{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho dung d\u1ecbch axit sunfuric lo\u00e3ng t\u00e1c d\u1ee5ng v\u1edbi mu\u1ed1i natrisunfit ($Na_2SO_3$). Ch\u1ea5t kh\u00ed n\u00e0o sinh ra ","select":["A. Kh\u00ed hi\u0111ro ","B. Kh\u00ed oxi","C. Kh\u00ed l\u01b0u hu\u1ef3nh \u0111ioxit ","D. Kh\u00ed hi\u0111ro sunfua"],"hint":"","explain":"<span class='basic_left'>$N{{a}_{2}}S{{O}_{3}}+{{H}_{2}}S{{O}_{4}}\\to N{{a}_{2}}S{{O}_{4}}+S{{O}_{2}}\\uparrow +{{H}_{2}}O$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1891},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" \u0110i\u1ec7n ph\u00e2n dung d\u1ecbch $NaCl$ b\u00e3o ho\u00e0, c\u00f3 m\u00e0ng ng\u0103n gi\u1eefa hai \u0111i\u1ec7n c\u1ef1c, s\u1ea3n ph\u1ea9m thu \u0111\u01b0\u1ee3c l\u00e0","select":["A. $NaOH, H_2, Cl_2$ ","B. $NaCl, NaClO, H_2, Cl_2$","C. $NaCl, NaClO, Cl_2$ ","D. $NaClO, H_2 v\u00e0 Cl_2$"],"hint":"","explain":"<span class='basic_left'>$NaCl+{{H}_{2}}O\\xrightarrow[cmn]{\\text{\u0111pdd}}\\text{ NaOH}+C{{l}_{2}}\\uparrow +{{H}_{2}}\\uparrow $ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1892},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Cho 50 g $CaCO_3$ v\u00e0o dung d\u1ecbch $HCl$ d\u01b0 th\u1ec3 t\u00edch $CO_2$ thu \u0111\u01b0\u1ee3c \u1edf \u0111ktc l\u00e0:","select":["A. 11,2 l\u00edt","B. 1,12 l\u00edt","C. 2,24 l\u00edt ","D. 22,4 l\u00edt"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{CaC{{O}_{3}}}}=0,5\\,mol \\\\ & CaC{{O}_{3}}+2HCl\\to CaC{{l}_{2}}+C{{O}_{2}}\\uparrow +{{H}_{2}}O \\\\ & {{n}_{C{{O}_{2}}}}={{n}_{CaC{{O}_{3}}}}=0,5\\,mol \\\\ & {{V}_{C{{O}_{2}}}}=11,2\\,l \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1893},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho dung d\u1ecbch $KOH$ v\u00e0o \u1ed1ng nghi\u1ec7m \u0111\u1ef1ng dung d\u1ecbch $FeCl_3$, hi\u1ec7n t\u01b0\u1ee3ng quan s\u00e1t \u0111\u01b0\u1ee3c l\u00e0","select":["A. C\u00f3 k\u1ebft t\u1ee7a tr\u1eafng xanh ","B. C\u00f3 kh\u00ed tho\u00e1t ra ","C. C\u00f3 k\u1ebft t\u1ee7a n\u00e2u \u0111\u1ecf ","D. C\u00f3 k\u1ebft t\u1ee7a m\u00e0u tr\u1eafng"],"hint":"","explain":"<span class='basic_left'>$3KOH+FeC{{l}_{3}}\\to 3KCl+Fe{{(OH)}_{3}}\\downarrow$ <br\/>$Fe(OH)_3$ m\u00e0u n\u00e2u \u0111\u1ecf.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1894},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" D\u00e3y ch\u1ea5t n\u00e0o sau \u0111\u00e2y b\u1ecb nhi\u1ec7t ph\u00e2n h\u1ee7y \u1edf nhi\u1ec7t \u0111\u1ed9 cao:","select":["A. $BaSO_3, BaCl_2, KOH, Na_2SO_4$","B. $AgNO_3, Na_2CO_3, KCl, BaSO_4$","C. $CaCO_3, Zn(OH)_2, KNO_3, KMnO_4$","D. $Fe(OH)_3, Na_2SO_4, BaSO_4, KCl$"],"hint":"","explain":"<span class='basic_left'>CH\u00da \u00dd: C\u00e1c mu\u1ed1i sunfat kh\u00f4ng b\u1ecb nhi\u1ec7t ph\u00e2n. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n C th\u1ecfa m\u00e3n kh\u00f4ng c\u00f3 mu\u1ed1i sunfat.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":1895},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" H\u1ee3p ch\u1ea5t n\u00e0o sau \u0111\u00e2y b\u1ecb nhi\u1ec7t ph\u00e2n h\u1ee7y t\u1ea1o ra h\u1ee3p ch\u1ea5t oxit v\u00e0 m\u1ed9t ch\u1ea5t kh\u00ed l\u00e0m \u0111\u1ee5c n\u01b0\u1edbc v\u00f4i trong","select":["A. Mu\u1ed1i sufat","B. Mu\u1ed1i cacbonat kh\u00f4ng tan","C. Mu\u1ed1i clorua ","D. Mu\u1ed1i nitrat"],"hint":"","explain":"<span class='basic_left'>C\u00e1c mu\u1ed1i cacbonat kh\u00f4ng tan b\u1ecb nhi\u1ec7t ph\u00e2n t\u1ea1o oxit v\u00e0 kh\u00ed cacbonic (l\u00e0m \u0111\u1ee5c n\u01b0\u1edbc v\u00f4i trong)<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1896},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Ng\u00e2m 1 thanh s\u1eaft v\u00e0o dung d\u1ecbch $CuSO_4$. Hi\u1ec7n t\u01b0\u1ee3ng quan s\u00e1t \u0111\u01b0\u1ee3c l\u00e0:","select":["A. Thanh s\u1eaft tan ra, dung d\u1ecbch m\u00e0u xanh","B. Thanh s\u1eaft kh\u00f4ng tan, kh\u00f4ng c\u00f3 hi\u1ec7n t\u01b0\u1ee3ng g\u00ec","C. Thanh s\u1eaft tan ra, dung d\u1ecbch m\u00e0u xanh nh\u1ea1t d\u1ea7n","D. Thanh s\u1eaft kh\u00f4ng tan, c\u00f3 kh\u00ed tho\u00e1t ra"],"hint":"","explain":"<span class='basic_left'>$Fe+CuS{{O}_{4}}\\to FeS{{O}_{4}}+Cu$ <br\/>Fe tham gia ph\u1ea3n \u1ee9ng v\u00e0 tan d\u1ea7n ra. Dung d\u1ecbch $CuSO_4$ c\u00f3 m\u00e0u xanh tham gia ph\u1ea3n \u1ee9ng n\u00ean nh\u1ea1t m\u00e0u d\u1ea7n.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":1897},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Dung d\u1ecbch mu\u1ed1i KH\u00d4NG th\u1ec3 t\u00e1c d\u1ee5ng v\u1edbi","select":["A. Axit ","B. Baz\u01a1","C. Oxit ","D. Mu\u1ed1i"],"hint":"","explain":"<span class='basic_left'>Dung d\u1ecbch mu\u1ed1i kh\u00f4ng th\u1ec3 t\u00e1c d\u1ee5ng v\u1edbi oxit.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1898},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Ph\u1ea3n \u1ee9ng trao \u0111\u1ed5i:","select":["A. Lu\u00f4n x\u1ea3y ra ","B. Ch\u1ec9 x\u1ea3y ra khi t\u1ea1o th\u00e0nh ch\u1ea5t kh\u00ed","C. Ch\u1ec9 x\u1ea3y ra khi t\u1ea1o th\u00e0nh k\u1ebft t\u1ee7a. ","D. X\u1ea3y ra khi t\u1ea1o th\u00e0nh ch\u1ea5t k\u1ebft t\u1ee7a ho\u1eb7c ch\u1ea5t kh\u00ed ho\u1eb7c c\u1ea3 k\u1ebft t\u1ee7a c\u1ea3 kh\u00ed."],"hint":"","explain":"<span class='basic_left'>Ph\u1ea3n \u1ee9ng trao \u0111\u1ed5i x\u1ea3y ra khi s\u1ea3n ph\u1ea9m t\u1ea1o th\u00e0nh c\u00f3 ch\u1ea5t kh\u00ed ho\u1eb7c ch\u1ea5t k\u1ebft t\u1ee7a ho\u1eb7c c\u1ea3 ch\u1ea5t k\u1ebft t\u1ee7a c\u1ea3 ch\u1ea5t kh\u00ed.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":1899},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Mu\u1ed1i n\u00e0o sau \u0111\u00e2y kh\u00f4ng tan l\u00e0:","select":["A. $BaCl_2$ ","B. $CaCO_3$ ","C. $MgSO_4$ ","D. $AgNO_3$"],"hint":"","explain":"<span class='basic_left'>C\u00e1c mu\u1ed1i cacbonat h\u1ea7u h\u1ebft \u0111\u1ec1u kh\u00f4ng tan (tr\u1eeb mu\u1ed1i c\u1ee7a kim lo\u1ea1i ki\u1ec1m)<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1900},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" \u0110\u1ec3 l\u00e0m s\u1ea1ch dung d\u1ecbch \u0111\u1ed3ng nitrat $Cu(NO_3)_2$ c\u00f3 l\u1eabn t\u1ea1p ch\u1ea5t b\u1ea1c nitrat $AgNO_3$. Ta d\u00f9ng kim lo\u1ea1i:","select":["A. Mg ","B. Cu ","C. Fe ","D. Au"],"hint":"","explain":"<span class='basic_left'>\u0110\u1ec3 l\u00e0m s\u1ea1ch $Cu(NO_3)_2$ ta cho h\u1ed7n h\u1ee3p t\u00e1c d\u1ee5ng v\u1edbi ch\u1ea5t m\u00e0 kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi $Cu(NO_3)_2$, t\u00e1c d\u1ee5ng v\u1edbi $AgNO_3$. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n B th\u1ecfa m\u00e3n<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1901},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Dung d\u1ecbch c\u1ee7a ch\u1ea5t X c\u00f3 pH>7 v\u00e0 khi cho t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch kali sunfat($K_2SO_4$) t\u1ea1o ra ch\u1ea5t kh\u00f4ng tan (k\u1ebft t\u1ee7a). Ch\u1ea5t X l\u00e0:","select":["A. $BaCl_2$ ","B. $NaOH$","C. $Ba(OH)_2$ ","D. $H_2SO_4$"],"hint":"","explain":"<span class='basic_left'>Ch\u1ea5t c\u00f3 pH > 7 l\u00e0 baz\u01a1, khi t\u00e1c d\u1ee5ng v\u1edbi kalisunfat t\u1ea1o k\u1ebft t\u1ee7a n\u00ean ch\u1ec9 c\u00f3 $Ba(OH)_2$ th\u1ecfa m\u00e3n<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1902},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Khi cho 200g dung d\u1ecbch $Na_2CO_3$ 10,6% v\u00e0o dung d\u1ecbch $HCl$ d\u01b0, kh\u1ed1i l\u01b0\u1ee3ng kh\u00ed sinh ra","select":["A. 4,6 g ","B.8 g","C. 8,8 g ","D. 10 g"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{m}_{N{{a}_{2}}C{{O}_{3}}}}=\\dfrac{200.10,6%}{100%}=21,2\\,gam \\\\ & {{n}_{N{{a}_{2}}C{{O}_{3}}}}=0,2\\,mol \\\\ & N{{a}_{2}}C{{O}_{3}}+2HCl\\to 2NaCl+C{{O}_{2}}+{{H}_{2}}O \\\\ & {{n}_{C{{O}_{2}}}}={{n}_{N{{a}_{2}}C{{O}_{3}}}}=0,2 \\\\ & {{m}_{C{{O}_{2}}}}=8,8\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1903},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho a g $Na_2CO_3$ v\u00e0o dung d\u1ecbch $HCl$, sau ph\u1ea3n \u1ee9ng thu \u0111\u01b0\u1ee3c 3,36 l\u00edt kh\u00ed \u1edf \u0111ktc. V\u1eady a c\u00f3 gi\u00e1 tr\u1ecb:","select":["A. 34,8 g","B.18,2 g","C. 15,9 g ","D. 10,5 g"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{C{{O}_{2}}}}=0,15\\,mol \\\\ & N{{a}_{2}}C{{O}_{3}}+2HCl\\to 2NaCl+C{{O}_{2}}+{{H}_{2}}O \\\\ & {{n}_{N{{a}_{2}}C{{O}_{3}}}}={{n}_{C{{O}_{2}}}}=0,15\\,mol \\\\ & {{m}_{N{{a}_{2}}C{{O}_{3}}}}=15,9gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1904},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Tr\u01b0\u1eddng h\u1ee3p n\u00e0o t\u1ea1o ra ch\u1ea5t k\u1ebft t\u1ee7a khi tr\u1ed9n 2 dung d\u1ecbch sau:","select":["A. $NaCl$ v\u00e0 $AgNO_3$","B. $NaCl$ v\u00e0 $Ba(NO_3)_2$","C. $KNO_3$ v\u00e0 $BaCl_2$ ","D. $CaCl_2$ v\u00e0 $NaNO_3$"],"hint":"","explain":"<span class='basic_left'>$NaCl+AgN{{O}_{3}}\\to NaN{{O}_{3}}+AgCl\\downarrow $ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1905},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Cho 200g dung d\u1ecbch $KOH$ 5,6% v\u00e0o dung d\u1ecbch $CuCl_2$ d\u01b0, sau ph\u1ea3n \u1ee9ng thu \u0111\u01b0\u1ee3c l\u01b0\u1ee3ng ch\u1ea5t k\u1ebft t\u1ee7a l\u00e0:","select":["A. 4,9 g ","B. 9,8 g","C. 17,4 g ","D. 19,6 g"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{m}_{KOH}}=\\dfrac{200.5,6%}{100%}=11,2\\,gam \\\\ & {{n}_{KOH}}=0,2\\,mol \\\\ & CuC{{l}_{2}}+2KOH\\to Cu{{(OH)}_{2}}\\downarrow +2KCl \\\\ & {{n}_{Cu{{(OH)}_{2}}}}=\\dfrac{1}{2}{{n}_{KOH}}=0,1\\,mol \\\\ & {{m}_{Cu{{(OH)}_{2}}}}=9,8\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1906},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" S\u1ed1 mol c\u1ee7a 200 gam dung d\u1ecbch $CuSO_4$ 32% l\u00e0:","select":["A. 0,4 mol ","B. 0,3 mol","C. 0,2 mol ","D. 0,25 mol"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{m}_{CuS{{O}_{4}}}}=\\dfrac{200.32%}{100%}=64\\,gam \\\\ & {{n}_{CuS{{O}_{4}}}}=\\dfrac{64}{160}=0,4\\,mol \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1907},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho $Mg$ t\u00e1c d\u1ee5ng v\u1eeba \u0111\u1ee7 v\u1edbi 0,1mol $CuSO_4$. Kh\u1ed1i l\u01b0\u1ee3ng $Cu$ sinh ra l\u00e0:","select":["A. 4 g ","B. 6 g","C. 6,4 g ","D. 12,8 g"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& CuS{{O}_{4}}+Mg\\to MgS{{O}_{4}}+Cu \\\\ & {{n}_{Cu}}={{n}_{CuS{{O}_{4}}}}=0,1\\,mol \\\\ & {{m}_{Cu}}=6,4\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1908},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Cho 500 ml dung d\u1ecbch $NaCl$ 2M t\u00e1c d\u1ee5ng v\u1edbi 600 ml dung d\u1ecbch $AgNO_3$ 2M. Kh\u1ed1i l\u01b0\u1ee3ng k\u1ebft t\u1ee7a thu \u0111\u01b0\u1ee3c l\u00e0","select":["A. 143,5 g","B. 14,35 g","C. 157,85 g","D. 15,785 g"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{NaCl}}=1\\,mol \\\\ & {{n}_{AgN{{O}_{3}}}}=1,2\\,mol \\\\ & NaCl+AgN{{O}_{3}}\\to AgCl\\downarrow +NaN{{O}_{3}} \\\\ & \\dfrac{{{n}_{NaCl}}}{1}<\\dfrac{{{n}_{AgN{{O}_{3}}}}}{1} \\\\ \\end{aligned}$<br\/>NaCl h\u1ebft. T\u00ednh theo NaCl <br\/>$\\begin{aligned}& {{n}_{AgN{{O}_{3}}}}={{n}_{NaCl}}=1\\,mol \\\\ & {{m}_{AgN{{O}_{3}}}}=143,5\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1909},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Khi ph\u00e2n h\u1ee7y b\u1eb1ng nhi\u1ec7t 14,2 g $CaCO_3$ v\u00e0 $MgCO_3$ ta thu \u0111\u01b0\u1ee3c 3,36 l\u00edt $CO_2$ \u1edf \u0111ktc. Th\u00e0nh ph\u1ea7n ph\u1ea7n tr\u0103m v\u1ec1 kh\u1ed1i l\u01b0\u1ee3ng c\u00e1c ch\u1ea5t trong h\u1ed7n h\u1ee3p \u0111\u1ea7u l\u00e0:","select":["A. 29,58% v\u00e0 70,42% ","B. 65% v\u00e0 35%","C. 70,42% v\u00e0 29,58% ","D. 35% v\u00e0 65%"],"hint":"","explain":"<span class='basic_left'>\u0110\u1eb7t x, y l\u1ea7n l\u01b0\u1ee3t l\u00e0 s\u1ed1 mol c\u1ee7a $CaCO_3$ v\u00e0 $MgCO_3$ trong h\u1ed7n h\u1ee3p <br\/>V\u00ec kh\u1ed1i l\u01b0\u1ee3ng h\u1ed7n h\u1ee3p b\u1eb1ng 14,2 gam => 100x + 84y = 14,2 (1) <br\/>$\\begin{aligned}& CaC{{O}_{3}}+2HCl\\to CaC{{l}_{2}}+C{{O}_{2}}+{{H}_{2}}O \\\\ & MgC{{O}_{3}}+2HCl\\to MgC{{l}_{2}}+C{{O}_{2}}+{{H}_{2}}O \\\\ & {{n}_{C{{O}_{2}}}}=0,15\\,\\,mol \\\\ & =>\\,x+y=0,15\\,mol(2) \\\\ \\end{aligned}$<br\/> Gi\u1ea3i (1) v\u00e0 (2) ta \u0111\u01b0\u1ee3c x = 0,1 mol; y = 0,05 mol<br\/>$\\begin{aligned}& \\%{{m}_{CaC{{O}_{3}}}}=\\dfrac{0,1.100}{14,2}.100\\%\\approx 70,42\\% \\\\ & \\%{{m}_{MgC{{O}_{3}}}}=100\\%-70,42\\%=29,58\\% \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1910}],"lesson":{"save":0,"level":2}}