{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" D\u00e3y c\u00e1c ch\u1ea5t n\u00e0o sau \u0111\u00e2y l\u00e0 mu\u1ed1i axit ","select":["A. $KHCO_3, CaCO_3, Na_2CO_3$. ","B. $Ba(HCO_3)_2, NaHCO_3, Ca(HCO_3)_2$.","C. $Ca(HCO_3)_2, Ba(HCO_3)_2, BaCO_3$. ","D. $Mg(HCO_3)_2, Ba(HCO_3)_2, CaCO_3$."],"hint":"","explain":"<span class='basic_left'>Mu\u1ed1i axit l\u00e0 mu\u1ed1i trong ph\u00e2n t\u1eed g\u1ed1c axit c\u00f2n H. 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":2251},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" D\u00e3y g\u1ed3m c\u00e1c mu\u1ed1i \u0111\u1ec1u tan trong n\u01b0\u1edbc l\u00e0 ","select":["A. $CaCO_3, BaCO_3, Mg(HCO_3)_2, K_2CO_3$.","B. $BaCO_3, NaHCO_3, Mg(HCO_3)_2, Na_2CO_3$.","C. $CaCO_3, BaCO_3, NaHCO_3, MgCO_3$.","D. $Na_2CO_3, Ca(HCO_3)_2, Ba(HCO_3)_2, K_2CO_3$."],"hint":"","explain":"<span class='basic_left'>Mu\u1ed1i cacbonat c\u1ee7a kim lo\u1ea1i ki\u1ec1m tan trong n\u01b0\u1edbc. Mu\u1ed1i hi\u0111rocacbonat tan trong n\u01b0\u1edbc. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n D th\u1ecfa m\u00e3n<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":2252},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" D\u00e3y g\u1ed3m c\u00e1c ch\u1ea5t b\u1ecb ph\u00e2n h\u1ee7y b\u1edfi nhi\u1ec7t l\u00e0 ","select":["A. $Na_2CO_3, MgCO_3, Ca(HCO_3)_2, BaCO_3$.","B. $MgCO_3, BaCO_3, Ca(HCO_3)_2, NaHCO_3$.","C. $K_2CO_3, KHCO_3, MgCO_3, Ca(HCO_3)_2$.","D. $NaHCO_3, KHCO_3, Na_2CO_3, K_2CO_3$."],"hint":"","explain":"<span class='basic_left'>C\u00e1c mu\u1ed1i cacbonat kh\u00f4ng tan \u0111\u1ec1u b\u1ecb nhi\u1ec7t ph\u00e2n. C\u00e1c mu\u1ed1i hi\u0111rocacbonat \u0111\u1ec1u b\u1ecb nhi\u1ec7t ph\u00e2n h\u1ee7y. 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":1}]}],"id_ques":2253},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" D\u00e3y g\u1ed3m c\u00e1c ch\u1ea5t \u0111\u1ec1u ph\u1ea3n \u1ee9ng v\u1edbi dung d\u1ecbch HCl l\u00e0 ","select":["A. $Na_2CO_3, CaCO_3$.","B. $K_2SO_4, Na_2CO_3$.","C. $Na_2SO_4, MgCO_3$. ","D. $Na_2SO_3, KNO_3$."],"hint":"","explain":"<span class='basic_left'>C\u00e1c mu\u1ed1i cacbonat \u0111\u1ec1u c\u00f3 ph\u1ea3n \u1ee9ng v\u1edbi dung d\u1ecbch HCl<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2254},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" C\u1eb7p ch\u1ea5t n\u00e0o sau \u0111\u00e2y c\u00f3 th\u1ec3 c\u00f9ng t\u1ed3n t\u1ea1i trong dung d\u1ecbch ","select":["A. $HNO_3$ v\u00e0 $KHCO_3$. ","B. $Ba(OH)_2$ v\u00e0 $Ca(HCO_3)_2$","C. $Na_2CO_3$ v\u00e0 $CaCl_2$. ","D. $K_2CO_3$ v\u00e0 $Na_2SO_4$."],"hint":"","explain":"<span class='basic_left'>C\u00e1c ch\u1ea5t c\u00f3 th\u1ec3 t\u1ed3n t\u1ea1i trong c\u00f9ng 1 dung d\u1ecbch khi ch\u00fang kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi nhau. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n D th\u1ecfa m\u00e3n<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2255},{"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 $K_2CO_3$ v\u00e0o dung d\u1ecbch $Ca(OH)_2$ quan s\u00e1t hi\u1ec7n t\u01b0\u1ee3ng th\u1ea5y","select":["A. B\u1ecdt kh\u00ed tho\u00e1t ra, dung d\u1ecbch kh\u00f4ng m\u00e0u","B. Kh\u00f4ng c\u00f3 hi\u1ec7n t\u01b0\u1ee3ng g\u00ec","C. Dung d\u1ecbch v\u1ea9n \u0111\u1ee5c, kh\u00f4ng c\u00f3 b\u1ecdt kh\u00ed ","D. Dung d\u1ecbch v\u1ea9n \u0111\u1ee5c, c\u00f3 b\u1ecdt kh\u00ed"],"hint":"","explain":"<span class='basic_left'>${{K}_{2}}C{{O}_{3}}+Ca{{(OH)}_{2}}\\to 2KOH+CaC{{O}_{3}}\\downarrow $ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2256},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" C\u1eb7p ch\u1ea5t n\u00e0o sau \u0111\u00e2y c\u00f3 th\u1ec3 t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi nhau ","select":["A. $HCl$ v\u00e0 $KHCO_3$.","B. $Na_2CO_3$ v\u00e0 $K_2CO_3$.","C. $K_2CO_3$ v\u00e0 $NaCl$.","D. $CaCO_3$ v\u00e0 $NaHCO_3$."],"hint":"","explain":"<span class='basic_left'>$KHC{{O}_{3}}+HCl\\to KCl+C{{O}_{2}}\\uparrow +{{H}_{2}}O$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2257},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" S\u1ea3n ph\u1ea9m nhi\u1ec7t ph\u00e2n mu\u1ed1i hi\u0111rocacbonat l\u00e0 ","select":["A. $CO_2$.","B. $Cl_2$. ","C. $CO$. ","D. $Na_2O$."],"hint":"","explain":"<span class='basic_left'>Khi nhi\u1ec7t ph\u00e2n mu\u1ed1i hi\u0111rocacbonat thu \u0111\u01b0\u1ee3c mu\u1ed1i cacbonat v\u00e0 kh\u00ed cacbonic<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2258},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" D\u00e3y g\u1ed3m c\u00e1c mu\u1ed1i \u0111\u1ec1u ph\u1ea3n \u1ee9ng \u0111\u01b0\u1ee3c v\u1edbi dung d\u1ecbch $NaOH$ l\u00e0 ","select":["A. $Na_2CO_3, NaHCO_3, MgCO_3, K_2CO_3$.","B. $NaHCO_3, Ca(HCO_3)_2, Mg(HCO_3)_2, Ba(HCO_3)_2$.","C. $Ca(HCO_3)_2, Mg(HCO_3)_2, BaCO_3, Ba(HCO_3)_2$.","D. $CaCO_3, BaCO_3, Na_2CO_3, MgCO_3$."],"hint":"","explain":"<span class='basic_left'>T\u1ea5t c\u1ea3 c\u00e1c mu\u1ed1i hi\u0111rocacbonat \u0111\u1ec1u ph\u1ea3n \u1ee9ng \u0111\u01b0\u1ee3c v\u1edbi dung d\u1ecbch ki\u1ec1m<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":2259},{"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 $H_2CO_3$ nh\u00fang qu\u1ef3 t\u00edm v\u00e0o hi\u1ec7n t\u01b0\u1ee3ng x\u1ea3y ra l\u00e0:","select":["A. Qu\u1ef3 kh\u00f4ng \u0111\u1ed5i m\u00e0u","B. Qu\u1ef3 chuy\u1ec3n m\u00e0u xanh ","C. Qu\u1ef3 chuy\u1ec3n m\u00e0u \u0111\u1ecf nh\u1ea1t ","D. Qu\u1ef3 m\u1ea5t m\u00e0u "],"hint":"","explain":"<span class='basic_left'>Dung d\u1ecbch $H_2CO_3$ c\u00f3 m\u00f4i tr\u01b0\u1eddng axit y\u1ebfu n\u00ean l\u00e0m qu\u1ef3 chuy\u1ec3n m\u00e0u \u0111\u1ecf nh\u1ea1t<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2260},{"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 $K_2CO_3$ t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch $BaCl_2$. Hi\u1ec7n t\u01b0\u1ee3ng quan s\u00e1t \u0111\u01b0\u1ee3c l\u00e0","select":["A. Kh\u00f4ng c\u00f3 hi\u1ec7n t\u01b0\u1ee3ng g\u00ec","B. Dung d\u1ecbch chuy\u1ec3n m\u00e0u xanh","C. Dung d\u1ecbch c\u00f3 v\u1ea9n \u0111\u1ee5c ","D. C\u00f3 kh\u00ed kh\u00f4ng m\u00e0u tho\u00e1t ra."],"hint":"","explain":"<span class='basic_left'>${{K}_{2}}C{{O}_{3}}+BaC{{l}_{2}}\\to 2KCl+BaC{{O}_{3}}\\downarrow $ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2261},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Cho dung d\u1ecbch $AgNO_3$ ph\u1ea3n \u1ee9ng v\u1eeba \u0111\u1ee7 v\u1edbi dung d\u1ecbch g\u1ed3m $Na_2CO_3$ v\u00e0 $NaCl$. C\u00e1c ch\u1ea5t thu \u0111\u01b0\u1ee3c sau ph\u1ea3n \u1ee9ng ","select":["A. $AgCl, AgNO_3, Na_2CO_3$.","B. $Ag_2CO_3, AgCl, AgNO_3$","C. $Ag_2CO_3, AgNO_3, Na_2CO_3$.","D. $AgCl, Ag_2CO_3, NaNO_3$"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& 2AgN{{O}_{3}}+N{{a}_{2}}C{{O}_{3}}\\to 2NaN{{O}_{3}}+A{{g}_{2}}C{{O}_{3}}\\downarrow \\\\ & AgN{{O}_{3}}+NaCl\\to NaN{{O}_{3}}+AgCl\\downarrow \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2262},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Cho 21 gam $MgCO_3$ t\u00e1c d\u1ee5ng v\u1edbi m\u1ed9t l\u01b0\u1ee3ng v\u1eeba \u0111\u1ee7 dung d\u1ecbch $HCl$ 2M. Th\u1ec3 t\u00edch dung d\u1ecbch $HCl$ \u0111\u00e3 d\u00f9ng l\u00e0 ","select":["A. 0,50 l\u00edt.","B. 0,25 l\u00edt. ","C. 0,75 l\u00edt. ","D. 0,15 l\u00edt"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & MgC{{O}_{3}}+2HCl\\to MgC{{l}_{2}}+C{{O}_{2}}+{{H}_{2}}O \\\\ & {{n}_{HCl}}=2{{n}_{MgC{{O}_{3}}}}=0,5\\,mol \\\\ & {{V}_{HCl}}=\\dfrac{0,5}{2}=0,25\\,l \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2263},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Th\u00ed nghi\u1ec7m n\u00e0o sau \u0111\u00e2y c\u00f3 hi\u1ec7n t\u01b0\u1ee3ng sinh ra k\u1ebft t\u1ee7a tr\u1eafng v\u00e0 b\u1ecdt kh\u00ed tho\u00e1t ra kh\u1ecfi dung d\u1ecbch ","select":["A. Nh\u1ecf t\u1eebng gi\u1ecdt dung d\u1ecbch $NaOH$ v\u00e0o \u1ed1ng nghi\u1ec7m \u0111\u1ef1ng dung d\u1ecbch $CuCl_2$.","B. Nh\u1ecf t\u1eeb t\u1eeb dung d\u1ecbch $H_2SO_4$ v\u00e0o \u1ed1ng nghi\u1ec7m c\u00f3 s\u1eb5n m\u1ed9t m\u1eabu $BaCO_3$.","C. Nh\u1ecf t\u1eeb t\u1eeb dung d\u1ecbch $BaCl_2$ v\u00e0o \u1ed1ng nghi\u1ec7m \u0111\u1ef1ng dung d\u1ecbch $AgNO_3$.","D. Nh\u1ecf t\u1eeb t\u1eeb dung d\u1ecbch $HCl$ v\u00e0o \u1ed1ng nghi\u1ec7m \u0111\u1ef1ng dung d\u1ecbch $Na_2CO_3$."],"hint":"","explain":"<span class='basic_left'>${{H}_{2}}S{{O}_{4}}+BaC{{O}_{3}}\\to BaS{{O}_{4}}\\downarrow +C{{O}_{2}}\\uparrow +{{H}_{2}}O$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":2264},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Kh\u1ed1i l\u01b0\u1ee3ng k\u1ebft t\u1ee7a t\u1ea1o ra, khi cho 21,2 gam $Na_2CO_3$ t\u00e1c d\u1ee5ng v\u1eeba \u0111\u1ee7 v\u1edbi dung d\u1ecbch $Ba(OH)_2$ l\u00e0 ","select":["A. 3,94 gam","B. 39,4 gam","C. 25,7 gam","D. 51,4 gam"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & N{{a}_{2}}C{{O}_{3}}+Ba{{(OH)}_{2}}\\to 2NaOH+BaC{{O}_{3}}\\downarrow \\\\ & {{n}_{\\downarrow }}={{n}_{N{{a}_{2}}C{{O}_{3}}}}=0,2\\,mol \\\\ & {{m}_{\\downarrow }}=39,4\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2265},{"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 3 l\u1ecd \u0111\u1ef1ng 3 h\u00f3a ch\u1ea5t: $Cu(OH)_2, BaCl_2, KHCO_3$ \u0111\u1ec3 nh\u1eadn bi\u1ebft 3 l\u1ecd tr\u00ean c\u1ea7n d\u00f9ng h\u00f3a ch\u1ea5t n\u00e0o ","select":["A. $NaCl$.","B. $NaOH$.","C. $H_2SO_4$. ","D. $CaCl_2$"],"hint":"","explain":"<span class='basic_left'>Cho 3 h\u00f3a ch\u1ea5t l\u1ea7n l\u01b0\u1ee3t ph\u1ea3n \u1ee9ng v\u1edbi dung d\u1ecbch $H_2SO_4$.. Ch\u1ea5t n\u00e0o t\u1ea1o k\u1ebft t\u1ee7a tr\u1eafng l\u00e0 $BaCl_2$.<br\/> $BaC{{l}_{2}}+{{H}_{2}}S{{O}_{4}}\\to 2HCl+BaS{{O}_{4}}\\downarrow $ <br\/>Ch\u1ea5t n\u00e0o t\u1ea1o kh\u00ed kh\u00f4ng m\u00e0u tho\u00e1t ra l\u00e0 $KHCO_3$ <br\/> $2KHC{{O}_{3}}+{{H}_{2}}S{{O}_{4}}\\to {{K}_{2}}S{{O}_{4}}+2C{{O}_{2}}+2{{H}_{2}}O$ . <br\/>Ch\u1ea5t c\u00f2n l\u1ea1i t\u1ea1o dung d\u1ecbch m\u00e0u xanh l\u00e0 $Cu(OH)_2$. <br\/> $Cu{{(OH)}_{2}}+{{H}_{2}}S{{O}_{4}}\\to CuS{{O}_{4}}+2{{H}_{2}}O$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2266},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Nung ho\u00e0n to\u00e0n h\u1ed7n h\u1ee3p 2 mu\u1ed1i $CaCO_3$ v\u00e0 $MgCO_3$ thu \u0111\u01b0\u1ee3c 76 gam hai oxit v\u00e0 33,6 l\u00edt $CO_2$ (\u0111ktc). Kh\u1ed1i l\u01b0\u1ee3ng h\u1ed7n h\u1ee3p mu\u1ed1i ban \u0111\u1ea7u l\u00e0 ","select":["A. 142 gam","B. 124 gam. ","C. 141 gam. ","D. 140 gam."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & CaC{{O}_{3}}\\xrightarrow{{{t}^{0}}}CaO+C{{O}_{2}} \\\\ & MgC{{O}_{3}}\\xrightarrow{{{t}^{0}}}MgO+C{{O}_{2}} \\\\ & {{n}_{C{{O}_{2}}}}=1,5\\,mol \\\\ & {{m}_{hh\\,\\text{?}}}={{m}_{\\text{ox}it}}+{{m}_{C{{O}_{2}}}}=76+1,5.44=142\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2267},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Tr\u1ed9n 2 l\u00edt dung d\u1ecbch $Na_2CO_3$ 0,1M v\u1edbi 3 l\u00edt dung d\u1ecbch $Na_2CO_3$ 0,5M. N\u1ed3ng \u0111\u1ed9 mol c\u1ee7a dung d\u1ecbch thu \u0111\u01b0\u1ee3c l\u00e0","select":["A. 0,33M","B. 0,34M","C. 0,35M","D. 0,4M"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & \\text{T\u1ed5ng}\\,{{n}_{N{{a}_{2}}C{{O}_{3}}}}=0,1.2+0,5.3=1,7\\,mol \\\\ & {{V}_{\\text{t\u1ed5ng}}}=2+3=5\\,l \\\\ & {{C}_{M\\,\\,\\text{dd}}}=\\dfrac{1,7}{5}=0,34\\,M \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2268},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Cho 19 gam h\u1ed7n h\u1ee3p $Na_2CO_3$ v\u00e0 $NaHCO_3$ t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch $HCl$ d\u01b0, sinh ra 4,48 l\u00edt kh\u00ed (\u0111ktc). Kh\u1ed1i l\u01b0\u1ee3ng m\u1ed7i mu\u1ed1i trong h\u1ed7n h\u1ee3p ban \u0111\u1ea7u l\u00e0","select":["A. 10,6 gam v\u00e0 8,4 gam. ","B. 16 gam v\u00e0 3 gam. ","C. 10,5 gam v\u00e0 8,5 gam. ","D. 16 gam v\u00e0 4,8 gam."],"hint":"","explain":"<span class='basic_left'>G\u1ecdi x, y l\u1ea7n l\u01b0\u1ee3t l\u00e0 s\u1ed1 mol c\u1ee7a $Na_2CO_3$ v\u00e0 $NaHCO_3$ trong h\u1ed7n h\u1ee3p ban \u0111\u1ea7u. <br\/>Kh\u1ed1i l\u01b0\u1ee3ng h\u1ed7n h\u1ee3p b\u1eb1ng 19 gam n\u00ean ta c\u00f3: 106x + 84y = 19 (1). <br\/>$\\begin{aligned} & N{{a}_{2}}C{{O}_{3}}+2HCl\\to 2NaCl+C{{O}_{2}}\\uparrow +{{H}_{2}}O \\\\ & x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x \\\\ & NaHC{{O}_{3}}+HCl\\to NaCl+C{{O}_{2}}\\uparrow +{{H}_{2}}O \\\\ & y\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,y \\\\ & {{n}_{C{{O}_{2}}}}=0,2\\,mol\\Rightarrow x+y=0,2\\,\\,\\,\\,\\,\\,\\,(2) \\\\ \\end{aligned}$<br\/>Gi\u1ea3i (1) v\u00e0 (2) ta c\u00f3 x = y = 0,1 mol. <br\/>${{m}_{N{{a}_{2}}C{{O}_{3}}}}=10,6\\,gam;\\,\\,\\,{{m}_{NaHC{{O}_{3}}}}=8,4\\,gam$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2269},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Nung h\u1ed7n h\u1ee3p 18,4 gam g\u1ed3m $MgCO_3$ v\u00e0 $CaCO_3$. S\u1ea3n ph\u1ea9m kh\u00ed thu \u0111\u01b0\u1ee3c d\u1eabn qua dung d\u1ecbch n\u01b0\u1edbc v\u00f4i trong d\u01b0 th\u1ea5y c\u00f3 20 gam k\u1ebft t\u1ee7a. Kh\u1ed1i l\u01b0\u1ee3ng mu\u1ed1i canxi cacbonat c\u00f3 trong h\u1ed7n h\u1ee3p ban \u0111\u1ea7u l\u00e0","select":["A. 8,4 gam","B. 12 gam ","C. 10 gam ","D. 15 gam"],"hint":"","explain":"<span class='basic_left'>G\u1ecdi x, y l\u1ea7n l\u01b0\u1ee3t l\u00e0 s\u1ed1 mol c\u1ee7a $MgCO_3$ v\u00e0 $CaCO_3$ trong h\u1ed7n h\u1ee3p ban \u0111\u1ea7u.<br\/> Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a h\u1ed7n h\u1ee3p l\u00e0 18,4 gam n\u00ean ta c\u00f3: 84x + 100y = 18,4 (1)<br\/>$\\begin{aligned} & MgC{{O}_{3}}\\xrightarrow{{{t}^{0}}}MgO+C{{O}_{2}} \\\\ & x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x \\\\ & CaC{{O}_{3}}\\xrightarrow{{{t}^{0}}}CaO+C{{O}_{2}} \\\\ & y\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,y \\\\ & C{{O}_{2}}+CaC{{O}_{3}}\\to CaC{{O}_{3}}\\downarrow +{{H}_{2}}O \\\\ & {{n}_{C{{O}_{2}}}}={{n}_{\\downarrow }}=0,2\\,mol\\Rightarrow x+y=0,2\\,mol\\,\\,\\,\\,(2) \\\\ \\end{aligned}$ <br\/>Gi\u1ea3i (1) v\u00e0 (2) ta c\u00f3 x = y = 0,1 mol <br\/>${{m}_{CaC{{O}_{3}}}}=10\\,gam$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2270}],"lesson":{"save":0,"level":2}}