{"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":"Gi\u1ea3 s\u1eed c\u00f3 ph\u1ea3n \u1ee9ng gi\u1eefa A + B t\u1ea1o ra C + D. Bi\u1ec3u th\u1ee9c n\u00e0o sau \u0111\u00e2y th\u1ec3 hi\u1ec7n \u0111\u00fang \u0111\u1ecbnh lu\u1eadt b\u1ea3o to\u00e0n kh\u1ed1i l\u01b0\u1ee3ng ","select":["A. ${{m}_{A}}+{{m}_{C}}={{m}_{A}}+{{m}_{D}}$ ","B. ${{m}_{A}}+{{m}_{B}}={{m}_{C}}+{{m}_{D}}$ ","C. ${{m}_{D}}+{{m}_{B}}={{m}_{C}}+{{m}_{A}}$ ","D. ${{m}_{A}}+{{m}_{D}}={{m}_{C}}+{{m}_{B}}$ "],"hint":"","explain":"<span class='basic_left'>\u0110\u1ecbnh lu\u1eadt: Trong m\u1ed9t ph\u1ea3n \u1ee9ng h\u00f3a h\u1ecdc, t\u1ed5ng kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a c\u00e1c s\u1ea3n ph\u1ea9m b\u1eb1ng t\u1ed5ng kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a c\u00e1c ch\u1ea5t tham gia ph\u1ea3n \u1ee9ng<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2040},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Trong ph\u1ea3n \u1ee9ng h\u00f3a h\u1ecdc : bari clorua + natri sunphat $\\to $ bari sunphat + natri clorua. Cho bi\u1ebft kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a natri sunphat $N{{a}_{2}}S{{O}_{4}}$ l\u00e0 14,2 gam, kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a bari sunphat $BaS{{O}_{4}}$ v\u00e0 natri clorua $NaCl$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 : 23,3 g v\u00e0 11,7 g. Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a Bari clorua $BaC{{l}_{2}}$ \u0111\u00e3 ph\u1ea3n \u1ee9ng l\u00e0 ","select":["A. $20,8$ g ","B. $25,8$ g ","C. $35$ g ","D. $21,8$ g "],"hint":"","explain":"<span class='basic_left'>\u00c1p d\u1ee5ng \u0111\u1ecbnh lu\u1eadt b\u1ea3o to\u00e0n kh\u1ed1i l\u01b0\u1ee3ng : <br\/>${{m}_{bari clorua}}+{{m}_{natri sunfat}}={{m}_{bari sunfat}}+{{m}_{natri clorua}}$<br\/>$\\to {{m}_{BaC{{l}_{2}}}}+{{m}_{N{{a}_{2}}S{{O}_{4}}}}={{m}_{BaS{{O}_{4}}}}+{{m}_{NaCl}}$<br\/>$\\to {{m}_{BaC{{l}_{2}}}}=23,3+11,7-14,2=20,8$ g<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2041},{"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 kim lo\u1ea1i magie $Mg$ trong kh\u00f4ng kh\u00ed thu \u0111\u01b0\u1ee3c 8g h\u1ee3p ch\u1ea5t magie oxit $(MgO)$. Bi\u1ebft l\u01b0\u1ee3ng kh\u1ed1i l\u01b0\u1ee3ng oxi trong kh\u00f4ng kh\u00ed tham gia ph\u1ea3n \u1ee9ng l\u00e0 3,2 g. Kh\u1ed1i l\u01b0\u1ee3ng $Mg$ tham gia ph\u1ea3n \u1ee9ng b\u1eb1ng bao nhi\u00eau l\u1ea7n l\u01b0\u1ee3ng oxi tham gia ph\u1ea3n \u1ee9ng. ","select":["A. 2 l\u1ea7n ","B. 2,5 l\u1ea7n ","C. 3 l\u1ea7n ","D. 1,5 l\u1ea7n "],"hint":"","explain":"<span class='basic_left'>\u00c1p d\u1ee5ng \u0111\u1ecbnh lu\u1eadt b\u1ea3o to\u00e0n kh\u1ed1i l\u01b0\u1ee3ng :<br\/>${{m}_{Mg}}+{{m}_{Oxi}}={{m}_{MgO}}$<br\/>$\\to {{m}_{Mg}} = 8 - 3,2 = 4,8$ g<br\/> Suy ra t\u1ef7 l\u1ec7 : $\\dfrac{{{m}_{Mg}}}{{{m}_{O}}}=\\dfrac{4,8}{3,2}=1,5$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2042},{"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\u00f2a tan canxi cacbua ($Ca{{C}_{2}}$) v\u00e0o n\u01b0\u1edbc (${{H}_{2}}O$) ta thu \u0111\u01b0\u1ee3c kh\u00ed axetylen (${{C}_{2}}{{H}_{2}}$) v\u00e0 canxi hi\u0111roxit ($Ca{{(OH)}_{2}}$). Ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc cho qu\u00e1 tr\u00ecnh tr\u00ean l\u00e0 ","select":["A. Canxi cacbua + n\u01b0\u1edbc $\\to $ axetylen + canxi hi\u0111roxit ","B. Canxi cacbua + axetylen $\\to $ n\u01b0\u1edbc + canxi hi\u0111roxit ","C. Axetylen + canxi hi\u0111roxit $\\to $ Canxi cacbua + n\u01b0\u1edbc","D. Canxi cacbua + canxi hi\u0111roxit $\\to $ axetylen + n\u01b0\u1edbc "],"hint":"","explain":"<span class='basic_left'><br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":1}]}],"id_ques":2043},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"Ph\u1ea3n \u1ee9ng \u0111i\u1ec1u ch\u1ebf axetylen l\u00e0 Canxi cacbua + n\u01b0\u1edbc $\\to $ axetylen + canxi hi\u0111roxit. N\u1ebfu d\u00f9ng 41 g $Ca{{C}_{2}}$ th\u00ec thu \u0111\u01b0\u1ee3c 17 g ${{C}_{2}}{{H}_{2}}$ v\u00e0 37 g $Ca{{(OH)}_{2}}$. S\u1ed1 mililit n\u01b0\u1edbc ph\u1ea3i d\u00f9ng l\u00e0( kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a n\u01b0\u1edbc l\u00e0 1 g\/ml ) ","select":["A. 12 ml ","B. 14 ml ","C. 13 ml ","D. 15 ml "],"hint":"","explain":"<span class='basic_left'>\u00c1p d\u1ee5ng \u0111\u1ecbnh lu\u1eadt b\u1ea3o to\u00e0n kh\u1ed1i l\u01b0\u1ee3ng : <br\/>$\\begin{aligned} & {{m}_{Ca{{C}_{2}}}}+{{m}_{{{H}_{2}}O}}={{m}_{{{C}_{2}}{{H}_{2}}}}+{{m}_{Ca{{(OH)}_{2}}}} \\\\ & \\to {{m}_{{{H}_{2}}O}}=17+37-41=13 \\\\ \\end{aligned}$ gam<br\/>V\u1eady th\u1ec3 t\u00edch n\u01b0\u1edbc c\u1ea7n d\u00f9ng l\u00e0 13 mililit<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2044},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"\u0110un n\u00f3ng m\u1ea1nh h\u1ed7n h\u1ee3p g\u1ed3m 30 gam b\u1ed9t s\u1eaft v\u00e0 22 gam b\u1ed9t l\u01b0u hu\u1ef3nh thu \u0111\u01b0\u1ee3c 46 gam ch\u1ea5t s\u1eaft($II$) sunfua ($FeS$) m\u00e0u x\u00e1m. \u0110\u1ec3 ph\u1ea3n \u1ee9ng h\u00f3a h\u1ee3p x\u1ea3y ra h\u1ebft, ng\u01b0\u1eddi ta l\u1ea5y d\u01b0 l\u01b0u hu\u1ef3nh. Kh\u1ed1i l\u01b0\u1ee3ng d\u01b0 c\u1ee7a l\u01b0u hu\u1ef3nh l\u00e0","select":["A. 4 gam ","B. 5 gam ","C. 3 gam ","D. 6 gam "],"hint":"","explain":"<span class='basic_left'>\u00c2p d\u1ee5ng \u0111\u1ecbnh lu\u1eadt b\u1ea3o to\u00e0n kh\u1ed1i l\u01b0\u1ee3ng : <br\/>${{m}_{Fe}}+{{m}_{S}}={{m}_{FeS}}$<br\/>$\\to {{m}_{S}} = 46 - 30 = 16$ gam<br\/>V\u1eady l\u01b0\u1ee3ng l\u01b0u hu\u1ef3nh c\u00f2n d\u01b0 l\u00e0 $\\to {{m}_{S}} = 22 - 16 = 6$ gam<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2045},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"\u0110\u1ed1t ch\u00e1y 20 gam h\u1ed7n h\u1ee3p $C{{H}_{4}},{{C}_{2}}{{H}_{2}}$, thu \u0111\u01b0\u1ee3c 32 gam kh\u00ed cacbon \u0111ioxit v\u00e0 28 gam n\u01b0\u1edbc. T\u1ef7 l\u1ec7 gi\u1eefa kh\u1ed1i l\u01b0\u1ee3ng kh\u00ed cacbon \u0111ioxit v\u1edbi kh\u1ed1i l\u01b0\u1ee3ng Oxi tham gia ph\u1ea3n \u1ee9ng l\u00e0 ","select":["A. $\\dfrac{4}{5}$ ","B. $\\dfrac{3}{5}$ ","C. $\\dfrac{4}{3}$ ","D. $\\dfrac{2}{3}$ "],"hint":"","explain":"<span class='basic_left'>\u00c1p d\u1ee5ng \u0111\u1ecbnh lu\u1eadt b\u1ea3o to\u00e0n kh\u1ed1i l\u01b0\u1ee3ng:<br\/>$\\begin{aligned} & {{m}_{hh}}+{{m}_{Oxi}}={{m}_{C{{O}_{2}}}}+{{m}_{{{H}_{2}}O}} \\\\ & \\to {{m}_{Oxi}}=32+28-20=40 \\\\ \\end{aligned}$ gam<br\/>V\u1eady t\u1ef7 l\u1ec7 kh\u1ed1i l\u01b0\u1ee3ng l\u00e0<br\/>$\\dfrac{{{m}_{c{{o}_{2}}}}}{{{m}_{Oxi}}}=\\dfrac{32}{40}=\\dfrac{4}{5}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2046},{"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 n\u00f3ng 15,8 gam kali pemanganat ( thu\u1ed1c t\u00edm ) $KMn{{O}_{4}}$ trong \u1ed1ng nghi\u1ec7m \u0111\u1ec3 \u0111i\u1ec1u ch\u1ebf kh\u00ed oxi. Bi\u1ebft r\u1eb1ng, ch\u1ea5t r\u1eafn c\u00f2n l\u1ea1i trong \u1ed1ng nghi\u1ec7m c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng 12,6 gam; kh\u1ed1i l\u01b0\u1ee3ng kh\u00ed oxi thu \u0111\u01b0\u1ee3c l\u00e0 2,8 gam. Hi\u1ec7u su\u1ea5t c\u1ee7a ph\u1ea3n \u1ee9ng ph\u00e2n h\u1ee7y l\u00e0","select":["A. $85\\%$ ","B. $87,5\\%$ ","C. $86,5\\%$ ","D. $88,5\\%$ "],"hint":"","explain":"<span class='basic_left'>Hi\u1ec7u su\u1ea5t b\u1eb1ng l\u01b0\u1ee3ng th\u1ef1c t\u1ebf thu \u0111\u01b0\u1ee3c chia cho l\u01b0\u1ee3ng thu \u0111\u01b0\u1ee3c theo l\u00fd thuy\u1ebft r\u1ed3i nh\u00e2n v\u1edbi $100\\%$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh lu\u1eadt b\u1ea3o to\u00e0n kh\u1ed1i l\u01b0\u1ee3ng: <br\/>$\\begin{aligned} & {{m}_{o}}={{m}_{KMn{{O}_{4}}}}-{{m}_{chatranconlai}} \\\\ & \\to {{m}_{o}}=15,8-12,6=3,2g \\\\ \\end{aligned}$<br\/>Hi\u1ec7u su\u1ea5t c\u1ee7a ph\u1ea3n \u1ee9ng l\u00e0 <br\/>$Hs=\\dfrac{2,8}{3,2}.100\\%=87,5\\%$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2047},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"Bi\u1ebft r\u1eb1ng canxi oxit ( v\u00f4i s\u1ed1ng ) $CaO$ h\u00f3a h\u1ee3p v\u1edbi n\u01b0\u1edbc t\u1ea1o ra canxi hi\u0111roxit ( v\u00f4i t\u00f4i ) $Ca{{(OH)}_{2}}$, ch\u1ea5t n\u00e0y tan \u0111\u01b0\u1ee3c trong n\u01b0\u1edbc, c\u1ee9 56 gam $CaO$ h\u00f3a h\u1ee3p v\u1eeba \u0111\u1ee7 v\u1edbi 18 gam n\u01b0\u1edbc. T\u00ednh kh\u1ed1i l\u01b0\u1ee3ng th\u1ef1c t\u1ebf $Ca{{(OH)}_{2}}$ thu \u0111\u01b0\u1ee3c, bi\u1ebft hi\u1ec7u su\u1ea5t c\u1ee7a ph\u1ea3n \u1ee9ng l\u00e0 $80\\%$ ","select":["A. $74$ gam ","B. $58,2$ gam ","C. $75$ gam ","D. $59,2$ gam "],"hint":"","explain":"<span class='basic_left'>\u00c1p d\u1ee5ng \u0111\u1ecbnh lu\u1eadt b\u1ea3o to\u00e0n kh\u1ed1i l\u01b0\u1ee3ng:<br\/>$\\begin{aligned} & {{m}_{CaO}}+{{m}_{{{H}_{2}}O}}={{m}_{Ca{{(OH)}_{2}}}} \\\\ & \\to {{m}_{Ca{{(OH)}_{2}}}}=56+18=74g \\\\ \\end{aligned}$<br\/>Kh\u1ed1i l\u01b0\u1ee3ng $Ca{{(OH)}_{2}}$ th\u1ef1c t\u1ebf thu \u0111\u01b0\u1ee3c l\u00e0<br\/>${{m}_{Ca{{(OH)}_{2}}}}=\\dfrac{74.80}{100}=59,2g$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2048},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Khi nung n\u00f3ng c\u1ee5c \u0111\u00e1 v\u00f4i trong l\u00f2 nung v\u00f4i th\u00ec kh\u1ed1i l\u01b0\u1ee3ng \u0111\u00e1 v\u00f4i s\u1ebd thay \u0111\u1ed5i nh\u01b0 th\u1ebf n\u00e0o? ","select":["A. Gi\u1ea3m ","B. T\u0103ng ","C. Kh\u00f4ng thay \u0111\u1ed5i ","D. L\u00fac t\u0103ng l\u00fac gi\u1ea3m "],"hint":"","explain":"<span class='basic_left'>Khi nung n\u00f3ng c\u1ee5c \u0111\u00e1 v\u00f4i th\u00ec ch\u1ea5t canxi cacbonat b\u1ecb ph\u00e2n h\u1ee7y th\u00e0nh ch\u1ea5t canxi oxit v\u00e0 kh\u00ed cacbon \u0111ioxit tho\u00e1t ra n\u00ean kh\u1ed1i l\u01b0\u1ee3ng gi\u1ea3m \u0111i<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2049}],"lesson":{"save":0,"level":2}}