Bài tập trung bình bài Bài 8: Một số bazơ quan trọng

Home Lớp 9 Hóa học 9 Bài 8: Một số bazơ quan trọng Bài tập luyện

{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Thu\u1ed1c th\u1eed \u0111\u1ec3 nh\u1eadn bi\u1ebft dung d\u1ecbch $Ca(OH)_2$ l\u00e0:","select":["A. $Na_2CO_3$ ","B. $KCl$","C. $NaOH$","D. $NaNO_3$"],"hint":"","explain":"Trong 4 \u0111\u00e1p \u00e1n ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n A ph\u1ea3n \u1ee9ng v\u1edbi dung d\u1ecbch $Na_2CO_3$ t\u1ea1o th\u00e0nh k\u1ebft t\u1ee7a theo ph\u1ea3n \u1ee9ng: <br\/>$Ca{{(OH)}_{2}}+N{{a}_{2}}C{{O}_{3}}\\to CaC{{O}_{3}}\\downarrow +2NaOH$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1871},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Dung d\u1ecbch c\u00f3 \u0111\u1ed9 baz\u01a1 m\u1ea1nh nh\u1ea5t trong c\u00e1c dung d\u1ecbch c\u00f3 gi\u00e1 tr\u1ecb pH sau:","select":["A. pH = 8 ","B. pH = 12","C. pH = 10 ","D. pH = 14"],"hint":"","explain":"pH c\u00e0ng l\u1edbn t\u00ednh baz\u01a1 c\u00e0ng m\u1ea1nh<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":1872},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" NaOH c\u00f3 t\u00ednh ch\u1ea5t v\u1eadt l\u00fd n\u00e0o sau \u0111\u00e2y ","select":["A. Natri hi\u0111roxit l\u00e0 ch\u1ea5t r\u1eafn kh\u00f4ng m\u00e0u, \u00edt tan trong n\u01b0\u1edbc ","B. Natri hi\u0111roxit l\u00e0 ch\u1ea5t r\u1eafn kh\u00f4ng m\u00e0u, h\u00fat \u1ea9m m\u1ea1nh, tan nhi\u1ec1u trong n\u01b0\u1edbc v\u00e0 t\u1ecfa nhi\u1ec7t","C. Natri hi\u0111roxit l\u00e0 ch\u1ea5t r\u1eafn kh\u00f4ng m\u00e0u, h\u00fat \u1ea9m m\u1ea1nh v\u00e0 kh\u00f4ng t\u1ecfa nhi\u1ec7t ","D. Natri hi\u0111roxit l\u00e0 ch\u1ea5t r\u1eafn kh\u00f4ng m\u00e0u, kh\u00f4ng tan trong n\u01b0\u1edbc, kh\u00f4ng t\u1ecfa nhi\u1ec7t"],"hint":"","explain":"Natri hi\u0111roxit l\u00e0 ch\u1ea5t r\u1eafn kh\u00f4ng m\u00e0u, h\u00fat \u1ea9m m\u1ea1nh, tan nhi\u1ec1u trong n\u01b0\u1edbc v\u00e0 t\u1ecfa nhi\u1ec7t<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":1873},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Nh\u00f3m c\u00e1c dung d\u1ecbch c\u00f3 pH > 7 l\u00e0","select":["A. $HCl, NaOH$ ","B. $H_2SO_4, HNO_3$","C. $NaOH, Ca(OH)_2$ ","D. $BaCl_2, NaNO_3$"],"hint":"","explain":"C\u00e1c dung d\u1ecbch baz\u01a1 c\u00f3 pH > 7. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n C th\u1ecfa m\u00e3n<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1874},{"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 ho\u00e1 h\u1ecdc n\u00e0o sau \u0111\u00e2y t\u1ea1o ra oxit baz\u01a1 ","select":["A. Cho dd $Ca(OH)_2$ ph\u1ea3n \u1ee9ng v\u1edbi $SO_2$","B. Cho dd $NaOH$ ph\u1ea3n \u1ee9ng v\u1edbi dd $H_2SO_4$","C. Cho dd $Cu(OH)_2$ ph\u1ea3n \u1ee9ng v\u1edbi $HCl$","D. Nung n\u00f3ng $Cu(OH)_2$"],"hint":"","explain":"C\u00e1c baz\u01a1 kh\u00f4ng tan \u0111\u1ec1u b\u1ecb nhi\u1ec7t ph\u00e2n h\u1ee7y t\u1ea1o oxit baz\u01a1. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n D th\u1ecfa m\u00e3n<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":1875},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Dung d\u1ecbch $KOH$ t\u00e1c d\u1ee5ng v\u1edbi nh\u00f3m ch\u1ea5t n\u00e0o sau \u0111\u00e2y \u0111\u1ec1u t\u1ea1o th\u00e0nh mu\u1ed1i v\u00e0 n\u01b0\u1edbc","select":["A. $Ca(OH)_2,CO_2, CuCl_2$ ","B. $P_2O_5; H_2SO_4, SO_3$","C. $CO_2; Na_2CO_3, HNO_3$ ","D. $Na_2O; Fe(OH)_3, FeCl_3$."],"hint":"","explain":"Dung d\u1ecbch ki\u1ec1m t\u00e1c d\u1ee5ng v\u1edbi oxit axit, axit \u0111\u1ec1u t\u1ea1o th\u00e0nh mu\u1ed1i v\u00e0 n\u01b0\u1edbc. ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n B th\u1ecfa m\u00e3n<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1876},{"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 khi ph\u1ea3n \u1ee9ng v\u1edbi nhau t\u1ea1o th\u00e0nh ch\u1ea5t k\u1ebft t\u1ee7a tr\u1eafng :","select":["A. $Ca(OH)_2$ v\u00e0 $Na_2CO_3$ ","B. $NaOH$ v\u00e0 $Na_2CO_3$","C. $KOH$ v\u00e0 $NaNO_3$ ","D. $Ca(OH)_2$ v\u00e0 $NaCl$"],"hint":"","explain":"$Ca{{(OH)}_{2}}+N{{a}_{2}}C{{O}_{3}}\\to 2NaOH+CaC{{O}_{3}}\\downarrow $ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1877},{"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 KH\u00d4NG th\u1ec3 t\u1ed3n t\u1ea1i trong m\u1ed9t dung d\u1ecbch (t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi nhau) l\u00e0","select":["A. $Ca(OH)_2$ v\u00e0 $Na_2CO_3$ ","B. $Ca(OH)_2$ v\u00e0 $NaCl$","C. $Ca(OH)_2$ v\u00e0 $KNO_3$ ","D. $NaOH$ v\u00e0 $NaNO_3$"],"hint":"","explain":"C\u00e1c ch\u1ea5t kh\u00f4ng th\u1ec3 t\u1ed3n t\u1ea1i trong 1 dung d\u1ecbch khi ch\u00fang c\u00f3 t\u00e1c d\u1ee5ng v\u1edbi nhau. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n A th\u1ecfa m\u00e3n <br\/>$Ca{{(OH)}_{2}}+N{{a}_{2}}C{{O}_{3}}\\to 2NaOH+CaC{{O}_{3}}\\downarrow $<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1878},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Dung d\u1ecbch $NaOH$ v\u00e0 dung d\u1ecbch $KOH$ KH\u00d4NG c\u00f3 t\u00ednh ch\u1ea5t n\u00e0o sau \u0111\u00e2y","select":["A. L\u00e0m \u0111\u1ed5i m\u00e0u qu\u1ef3 t\u00edm v\u00e0 phenophtalein ","B. B\u1ecb nhi\u1ec7t ph\u00e2n h\u1ee7y khi \u0111un n\u00f3ng t\u1ea1o th\u00e0nh oxit baz\u01a1 v\u00e0 n\u01b0\u1edbc","C. T\u00e1c d\u1ee5ng v\u1edbi oxit axit t\u1ea1o th\u00e0nh mu\u1ed1i v\u00e0 n\u01b0\u1edbc ","D. T\u00e1c d\u1ee5ng v\u1edbi axit t\u1ea1o th\u00e0nh mu\u1ed1i v\u00e0 n\u01b0\u1edbc"],"hint":"","explain":"C\u00e1c baz\u01a1 kh\u00f4ng tan trong n\u01b0\u1edbc b\u1ecb nhi\u1ec7t ph\u00e2n h\u1ee7y. C\u00f2n c\u00e1c baz\u01a1 tan kh\u00f4ng b\u1ecb ph\u00e2n h\u1ee7y b\u1edfi nhi\u1ec7t.<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":1879},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Dung d\u1ecbch $Ca(OH)_2$ kh\u00f4ng ph\u1ea3n \u1ee9ng \u0111\u01b0\u1ee3c v\u1edbi","select":["A. Dung d\u1ecbch $Na_2CO_3$","B. Dung d\u1ecbch $MgSO_4$","C. Dung d\u1ecbch $CuCl_2$ ","D. Dung d\u1ecbch $KNO_3$"],"hint":"","explain":"Dung d\u1ecbch baz\u01a1 t\u00e1c d\u1ee5ng v\u1edbi mu\u1ed1i th\u00ec s\u1ea3n ph\u1ea9m pahir c\u00f3 ch\u1ea5t k\u1ebft t\u1ee7a ho\u1eb7c ch\u1ea5t kh\u00ed. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n D l\u00e0 kh\u00f4ng th\u1ecfa m\u00e3n v\u00e0 kh\u00f4ng c\u00f3 ph\u1ea3n \u1ee9ng x\u1ea3y ra<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":1880},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Sau khi l\u00e0m th\u00ed nghi\u1ec7m, c\u00f3 nh\u1eefng kh\u00ed th\u1ea3i \u0111\u1ed9c h\u1ea1i: $HCl, H_2S, CO_2, SO_2$. D\u00f9ng ch\u1ea5t n\u00e0o sau \u0111\u00e2y \u0111\u1ec3 lo\u1ea1i b\u1ecf ch\u00fang l\u00e0 t\u1ed1t nh\u1ea5t","select":["A. Mu\u1ed1i $NaCl$ ","B. N\u01b0\u1edbc v\u00f4i trong","C. Dung d\u1ecbch $HCl$ ","D. Dung d\u1ecbch $NaNO_3$"],"hint":"","explain":"Lo\u1ea1i b\u1ecf c\u00e1c kh\u00ed b\u1eb1ng c\u00e1ch cho 1 ch\u1ea5t t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi t\u1ea5t c\u1ea3 c\u00e1c ch\u1ea5t \u0111\u00f3. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n B dung d\u1ecbch n\u01b0\u1edbc v\u00f4i trong ($Ca(OH)_2$) th\u1ecfa m\u00e3n<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1881},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Th\u00e0nh ph\u1ea7n ph\u1ea7n tr\u0103m c\u1ee7a $Na$ v\u00e0 $Ca$ trong h\u1ee3p ch\u1ea5t $NaOH$ v\u00e0 $Ca(OH)_2$ l\u1ea7n l\u01b0\u1ee3t l\u00e0","select":["A. 50,0%, 54,0% ","B. 52,0%, 56,0%","C. 54,1%, 57,5% ","D. 57,5%, 54,1%"],"hint":"","explain":"$\\begin{aligned}& \\%{{m}_{Na}}=\\dfrac{23}{40}.100\\%=57,5\\% \\\\ & \\%{{m}_{Ca}}=\\dfrac{40}{74}.100\\%\\approx 54,1\\% \\\\ \\end{aligned}$<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":1882},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Cho 2,24 l\u00edt kh\u00ed $CO_2$ ( \u0111ktc) h\u1ea5p th\u1ee5 ho\u00e0n to\u00e0n b\u1edfi 200 ml dung d\u1ecbch $Ca(OH)_2$, ch\u1ec9 thu \u0111\u01b0\u1ee3c mu\u1ed1i $CaCO_3$. N\u1ed3ng \u0111\u1ed9 mol c\u1ee7a dung d\u1ecbch $Ca(OH)_2$ c\u1ea7n d\u00f9ng l\u00e0:","select":["A. 0,5 M ","B. 0,25 M","C. 0,1 M ","D. 0,05 M"],"hint":"","explain":"$\\begin{aligned}& C{{O}_{2}}+Ca{{(OH)}_{2}}\\to CaC{{O}_{3}}+{{H}_{2}}O \\\\ & {{n}_{C{{O}_{2}}}}=0,1\\,mol \\\\ & {{n}_{Ca{{(OH)}_{2}}}}={{n}_{C{{O}_{2}}}}=0,1\\,mol \\\\ & {{C}_{M\\,\\,\\,}}(Ca{{(OH)}_{2}})=\\dfrac{0,1}{0,2}=0,5\\,M \\\\ \\end{aligned}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A<\/span><\/span> ","column":2}]}],"id_ques":1883},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" N\u1ebfu r\u00f3t 200 ml dung d\u1ecbch $NaOH$ 1M v\u00e0o \u1ed1ng nghi\u1ec7m \u0111\u1ef1ng 100 ml dung d\u1ecbch $H_2SO_4$ 1M th\u00ec dung d\u1ecbch t\u1ea1o th\u00e0nh sau ph\u1ea3n \u1ee9ng s\u1ebd","select":["A. L\u00e0m qu\u1ef3 t\u00edm chuy\u1ec3n \u0111\u1ecf","B. L\u00e0m qu\u1ef3 t\u00edm chuy\u1ec3n xanh","C. L\u00e0m dung d\u1ecbch phenolphtalein kh\u00f4ng m\u00e0u chuy\u1ec3n \u0111\u1ecf ","D. Kh\u00f4ng l\u00e0m thay \u0111\u1ed5i m\u00e0u qu\u1ef3 t\u00edm."],"hint":"","explain":"$\\begin{aligned}& {{n}_{NaOH}}=0,2\\,mol \\\\ & {{n}_{{{H}_{2}}S{{O}_{4}}}}=0,1\\,mol \\\\ & 2NaOH+{{H}_{2}}S{{O}_{4}}\\to N{{a}_{2}}S{{O}_{4}}+{{H}_{2}}O \\\\ & \\dfrac{{{n}_{NaOH}}}{2}=\\dfrac{{{n}_{{{H}_{2}}S{{O}_{4}}}}}{1}\\, \\\\ \\end{aligned}$ <br\/>2 ch\u1ea5t ph\u1ea3n \u1ee9ng v\u1eeba \u0111\u1ee7, dung d\u1ecbch t\u1ea1o th\u00e0nh c\u00f3 m\u00f4i tr\u01b0\u1eddng trung t\u00ednh n\u00ean kh\u00f4ng l\u00e0m \u0111\u1ed5i m\u00e0u qu\u1ef3 t\u00edm v\u00e0 phenolphtalein.<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":1884},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" H\u00f2a tan 30 g NaOH v\u00e0o 170 g n\u01b0\u1edbc th\u00ec thu \u0111\u01b0\u1ee3c dung d\u1ecbch NaOH c\u00f3 n\u1ed3ng \u0111\u1ed9 l\u00e0:","select":["A. 18% ","B. 16 %","C. 15 % ","D. 17 %"],"hint":"","explain":"$C{{\\%}_{\\text{dd}}}=\\dfrac{30}{30+170}.100\\%=15\\%$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1885},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" $NaOH$ c\u00f3 th\u1ec3 l\u00e0m kh\u00f4 ch\u1ea5t kh\u00ed \u1ea9m sau","select":["A. $CO_2$ ","B. $SO_2$","C. $N_2$ ","D. $HCl$"],"hint":"","explain":"Kh\u00ed \u0111\u01b0\u1ee3c l\u00e0m kh\u00f4 ph\u1ea3i kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi NaOH. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n C th\u1ecfa m\u00e3n<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1886},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Trung h\u00f2a 200 g dung d\u1ecbch NaOH 10% b\u1eb1ng dung d\u1ecbch HCl 3,65%. Kh\u1ed1i l\u01b0\u1ee3ng dung d\u1ecbch HCl c\u1ea7n d\u00f9ng l\u00e0","select":["A. 200g ","B. 300g","C. 400g ","D. 500g"],"hint":"","explain":"$\\begin{aligned}& {{m}_{NaOH}}=\\dfrac{200.10\\%}{100\\%}=20\\,gam \\\\ & {{n}_{NaOH}}=0,5\\,mol \\\\ & NaOH+HCl\\to NaCl+{{H}_{2}}O \\\\ & {{n}_{HCl}}={{n}_{NaOH}}=0,5\\,mol \\\\ & {{m}_{HCl}}=18,25\\,gam \\\\ & {{m}_{\\text{dd}\\,HCl}}=\\dfrac{18,25.100\\%}{3,65\\%}=500\\,gam \\\\ \\end{aligned}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":1887},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" H\u00f2a tan 80 g NaOH v\u00e0o n\u01b0\u1edbc thu \u0111\u01b0\u1ee3c dung d\u1ecbch c\u00f3 n\u1ed3ng \u0111\u1ed9 1M. Th\u1ec3 t\u00edch dung d\u1ecbch NaOH l\u00e0","select":["A. 1 l\u00edt","B. 1,5 l\u00edt","C. 2 l\u00edt ","D. 3 l\u00edt"],"hint":"","explain":"$\\begin{aligned}& {{n}_{NaOH}}=2\\,mol \\\\ & {{V}_{\\,\\,\\text{dd}}}=\\dfrac{2}{1}=2\\,l \\\\ \\end{aligned}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1888},{"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 22,4 l\u00edt kh\u00ed $CO_2$ ( \u0111ktc) v\u00e0o 200g dung d\u1ecbch $NaOH$ 20%. Sau ph\u1ea3n \u1ee9ng t\u1ea1o ra s\u1ea3n ph\u1ea9m n\u00e0o trong s\u1ed1 c\u00e1c s\u1ea3n ph\u1ea9m sau","select":["A. Mu\u1ed1i natricacbonat v\u00e0 n\u01b0\u1edbc ","B. Mu\u1ed1i natri hidrocacbonat","C. Mu\u1ed1i natricacbonat ","D. Mu\u1ed1i natrihi\u0111rocacbonat v\u00e0 natricacbonat"],"hint":"","explain":"$\\begin{aligned}& {{n}_{C{{O}_{2}}}}=1\\,mol \\\\ & {{n}_{NaOH}}=\\dfrac{200.20}{100.40}=1\\,mol \\\\ & \\dfrac{{{n}_{NaOH}}}{{{n}_{C{{O}_{2}}}}}=\\dfrac{1}{1} \\\\ \\end{aligned}$ => Ph\u1ea3n \u1ee9ng t\u1ea1o mu\u1ed1i $NaHCO_3$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":1889},{"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 2,24 l\u00edt kh\u00ed $CO_2$ ( \u0111ktc) v\u00e0o 300 dung d\u1ecbch $NaOH$ 1M. Sau ph\u1ea3n \u1ee9ng c\u00f4 c\u1ea1n dung d\u1ecbch thu \u0111\u01b0\u1ee3c ch\u1ea5t r\u1eafn c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng l\u00e0","select":["A. 10,6g ","B. 14,6g","C. 20g ","D. 25g"],"hint":"","explain":"$\\begin{aligned}& {{n}_{C{{O}_{2}}}}=0,1\\,mol \\\\ & {{n}_{NaOH}}=0,3\\,mol \\\\ & 2NaOH+C{{O}_{2}}\\to N{{a}_{2}}C{{O}_{3}}+{{H}_{2}}O \\\\ & \\dfrac{{{n}_{NaOH}}}{2}>\\dfrac{{{n}_{C{{O}_{2}}}}}{1} \\\\ & =>\\,\\text{NaOH d\u01b0}\\text{.}\\,{{\\text{n}}_{NaOH}}\\,\\text{(d\u01b0)=}0,3-0,1.2=0,1\\,mol \\\\ & {{n}_{N{{a}_{2}}C{{O}_{3}}}}={{n}_{C{{O}_{2}}}}=0,1\\,mol \\\\ & {{m}_{\\text{ch\u1ea5t r\u1eafn}}}=0,1.106+0,1.40=14,6\\,gam \\\\ \\end{aligned}$<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1890}],"lesson":{"save":0,"level":2}}

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Câu hỏi này theo dạng chọn đáp án đúng, sau khi đọc xong câu hỏi, bạn bấm vào một trong số các đáp án mà chương trình đưa ra bên dưới, sau đó bấm vào nút gửi để kiểm tra đáp án và sẵn sàng chuyển sang câu hỏi kế tiếp

Trả lời đúng trong khoảng thời gian quy định bạn sẽ được + số điểm như sau:
Trong khoảng 1 phút đầu tiên	+ 5 điểm
Trong khoảng 1 phút -> 2 phút	+ 4 điểm
Trong khoảng 2 phút -> 3 phút	+ 3 điểm
Trong khoảng 3 phút -> 4 phút	+ 2 điểm
Trong khoảng 4 phút -> 5 phút	+ 1 điểm

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