·ÖÎö»¯Ñ§£¨¹¤¿ÆÀࣩÆÚÄ©¸´Ï°Ìâ

·ÖÎö»¯Ñ§¸´Ï°Ìâ

µÚ 21 Ò³ ¹² 56 Ò³

13-4-7. ÈôÅäÖÆÊÔÑùÈÜÒºµÄÕôÁóË®Öк¬ÓÐÉÙÁ¿Ca,ÔÚpH=5.5»òÔÚpH=10(°±ÐÔ»º³åÈÜÒº)µÎ¶¨Zn,ËùÏûºÄEDTAµÄÌå»ýÊÇ·ñÏàͬ?ÄÄÖÖÇé¿ö²úÉúµÄÎó²î´ó?

13-4-8. ÈôÈÜÒºµÄpH=11¡£00£¬ÓÎÀëCNµÄŨ¶ÈΪ1.0¡Á10 mol.L£¬¼ÆËãHgYÂçºÏÎïµÄlgK¡¯

HgY

--2

-1

2+

2+

Öµ£¬ÒÑÖªHg-CNÂçºÏÎïµÄÖð¼¶Ðγɳ£ÊýlgK1~K4·Ö±ðΪ£º18.00,16.70,3.83ºÍ2.98¡£

2+-

Ò»¡¢ÅжÏÌâ

14£­1£­1. KMnO·¨Ó¦ÔÚËáÐÔÈÜÒºÖе樣¬µ÷½ÚËá¶ÈÖ»ÄÜÓÃHSO¡£¡Á

4

2

4

14£­1£­2. KMnOµÎ¶¨Fe£¬»¯Ñ§¼ÆÁ¿µãÔÚͻԾ·¶Î§µÄÖе㡣¡Á

4

2+

14£­1£­3. Ñõ»¯»¹Ô­Ö¸Ê¾¼ÁÖ»ÄÜÓÃÓÚKCrO·¨£¬²»ÄÜÓÃÓÚKMnO·¨¡£¡Á

2

2

7

4

14£­1£­4. Ó°ÏìÑõ»¯»¹Ô­µÎ¶¨Í»Ô¾´óСµÄÒòËØÓÐÁ½µç¶ÔÌõ¼þµç¼«µçÊÆµÄ²îÖµ¡¢½éÖʼ° Ëá¶È¡£¡Ì

14£­1£­5. ijÑõ»¯»¹Ô­µÎ¶¨»¯Ñ§¼ÆÁ¿µãʱµçÊÆÎª0.94V£¬ÄÇôѡÔñ±äÉ«µãµçÊÆÎª0.94VµÄÑõ»¯»¹Ô­Ö¸Ê¾¼ÁÊǺÏÊʵġ£.¡Ì

14£­1£­6. KCrO·¨²âÌú£¬ÖÕµãΪÀ¶×ÏÉ«£¬´ËÉ«ÊǶþ±½°·»ÇËáÄÆÑõ»¯Ì¬É«ÓëCr֮ɫµÄ»ì

2

2

7

3+

ºÏÉ«¡Ì

14£­1£­7. ÈÜÒºµçÊÆ´óÓÚָʾ¼Á±äÉ«µãµçÊÆ¦Õ£§(In)ʱ£¬ÈÜÒº³Êָʾ¼ÁÑõ»¯Ì¬É«¡£¡Á 14£­1£­8. Ìõ¼þµç¼«µçÊÆ¦Õ£§ÊÇËæÈÜÒºÌõ¼þ¶øÒìµÄ±ê×¼µç¼«µçÊÆ¡£¡Ì

14£­1£­9£®µâÁ¿·¨¿ÉÒÔÔÚÈõËáÐÔ¡¢ÖÐÐÔ»òÈõ¼õÐÔÈÜÒºÖнøÐУ¬µ«²»ÒËÔÚÇ¿¼îÐÔÈÜÒºÖÐ ½øÐС£¡Ì

14£­1£­10£®µâÁ¿·¨ÖУ¬µí·Ûָʾ¼Á¶¼±ØÐë½üÖÕµãʱ¼ÓÈë¡£¡Á

14£­1£­11£®±ê¶¨¸ßÃÌËá¼ØÈÜҺʱ£¬ÎªÊ¹·´Ó¦½Ï¿ì½øÐУ¬¿ÉÒÔ¼ÓÈëMn¡£ (¡Ì )

14£­1£­12ÔÚµâÁ¿·¨ÖУ¬¿ÉÒÔÓÃÖØ¸õËá¼Ø×÷Ϊ»ù×¼ÎïÖʱ궨Áò´úÁòËáÄÆÈÜÒºµÄŨ¶È¡££¨¡Ì£© 14£­1£­13.Ñõ»¯»¹Ô­µÎ¶¨µÄͻԾ·¶Î§Ö»ÓëÑõ»¯ºÍ»¹Ô­µç¶ÔµÄµçÊÆ²îÓйأ¬ÓëÈÜÒºµÄŨ¶ÈÎ޹ء££¨¡Ì£©

14£­1£­14. ÖØ¸õËá¼Ø±ê×¼ÈÜÒº¿ÉÒÔÓÃÖ±½Ó·¨ÅäÖÆ¡££¨¡Ì£© 14£­1£­15.µâÁ¿·¨µÄÊÊÒËÌõ¼þΪÖÐÐÔ»òÈõ¼îÐÔÌõ¼þ¡££¨¡Ì£© 14£­1£­16.ÖØ¸õËá¼ØµÎ¶¨·¨Ó¦¿ØÖÆÈÜҺΪËáÐÔ¡££¨¡Ì£©

14£­1£­17£®ÌØÊâָʾ¼ÁÊÇÖ¸ËüµÄÑõ»¯Ì¬Ó뻹ԭ̬Äܹ»³ÊÏÖ¸÷×Ô²»Í¬µÄ´ýÊâµÄÑÕÉ«¡££¨¡Á£© 14£­1£­18£®ÁÚ¶þµª·Æ¡ªÑÇÌú¼È¿É×öΪCeµÎ¶¨FeµÄָʾ¼Á£¬Ò²¿É×öCr2O72 µÎ¶¨FeµÄÖ¸

4+

2+

£­

2+

2+

¦È

¦È

·ÖÎö»¯Ñ§¸´Ï°Ìâ

µÚ 22 Ò³ ¹² 56 Ò³

ʾ¼Á£¨¡Á£©

14£­1£­19£®K2Cr2O7ÈÜÒººÍKMnO4ÈÜÒº¶¼¿ÉÒÔ³¤ÆÚÖü´æ¡££¨¡Á£© 14£­1£­20£®KMnO4µÎ¶¨·¨²»ÄÜÔÚHCl½éÖÊÖнøÐС££¨¡Ì£©

2?14£­1£­21£®ÔÚK2Cr2O7·¨ÖУ¬³ÈºìÉ«µÄCr2O7±»»¹Ô­³É²ÝÂÌÉ«µÄCr3+£¬Òò´Ë,¿É¸ù¾ÝËü±¾

ÉíÑÕÉ«µÄ±ä»¯À´È·¶¨Öյ㡣£¨¡Á£©

¶þ¡¢Ñ¡ÔñÌâ

14£­2£­1£®Ìõ¼þµçλÊÇ£¨ D £©

£Á£®±ê×¼µç¼«µçλ £Â£®ÈÎÒâζÈϵĵ缫µçλ

C£®ÈÎÒâŨ¶Èϵĵ缫µçλ £Ä£®µç¶ÔµÄÑõ»¯Ðͺͻ¹Ô­ÐÍŨ¶È¶¼µÈÓÚ1 mol¡¤LµÄµç¼«µçλ

14£­2£­2£®I3-1 (0.01mol/L) + 2e = 3I- (0.100mol/L)£»??0.545V£¬Õâ¸öµç¶ÔµÄµç¼«µçλΪ£¨ C £©¡£

£Á£®0.545V £Â£®0.507V £Ã£®0.574V £Ä£®0.486V

14£­2£­3£®ÔÚ1mol/LµÄHClÖУ¬?Sn4?/Sn2??0.14V£¬?Fe3?/Fe2??0.70V£¬ÔÚ´ËÌõ¼þÏ£¬ÒÔFe3+µÎ¶¨Sn2+£¬¼ÆÁ¿µãµÄµçλΪ£¨ C £©¡£

£Á£®0.25V £Â£®0.23V £Ã£®0.33V £Ä£®0.52V

14£­2£­4£®¶ÔÓÚ2A+ + 3B4+ = 2A4+ + 3B2+ Õâ¸öµÎ¶¨·´Ó¦£¬µÈÁ¿µãʱµÄµç¼«µçλÊÇ£¨ D £©

¦¨¦¨¦¨¦¨¦¨¦¨¦¨3?¦¨??3??2?3??2?3??2?BBBB£Á£®A B£®A £Ã£®A D£®A

5655-1

¦¨??14£­2£­5£®ÔÚÑõ»¯»¹Ô­µÎ¶¨ÖУ¬¿É×öÎªÌØÊâָʾ¼ÁʹÓõÄÊÇ£¨ B £©¡£ A£®KMnO4 B£®I2 C£®´Î¼×»ùÀ¶ £Ä£®¶þ±½°·»ÇËáÄÆ

14£­2£­6£®ÔÚ1mol/LµÄH2SO4ÈÜÒºÖУ¬ÓÃ0.1000 mol/LCe4+µÎ¶¨0.1000mol/L Fe2+ÈÜÒº£¬×îÇ¡µ±µÄÑõ»¯»¹Ô­Ö¸Ê¾¼ÁÊÇ£¨ C £©¡£

£Á£®´Î¼×»ùÀ¶ £Â£®ÁÚ±½°±»ù±½¼×Ëá £Ã£®ÁÚ¶þµª·Æ¡ªÑÇÌú D£®KSCN 14£­2£­7£®¿ÉÓÃÓڵ樵âµÄ±ê×¼ÈÜÒºÊÇ£¨ C £© £Á£®H2SO4 £Â£®KBrO3 £Ã£®Na2S2O3 £Ä£®K2Cr2O7 14£­2£­8. ijÑõ»¯»¹Ô­Ö¸Ê¾¼ÁµÄ±äÉ«·¶Î§ÊÇ£¨ C £© A£®pK£§(HIn)¡À1 B. ¦Õ£§(In)¡À1

C. ¦Õ£§(In)¡À0.0592V/n D. ¦Õ£§ (O/R)¡À0.0592V/n lg {c(O)/c(R) }

r

r

¦È

¦È

¦È

¦È

·ÖÎö»¯Ñ§¸´Ï°Ìâ

µÚ 23 Ò³ ¹² 56 Ò³

14£­2£­9 ¶Ô³ÆÐÔÑõ»¯»¹Ô­µÎ¶¨·´Ó¦£¬n=n=1ʱ£¬¶Ô·´Ó¦Æ½ºâ³£ÊýµÄÒªÇóÊÇ(A )

1

2

¦È6 ¦È¦È¦È9 A. K£§¡Ý10 B. K£§¡Ý6 C. K£§¡Ý 8 D. K£§¡Ý10

14£­2£­10. KMnO·¨²â¶¨HO£¬Îª¼ÓËÙ·´Ó¦£¬¿É( C )

4

2

2

A.¼ÓÈÈ B.Ôö´óŨ¶È C.¼ÓMnD.¿ªÊ¼¶à¼ÓKMnO.

4

2+

14£­2£­11.NaCO±ê¶¨KMnO£¬Î¶ȿØÖÆ75~85¡æÎ¶ȣ¬¹ý¸ßÔò( B )

2

2

4

4

A£®Éú³ÉMnO

2

2

2

B. HCO·Ö½â C. ·´Ó¦Ì«¿ì D. Öյ㲻Ã÷ÏÔ

2

2

2

4

4

4

14£­2£­12. ÓÃNaCO±ê¶¨KMnOʱ£¬µÎÈëµÚÒ»µÎKMnO£¬ÑÕÉ«ÏûÍʺó·´Ó¦¼Ó¿ì£¬Ô­ÒòÊÇ(C)

A. ÈÜÒºËá¶ÈÔö´ó B.·´Ó¦·ÅÈÈ C. ²úÉúµÄMnÆð´ß»¯×÷Óà D.ÓÕµ¼Ð§Ó¦ 14£­2£­13. ÔÚ0.5mol¡¤LHSOÖУ¬ÓÃKMnOµÎ¶¨ÏàͬŨ¶ÈµÄÏÂÁÐÈÜÒº£¬Í»Ô¾·¶Î§×î´óµÄ

2

4

4

-1

2+

ÊÇ(C )

A. HO(¦Õ=0.68V) B. Fe

2

2

¦È

2+

(¦Õ=0.77V) C. HCO(¦Õ= - 0.49V) D. Sn

2

2

4

¦È ¦È 2+

(¦Õ

¦È

=0.15V)

14£­2£­14. KCrO·¨ÔÚËáÐÔÌõ¼þÏµζ¨£¬ÆäËáΪ( D )

2

2

7

A. Ö»ÄÜÓÃHNO B.Ö»ÄÜÓÃHCl C.Ö»ÄÜÓÃHSOD.HCl¡¢HSO¾ù¿É

3 24 2414£­2£­15. ÔÚHPO´æÔÚϵÄHClÈÜÒºÖУ¬ÓÃ0.02mol¡¤LKCrOµÎ¶¨0.1mol¡¤LFe£¬»¯

3

4

2

2

7

-1

-1

2+

ѧ¼ÆÁ¿µãµçÊÆÎª0.86V, ×îºÏÊʵÄָʾ¼ÁΪ( C )

A. ´Î¼×»ùÀ¶(¦Õ£§=0.36V) B. ¶þ±½°·(¦Õ£§= 0.76V) C. ¶þ±½°·»ÇËáÄÆ(¦Õ£§=0.85V) D. ÁÚ¶þµª·ÆÑÇÌú( ¦Õ£§=1.06V) 14£­2£­16£®¼ä½ÓµâÁ¿·¨ÖУ¬¼ÓÈëµí·Ûָʾ¼ÁµÄÊÊÒËʱ¼äÊÇ £¨B£©

A£®µÎ¶¨¿ªÊ¼Ê± B£®µÎ¶¨½Ó½üÖÕµãʱ C£®±ê×¼ÈÜÒºµÎ¶¨ÖÁ50%ʱ D£®ÈκÎʱºò¶¼¿ÉÒÔ 14£­2£­17£®µâÁ¿·¨²âÍ­£¬¼ÓÈëµí·ÛµÄÊÊÒËʱ¼äΪ £¨B£© A£®µÎ¶¨¿ªÊ¼ B£®µÎ¶¨µ½ÈÜÒº³Êdz»ÆÉ« C£®µÎ¶¨µ½ÈÜÒº³öÏÖ°×É«³Áµí D£®µÎ¶¨µ½ÈÜÒº³ÊdzÀ¶É«

14£­2£­18£®ÓÃ0.1000 mol¡¤LCe£¨SO4£©2µÎ¶¨0.1000 mol¡¤LFe2+ÈÜÒº£¬ÒÑÖª

-1

-1

¦È¦È

¦È¦È

? , ? ? ? 0 .69 £¬ÔòÆäµçÊÆÍ»Ô¾·¶Î§Îª( A )¡£ ? /Ce?Ce1 .44VVFe/Fe4?3?3?2?a.0.86-1.26V b.0.86-1.44V c.0.68-1.26V d.0.68-1.44V

14£­2£­19£®ÔÚÒ»¶¨Ìõ¼þÏÂÓÃCe£¨SO4£©2±ê×¼ ÈÜÒºµÎ¶¨ÏàͬŨ¶ÈµÄFe2+ÈÜÒº£¬µ±µÎ¶¨°Ù·Ö

·ÖÎö»¯Ñ§¸´Ï°Ìâ

µÚ 24 Ò³ ¹² 56 Ò³

ÊýΪ50%ʱ£¬ÈÜÒºµÄµçλµÈÓÚ(B)

A.µÎ¶¨¼Áµç¶ÔµÄÌõ¼þµç¼«µçÊÆ B.±»µÎÎïµç¶ÔµÄÌõ¼þµç¼«µçÊÆ C.µÎ¶¨¼Áµç¶ÔµÄÌõ¼þµç¼«µçÊÆµÄÒ»°ë D.±»µÎÎïµç¶ÔµÄÌõ¼þµç¼«µçÊÆµÄÒ»°ë 14£­2£­20£®ÏÂÁлù×¼ÎïÖÊÖУ¬¼È¿É±ê¶¨NaOH£¬Óֿɱ궨KMnO4ÈÜÒºµÄÊÇ(C)¡£ B.Na2C2O4 C.H2C2O4¡¤2H2O D.Na2B4O7¡¤10H2O

14£­2£­21£®ÓÃNa2C2O4±ê¶¨KMnO4ÈÜҺʱ£¬µÎ¶¨¿ªÊ¼Ç°²»É÷½«±»µÎ¶¨ÈÜÒº¼ÓÈÈÖÁ·Ð£¬Èç¹û¼ÌÐøµÎ¶¨£¬Ôò×îºó±ê¶¨µÄ½á¹û( A )

A.Æ«¸ß B.Æ«µÍ C.׼ȷÎÞÎó D.²»È·¶¨

14£­2£­22£®ÔÚH2SO4½éÖÊÖУ¬ÓÃNa2C2O4±ê¶¨KMnO4ÈÜÒº£¬Èç¹ûÈÜÒºÖÐH2SO4Ũ¶ÈΪ5 mol¡¤L£¬Ôò×îºó±ê¶¨µÄ½á¹û(A)

A.Æ«¸ß B.Æ«µÍ C.ÕýÈ· D.²»È·¶¨

14£­2£­23£®ÓÃNa2C2O4±ê¶¨KMnO4ÈÜÒº£¬Èô²»Ê¹ÓÃMn2+´ß»¯¼Á£¬ÔòµÎ¶¨ÈÜҺζÈÓ¦¿ØÖÆÔÚ ¡£

A.ÊÒÎÂ B.50¡æÒÔÏÂ C.75-80¡æ D.100¡æ

14£­2£­24£®ÓÃͬһÖÖKMnO4±ê×¼ÈÜÒº·Ö±ðµÎ¶¨µÈÌå»ýµÄFeSO4ºÍH2C2O4ÈÜÒº£¬ÈôÏûºÄµÄKMnO4ÈÜÒºµÄÌå»ýÏàµÈ£¬Ôò(B)

C.cH2C2 Ac.FeSO4?cH2C2O4 B.cFeSO4?2cH2O4?2cFeSO 4 D.cFeSO4?4cH2COCO242414£­2£­25£®ÓÃKMnO4·¨ ²â¶¨FeSO4ÑùÆ·ÖÐFe2+µÄº¬Á¿£¬½éÖʳý¼ÓH2SO4Í⣬×îºÃ»¹¼Ó(B) A.HAc B.H3PO4 C.HCl D.HNO3

14£­2£­26£®K2Cr2O7·¨²âÌú¿óʯÖÐFeº¬Á¿Ê±£¬¼ÓÈëH3PO4µÄÖ÷ҪĿµÄÖ®Ò»ÊÇ(C) A.¼Ó¿ì·´Ó¦µÄËÙ¶È B.·ÀÖ¹³öÏÖFe£¨OH£©3³Áµí C.ʹFe3+ת»¯ÎªÎÞÉ«ÅäÀë×Ó D.³ÁµíCr3+

14£­2£­27£®±ê¶¨Na2S2O3ÈÜÒºÖУ¬¿ÉÑ¡ÓõĻù×¼ÎïÖÊÊÇ(C) A.KMnO4 B.´¿Fe C. K2Cr2O7 D.Vc

14£­2£­28£®µâÁ¿·¨²âVcº¬Á¿£¬±ØÐë²ÉÓúÎÖÖËá¿ØÖÆµÎ¶¨Ê±ÈÜÒºµÄËá¶È¡££¨C£© A.H2SO4 B.HCl C.HAc D.H3PO4

14£­2£­29£®24. ÓÃÁÚ¶þµª·Æ·¨²â¶¨Î¢Á¿Ìúʱ£¬¼ÓÈëÑÎËáôǰ·µÄ×÷ÓÃÊÇ£¨ A £©¡£ £¨A£©»¹Ô­¼Á £¨B£©Ñõ»¯¼Á £¨C£©Îȶ¨¼Á £¨D£©µ÷½ÚËá¶È

14£­2£­30. ÏÂÁÐÄÄÖÖÇé¿öÓ¦²ÉÓÃÖû»µÎ¶¨·¨£¨ D £©

-1

·ÖÎö»¯Ñ§¸´Ï°Ìâ

µÚ 25 Ò³ ¹² 56 Ò³

£¨A£©. ÓÃHCl±ê×¼ÈÜÒº²â¶¨NaOHÊÔÑùº¬Á¿ £¨B£©. ÓÃHCl±ê×¼ÈÜÒº²â¶¨CaCO3ÊÔÑùº¬Á¿ £¨C£©. ÓÃKMnO4±ê×¼ÈÜÒº²â¶¨CaCO3ÊÔÑùº¬Á¿ £¨D£©. ÓÃNa2S2O3±ê×¼ÈÜÒº²â¶¨K2Cr2O7ÊÔÑùº¬Á¿

14£­2£­31.ÔÚËáÐÔ½éÖÊÖÐ,ÓÃKMnO4ÈÜÒºµÎ¶¨²ÝËáÑÎʱ,µÎ¶¨ËÙ¶ÈÓ¦¿ØÖÆ£¨B£© £¨A£©.ÏóËá¼îµÎ¶¨ÄÇÑù¿ìËÙ½øÐÐ £¨B£©.ÔÚ¿ªÊ¼ÊÇ»ºÂý½øÐÐ,ÒÔºóÖð½¥¼Ó¿ì £¨C£©.ʼÖÕ»ºÂý½øÐÐ £¨D£©.¿ªÊ¼Ê±¿ì,È»ºó»ºÂý 14£­2£­32.ÖØ¸õËá¼ØµÎ¶¨ÊÇΪÁË..( E )

(A) ±ÜÃâÓÕµ¼·´Ó¦µÄ·¢Éú (B) ¼Ó¿ì·´Ó¦µÄËÙ¶È

(C) ʹָʾ¼ÁµÄ±äÉ«µãµÄµçλ´¦Ôڵζ¨ÌåϵµÄµçλͻԾ·¶Î§ÄÚ¡£ (D) ÖÕµãÒ×ÓÚ¹Û²ì (E) ¼æÓÐCÓëDµÄ×÷ÓÃ

14£­2£­33. µâ·¨ÖеÄÖ÷ÒªÎó²îÀ´ÊÇ?????????????????.( C ) £¨A£©I2»Ó·¢ £¨B£© IÑõ»¯ £¨C£©.I2»Ó·¢,IÑõ»¯ £¨D£©.ÖÕµãÎó²î 14£­2£­34.¶þ±½°·»ÇËáÄÆÊÇ

µÎ¶¨

µÄ³£ÓÃָʾ¼Á¡£ËüÊÇÊôÓÚ???

--

£¬ÈôÑ¡Óöþ±½°·»ÇËáÄÆ×÷ָʾ¼Á£¬ÐèÔÚÁòÁ×»ìËá½éÖÊÖнøÐУ¬

(A) ×ÔÉíָʾ¼Á (B) ÌØÊâָʾ¼Á (C) Ñõ»¯»¹Ô­Ö¸Ê¾¼Á (D) ÆäËûָʾ¼Á

14£­2£­35. ÒÑÖªÔÚ1mol/L H2SO4ÈÜÒºÖÐ,EMnO4/Mn=1.45V ,EFe/Fe=0.68V ,ÔÚ´ËÌõ¼þÏÂÓÃKMnO4±ê×¼ÈÜÒºµÎ¶¨Fe,Æä»¯Ñ§¼ÆÁ¿µãʱµçλֵÊÇ(C)

(A).0.77V (B).1.06V (C).1.32V (D).1.45V 14£­2£­36£®ÓÃK2Cr2O7±êÒºµÎ¶¨Fe ,¼ÆÁ¿µãʱŨ¶ÈµÄ¹ØÏµÎª(B)

(A).[Fe]=[Cr],[Fe]=[Cr2O7]; (B).[Fe]=3[Cr],[Fe]=6[Cr2O7]; (C) .3[Fe]=[Cr],[Fe]=6[Cr2O7]; (D).[Fe]=3[Cr],6[Fe]=[Cr2O7]; 14£­2£­37.µâÁ¿·¨ÖÐËùÐè

±ê×¼ÈÜÒºÖÐÔÚ±£´æÖÐÎüÊÕÁËÈôÓøÃ

½«???

(A) Æ«¸ß (B) Æ«µÍ (C) ÎÞÓ°Ïì (D) ÎÞ·¨ÅжÏ

14£­2£­38.ÈôÁ½µç¶ÔÔÚ·´Ó¦Öеç×Ó×ªÒÆÊý·Ö±ðΪ1ºÍ2£¬ÎªÊ¹·´Ó¦µÄÍêÈ«³Ì¶È´ïµ½99.9£¥£¬Á½µç¶ÔµÄÌõ¼þµçÊÆÖ®²î¡÷E

¦È¡¯

3+

3+

2+

2-3+

3+

2+

2-3+

3+

2+

2-3+

3+

2+

2-2+

2+

0

-2+

0

3+

2+

¶ø·¢ÉúÏÂÊö·´Ó¦£º

µÄÁ¿

µÎ¶¨I2ÈÜÒºÔòÏûºÄ

ÖÁÉÙÓ¦´óÓÚ_____

A. 0.09V B.0.27V C.0.36V D.0.18V

ÁªÏµ¿Í·þ£º779662525#qq.com(#Ìæ»»Îª@) ËÕICP±¸20003344ºÅ-4