10}^{24}}}{{{6.10}^{23}}}=13,37(mol)$<br\/>$\\to {{m}_{{{O}_{2}}}}=13,37.32=427,7(g)$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2224},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"oxit kim lo\u1ea1i n\u00e0o d\u01b0\u1edbi \u0111\u00e2y l\u00e0 oxit axit? ","select":["A. $Cr{{O}_{3}}$ ","B. $C{{r}_{2}}{{O}_{3}}$ ","C. $N{{a}_{2}}O$ ","D. $CuO$ "],"hint":"","explain":"<span class='basic_left'>Nh\u1eefng kim lo\u1ea1i h\u00f3a tr\u1ecb I, II, III k\u1ebft h\u1ee3p v\u1edbi oxi \u0111\u1ec1u l\u00e0 oxit baz\u01a1<br\/>Nh\u1eefng kim lo\u1ea1i h\u00f3a tr\u1ecb V, VI, VII k\u1ebft h\u1ee3p v\u1edbi oxi \u0111\u1ec1u l\u00e0 oxit axit<br\/>Nh\u1eefng kim lo\u1ea1i h\u00f3a tr\u1ecb IVk\u1ebft h\u1ee3p v\u1edbi oxi v\u1eeba l\u00e0 oxit baz\u01a1 v\u1eeba l\u00e0 oxit axit<br\/>Theo quy t\u1eafc h\u00f3a tr\u1ecb :<br\/>Cr trong $Cr{{O}_{3}}$ c\u00f3 h\u00f3a tr\u1ecb VI<br\/>Cr trong $C{{r}_{2}}{{O}_{3}}$ c\u00f3 h\u00f3a tr\u1ecb III<br\/>Na trong $N{{a}_{2}}O$ c\u00f3 h\u00f3a tr\u1ecb I<br\/>Cu trong $CuO$ c\u00f3 h\u00f3a tr\u1ecb II<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2225},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Oxit n\u00e0o d\u01b0\u1edbi \u0111\u00e2y c\u00f3 ph\u1ea7n tr\u0103m kh\u1ed1i l\u01b0\u1ee3ng oxi cao nh\u1ea5t ? ","select":["A. $CuO$ ","B. $MgO$ ","C. $A{{l}_{2}}{{O}_{3}}$ ","D. $F{{e}_{3}}{{O}_{4}}$ "],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3 : <br\/>$\\begin{aligned} & \\%{{O}_{A{{l}_{2}}{{O}_{3}}}}=\\dfrac{16.3}{27.2+16.3}.100=47,06\\% \\\\ & \\%{{O}_{CuO}}=\\dfrac{16}{64+16}.100=20\\% \\\\ & \\%{{O}_{MgO}}=\\dfrac{16}{24+16}.100=40\\% \\\\ & \\%{{O}_{F{{e}_{3}}{{O}_{4}}}}=\\dfrac{16.4}{56.3+16.4}.100=27,58\\% \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2226},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Thi\u1ebfc c\u00f3 th\u1ec3 c\u00f3 h\u00f3a tr\u1ecb II ho\u1eb7c IV. H\u1ee3p ch\u1ea5t c\u00f3 c\u00f4ng th\u1ee9c $Sn{{O}_{2}}$ c\u00f3 t\u00ean l\u00e0:","select":["A. Thi\u1ebfc(IV) oxit","B. Thi\u1ebfc(II) oxit ","C. Thi\u1ebfc oxit ","D. Thi\u1ebfc pentaoxit "],"hint":"","explain":"<span class='basic_left'>Theo quy t\u1eafc h\u00f3a tr\u1ecb : thi\u1ebfc trong $Sn{{O}_{2}}$ c\u00f3 h\u00f3a tr\u1ecb IV <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2227},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Trong ph\u00f2ng th\u00ed nghi\u1ec7m, Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a mu\u1ed1i n\u00e0o sau \u0111\u00e2y l\u00e0 nh\u1ecf nh\u1ea5t d\u00f9ng \u0111\u1ec3 \u0111i\u1ec1u ch\u1ebf $4,48$ l\u00edt kh\u00ed oxi \u1edf \u0111ktc ?","select":["A. ${{H}_{2}}O$ ( \u0111i\u1ec7n ph\u00e2n n\u01b0\u1edbc )","B. $KN{{O}_{3}}$ ","C. $KMn{{O}_{4}}$ ","D. $KCl{{O}_{3}}$ "],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec7n ph\u00e2n n\u01b0\u1edbc l\u00e0 \u0111i\u1ec1u ch\u1ebf oxi trong c\u00f4ng nghi\u1ec7p<br\/>${{n}_{{{O}_{2}}}}=\\dfrac{4,48}{22,4}=0,2(mol)$<br\/>$\\begin{aligned} & 2KMn{{O}_{4}}\\xrightarrow{{{t}^{o}}}{{K}_{2}}Mn{{O}_{4}}+Mn{{O}_{2}}+{{O}_{2}} \\\\ & \\begin{matrix} 0,4mol & {} & {} & {} & \\leftarrow & {} & {} & {} & 0,2mol & {} \\\\\\end{matrix} \\\\ & \\to {{m}_{KMn{{O}_{4}}}}=0,4.158=63,2(g) \\\\ \\end{aligned}$<br\/>$\\begin{aligned} & 2KCl{{O}_{3}}\\xrightarrow{{{t}^{o}}}2KCl+3{{O}_{2}} \\\\ & \\begin{matrix} \\dfrac{2}{3}.0,2mol & \\leftarrow & {} & {} & 0,2mol & {} & {} \\\\\\end{matrix} \\\\ & {{m}_{KCl{{O}_{3}}}}=\\dfrac{2}{3}.0,2.122,5=16,33(g) \\\\ \\end{aligned}$ <br\/>$\\begin{aligned} & 2KN{{O}_{3}}\\xrightarrow{{{t}^{o}}}2KN{{O}_{2}}+{{O}_{2}} \\\\ & \\begin{matrix} 0,4mol & \\leftarrow & {} & {} & {} & 0,2mol & {} \\\\\\end{matrix} \\\\ & \\to {{m}_{KN{{O}_{3}}}}=0,4.101=40,4(g) \\\\ \\end{aligned}$ <br\/>V\u1eady kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a $KCl{{O}_{3}}$ l\u00e0 nh\u1ecf nh\u1ea5t<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2228},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Ch\u1ecdn \u0111\u1ecbnh ngh\u0129a ph\u1ea3n \u1ee9ng ph\u00e2n h\u1ee7y \u0111\u1ea7y \u0111\u1ee7 nh\u1ea5t : ","select":["A. Ph\u1ea3n \u1ee9ng ph\u00e2n h\u1ee7y l\u00e0 ph\u1ea3n \u1ee9ng h\u00f3a h\u1ecdc trong \u0111\u00f3 m\u1ed9t ch\u1ea5t sinh ra m\u1ed9t ch\u1ea5t m\u1edbi ","B. Ph\u1ea3n \u1ee9ng ph\u00e2n h\u1ee7y l\u00e0 ph\u1ea3n \u1ee9ng h\u00f3a h\u1ecdc trong \u0111\u00f3 m\u1ed9t ch\u1ea5t sinh ra hai hay nhi\u1ec1u ch\u1ea5t m\u1edbi ","C. Ph\u1ea3n \u1ee9ng ph\u00e2n h\u1ee7y l\u00e0 ph\u1ea3n \u1ee9ng h\u00f3a h\u1ecdc c\u00f3 ch\u1ea5t kh\u00ed tho\u00e1t ra ","D. Ph\u1ea3n \u1ee9ng ph\u00e2n h\u1ee7y l\u00e0 ph\u1ea3n \u1ee9ng h\u00f3a h\u1ecdc trong \u0111\u00f3 m\u1ed9t ch\u1ea5t sinh ra hai ch\u1ea5t m\u1edbi "],"hint":"","explain":"<span class='basic_left'>Ph\u1ea3n \u1ee9ng ph\u00e2n h\u1ee7y l\u00e0 ph\u1ea3n \u1ee9ng h\u00f3a h\u1ecdc trong \u0111\u00f3 m\u1ed9t ch\u1ea5t sinh ra hai hay nhi\u1ec1u ch\u1ea5t m\u1edbi<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":2229},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Ph\u1ea3n \u1ee9ng n\u00e0o sau \u0111\u00e2u kh\u00f4ng ph\u1ea3i l\u00e0 ph\u1ea3n \u1ee9ng h\u00f3a h\u1ee3p ? ","select":["A. ${{N}_{2}}+3{{H}_{2}}\\xrightarrow{xt,{{t}^{o}},p}2N{{H}_{3}}$ ","B. $C{{O}_{2}}+Ca{{(OH)}_{2}}\\to CaC{{O}_{3}}+{{H}_{2}}O$ ","C. ${{P}_{2}}{{O}_{5}}+3{{H}_{2}}O\\to 2{{H}_{3}}P{{O}_{4}}$ ","D. $S+{{O}_{2}}\\to S{{O}_{2}}$ "],"hint":"","explain":"<span class='basic_left'>Ph\u1ea3n \u1ee9ng h\u00f3a h\u1ee3p l\u00e0 ph\u1ea3n \u1ee9ng h\u00f3a h\u1ecdc trong \u0111\u00f3 ch\u1ec9 c\u00f3 m\u1ed9t ch\u1ea5t m\u1edbi \u0111\u01b0\u1ee3c t\u1ea1o th\u00e0nh t\u1eeb hai hay nhi\u1ec1u ch\u1ea5t ban \u0111\u1ea7u<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2230},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"C\u00e2u n\u00e0o \u0111\u00fang khi n\u00f3i v\u1ec1 kh\u00f4ng kh\u00ed trong c\u00e1c c\u00e2u sau \u0111\u00e2y ? ","select":["A. Kh\u00f4ng kh\u00ed l\u00e0 m\u1ed9t h\u1ee3p ch\u1ea5t ","B. Kh\u00f4ng kh\u00ed l\u00e0 m\u1ed9t nguy\u00ean t\u1ed1 h\u00f3a h\u1ecdc ","C. Kh\u00f4ng kh\u00ed l\u00e0 h\u1ed7n h\u1ee3p kh\u00ed trong \u0111\u00f3 c\u00f3 $78\\%$ ${{N}_{2}}$, $21\\%$ ${{O}_{2}}$, $1\\%$ kh\u00ed kh\u00e1c v\u1ec1 th\u1ec3 t\u00edch","D. Kh\u00f4ng kh\u00ed l\u00e0 m\u1ed9t h\u1ed7n h\u1ee3p ch\u1ea5t c\u1ee7a 2 nguy\u00ean t\u1ed1 l\u00e0 oxi v\u00e0 nit\u01a1 "],"hint":"","explain":"<span class='basic_left'>Kh\u00f4ng kh\u00ed l\u00e0 m\u1ed9t h\u1ed7n h\u1ee3p kh\u00ed trong \u0111\u00f3 c\u00f3 $78\\%$ ${{N}_{2}}$, $21\\%$ ${{O}_{2}}$, $1\\%$ kh\u00ed kh\u00e1c v\u1ec1 th\u1ec3 t\u00edch<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2231},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"N\u1ebfu \u0111\u1ed1t ch\u00e1y ho\u00e0n to\u00e0n $2,4$ gam cacbon trong $4,8$ gam oxi th\u00ec thu \u0111\u01b0\u1ee3c t\u1ed1i \u0111a bao nhi\u00eau gam kh\u00ed $C{{O}_{2}}$ ? ","select":["A. $6,6(g)$ ","B. $6,8(g)$ ","C. $6,4(g)$ ","D. $6,3(g)$ "],"hint":"","explain":"<span class='basic_left'>PTHH : $C+{{O}_{2}}\\to C{{O}_{2}}$<br\/>$\\begin{aligned} & {{n}_{C}}=\\dfrac{2,4}{12}=0,2(mol) \\\\ & {{n}_{{{O}_{2}}}}=\\dfrac{4,8}{32}=0,15(mol) \\\\ \\end{aligned}$<br\/>Trong ph\u1ea3n \u1ee9ng n\u00e0y l\u01b0\u1ee3ng oxi tham gia ph\u1ea3n \u1ee9ng \u00edt h\u01a1n cacbon n\u00ean \u0111\u1ec3 thu \u0111\u01b0\u1ee3c l\u01b0\u1ee3ng $C{{O}_{2}}$ t\u1ed1i \u0111a th\u00ec oxi ph\u1ea3i ph\u1ea3n \u1ee9ng h\u1ebft v\u1edbi cacbon<br\/>${{n}_{C{{O}_{2}}}}={{n}_{{{O}_{2}}}}=0,15(mol)\\to {{m}_{C{{O}_{2}}}}=0,15.44=6,6(g)$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2232},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"M\u1ed9t lo\u1ea1i \u0111\u1ed3ng oxit c\u00f3 t\u1ef7 l\u1ec7 kh\u1ed1i l\u01b0\u1ee3ng gi\u1eefa $Cu$ v\u00e0 $O$ l\u00e0 $\\dfrac{{{m}_{Cu}}}{{{m}_{O}}}=8$.C\u00f4ng th\u1ee9c h\u00f3a h\u1ecdc c\u1ee7a oxit n\u00e0y l\u00e0 ","select":["A. $CuO$ ","B. $Cu{{O}_{2}}$ ","C. $C{{u}_{2}}{{O}_{3}}$","D. $C{{u}_{2}}O$ "],"hint":"","explain":"<span class='basic_left'>G\u1ecdi c\u00f4ng th\u1ee9c oxit c\u00f3 d\u1ea1ng : $C{{u}_{x}}{{O}_{y}}$<br\/>Theo \u0111\u1ea7u b\u00e0i ta c\u00f3 :<br\/>$\\dfrac{{{m}_{Cu}}}{{{m}_{O}}}=8=\\dfrac{64x}{16y}\\to \\dfrac{x}{y}=2\\to \\left\\{ \\begin{align} & x=2 \\\\ & y=1 \\\\ \\end{align} \\right.$<br\/>V\u1eady oxit c\u00f3 c\u00f4ng th\u1ee9c l\u00e0 $C{{u}_{2}}O$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2233},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"\u0110\u1ed1t ch\u00e1y ho\u00e0n to\u00e0n $2,4$ gam magie $Mg$ trong kh\u00ed oxi thu \u0111\u01b0\u1ee3c magie oxit $MgO$. S\u1ed1 gam $KCl{{O}_{3}}$ c\u1ea7n d\u00f9ng \u0111\u1ec3 \u0111i\u1ec1u ch\u1ebf l\u01b0\u1ee3ng oxi tr\u00ean l\u00e0 ","select":["A. $5,083(g)$ ","B. $3,083(g)$ ","C. $2,083(g)$ ","D. $4,083(g)$ "],"hint":"","explain":"<span class='basic_left'>${{n}_{Mg}}=\\dfrac{2,4}{24}=0,1(mol)$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ed1t ch\u00e1y l\u00e0<br\/> $\\begin{aligned} & 2Mg+{{O}_{2}}\\to 2MgO \\\\ & \\begin{matrix} 0,1\\to 0,05 & {} & {} & {} & mol & {} & {} \\\\\\end{matrix} \\\\ & 2KCl{{O}_{3}}\\xrightarrow{{{t}^{o}}}KCl+3{{O}_{2}} \\\\ & \\begin{matrix} \\dfrac{2}{3}.0,05 & \\leftarrow & {} & {} & 0,05 & {} & mol \\\\\\end{matrix} \\\\ & \\to {{m}_{KCl{{O}_{3}}}}=\\dfrac{2}{3}.0,05.122,5=4,083(g) \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2234},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Nhi\u1ec7t ph\u00e2n $15,8$ gam $KMn{{O}_{4}}$ thu \u0111\u01b0\u1ee3c l\u01b0\u1ee3ng kh\u00ed oxi ${{O}_{2}}$. \u0110\u1ed1t ch\u00e1y $5,6$ gam $Fe$ trong l\u01b0\u1ee3ng ${{O}_{2}}$ v\u1eeba thu \u0111\u01b0\u1ee3c th\u00ec c\u00e1c ch\u1ea5t sau ph\u1ea3n \u1ee9ng l\u00e0 ","select":["A. $F{{e}_{3}}{{O}_{4}}$ ","B. ${{O}_{2}}$ d\u01b0 v\u00e0 $F{{e}_{3}}{{O}_{4}}$ ","C. $Fe$ d\u01b0 v\u00e0 $F{{e}_{3}}{{O}_{4}}$ ","D. $Fe$ d\u01b0 "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{KMn{{O}_{4}}}}=\\dfrac{15,8}{158}=0,1(mol) \\\\ & {{n}_{Fe}}=\\dfrac{5,6}{56}=0,1(mol) \\\\ & 2KMn{{O}_{4}}\\xrightarrow{{{t}^{o}}}{{K}_{2}}Mn{{O}_{4}}+Mn{{O}_{2}}+{{O}_{2}} \\\\ & \\begin{matrix} 0,1mol & {} & {} & {} & \\to & {} & {} & {} & 0,05mol & {} \\\\\\end{matrix} \\\\ & 3Fe+2{{O}_{2}}\\to F{{e}_{3}}{{O}_{4}} \\\\ & \\begin{matrix} 3mol & 2mol & {} & {} & {} & {} & {} & {} & {} & {} \\\\\\end{matrix} \\\\ \\end{aligned}$<br\/>L\u1eadp t\u1ef7 s\u1ed1 : $\\dfrac{0,1}{3}>\\dfrac{0,05}{2}$<br\/>V\u1eady Fe sau ph\u1ea3n \u1ee9ng c\u00f2n d\u01b0<br\/>$\\to$ sau ph\u1ea3n \u1ee9ng c\u00f3 Fe d\u01b0 v\u00e0 $F{{e}_{3}}{{O}_{4}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2235},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"\u0110\u1ed1t ch\u00e1y $12,4$ gam photpho trong b\u00ecnh ch\u1ee9a kh\u00ed oxi t\u1ea1o th\u00e0nh \u0111iphotpho pentaoxit $({{P}_{2}}{{O}_{5}})$. Kh\u1ed1i l\u01b0\u1ee3ng ch\u1ea5t t\u1ea1o th\u00e0nh l\u00e0 ","select":["A. $29,4(g)$ ","B. $28,4(g)$ ","C. $30,4(g)$ ","D. $43,4(g)$ "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & {{n}_{P}}=\\dfrac{12,4}{31}=0,4(mol) \\\\ & 4P+5{{O}_{2}}\\to 2{{P}_{2}}{{O}_{5}} \\\\ & \\begin{matrix} 0,4 & \\to & {} & 0,2 & {} \\\\\\end{matrix}mol \\\\ & \\to {{m}_{{{P}_{2}}{{O}_{5}}}}=0,2.142=28,4(g) \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2236},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"\u0110\u1ec3 s\u1ea3n xu\u1ea5t v\u00f4i, trong l\u00f2 v\u00f4i ng\u01b0\u1eddi ta th\u01b0\u1eddng s\u1eafp x\u1ebfp m\u1ed9t l\u1edbp than, m\u1ed9t l\u1edbp \u0111\u00e1 v\u00f4i, sau \u0111\u00f3 \u0111\u1ed1t l\u00f2. C\u00f3 nh\u1eefng ph\u1ea3n \u1ee9ng h\u00f3a h\u1ecdc n\u00e0o x\u1ea3y ra trong l\u00f2 v\u00f4i? ","select":["A. $C+{{O}_{2}}\\xrightarrow{{{t}^{o}}}C{{O}_{2}}$ ","B. $CaC{{O}_{3}}\\xrightarrow{{{t}^{o}}}CaO+C{{O}_{2}}$ ","C. $S+{{O}_{2}}\\xrightarrow{{{t}^{o}}}S{{O}_{2}}$ ","D. C\u1ea3 $A$ v\u00e0 $B$ "],"hint":"","explain":"<span class='basic_left'>M\u1ed9t l\u1edbp than x\u1ea3y ra ph\u1ea3n \u1ee9ng: $C+{{O}_{2}}\\xrightarrow{{{t}^{o}}}C{{O}_{2}}$<br\/>M\u1ed9t l\u1edbp \u0111\u00e1 v\u00f4i x\u1ea3y ra ph\u1ea3n \u1ee9ng: $CaC{{O}_{3}}\\xrightarrow{{{t}^{o}}}CaO+C{{O}_{2}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2237},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Trong qu\u00e1 tr\u00ecnh quang h\u1ee3p, c\u00e2y c\u1ed1i tr\u00ean m\u1ed7i hecta \u0111\u1ea5t trong m\u1ed9t ng\u00e0y h\u1ea5p th\u1ee5 kho\u1ea3ng $100$ kg kh\u00ed cacbonic v\u00e0 sau khi \u0111\u1ed3ng h\u00f3a c\u00e2y c\u1ed1i nh\u1ea3 ra kh\u00ed oxi. Kh\u1ed1i l\u01b0\u1ee3ng kh\u00ed oxi do c\u00e2y c\u1ed1i tr\u00ean 5 hecta \u0111\u1ea5t tr\u1ed3ng sinh ra m\u1ed7i ng\u00e0y. Bi\u1ebft r\u1eb1ng s\u1ed1 mol kh\u00ed c\u00e2y sinh ra b\u1eb1ng s\u1ed1 mol kh\u00ed cacbonic h\u1ea5p th\u1ee5. ","select":["A. $563636(g)$ ","B. $483636(g)$ ","C. $363636(g)$ ","D. $367895(g)$ "],"hint":"","explain":"<span class='basic_left'>\u0110\u1ed5i 100 kg = 100000 gam<br\/>S\u1ed1 mol kh\u00ed oxi sinh ra tr\u00ean m\u1ed7i hecta trong m\u1ed9t ng\u00e0y : ${{n}_{C{{O}_{2}}}}=\\dfrac{m}{M}=\\dfrac{100000}{44}={{n}_{{{O}_{2}}}}$<br\/>Kh\u1ed1i l\u01b0\u1ee3ng kh\u00ed oxi sinh ra tr\u00ean 5 hecta trong 1 ng\u00e0y l\u00e0 ${{m}_{{{O}_{2}}}}=\\dfrac{100000}{44}.5.32=363636(g)$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2238},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Cho ph\u1ea3n \u1ee9ng sau : $2KMn{{O}_{4}}\\xrightarrow{{{t}^{o}}}{{K}_{2}}Mn{{O}_{4}}+Mn{{O}_{2}}+{{O}_{2}}$. Ph\u1ea3n \u1ee9ng tr\u00ean thu\u1ed9c lo\u1ea1i ph\u1ea3n \u1ee9ng n\u00e0o? ","select":["A. Ph\u1ea3n \u1ee9ng ph\u00e2n h\u1ee7y ","B. Ph\u1ea3n \u1ee9ng h\u00f3a h\u1ee3p ","C. Ph\u1ea3n \u1ee9ng th\u1ebf ","D. Ph\u1ea3n \u1ee9ng trao \u0111\u1ed5i "],"hint":"","explain":"<span class='basic_left'>Ph\u1ea3n \u1ee9ng ph\u00e2n h\u1ee7y l\u00e0 ph\u1ea3n \u1ee9ng h\u00f3a h\u1ecdc trong \u0111\u00f3 m\u1ed9t ch\u1ea5t sinh ra hai hay nhi\u1ec1u ch\u1ea5t m\u1edbi<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2239}],"lesson":{"save":0,"level":2}}