{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" S\u1eaft (II) oxit t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi","select":["A. Dung d\u1ecbch $Ca(OH)_2$","B. Dung d\u1ecbch $Na_2SO_4$","C. N\u01b0\u1edbc ","D. Dung d\u1ecbch $H_2SO_4$"],"hint":"","explain":"<span class='basic_left'>FeO l\u00e0 oxit baz\u01a1 n\u00ean c\u00f3 th\u1ec3 t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi dung d\u1ecbch axit.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":1951},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" B\u1eb1ng ph\u01b0\u01a1ng ph\u00e1p n\u00e0o kh\u1eb3ng \u0111\u1ecbnh \u0111\u01b0\u1ee3c trong kh\u00ed oxi c\u00f3 l\u1eabn kh\u00ed $CO_2$ v\u00e0 kh\u00ed $SO_2$","select":["A. Cho kh\u00ed oxi \u0111i qua dung d\u1ecbch $KCl$","B. Cho kh\u00ed oxi \u0111i qua dung d\u1ecbch $Ca(OH)_2$","C. Cho kh\u00ed oxi \u0111i qua dung d\u1ecbch $HCl$ ","D. C\u1ea3 3 ph\u01b0\u01a1ng ph\u00e1p tr\u00ean \u0111\u1ec1u \u0111\u00fang"],"hint":"","explain":"<span class='basic_left'> $CO_2$ v\u00e0 $SO_2$ \u0111\u1ec1u l\u00e0 c\u00e1c kh\u00ed l\u00e0m \u0111\u1ee5c n\u01b0\u1edbc v\u00f4i trong <br\/>$\\begin{aligned}& C{{O}_{2}}+Ca{{(OH)}_{2}}\\to CaC{{O}_{3}}\\downarrow +{{H}_{2}}O \\\\ & S{{O}_{2}}+Ca{{(OH)}_{2}}\\to CaS{{O}_{3}}\\downarrow +{{H}_{2}}O \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1952},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Cho c\u00e1c ch\u1ea5t c\u00f3 t\u00ean g\u1ecdi sau: Natri oxit, canxi hi\u0111roxit, axit sunfuric.D\u00e3y c\u00f4ng th\u1ee9c ho\u00e1 h\u1ecdc c\u1ee7a c\u00e1c ch\u1ea5t \u1ee9ng v\u1edbi t\u00ean g\u1ecdi tr\u00ean l\u1ea7n l\u01b0\u1ee3t l\u00e0","select":["A. $Na_2O, CaO, H_2SO_4$ ","B. $Na_2O, Ca(OH)_2, H_2SO_4$","C. $Na_2O, CaO, H_2SO_3$","D. $Na_2O, Ca(OH)_2, H_2SO_3$"],"hint":"","explain":"<span class='basic_left'>\u0110\u00e1p \u00e1n: B<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1953},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Cho ph\u01b0\u01a1ng tr\u00ecnh ph\u1ea3n \u1ee9ng sau:$Fe+HCl\\to X+{{H}_{2}}$ . X l\u00e0:","select":["A. $FeCl$ ","B. $FeCl_2$","C. $FeCl_3$ ","D. $FeCl_4$"],"hint":"","explain":"<span class='basic_left'>$Fe+2HCl\\to FeC{{l}_{2}}+{{H}_{2}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1954},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho $CuSO_4$ ph\u1ea3n \u1ee9ng v\u1edbi $NaOH, HCl, BaCl_2$. S\u1ed1 ph\u1ea3n \u1ee9ng x\u1ea3y ra l\u00e0:","select":["A. 0 ","B. 1","C. 2 ","D. 3"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& CuS{{O}_{4}}+2NaOH\\to N{{a}_{2}}S{{O}_{4}}+Cu{{(OH)}_{2}}\\downarrow \\\\ & CuS{{O}_{4}}+BaC{{l}_{2}}\\to BaS{{O}_{4}}\\downarrow +CuC{{l}_{2}} \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1955},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Ph\u1ea3n \u1ee9ng gi\u1eefa HCl v\u00e0 NaOH thu\u1ed9c lo\u1ea1i ph\u1ea3n \u1ee9ng","select":["A. Trung h\u00f2a ","B. Nhi\u1ec7t ph\u00e2n","C. Trao \u0111\u1ed5i ","D. Th\u1ebf"],"hint":"","explain":"<span class='basic_left'>Ph\u1ea3n \u1ee9ng gi\u1eefa dung d\u1ecbch axit v\u00e0 dung d\u1ecbch baz\u01a1 thu\u1ed9c ph\u1ea3n \u1ee9ng trung h\u00f2a.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1956},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Khi cho $SO_2$ t\u00e1c d\u1ee5ng v\u1edbi n\u01b0\u1edbc axit t\u01b0\u01a1ng \u1ee9ng sinh ra l\u00e0:","select":["A. $H_2S$ ","B. $H_2SO_3$","C. $H_2SO_4$ ","D. $SO_3$"],"hint":"","explain":"<span class='basic_left'>$S{{O}_{2}}+{{H}_{2}}O\\to {{H}_{2}}S{{O}_{3}}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1957},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" H\u00f2a tan 40 gam NaOH v\u00e0o 120 gam n\u01b0\u1edbc thu \u0111\u01b0\u1ee3c dung d\u1ecbch c\u00f3 n\u1ed3ng \u0111\u1ed9 l\u00e0:","select":["A. 20% ","B. 25%","C. 30% ","D. 35%"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{m}_{\\text{dd}}}=40+120=160\\,gam \\\\ & C{{\\%}_{\\text{dd}}}=\\dfrac{40}{160}.100\\%=25\\% \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1958},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Oxit axit c\u00f3 th\u1ec3 t\u00e1c d\u1ee5ng v\u1edbi oxit baz\u01a1 c\u1ee7a kim lo\u1ea1i ki\u1ec1m, ki\u1ec1m th\u1ed5 \u0111\u1ec3 t\u1ea1o th\u00e0nh:","select":["A. Axit v\u00e0 n\u01b0\u1edbc ","B. Baz\u01a1 v\u00e0 n\u01b0\u1edbc","C. Mu\u1ed1i","D. Mu\u1ed1i v\u00e0 n\u01b0\u1edbc"],"hint":"","explain":"<span class='basic_left'>$VD:\\,C{{O}_{2}}+N{{a}_{2}}O\\to N{{a}_{2}}C{{O}_{3}}\\,$ <br\/> $Na_2CO_3$ l\u00e0 mu\u1ed1i<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1959},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" H\u00f2a tan ho\u00e0n to\u00e0n 1,6 gam $CuSO_4$ trong 200 ml n\u01b0\u1edbc thu \u0111\u01b0\u1ee3c dung d\u1ecbch c\u00f3 n\u1ed3ng \u0111\u1ed9 l\u00e0 (coi th\u1ec3 t\u00edch dung d\u1ecbch ch\u00ednh l\u00e0 th\u1ec3 t\u00edch c\u1ee7a n\u01b0\u1edbc)","select":["A. 0,02M ","B. 0,05M","C. 0,2M ","D. 0,5M"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{CuS{{O}_{4}}}}=0,01\\,mol \\\\ & {{C}_{M\\,dd}}=\\dfrac{0,01}{0,2}=0,05M \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1960},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Ph\u1ea3n \u1ee9ng n\u00e0o sau \u0111\u00e2y KH\u00d4NG x\u1ea3y ra \u1edf \u0111i\u1ec1u ki\u1ec7n th\u01b0\u1eddng:","select":["A. $Fe$ + $H_2O$ ","B. $CaO$ + $H_2O$","C. $Na$ + $H_2O$ ","D. $BaO$ + $CO_2$"],"hint":"","explain":"<span class='basic_left'>Ch\u1ec9 c\u00f3 kim lo\u1ea1i ki\u1ec1m, ki\u1ec1m th\u1ed5 m\u1edbi t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi n\u01b0\u1edbc \u1edf \u0111i\u1ec1u ki\u1ec7n th\u01b0\u1eddng.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1961},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" 0,2 mol dung d\u1ecbch $FeCl_2$ c\u00f3 th\u1ec3 t\u00e1c d\u1ee5ng t\u1ed1i \u0111a v\u1edbi x mol dung d\u1ecbch $Ba(OH)_2$. Gi\u00e1 tr\u1ecb c\u1ee7a x l\u00e0","select":["A. 0,1 ","B. 0,2","C. 0,3 ","D. 0,4"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& FeC{{l}_{2}}+Ba{{(OH)}_{2}}\\to BaC{{l}_{2}}+Fe{{(OH)}_{2}}\\downarrow \\\\ & {{n}_{Ba{{(OH)}_{2}}}}={{n}_{FeC{{l}_{2}}}}=0,2\\,mol \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1962},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" D\u1eabn 4,48 l\u00edt kh\u00ed $CO_2$ v\u00e0o dung d\u1ecbch $KOH$ d\u01b0 kh\u1ed1i l\u01b0\u1ee3ng mu\u1ed1i thu \u0111\u01b0\u1ee3c l\u00e0","select":["A. 13,8 g ","B. 27,6 g ","C. 34,5 g ","D. 41,4 g"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& C{{O}_{2}}+2KOH\\to {{K}_{2}}C{{O}_{3}}+{{H}_{2}}O \\\\ & {{n}_{{{K}_{2}}C{{O}_{3}}}}={{n}_{C{{O}_{2}}}}=0,2\\,mol \\\\ & {{m}_{{{K}_{2}}C{{O}_{3}}}}=138.0,2=27,6\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1963},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Khi cho dung d\u1ecbch NaOH t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch $H_2S$ c\u00f3 th\u1ec3 thu \u0111\u01b0\u1ee3c t\u1ed1i \u0111a bao nhi\u00eau Mu\u1ed1i","select":["A. 1 ","B. 2","C. 3 ","D. 4"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& 2NaOH+{{H}_{2}}S\\to N{{a}_{2}}S+2{{H}_{2}}O \\\\ & NaOH+{{H}_{2}}S\\to NaHS+{{H}_{2}}O \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1964},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" C\u00e1c baz\u01a1 c\u00f3 th\u1ec3 b\u1ecb nhi\u1ec7t ph\u00e2n l\u00e0","select":["A. C\u00e1c baz\u01a1 c\u1ee7a kim lo\u1ea1i ki\u1ec1m","B. C\u00e1c baz\u01a1 c\u1ee7a kim lo\u1ea1i ki\u1ec1m th\u1ed5","C. C\u00e1c baz\u01a1 tan trong n\u01b0\u1edbc ","D. C\u00e1c baz\u01a1 kh\u00f4ng tan trong n\u01b0\u1edbc"],"hint":"","explain":"<span class='basic_left'>C\u00e1c baz\u01a1 kh\u00f4ng tan b\u1ecb nhi\u1ec7t ph\u00e2n h\u1ee7y t\u1ea1o oxit v\u00e0 n\u01b0\u1edbc<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":1965},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Th\u1ec3 t\u00edch kh\u00ed $CO_2$ \u0111\u1ec3 ph\u1ea3n \u1ee9ng ho\u00e0n to\u00e0n v\u1edbi 15,3 gam $BaO$ l\u00e0:","select":["A. 1.12 l\u00edt ","B. 2,24 l\u00edt","C. 3,36 l\u00edt ","D. 4,48 l\u00edt"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& C{{O}_{2}}+BaO\\to BaC{{O}_{3}} \\\\ & {{n}_{C{{O}_{2}}}}={{n}_{BaO}}=0,1\\,mol \\\\ & {{V}_{C{{O}_{2}}}}=2,24\\,l \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1966},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Mu\u1ed1i n\u00e0o sau \u0111\u00e2y l\u00e0 mu\u1ed1i axit","select":["A. $Na_2CO_3$","B. $K_2SO_4$","C. $KHCO_3$","D. $BaCl_2$"],"hint":"","explain":"<span class='basic_left'>Mu\u1ed1i axit l\u00e0 mu\u1ed1i cuar kim lo\u1ea1i v\u00e0 g\u1ed1c axit v\u1eabn c\u00f2n H. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n C th\u1ecfa m\u00e3n<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1967},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Kh\u1ed1i l\u01b0\u1ee3ng k\u1ebft t\u1ee7a thu \u0111\u01b0\u1ee3c khi cho 200 ml dung d\u1ecbch $H_2SO_4$ 1M t\u00e1c d\u1ee5ng v\u1edbi l\u01b0\u1ee3ng d\u01b0 dung d\u1ecbch $BaCl_2$ l\u00e0","select":["A. 23,3 gam ","B. 34,95 gam","C. 46,6 gam ","D. 58,25 gam"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& Ba{{(OH)}_{2}}+{{H}_{2}}S{{O}_{4}}\\to BaS{{O}_{4}}\\downarrow +2{{H}_{2}}O \\\\ & {{n}_{BaS{{O}_{4}}}}={{n}_{{{H}_{2}}S{{O}_{4}}}}=0,2\\,mol \\\\ & {{m}_{\\downarrow }}=0,2.233=46,6\\,gam \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1968},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho 23,2 gam $Fe_3O_4$ t\u00e1c d\u1ee5ng ho\u00e0n to\u00e0n v\u1edbi l\u01b0\u1ee3ng d\u01b0 dung d\u1ecbch $HCl$ thu \u0111\u01b0\u1ee3c m gam mu\u1ed1i. Gi\u00e1 tr\u1ecb c\u1ee7a m l\u00e0","select":["B. 12,7 ","B. 32,5 ","C. 45,2","D. 57,9"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{F{{e}_{3}}{{O}_{4}}}}=\\dfrac{23,2}{232}=0,1\\,mol \\\\ & F{{e}_{3}}{{O}_{4}}+8HCl\\to FeC{{l}_{2}}+2FeC{{l}_{3}}+4{{H}_{2}}O \\\\ & {{n}_{FeC{{l}_{2}}}}={{n}_{F{{e}_{3}}{{O}_{4}}}}=0,1\\,mol \\\\ & {{n}_{FeC{{l}_{3}}}}=2{{n}_{F{{e}_{3}}{{O}_{4}}}}=0,2\\,mol \\\\ & m=0,1.127+0,2.162,5=45,2\\,gam \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1969},{"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 0,2 mol $CuCl_2$ v\u1edbi 20 gam $NaOH$ thu \u0111\u01b0\u1ee3c k\u1ebft t\u1ee7a X. Nung X \u0111\u1ebfn kh\u1ed1i l\u01b0\u1ee3ng kh\u00f4ng \u0111\u1ed5i thu \u0111\u01b0\u1ee3c ch\u1ea5t r\u1eafn c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng m gam. Gi\u00e1 tr\u1ecb c\u1ee7a m l\u00e0","select":["A. 8 ","B. 16","C. 20 ","D. 24"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{NaOH}}=0,5\\,mol \\\\ & CuC{{l}_{2}}+2NaOH\\to 2NaCl+Cu{{(OH)}_{2}}\\downarrow \\\\ & \\dfrac{{{n}_{CuC{{l}_{2}}}}}{1}<\\dfrac{{{n}_{NaOH}}}{2} \\\\ & =>\\,NaOH\\text{ d\u01b0}\\text{.} \\\\ \\end{aligned}$ <br\/>T\u00ednh theo $CuCl_2$ <br\/>$\\begin{aligned}& {{n}_{Cu{{(OH)}_{2}}}}={{n}_{CuC{{l}_{2}}}}=0,2\\,mol \\\\ & Cu{{(OH)}_{2}}\\xrightarrow{{{t}^{0}}}CuO+{{H}_{2}}O \\\\ & {{n}_{Cuo}}={{n}_{Cu{{(OH)}_{2}}}}=0,2\\,mol \\\\ & {{m}_{CuO}}=0,2.80=16\\,gam \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1970}],"lesson":{"save":0,"level":2}}