Bài tập trung bình bài Bài 1: Tính chất hóa học của oxit. Khái quát về sự phân loại oxit

Home Lớp 9 Hóa học 9 Bài 1: Tính chất hóa học của oxit. Khái quát về sự phân loại oxit Bài tập luyện

{"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 ch\u1ea5t sau \u0111\u00e2y ch\u1ec9 g\u1ed3m c\u00e1c oxit:","select":["A. $MgO, Ba(OH)_2, CaSO_4, HCl$.","B. $MgO, CaO, CuO, FeO$.","C. $SO_2, CO_2, NaOH, CaSO_4$.","D. $CaO, Ba(OH)_2, MgSO_4, BaO$."],"hint":"","explain":"oxit l\u00e0 h\u1ee3p ch\u1ea5t c\u1ee7a oxi v\u00e0 1 nguy\u00ean t\u1ed1 kh\u00e1c. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n B th\u1ecfa m\u00e3n<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":1751},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" C\u00f4ng th\u1ee9c ho\u00e1 h\u1ecdc c\u1ee7a s\u1eaft (III) oxit l\u00e0:","select":["A.$ Fe_2O_3$.","B.$ Fe_3O_4$.","C.$ FeO$. ","D.$ Fe_3O_2$."],"hint":"","explain":" $Fe$ c\u00f3 h\u00f3a tr\u1ecb III, oxi c\u00f3 h\u00f3a tr\u1ecb II. N\u00ean c\u00f4ng th\u1ee9c c\u1ee7a s\u1eaft (III) oxit l\u00e0 $Fe_2O_3$. <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1752},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Ch\u1ea5t t\u00e1c d\u1ee5ng v\u1edbi n\u01b0\u1edbc t\u1ea1o ra dung d\u1ecbch baz\u01a1 l\u00e0:","select":["A. $CO_2$. ","B. $Na_2O$.","C. $SO_2$.","D. $P_2O_5$."],"hint":"","explain":"C\u00e1c oxit c\u1ee7a kim lo\u1ea1i ki\u1ec1m, ki\u1ec1m th\u1ed5 t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi n\u01b0\u1edbc t\u1ea1o dung d\u1ecbch baz\u01a1. Ch\u1ec9 c\u00f3 $Na_2O$ th\u1ecfa m\u00e3n<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1753},{"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 t\u00e1c d\u1ee5ng v\u1edbi n\u01b0\u1edbc t\u1ea1o ra dung d\u1ecbch axit l\u00e0:","select":["A. $CaO$.","B. $BaO$.","C. $Na_2O$. ","D. $SO_3$. "],"hint":"","explain":"C\u00e1c oxit axit t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi n\u01b0\u1edbc t\u1ea1o dung d\u1ecbch axit. Ch\u1ec9 c\u00f3 $SO_3$ th\u1ecfa m\u00e3n<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":1754},{"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 g\u1ed3m c\u00e1c oxit axit l\u00e0:","select":["A. $CO_2, SO_2, NO, P_2O_5$.","B. $CO_2, SO_3, Na_2O, NO_2$.","C. $SO_2, P_2O_5, CO_2, SO_3$. ","D. $H_2O, CO, NO, Al_2O_3$."],"hint":"","explain":"Oxit axit l\u00e0 h\u1ee3p ch\u1ea5t c\u1ee7a oxi v\u00e0 1 nguy\u00ean t\u1ed1 phi kim.<br\/> Oxit axit t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi n\u01b0\u1edbc \u0111\u1ec3 t\u1ea1o axit t\u01b0\u01a1ng \u1ee9ng. <br\/>Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n C th\u1ecfa m\u00e3n<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":1755},{"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 ch\u1ea5t sau l\u00e0 oxit l\u01b0\u1ee1ng t\u00ednh:","select":["A. $Al_2O_3, ZnO, PbO_2, Cr_2O_3$. ","B. $Al_2O_3, MgO, PbO, SnO_2$.","C. $CaO, ZnO, Na_2O, Cr_2O_3$. ","D. $PbO_2, Al_2O_3, K_2O, SnO_2$."],"hint":"","explain":"oxit l\u01b0\u1ee1ng t\u00ednh l\u00e0 oxit v\u1eeba t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi axit, v\u1eeba t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi dung d\u1ecbch baz\u01a1. <br\/> C\u00e1c oxit l\u01b0\u1ee1ng t\u00ednh g\u1ed3m $Al_2O_3, ZnO, PbO_2, Cr_2O_3, SnO_2,$\u2026.<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":1}]}],"id_ques":1756},{"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 oxit kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch axit, kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch baz\u01a1 l\u00e0:","select":["A. $CuO, CaO$. ","B. $K_2O, BaO$.","C. $CuO, MnO$.","D. $CO, NO$."],"hint":"","explain":" C\u00e1c oxit trung t\u00ednh kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch axit, kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch baz\u01a1. <br\/> V\u00ed d\u1ee5: $NO, CO$<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":1757},{"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 l\u00e0m qu\u1ef3 \u1ea9m h\u00f3a \u0111\u1ecf l\u00e0:","select":["A. $CaO $","B. $Na_2O$","C. $CO $","D. $SO_2$"],"hint":"","explain":" C\u00e1c oxit axit s\u1ebd l\u00e0 qu\u1ef3 t\u00edm h\u00f3a \u0111\u1ecf do t\u00e1c d\u1ee5ng v\u1edbi n\u01b0\u1edbc t\u1ea1o th\u00e0nh axit<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":1758},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" C\u00e1c c\u1eb7p ch\u1ea5t t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi nhau l\u00e0:","select":["A. $H_2O$ v\u00e0 $PbO$, $BaO$ v\u00e0 $SO_2$","B. $Na_2O$ v\u00e0 $SO_3$, $CaO$ v\u00e0 $H_2SO_3$","C. $MgO$ v\u00e0 $H_2O$, $NaOH$ v\u00e0 $CO_2$","D. $CuO$ v\u00e0 $N_2O_5$, $CuO$ v\u00e0 $KOH$"],"hint":"","explain":"c\u00e1c oxit axit t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi n\u01b0\u1edbc, dung d\u1ecbch baz\u01a1, oxit baz\u01a1 c\u1ee7a kim lo\u1ea1i ki\u1ec1m, ki\u1ec1m th\u1ed5.<br\/> C\u00f2n c\u00e1c oxit baz\u01a1 t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi dung d\u1ecbch axit. <br\/> Ngo\u00e0i ra c\u00e1c oxit c\u1ee7a kim lo\u1ea1i ki\u1ec1m, ki\u1ec1m th\u1ed5 t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi n\u01b0\u1edbc t\u1ea1o dung d\u1ecbch baz\u01a1 t\u01b0\u01a1ng \u1ee9ng. <br\/> Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n B th\u1ecfa m\u00e3n <br\/> $\\begin{aligned}& N{{a}_{2}}O+S{{O}_{3}}\\to N{{a}_{2}}S{{O}_{4}} \\\\ & CaO+{{H}_{2}}S{{O}_{3}}\\to CaS{{O}_{3}}\\downarrow +{{H}_{2}}O \\\\ \\end{aligned}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":1759},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Cho v\u00e0o \u1ed1ng nghi\u1ec7m m\u1ed9t \u00edt b\u1ed9t $CuO$ th\u00eam v\u00e0i gi\u1ecdt dung d\u1ecbch $HCl$ l\u1eafc nh\u1eb9. Hi\u1ec7n t\u01b0\u1ee3ng quan s\u00e1t \u0111\u01b0\u1ee3c l\u00e0:","select":["A. B\u1ed9t $CuO$ tan d\u1ea7n, dung d\u1ecbch trong su\u1ed1t ","B. B\u1ed9t $CuO$ kh\u00f4ng tan","C. B\u1ed9t $CuO$ tan d\u1ea7n, dung d\u1ecbch c\u00f3 m\u00e0u v\u00e0ng","D. B\u1ed9t $CuO$ tan d\u1ea7n, dung d\u1ecbch c\u00f3 m\u00e0u xanh."],"hint":"","explain":" $CuO$ l\u00e0 oxit baz\u01a1 c\u00f3 ph\u1ea3n \u1ee9ng v\u1edbi dung d\u1ecbch axit theo ph\u1ea3n \u1ee9ng: <br\/> $CuO+HCl\\to CuC{{l}_{2}}+{{H}_{2}}O$ <br\/> C\u00e1c mu\u1ed1i c\u1ee7a \u0111\u1ed3ng c\u00f3 m\u00e0u xanh<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":1760},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho c\u00e1c oxit sau: $CaO, Fe_2O_3, P_2O_5$. S\u1ed1 oxit t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi axit clohi\u0111ric l\u00e0:","select":["A. 0 ","B. 1","C. 2 ","D. 3"],"hint":"","explain":"Oxit baz\u01a1 s\u1ebd t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi axit. <br\/> Ch\u1ec9 c\u00f3 $CaO$ v\u00e0 $Fe_2O_3$ th\u1ecfa m\u00e3n<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1761},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" 0,05 mol FeO t\u00e1c d\u1ee5ng v\u1eeba \u0111\u1ee7 v\u1edbi:","select":["A. $0,15$ mol HCl ","B. $0,1$ mol HCl.","C. $0,05$ mol HCl. ","D. $0,01$ mol HCl."],"hint":"","explain":"$FeO$ t\u00e1c d\u1ee5ng v\u1edbi $HCl$ theo ph\u1ea3n \u1ee9ng: <br\/> $\\begin{aligned}& FeO+2HCl\\to FeC{{l}_{2}}+{{H}_{2}} \\\\ & 0,05\\,\\,\\,\\,\\,\\,\\,0,05.2 \\\\ & {{n}_{HCl}}=0,1mol \\\\ \\end{aligned}$<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1762},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" M\u1ed9t oxit c\u1ee7a photpho c\u00f3 th\u00e0nh ph\u1ea7n ph\u1ea7n tr\u0103m c\u1ee7a $P$ b\u1eb1ng $43,66\\%$. Bi\u1ebft ph\u00e2n t\u1eed kh\u1ed1i c\u1ee7a oxit b\u1eb1ng $142$ \u0111vC. C\u00f4ng th\u1ee9c ho\u00e1 h\u1ecdc c\u1ee7a oxit l\u00e0:","select":["A. $P_2O_3$. ","B. $P_2O_5$.","C. $PO_2$. ","D. $P_2O_4$."],"hint":"","explain":"C\u00e1ch 1: Th\u1eed \u0111\u00e1p \u00e1n:<br\/>T\u00ednh ph\u00e2n t\u1eed kh\u1ed1i v\u00e0 $\\%$ kh\u1ed1i l\u01b0\u1ee3ng $P$ c\u1ee7a t\u1eebng \u0111\u00e1p \u00e1n ch\u1ec9 c\u00f3 $P_2O_5$ th\u1ecfa m\u00e3n.<br\/> C\u00e1ch 2: T\u00ednh<br\/> G\u1ecdi c\u00f4ng th\u1ee9c oxit c\u1ee7a photpho l\u00e0 $P_xO_y$ theo \u0111\u1ec1 b\u00e0i ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh:<br\/> $\\left\\{ \\begin{aligned} & \\dfrac{31x}{142}.100\\%=43,66\\% \\\\ & 31x+16y=142 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x=2 \\\\ & y=5 \\\\ \\end{aligned} \\right.$ <br\/>V\u1eady c\u00f4ng th\u1ee9c c\u1ee7a oxit l\u00e0 $P_2O_5$<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1763},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Cho $7,2$ gam m\u1ed9t lo\u1ea1i oxit s\u1eaft t\u00e1c d\u1ee5ng ho\u00e0n to\u00e0n v\u1edbi kh\u00ed hi\u0111ro cho $5,6$ gam s\u1eaft. C\u00f4ng th\u1ee9c oxit s\u1eaft l\u00e0:","select":["A. $FeO$. ","B. $Fe_2O_3$.","C. $Fe_3O_4$.","D. $FeO_2$."],"hint":"","explain":"$\\begin{aligned}& {{m}_{FeO}}=7,2\\,gam \\\\ & {{m}_{Fe}}=5,6\\,gam \\\\ & \\Rightarrow \\,{{m}_{O}}=7,2-5,6=1,6\\,gam \\\\ & {{n}_{Fe}}=\\dfrac{5,6}{56}=0,1\\,mol \\\\ & {{n}_{O}}=\\dfrac{1,6}{16}=0,1\\,mol \\\\ \\end{aligned}$<br\/>T\u1ef7 l\u1ec7: $\\dfrac{\\,{{n}_{Fe}}}{{{n}_{O}}}=\\dfrac{1}{1}$<br\/>C\u00f4ng th\u1ee9c l\u00e0: $FeO$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1764},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" H\u1ea5p th\u1ee5 ho\u00e0n to\u00e0n $2,24$ l\u00edt kh\u00ed $CO_2$ (\u0111ktc) v\u00e0o dung d\u1ecbch n\u01b0\u1edbc v\u00f4i trong c\u00f3 ch\u1ee9a $0,075$ mol $Ca(OH)_2$. Mu\u1ed1i thu \u0111\u01b0\u1ee3c sau ph\u1ea3n \u1ee9ng l\u00e0:","select":["A. $CaCO_3$. ","B. $Ca(HCO3)_2$","C. $CaCO_3$ v\u00e0 $Ca(HCO3)_2$ ","D. $CaCO_3$ v\u00e0 $CaHCO_3$."],"hint":"","explain":"$\\begin{aligned}& {{n}_{C{{O}_{2}}}}=0,1\\,mol \\\\ & {{n}_{Ca{{(OH)}_{2}}}}=0,075\\,mol \\\\ \\end{aligned}$<br\/>T\u1ef7 l\u1ec7: $1<\\dfrac{{{n}_{C{{O}_{2}}}}}{{{n}_{Ca{{(OH)}_{2}}}}}=\\dfrac{0,1}{0,075}=\\dfrac{4}{3}<2$ <br\/>Ph\u1ea3n \u1ee9ng t\u1ea1o $2$ mu\u1ed1i: $CaCO_3$ v\u00e0 $Ca(HCO_3)_2$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":1765},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Ho\u00e0 tan $6,2$ g natri oxit v\u00e0o $193,8$ g n\u01b0\u1edbc th\u00ec \u0111\u01b0\u1ee3c dung d\u1ecbch $A$. N\u1ed3ng \u0111\u1ed9 ph\u1ea7n tr\u0103m c\u1ee7a dung d\u1ecbch $A $ l\u00e0:","select":["A. $4\\%$. ","B. $6\\%$.","C. $4,5\\%$","D. $10\\%$"],"hint":"","explain":" $\\begin{aligned}& {{n}_{N{{a}_{2}}O}}=\\dfrac{6,2}{62}=0,1\\,mol \\\\ & N{{a}_{2}}O+{{H}_{2}}O\\to 2NaOH \\\\ & \\,0,1\\,mol\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0,1.2\\,mol \\\\ & {{n}_{NaOH}}=0,2\\,mol \\\\ & {{m}_{NaOH}}=0,2.40=8gam \\\\ & {{m}_{\\text{dd}}}=6,2+193,8=200\\,gam \\\\ & C{{\\%}_{dd\\,NaOH}}=\\dfrac{8}{200}.100\\%=4\\% \\\\ \\end{aligned}$<br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1766},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" \u0110\u1ec3 t\u00e1ch ri\u00eang $Fe_2O_3$ ra kh\u1ecfi h\u1ed7n h\u1ee3p $BaO$ v\u00e0 $Fe_2O_3$ ta d\u00f9ng:","select":["A. N\u01b0\u1edbc. ","B. Gi\u1ea5y qu\u00ec t\u00edm.","C. Dung d\u1ecbch $HCl$. ","D. Dung d\u1ecbch $NaOH$."],"hint":"","explain":" $BaO$ t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi n\u01b0\u1edbc, $Fe_2O_3$ kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi n\u01b0\u1edbc n\u00ean c\u00f3 th\u1ec3 h\u00f2a tan h\u1ed7n h\u1ee3p v\u00e0o n\u01b0\u1edbc \u0111\u1ec3 thu \u0111\u01b0\u1ee3c $Fe_2O_3$ <br\/>$BaO+{{H}_{2}}O\\to Ba{{(OH)}_{2}}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1767},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Cho $12g$ h\u1ed7n h\u1ee3p g\u1ed3m $MgO$ v\u00e0 $Ca$ t\u00e1c d\u1ee5ng h\u1ebft v\u1edbi dung d\u1ecbch $HCl$, thu \u0111\u01b0\u1ee3c $2,24$ l\u00edt kh\u00ed \u1edf \u0111ktc. Ph\u1ea7n tr\u0103m kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a $MgO$ v\u00e0 $Ca$ trong h\u1ed7n h\u1ee3p l\u1ea7n l\u01b0\u1ee3t l\u00e0:","select":["A. $23,7\\%$ v\u00e0 $76,3\\%$ ","B. $66,67\\%$ v\u00e0 $33,33\\%$","C. $53,3\\%$ v\u00e0 $46,7\\%$ ","D. $33,33\\%$ v\u00e0 $66,67\\%$ "],"hint":"","explain":"C\u00e1c ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc:<br\/>$\\begin{aligned}& MgO+2HCl\\to MgC{{l}_{2}}+{{H}_{2}}O \\\\ & Ca+2HCl\\to CaC{{l}_{2}}+{{H}_{2}} \\\\ \\end{aligned}$<br\/>$\\begin{aligned}& {{n}_{{{H}_{2}}}}=\\dfrac{2,24}{22,4}=0,1mol \\\\ & {{n}_{Ca}}={{n}_{{{H}_{2}}}}=0,1\\,mol \\\\ \\end{aligned}$ <br\/>$\\begin{aligned}& \\%{{m}_{Ca}}=\\dfrac{0,1.40}{12}.100\\%=33,33\\% \\\\ & \\%{{m}_{MgO}}=100\\%-33,33\\%=66,67\\% \\\\ \\end{aligned}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1768},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Ho\u00e0 tan $2,4$ g m\u1ed9t oxit kim lo\u1ea1i ho\u00e1 tr\u1ecb $II$ c\u1ea7n d\u00f9ng $30g$ dd $HCl $ $7,3\\%$. C\u00f4ng th\u1ee9c c\u1ee7a oxit kim lo\u1ea1i l\u00e0:","select":["A. $CaO$.","B. $CuO$.","C. $FeO$. ","D. $ZnO$."],"hint":"","explain":"Gi\u1ea3 s\u01b0 kim lo\u1ea1i l\u00e0 $M$ <br\/> $\\begin{aligned}& {{m}_{HCl}}=\\dfrac{30.7,3\\%}{100\\%}=2,19\\,gam \\\\ & {{n}_{HCl}}=\\dfrac{2,19}{36,5}=0,06\\,mol \\\\ & MO+2HCl\\to MC{{l}_{2}}+{{H}_{2}}O \\\\ & {{n}_{MO}}=\\dfrac{1}{2}{{n}_{HCl}}=0,03\\,mol \\\\ & {{M}_{MO}}=\\dfrac{2,4}{0,03}=80(CuO) \\\\ \\end{aligned}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1769},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Cho m\u1ed9t l\u01b0\u1ee3ng h\u1ed7n h\u1ee3p $CuO$ v\u00e0 $Fe_2O_3$ t\u00e1c d\u1ee5ng h\u1ebft v\u1edbi dung d\u1ecbch $HCl$ thu \u0111\u01b0\u1ee3c $2$ mu\u1ed1i c\u00f3 t\u1ec9 l\u1ec7 mol l\u00e0 1 : 1. Ph\u1ea7n tr\u0103m kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a $CuO$ v\u00e0 $Fe_2O_3$ trong h\u1ed7n h\u1ee3p l\u1ea7n l\u01b0\u1ee3t l\u00e0:","select":["A. 20% v\u00e0 80%","B. 30% v\u00e0 70%","C. 40% v\u00e0 60% ","D. 50% v\u00e0 50%"],"hint":"","explain":"Gi\u1ea3 s\u1eed s\u1ed1 mol c\u1ee7a m\u1ed7i mu\u1ed1i l\u00e0 1 mol <br\/> $\\begin{aligned}& CuO+2HCl\\to CuC{{l}_{2}}+{{H}_{2}}O \\\\ & 1\\,mol\\,\\,\\,\\,\\,\\,1\\,mol \\\\ & F{{e}_{2}}{{O}_{3}}+6HCl\\to 2FeC{{l}_{3}}+3{{H}_{2}}O \\\\ & \\dfrac{1}{2}\\,mol\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,1\\,mol \\\\ & {{m}_{CuO}}=1.80=80\\,gam \\\\ & {{m}_{F{{e}_{2}}{{O}_{3}}}}=160.\\dfrac{1}{2}\\,=80\\,gam \\\\ & =>\\,\\%{{m}_{CuO}}=\\%{{m}_{F{{e}_{2}}{{O}_{3}}}}=50\\% \\\\ \\end{aligned}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":1770}],"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

Quá 5 phút: không được cộng điểm

Tổng thời gian làm mỗi câu (không giới hạn)

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Bấm vào đây nếu phát hiện có lỗi hoặc muốn gửi góp ý

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