De thi hoc ki II khoi 12 (4 ma de)

Đề kiểm tra học kì II - Môn hoá học ban cơ bản Họ Tên: Lớp: 12B Thời gian: 45 phút Câu Nội Dung !"#$%&'()*++,! -#$%.$$+/0(,12* 304 " (5 - 0 06'(5 - 0 (06 " (5 - 0 704'(5 - " (8912+:$$2; $$2; < 30=91> 00=91?(5' " 0 (0=91@2+A*B@C(5' "D 070=91*8E0 - 6FGH2$$(I5 J 0K)L.M< 30#GHB$$@$N& 0#GHB$$& (0#GH>O$$@$N& 70#GH>O&$$& J IP8E*QR1C(#=#( " ' S 6( - ' T 6N. 30D-00!"0(0DJ070DU0 U (MHH>22$CE.Aỳng< 3 0"V GHW3>XJ) -$ S 0"V GH "Y W3>XJ) " -$ J ( 0"V GH "Y W3>X-$ J J) " 7 0"V GH -Y W3>X-$ U V Khụng $Z2)E[/C1@\< 306 " (5 - 0' " I5 J (0(5' " 706 - =5 J S =91/A+2@2$CE]8 88LL< 303 " 5 - "3Y " - 5 " 0( " Y( " (0(5Y=Q5 =QY(5 " 70GHY"365 - GH65 - " Y"30 ^ (2Q]_H2HL)891@^`0#2Q].(5 " )>28 %A2C,>2$.Va+A*0b> 30Va00Ja0(0"a070-a0 9 Nguyên tử kim loại có cấu hình electron 1s 2 2s 2 2p 6 3s 2 là: A. K (Z = 19) B. Ca (Z = 20) C. Mg (Z = 12) D. Al (Z = 13) a c8?2SDJH)H(' - (55(' - Qd$$65'04.65'e$Z 30JDa00^Da0(0VDa070-"Da0 (2H$%*"aaa$$'(aD-0+.)98f 30J^D-00J^DT0(0TJD^070^JDT0 12 Cho bột nhôm tác dụng với NaOH d thu đợc 6,72 lít khí H 2 (đktc). Khối lợng bột nhôm đã phản ứng là: A. 16,2 gam B. 10,8 gam C. 5,4 gam D. 8,1 gam - (2B)&&HD&H>2D>2H0g[8EQLB>_D[hN$Z 30Cij00C$$65'0(0$$65'070CQ>20 J (2$$GH( " D3( - $%C$$65'$D+A*.+>2+,+A +.+,kD>O. 30GH " 5 - 0GH5D3 " 5 - (0GH " 5 - D3 " 5 - 70GH5 U 4*8E8>2,>\+/M.*&Ql2 A. &H>2 B. 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(0(h).$EPQv? 704,L.M&9> ^ c8?2SDJH)H(' - (55(' - Qd$$65'04.65'e$Z 30JDa00^Da0(0VDa070-"Da0 T #*|3||3M< 30#2&A0 0#2&@0 (0#+A0 70#+@ 20 Cho bột nhôm tác dụng với NaOH d thu đợc 6,72 lít khí H 2 (đktc). Khối lợng bột nhôm đã phản ứng là: A. 16,2 gam B. 10,8 gam C. 5,4 gam D. 8,1 gam " I%2+(5 " 2$$(5' " .-+A*0u+A*D$$?@ _"+A*:0b>* 30aDaS0aDaV (0aDa^70aDaU "" #$[8EQL$$6' J (D6' J  " I5 J D3( - D( "  30$$65'00$$'(0(0$$5' " 070$$6' - 0 "- 6F(2$$1aDaJ2GH65 -  - 04GH65 -  - 891AM+.P 30+,k0 09"D"J0 (0v"D"J0 709D"^0 "J ]).[>_v*+Qd+2@0I]nD\++2@Qh0 '2$\E+9vE>L.>_< 3032HH207NB(07E, 703&2$> "U 422x.8H)H2D 1D@nM)98f. 30)2(5 " q)2' " 500)2(5 " y)2' " 50 (0)2(5 " z)2' " 5070+.(5 " q+.' " 50 "V (,1*8~D.$Z[>2C% 306' J  " I5 J 03 " I5 J  - 0"J' " 5 06 " I5 J 03 " I5 J  - 0"J' " 5 (04 " I5 J 03 " I5 J  - 0"J' " 5 70u " I5 J 03 " I5 J  - 0"J' " 5 "S (2a+2@+/k||3$%CC2>UDV++042@ 30I> 0( (0 70 "^ (/9$N+*_ 30•zGHz6z'z3z' 0GHy•y6y'y3y' (0•yGHy6y'y3y' 70GHz6z'z3z'z• "T =912)Ekhông&9> 30(I5 J Y( " 0(I5 J Y6 " (5 - (0((5 - Y6 " I5 J 70(5' " Y( " -a  !"#$%&'()*++,! -#$%.$$+/0(,12* 304 " (5 - 0 06'(5 - 0 (06 " (5 - 0 704'(5 - - (2$$GH( " D3( - $%C$$65'$D+A*.+>2+,+A +.+,kD>O. 30GH " 5 - 0GH5D3 " 5 - (0GH " 5 - D3 " 5 - 70GH5 -" 6FGH2$$(I5 J 0K)L.M< 30#GHB$$@$N& 0#GHB$$& (0#GH>O$$@$N& 70#GH>O&$$& (2]Q]GH2$$365 - $D$$.1 30GH65 " 365 - 0GH65 - GH65 " (0GH65 - DGH65 -  " 365 - 70GH65 - 365 - -J (MHH>22$CE.Ađúng< 3 0"V GHW3>XJ)  -$ S  0"V GH "Y W3>XJ) " -$ J ( 0"V GH "Y W3>X-$ J J) " 7 0"V GH -Y W3>X-$ U -U ]H)H 1@nc0#o8E22"DTcNp*Ua$$45'0I89 1.]]H0(,1@2*c 30(' - (55('q(' " 0( " ' U (55('q(' " (03/F0 70'(55('q('r(' - -V =91/A+2@2$CE]8 88LL< 303 " 5 -  → "3Y " - 5 " 0( "  → Y( " (0(5Y=Q5 → =QY(5 " 70GHY"365 -  → GH65 -  " Y"30 -S g22aD"2H(' - 6' " D)>s+6 " n+0b>*s 30D"00JDJ^0(0-D-V070"D"J0 38 Nguyªn tö kim lo¹i cã cÊu h×nh electron 1s 2 2s 2 2p 6 3s 2 lµ: A. K (Z = 19) B. Ca (Z = 20) C. Mg (Z = 12) D. 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-#$%.$$+/0(,12* 304 " (5 - 0 06'(5 - 0 (06 " (5 - 0 704'(5 - " g22aD"2H(' - 6' " D)>s+6 " n+0b>*s 30D"00JDJ^0(0-D-V070"D"J0 "" 6FGH2$$(I5 J 0K)L.M< 30#GHB$$@$N& 0#GHB$$& (0#GH>O$$@$N& 70#GH>O&$$& "- ]H)H 1@nc0#o8E22"DTcNp*Ua$$45'0I89 1.]]H0(,1@2*c 30(' - (55('q(' " 0( " ' U (55('q(' " (03/F0 70'(55('q('r(' - "J (MHH>22$CE.Aỳng< 3 0"V GHW3>XJ) -$ S 0"V GH "Y W3>XJ) " -$ J ( 0"V GH "Y W3>X-$ J J) " 7 0"V GH -Y W3>X-$ U "U (2]Q]GH2$$365 - $D$$.1 30GH65 " 365 - 0GH65 - GH65 " (0GH65 - DGH65 - " 365 - 70GH65 - 365 - "V =91/A+2@2$CE]8 88LL< 303 " 5 - "3Y " - 5 " 0( " Y( " (0(5Y=Q5 =QY(5 " 70GHY"365 - GH65 - " Y"30 "S (2$$GH( " D3( - $%C$$65'$D+A*.+>2+,+A +.+,kD>O. 30GH " 5 - 0GH5D3 " 5 - (0GH " 5 - D3 " 5 - 70GH5 28 Nguyên tử kim loại có cấu hình electron 1s 2 2s 2 2p 6 3s 2 là: A. K (Z = 19) B. Ca (Z = 20) C. Mg (Z = 12) D. Al (Z = 13) 29 Cho bột nhôm tác dụng với NaOH d thu đợc 6,72 lít khí H 2 (đktc). Khối lợng bột nhôm đã phản ứng là: A. 16,2 gam B. 10,8 gam C. 5,4 gam D. 8,1 gam -a (2H$%*"aaa$$'(aD-0+.)98f 30J^D-00J^DT0(0TJD^070^JDT0 - c8?2SDJH)H(' - (55(' - Qd$$65'04.65'e$Z 30JDa00^Da0(0VDa070-"Da0 -" (2B)&&HD&H>2D>2H0g[8EQLB>_D[hN$Z 30Cij00C$$65'0(0$$65'070CQ>20 (2Q]_H2HL)891@^`0#2Q].(5 " )>28 %A2C,>2$.Va+A*0b> 30Va00Ja0(0"a070-a0 -J 4*8E8>2,>\+/M.*&Ql2 A. &H>2 B. H)H 1 C. 8H2 D. 2 1 -U Khụng $Z2)E[/C1@\< 306 " (5 - 0' " I5 J (0(5' " 706 - =5 J -V Ma axit*A$2:)>>2i>M&9&,L8+,.&m>L[0g :2)E< 306' - '(0 0' " I( " 0 (0I5 " 65 " 0 70(5 " I5 " 0 -S IP8E*QR1C(#=#( " ' S 6( - ' T 6N. 30D-00!"0(0DJ070DU0 -^ (891>Z 30' " 6(' " (55'00( " ' U 5'0(0(' - (55'070(' " q('r(55'0 -T (8912+:$$2; $$2; < 30=91> 00=91?(5' " 0 (0=91@2+A*B@C(5' "D 070=91*8E0 Ja g[8EQL$$2e'(D'65 - D' " I5 J [$Z2)E< 3042@)OP007$5' " Q]P+2@0 (07$(5' " 0 7042@,)O Đề kiểm tra học kì II - Môn hoá học ban cơ bản Họ Tên: Lớp: 12B Thời gian: 45 phút Câu Nội Dung Ja !"#$%&'()*++,! -#$%.$$+/0(,12* 304 " (5 - 0 06'(5 - 0 (06 " (5 - 0 704'(5 - U (8912+:$$2; $$2; < 30=91> 00=91?(5' " 0 (0=91@2+A*B@C(5' "D 070=91*8E0 -^ 6FGH2$$(I5 J 0K)L.M< 30#GHB$$@$N& 0#GHB$$& (0#GH>O$$@$N& 70#GH>O&$$& - IP8E*QR1C(#=#( " ' S 6( - ' T 6N. 30D-00!"0(0DJ070DU0 -V (MHH>22$CE.Aỳng< 3 0"V GHW3>XJ) -$ S 0"V GH "Y W3>XJ) " -$ J ( 0"V GH "Y W3>X-$ J J) " 7 0"V GH -Y W3>X-$ U Khụng $Z2)E[/C1@\< 306 " (5 - 0' " I5 J (0(5' " 706 - =5 J -J =91/A+2@2$CE]8 88LL< 303 " 5 - "3Y " - 5 " 0( " Y( " (0(5Y=Q5 =QY(5 " 70GHY"365 - GH65 - " Y"30 T (2Q]_H2HL)891@^`0#2Q].(5 " )>28 %A2C,>2$.Va+A*0b> 30Va00Ja0(0"a070-a0 32 Nguyên tử kim loại có cấu hình electron 1s 2 2s 2 2p 6 3s 2 là: A. K (Z = 19) B. Ca (Z = 20) C. Mg (Z = 12) D. Al (Z = 13) S c8?2SDJH)H(' - (55(' - Qd$$65'04.65'e$Z 30JDa00^Da0(0VDa070-"Da0 -a (2H$%*"aaa$$'(aD-0+.)98f 30J^D-00J^DT0(0TJD^070^JDT0 5 Cho bột nhôm tác dụng với NaOH d thu đợc 6,72 lít khí H 2 (đktc). Khối lợng bột nhôm đã phản ứng là: A. 16,2 gam B. 10,8 gam C. 32,4 gam D. 8,1 gam "^ (2B)&&HD&H>2D>2H0g[8EQLB>_D[hN$Z 30Cij00C$$65'0(0$$65'070CQ>20 - (2$$GH( " D3( - $%C$$65'$D+A*.+>2+,+A +.+,kD>O. 30GH " 5 - 0GH5D3 " 5 - (0GH " 5 - D3 " 5 - 70GH5 "V 4*8E8>2,>\+/M.*&Ql2 A. &H>2 B. 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Đề ki m tra học kì II - Môn hoá học ban cơ bản Họ Tên: Lớp: 12B Thời gian: 45 phút Câu Nội Dung !"#$%&'()*++,! -#$%.$$+/0( ,12* 304 " (5 - 0 06'(5 - 0. 00=91?(5' " 0 (0=91@2+A*B@C(5' "D 070=91*8E0 Ja g[8EQL$$2e'(D'65 - D' " I5 J [$Z2)E< 3042@)OP007$5' " Q]P+2@0 (07$(5' " 0 7042@,)O Đề ki m tra học kì II - Môn hoá học ban cơ bản Họ Tên: Lớp: 12B Thời gian: 45 phút Câu Nội Dung Ja !"#$%&'()*++,! -#$%.$$+/0( ,12* 304 " (5 - 0 06'(5 - 0. (2H$%*"aaa$$'(aD-0+.)98f 30J^D-00J^DT0(0TJD^070^JDT0 §Ò ki m tra häc k× II - M«n ho¸ häc ban c¬ b¶n Hä Tªn: Líp: 12B Thêi gian: 45 phót C©u Néi Dung  (/9$N+*_ 30•zGHz6z'z3z'