Bài tập trung bình bài Bài 47: Chất béo

Home Lớp 9 Hóa học 9 Bài 47: Chất béo Bài tập luyện

{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Ch\u1ea5t b\u00e9o l\u00e0:","select":["A. H\u1ed7n h\u1ee3p c\u1ee7a nhi\u1ec1u axit b\u00e9o","B. H\u1ed7n h\u1ee3p c\u1ee7a nhi\u1ec1u este c\u1ee7a axit b\u00e9o v\u00e0 r\u01b0\u1ee3u etylic","C. H\u1ed7n h\u1ee3p c\u1ee7a nhi\u1ec1u este c\u1ee7a axit b\u00e9o v\u00e0 glixerol","D. H\u1ed7n h\u1ee3p c\u1ee7a glixerol v\u00e0 r\u01b0\u1ee3u etylic"],"hint":"","explain":"Ch\u1ea5t b\u00e9o l\u00e0 h\u1ee3p c\u1ee7a nhi\u1ec1u este c\u1ee7a axit b\u00e9o v\u00e0 glixerol c\u00f3 c\u00f4ng th\u1ee9c chung l\u00e0:$(RCOO)_3C_3H_5$<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2551},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" C\u00f4ng th\u1ee9c vi\u1ebft g\u1ecdn c\u1ee7a glixerol l\u00e0:","select":["A. $C_2H_4(OH)_2$","B. $CH_3OH$","C. $C_3H_5(OH)_3$","D. $C_2H_5OH$"],"hint":"","explain":"Glixerol: $C_3H_5(OH)_3$<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2552},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" M\u1ed9t ch\u1ea5t b\u00e9o c\u00f3 c\u00f4ng th\u1ee9c $(C_{17}H_{35}COO)_3C_3H_5$ c\u00f3 ph\u00e2n t\u1eed kh\u1ed1i l\u00e0 ","select":["A. 890 \u0111vC.","B. 422 \u0111vC.","C. 372 \u0111vC. ","D. 980 \u0111vC"],"hint":"","explain":"$(C_{17}H_{35}COO)_3C_3H_5$ c\u00f3 ph\u00e2n t\u1eed kh\u1ed1i l\u00e0: 890 \u0111vC<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2553},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Th\u1ee7y ph\u00e2n ch\u1ea5t b\u00e9o trong m\u00f4i tr\u01b0\u1eddng axit thu \u0111\u01b0\u1ee3c","select":["A. glixerol v\u00e0 m\u1ed9t lo\u1ea1i axit b\u00e9o ","B. glixerol v\u00e0 m\u1ed9t mu\u1ed1i c\u1ee7a axit b\u00e9o","C. glixerol v\u00e0 m\u1ed9t s\u1ed1 lo\u1ea1i axit b\u00e9o ","D. glixerol v\u00e0 x\u00e0 ph\u00f2ng."],"hint":"","explain":"${{(RCOO)}_{3}}{{C}_{3}}{{H}_{5}}+3{{H}_{2}}O\\xrightarrow{Axit}3RCOOH+{{C}_{3}}{{H}_{5}}{{(OH)}_{3}}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2554},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Ch\u1ea5t n\u00e0o sau \u0111\u00e2y kh\u00f4ng ph\u1ea3i l\u00e0 ch\u1ea5t b\u00e9o:","select":["A. $(C_{17}H_{35}COO)_3C_3H_5 $","B. $(C_{15}H_{31}COO)_3C_3H_5$.","C. $(C_{17}H_{33}COO)_3C_3H_5 $.","D. $(CH_3COO)_3C_3H_5$."],"hint":"","explain":"Ch\u1ea5t b\u00e9o l\u00e0 h\u1ee3p c\u1ee7a nhi\u1ec1u este c\u1ee7a axit b\u00e9o v\u00e0 glixerol c\u00f3 c\u00f4ng th\u1ee9c chung l\u00e0:$(RCOO)_3C_3H_5$. => $(CH_3COO)_3C_3H_5$ kh\u00f4ng l\u00e0 ch\u1ea5t b\u00e9o<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2555},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" H\u00e3y ch\u1ecdn c\u00e2u \u0111\u00fang trong c\u00e1c c\u00e2u sau. ","select":["A. D\u1ea7u \u0103n l\u00e0 h\u1ed7n h\u1ee3p nhi\u1ec1u este v\u1edbi dung d\u1ecbch axit.","B. D\u1ea7u \u0103n l\u00e0 h\u1ed7n h\u1ee3p c\u1ee7a glixerol v\u00e0 mu\u1ed1i c\u1ee7a axit b\u00e9o.","C. D\u1ea7u \u0103n l\u00e0 h\u1ed7n h\u1ee3p nhi\u1ec1u este c\u1ee7a glixerol v\u00e0 c\u00e1c axit b\u00e9o. ","D. D\u1ea7u \u0103n l\u00e0 h\u1ed7n h\u1ee3p dung d\u1ecbch ki\u1ec1m v\u00e0 glixerol."],"hint":"","explain":" D\u1ea7u \u0103n l\u00e0 ch\u1ea5t b\u00e9o n\u00ean l\u00e0 h\u1ed7n h\u1ee3p nhi\u1ec1u este c\u1ee7a glixerol v\u00e0 c\u00e1c axit b\u00e9o.<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2556},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y l\u00e0 \u0111\u00fang:","select":["A. D\u1ea7u m\u1ee1 lu\u00f4n s\u00f4i \u1edf nhi\u1ec7t \u0111\u1ed9 nh\u1ea5t \u0111\u1ecbnh l\u00e0 $220^0C$","B. D\u1ea7u m\u1ee1 tan nhi\u1ec1u trong n\u01b0\u1edbc","C. D\u1ea7u m\u1ee1 nh\u1eb9 h\u01a1n n\u01b0\u1edbc v\u00e0 kh\u00f4ng tan trong n\u01b0\u1edbc ","D. D\u1ea7u m\u1ee1 n\u1eb7ng h\u01a1n n\u01b0\u1edbc v\u00e0 kh\u00f4ng tan trong n\u01b0\u1edbc."],"hint":"","explain":"D\u1ea7u m\u1ee1 l\u00e0 ch\u1ea5t b\u00e9o n\u00ean kh\u00f4ng tan trong n\u01b0\u1edbc, nh\u1eb9 h\u01a1n n\u01b0\u1edbc.<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2557},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Th\u1ee7y ph\u00e2n ch\u1ea5t b\u00e9o trong m\u00f4i tr\u01b0\u1eddng ki\u1ec1m thu \u0111\u01b0\u1ee3c","select":["A. glixerol v\u00e0 mu\u1ed1i c\u1ee7a m\u1ed9t axit b\u00e9o.","B. glixerol v\u00e0 axit b\u00e9o.","C. glixerol v\u00e0 x\u00e0 ph\u00f2ng.","D. glixerol v\u00e0 mu\u1ed1i c\u1ee7a c\u00e1c axit b\u00e9o"],"hint":"","explain":"${{(RCOO)}_{3}}{{C}_{3}}{{H}_{5}}+3NaOH\\xrightarrow{{{t}^{0}}}3RCOONa+{{C}_{3}}{{H}_{5}}{{(OH)}_{3}}$<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":2558},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Ch\u1ea5t b\u00e9o c\u00f3 c\u00f4ng th\u1ee9c t\u1ed5ng qu\u00e1t l\u00e0:","select":["A. $(RCOO)_3C_3H_5$ ","B. $RCOOC_2H_5$ ","C. $RCOOH$ ","D. $(RCOO)_2C_2H_4$"],"hint":"","explain":"Ch\u1ea5t b\u00e9o l\u00e0 h\u1ee3p c\u1ee7a nhi\u1ec1u este c\u1ee7a axit b\u00e9o v\u00e0 glixerol c\u00f3 c\u00f4ng th\u1ee9c chung l\u00e0: $(RCOO)_3C_3H_5$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2559},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" X\u00e0 ph\u00f2ng \u0111\u01b0\u1ee3c \u0111i\u1ec1u ch\u1ebf b\u1eb1ng c\u00e1ch:","select":["A. Ph\u00e2n h\u1ee7y ch\u1ea5t b\u00e9o","B. Th\u1ee7y ph\u00e2n ch\u1ea5t b\u00e9o trong m\u00f4i tr\u01b0\u1eddng axit","C. Th\u1ee7y ph\u00e2n ch\u1ea5t b\u00e9o trong m\u00f4i tr\u01b0\u1eddng ki\u1ec1m ","D. H\u00f2a tan ch\u1ea5t b\u00e9o trong dung m\u00f4i h\u1eefu c\u01a1"],"hint":"","explain":"Ph\u1ea3n \u1ee9ng th\u1ee7y ph\u00e2n ch\u1ea5t b\u00e9o trong m\u00f4i tr\u01b0\u1eddng ki\u1ec1m thu \u0111\u01b0\u1ee3c x\u00e0 ph\u00f2ng. <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2560},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" C\u00f3 th\u1ec3 l\u00e0m s\u1ea1ch d\u1ea7u \u0103n d\u00ednh tr\u00ean qu\u1ea7n \u00e1o b\u1eb1ng c\u00e1ch","select":["A. Gi\u1eb7t b\u1eb1ng n\u01b0\u1edbc ","B. T\u1ea9y b\u1eb1ng gi\u1ea5m ","C. T\u1ea9y b\u1eb1ng x\u0103ng ","D. Gi\u1eb7t b\u1eb1ng n\u01b0\u1edbc c\u00f3 pha th\u00eam \u00edt mu\u1ed1i"],"hint":"","explain":"D\u1ea7u \u0103n l\u00e0 ch\u1ea5t b\u00e9o, tan trong benzen n\u00ean c\u00f3 th\u1ec3 \u0111\u01b0\u1ee3c l\u00e0m s\u1ea1ch b\u1eb1ng c\u00e1ch t\u1ea9y b\u1eb1ng x\u0103ng.<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2561},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho v\u00e0i gi\u1ecdt d\u1ea7u \u0103n v\u00e0o benzen l\u1eafc nh\u1eb9 quan s\u00e1t th\u1ea5y hi\u1ec7n t\u01b0\u1ee3ng l\u00e0:","select":["A. Ph\u00e2n 2 l\u1edbp d\u1ea7u \u0103n \u1edf tr\u00ean, benzen \u1edf d\u01b0\u1edbi","B. Ph\u00e2n 2 l\u1edbp d\u1ea7u \u0103n \u1edf d\u01b0\u1edbi, benzen \u1edf tr\u00ean","C. Thu \u0111\u01b0\u1ee3c dung d\u1ecbch \u0111\u1ed3ng nh\u1ea5t ","D. Benzen h\u00f2a tan 1 ph\u1ea7n d\u1ea7u \u0103n, 1 ph\u1ea7n c\u00f2n l\u1ea1i kh\u00f4ng tan n\u1ed5i l\u00ean tr\u00ean"],"hint":"","explain":"D\u1ea7u \u0103n tan trong benzen t\u1ea1o th\u00e0nh dung d\u1ecbch \u0111\u1ed3ng nh\u1ea5t.<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2562},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Cho m\u1ed9t lo\u1ea1i ch\u1ea5t b\u00e9o c\u00f3 c\u00f4ng th\u1ee9c chung l\u00e0 $(RCOO)_3C_3H_5$ ph\u1ea3n \u1ee9ng v\u1eeba \u0111\u1ee7 v\u1edbi a gam dung d\u1ecbch $NaOH$ 50% t\u1ea1o th\u00e0nh 9,2 gam glixerol. Gi\u00e1 tr\u1ecb c\u1ee7a a l\u00e0 ","select":["A. 23 gam.","B. 24 gam.","C. 20 gam.","D. 48 gam"],"hint":"","explain":"$\\begin{aligned} & {{(RCOO)}_{3}}{{C}_{3}}{{H}_{5}}+3NaOH\\xrightarrow{{{t}^{0}}}3RCOONa+{{C}_{3}}{{H}_{5}}{{(OH)}_{3}} \\\\ & {{n}_{NaOH}}=3{{n}_{{{C}_{3}}{{H}_{5}}{{(OH)}_{3}}}}=0,3\\,\\,mol\\Rightarrow {{m}_{NaOH}}=12\\,gam \\\\ & \\Rightarrow {{m}_{\\text{dd}}}=\\dfrac{12.100\\%}{50\\%}=24\\,gam \\\\ \\end{aligned}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2563},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Th\u1ee7y ph\u00e2n ho\u00e0n to\u00e0n 17,16 kg m\u1ed9t lo\u1ea1i ch\u1ea5t b\u00e9o c\u1ea7n v\u1eeba \u0111\u1ee7 2,4 kg NaOH, s\u1ea3n ph\u1ea9m thu \u0111\u01b0\u1ee3c g\u1ed3m 1,84 kg glixerol v\u00e0 h\u1ed7n h\u1ee3p mu\u1ed1i c\u1ee7a c\u00e1c axit b\u00e9o. Kh\u1ed1i l\u01b0\u1ee3ng h\u1ed7n h\u1ee3p c\u00e1c mu\u1ed1i l\u00e0 ","select":["A. 17,72 kg. ","B. 19,44 kg.","C. 11,92 kg","D. 12,77 kg."],"hint":"","explain":"G\u1ecdi x (kg) l\u00e0 kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a h\u1ed7n h\u1ee3p c\u00e1c mu\u1ed1i thu \u0111\u01b0\u1ee3c. <br\/> $\\begin{aligned} & {{(RCOO)}_{3}}{{C}_{3}}{{H}_{5}}+3NaOH\\xrightarrow{{{t}^{0}}}3RCOONa+{{C}_{3}}{{H}_{5}}{{(OH)}_{3}} \\\\ & 17,16\\,kg\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2,4\\,kg\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x\\,kg\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,1,84\\,kg \\\\ \\end{aligned}$ <br\/> \u00c1p d\u1ee5ng \u0111\u1ecbnh lu\u1eadt b\u1ea3o to\u00e0n kh\u1ed1i l\u01b0\u1ee3ng ta c\u00f3: <br\/> $17,16+2,4=x+1,84\\Rightarrow x=17,72\\,kg$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2564},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" \u0110un 26,7 kg ch\u1ea5t b\u00e9o $(C_{17}H_{35}COO)_3C_3H_5$ v\u1edbi dung d\u1ecbch $NaOH$ d\u01b0 (hi\u1ec7u su\u1ea5t 100%), kh\u1ed1i l\u01b0\u1ee3ng glixerol thu \u0111\u01b0\u1ee3c l\u00e0 ","select":["A. 1,2 kg. ","B. 2,76 kg.","C. 3,6 kg. ","D. 4,8 kg."],"hint":"","explain":"$\\begin{aligned} & {{({{C}_{17}}{{H}_{35}}COO)}_{3}}{{C}_{3}}{{H}_{5}}+3NaOH\\xrightarrow{{{t}^{0}}}3{{C}_{17}}{{H}_{35}}COONa+{{C}_{3}}{{H}_{5}}{{(OH)}_{3}} \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,890\\,kg\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,92\\,kg \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,26,7\\,kg\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x\\,kg\\,\\,\\,\\,\\,\\,\\,\\, \\\\ & \\Rightarrow x=\\dfrac{26,7.92}{890}=2,76\\,kg\\,\\,\\, \\\\ \\end{aligned}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2565},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" \u0110\u00e2u kh\u00f4ng ph\u1ea3i l\u00e0 ch\u1ea5t b\u00e9o trong c\u00e1c ch\u1ea5t sau:","select":["A. D\u1ea7u d\u1eeba","B. D\u1ea7u v\u1eebng","C. D\u1ea7u luyn ","D. D\u1ea7u l\u1ea1c"],"hint":"","explain":"D\u1ea7u luyn kh\u00f4ng ph\u1ea3i l\u00e0 ch\u1ea5t b\u00e9o.<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2566},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Th\u1ee7y ph\u00e2n ch\u1ea5t b\u00e9o trong n\u01b0\u1edbc c\u00f3 axit x\u00fac t\u00e1c thu \u0111\u01b0\u1ee3c 4,6 gam glixerol v\u00e0 70 gam axit b\u00e9o. T\u00ednh kh\u1ed1i l\u01b0\u1ee3ng ch\u1ea5t b\u00e9o bi\u1ebft hi\u1ec7u su\u1ea5t ph\u1ea3n \u1ee9ng \u0111\u1ea1t 100%","select":["A. 68,5 gam ","B. 70,2 gam","C. 71,9 gam ","D. 73,7 gam"],"hint":"","explain":"G\u1ecdi x l\u00e0 kh\u1ed1i l\u01b0\u1ee3ng ch\u1ea5t b\u00e9o c\u1ea7n t\u00ecm <br\/> $\\begin{aligned} & {{(RCOO)}_{3}}{{C}_{3}}{{H}_{5}}+3{{H}_{2}}O\\xrightarrow{Axit}3RCOOH+{{C}_{3}}{{H}_{5}}{{(OH)}_{3}} \\\\ & {{n}_{{{H}_{2}}O}}=3{{n}_{{{C}_{3}}{{H}_{5}}{{(OH)}_{3}}}}=3.0,05=0,15\\,\\,mol\\Rightarrow {{m}_{{{H}_{2}}O}}=2,7\\,gam \\\\ \\end{aligned}$ <br\/> \u00c1p d\u1ee5ng \u0111\u1ecbnh lu\u1eadt b\u1ea3o to\u00e0n kh\u1ed1i l\u01b0\u1ee3ng ta c\u00f3: <br\/> $\\begin{aligned} & {{m}_{{{(RCOO)}_{3}}{{C}_{3}}{{H}_{5}}}}+{{m}_{{{H}_{2}}O}}={{m}_{RCOOH}}+{{m}_{{{C}_{3}}{{H}_{5}}{{(OH)}_{3}}}} \\\\ & \\Rightarrow \\,\\,\\,\\,\\,\\,\\,\\,x\\,\\,\\,\\,\\,\\,\\,\\,\\,+\\,2,7\\,\\,\\,\\,\\,\\,=70\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+4,6 \\\\ & \\Rightarrow x=71,9\\,gam \\\\ \\end{aligned}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2567},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" T\u00ednh kh\u1ed1i l\u01b0\u1ee3ng $(C_{17}H_{35}COO)_3C_3H_5$ t\u1ed1i thi\u1ec3u \u0111\u1ec3 \u0111i\u1ec1u ch\u1ebf 1 t\u1ea5n $C_{17}H_{35}COONa$ d\u00f9ng l\u00e0m x\u00e0 ph\u00f2ng, bi\u1ebft r\u1eb1ng hi\u1ec7u su\u1ea5t ph\u1ea3n \u1ee9ng l\u00e0 80%. ","select":["A. 1,12 t\u1ea5n.","B. 1,25 t\u1ea5n.","C. 1,3 t\u1ea5n. ","D. 1,21 t\u1ea5n."],"hint":"","explain":"$\\begin{aligned} & {{({{C}_{17}}{{H}_{35}}COO)}_{3}}{{C}_{3}}{{H}_{5}}+3NaOH\\xrightarrow{{{t}^{0}}}3{{C}_{17}}{{H}_{35}}COONa+{{C}_{3}}{{H}_{5}}{{(OH)}_{3}} \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,890\\text{t\u1ea5n}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,918\\text{ t\u1ea5n } \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x\\,\\text{ t\u1ea5n }\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,1\\,\\text{ t\u1ea5n }\\,\\,\\,\\,\\,\\,\\, \\\\ & \\Rightarrow x=\\dfrac{890.1}{918}\\approx 0,97\\text{ t\u1ea5n } \\\\ & H=80\\%\\Rightarrow {{m}_{{{({{C}_{17}}{{H}_{35}}COO)}_{3}}{{C}_{3}}{{H}_{5}}\\,\\,(TT)}}=\\dfrac{0,97.100\\%}{80\\%}\\approx 1,21\\,\\text{ t\u1ea5n }\\,\\, \\\\ \\end{aligned}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2568},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" X\u00e0 ph\u00f2ng ho\u00e1 ho\u00e0n to\u00e0n 17,24 gam ch\u1ea5t b\u00e9o c\u1ea7n v\u1eeba \u0111\u1ee7 0,06 mol NaOH. C\u00f4 c\u1ea1n dung d\u1ecbch sau ph\u1ea3n \u1ee9ng thu \u0111\u01b0\u1ee3c kh\u1ed1i l\u01b0\u1ee3ng x\u00e0 ph\u00f2ng l\u00e0","select":["A. 17,80 gam","B. 18,24 gam","C. 16,68 gam ","D. 18,38 gam"],"hint":"","explain":"G\u1ecdi x l\u00e0 kh\u1ed1i l\u01b0\u1ee3ng x\u00e0 ph\u00f2ng thu \u0111\u01b0\u1ee3c <br\/> $\\begin{aligned} & {{(RCOO)}_{3}}{{C}_{3}}{{H}_{5}}+3NaOH\\xrightarrow{{{t}^{0}}}3RCOONa+{{C}_{3}}{{H}_{5}}{{(OH)}_{3}} \\\\ & {{n}_{{{C}_{3}}{{H}_{5}}{{(OH)}_{3}}}}=\\dfrac{1}{3}{{n}_{NaOH}}=0,02\\,mol\\Rightarrow {{m}_{{{C}_{3}}{{H}_{5}}{{(OH)}_{3}}}}=1,84\\,gam \\\\ & {{m}_{NaOH}}=2,4\\,gam \\\\ \\end{aligned}$ <br\/> \u00c1p d\u1ee5ng \u0111\u1ecbnh lu\u1eadt b\u1ea3o to\u00e0n kh\u1ed1i l\u01b0\u1ee3ng ta c\u00f3: <br\/> $\\begin{aligned} & {{m}_{{{(RCOO)}_{3}}{{C}_{3}}{{H}_{5}}}}+{{m}_{NaOH}}={{m}_{RCOONa}}+{{m}_{{{C}_{3}}{{H}_{5}}{{(OH)}_{3}}}} \\\\ & \\Rightarrow \\,\\,\\,\\,17,24\\,\\,+\\,2,4\\,\\,\\,\\,\\,\\,=\\,\\,\\,\\,\\,\\,\\,\\,\\,x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+1,84 \\\\ & \\Rightarrow x=17,8\\,gam \\\\ \\end{aligned}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2569},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Trung h\u00f2a ho\u00e0n to\u00e0n 85,2 gam axit b\u00e9o b\u1eb1ng KOH thu \u0111\u01b0\u1ee3c l\u01b0\u1ee3ng n\u01b0\u1edbc v\u1eeba \u0111\u1ee7 \u0111\u1ec3 th\u1ee7y ph\u00e2n ch\u1ea5t b\u00e9o thu \u0111\u01b0\u1ee3c 9,2 gam glixerol. Axit b\u00e9o ban \u0111\u1ea7u c\u00f3 c\u00f4ng th\u1ee9c l\u00e0:","select":["A. $C_{15}H_{31}COOH$","B. $C_{17}H_{33}COOH$","C. $C_{17}H_{35}COOH$","D. $C_{17}H_{31}COOH$ "],"hint":"","explain":"$\\begin{aligned} & RCOOH+KOH\\to RCOOK+{{H}_{2}}O \\\\ & {{(RCOO)}_{3}}{{C}_{3}}{{H}_{5}}+3{{H}_{2}}O\\xrightarrow{Axit}3RCOOH+{{C}_{3}}{{H}_{5}}{{(OH)}_{3}} \\\\ & {{n}_{{{H}_{2}}O}}=3{{n}_{{{C}_{3}}{{H}_{5}}{{(OH)}_{3}}}}=3.0,1=0,3\\,\\,mol \\\\ & {{n}_{RCOOH}}={{n}_{{{H}_{2}}O}}=0,3\\,mol \\\\ & \\Rightarrow {{M}_{RCOOH}}=\\dfrac{{{m}_{RCOOH}}}{{{n}_{RCOOH}}}=\\dfrac{85,2}{0,3}=284\\,({{C}_{17}}{{H}_{35}}COOH) \\\\ \\end{aligned}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2570}],"lesson":{"save":0,"level":2}}

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