Î人´óѧ°æÎÞ»ú»¯Ñ§ÊÔ¾í¼°¿Îºó¸÷Õ½ÚÊÔÌâ ÏÂÔØ±¾ÎÄ

´ð£ºÍ¨¹ý³àÈȵÄ̼·Û£»Í¨¹ý³àÈȵÄCuO·ÛÄ©¡£

?HCOÌáʾ£º

2?332[H?]K1[H?]2?;?HCO?;?2??2?[H]?[H]K1?K1K2[H]?[H]K1?K1K2?3?CO?K1K2[H?]2?[H?]K1?K1K2

2?8£® ¼ÆËãµ±ÈÜÒºµÄpH·Ö±ðµÈÓÚ4£¬8£¬12ʱ£¬H2CO3£¬HCO?3£¬CO3ËùÕ¼µÄ°Ù·ÖÊý¡£

½â£ºpH=4

2?CO2µÄ±¥ºÍ״̬Ï£¬H2CO3µÄpH=3.9£¬HCO?µÄpH=8.1£¬CO33µÄpH=12.6 2? pH=4ʱ£¬ÌåϵÖÐÖ÷Òª³É·ÖΪ[H2CO3]£¬[CO3]¿ÉºöÂÔ

H2CO3==H+ + HCO?3 m 10

?4 n (m+n=1)

n10?4xn?7?3 =4.2¡Á10 ËùÒÔ=4.2¡Á10

mm[H2CO3%]=99.4%£¬[ HCO?3%]=0.06%

2?+

HCO?3== H + CO3

10?4x 0.6% 10 x =Ka2

0.6%

?4 ËùÒÔx=2.4¡Á10

?10

%¡£

pH=8£¬12ͬÀí¡£

9£® Ï©ÌþÄÜÎȶ¨´æÔÚ£¬¶ø¹èÏ©ÌþH2Si=SiH2È´ÄÑÒÔ´æÔÚ¡£ÆäÔ­ÒòÊÇʲô£¿

´ð£ºÌ¼µÄÔ­×Ó°ë¾¶½ÏС£¬³ýÐγÉ?¼üÍ⣬p¹ìµÀ°´¼ç²¢¼çÔڽϴóÖØµþ£¬ÄÜÐγÉÎȶ¨µÄ?¼ü£¬Òò¶ø£¬Ì¼

Ô­×Ó¼ä¿ÉÐγɵ¥¼ü£¬Ë«¼ü£¬ÉõÖÁÈý¼ü£¬Ï©Ìþ£¬È²Ìþ¶¼ºÜÎȶ¨¡£

¶ø¹èÔ­×Ó°ë¾¶½Ï´ó£¬p¹ìµÀ¼ç²¢¼çÖØµþºÜÉÙ£¬ºÜÄÑÐγÉÎȶ¨µÄ?¼ü£¬Òò¶ø£¬¹èÔ­×Ó¼ä´ó¶àÊýÖ»ÄÜ

Ðγɵ¥¼ü£¬¶øÄÑÐγÉË«¼ü£¬¹èÏ©ÌþÄÑÒÔÎȶ¨´æÔÚ¡£ 10£® ´ÓË®²£Á§³ö·¢ÔõÑùÖÆÔì±äÉ«¹è½º£¿ ´ð£º½«Na2SiO3ÈÜÒºÓëËá»ìºÏÉú³ÉÄý½º¡£

Na2SiO3+2H+=SiO2.H2O+2Na+

µ÷½ÚËáºÍNa2SiO3µÄÓÃÁ¿£¬Ê¹Éú³ÉµÄ¹èÄý½ºÖк¬ÓÐ8%~10%µÄSiO2£¬½«Äý½º¾²Ö¹ÀÏ»¯Ò»Ììºó£¬ÓÃ

ÈÈˮϴȥ¿ÉÈÜÑΡ£½«Äý½ºÔÚ60~70¡æºæ¸Éºó£¬ÐìÐìÉýÎÂÖÁ300¡æ

½º¸ÉÔï»î»¯Ç°£¬ÓÃCOCl2ÈÜÒº½þÅÝ£¬ÔÙ¸ÉÔï»î»¯¼´µÃ±äÉ«¹è½º¡£ 11£® Óû¯Ñ§·´Ó¦·½³Ìʽ±íʾµ¥ÖʹèµÄÖÆ±¸·´Ó¦£¬µ¥ÖʹèµÄÖ÷Òª»¯Ñ§ÐÔÖÊ£¬¶þÑõ»¯¹èµÄÖÆ±¸ºÍÐÔÖÊ¡£ ´ð£º

SiO2?2C?2Cl2?SiCl4?2COSiCl4?2Zn?Si?2ZnCl2

17

12£® ʲôÊÇ·Ðʯ·Ö×Óɸ£¿ÊÔÊöÆä½á¹¹ÌصãºÍÓ¦Óã¿

´ð£º×ÔÈ»½çÖдæÔÚµÄÄ³Ð©ÍøÂç×´µÄ¹èËáÑκÍÂÁËáÑξßÓÐÁýÐνṹ£¬ÕâЩ¾ùÔȵÄÁý¿ÉÒÔÓÐÑ¡ÔñµØÎü¸½

Ò»¶¨´óСµÄ·Ö×Ó£¬ÕâÖÖ×÷ÓýÐ×ö·Ö×Óɸ×÷Óá£Í¨³£°ÑÕâÑùµÄÌìÈ»¹èËáÑκÍÂÁËáÑνÐ×ö·Ðʯ·Ö×Óɸ¡£

AÐÍ·Ö×ÓɸÖУ¬ËùÓйèÑõËÄÃæÌåºÍÂÁÑõËÄÃæÌåͨ¹ý¹²ÓÃÑõÔ­×ÓÁª½á³É¶àÔª»·¡£ 13£® ʵÑéÊÒÖÐÅäÖÆSnCl2ÈÜҺʱҪ²ÉÈ¡ÄÄЩ´ëÊ©£¿ÆäÄ¿µÄʱʲô£¿

4+2+

´ð£º¼ÓÈëһЩSnÁ££¬ÒÔ·ÀÖ¹±»¿ÕÆøÑõ»¯³ÉSn£»¼ÓÈëÏ¡ÑÎËáÊÇΪÁË·ÀÖ¹SnµÄË®½â¡£ 14£®

µ¥ÖÊÎýÓÐÄÄÐ©Í¬ËØÒìÐÎÌ壬ÐÔÖÊÓкβîÒ죿?-ÎýËáºÍ?-ÎýËáµÄÖÆÈ¡ºÍÐÔÖÊ·½ÃæÓкÎÒìͬ£¿

´ð£ºÎýÓУº»ÒÎý£¬°×Îý£¬´àÎý¡£

15£® ÊÔÊöǦÐîµç³ØµÄ»ù±¾Ô­Àí¡£

´ð£º ·Åµçʱ£º¸º¼« PbSO4 + 2e== PbSO4 + 2e

?Õý¼« PbO2 + 4H+ + SO24 + 2e == PbSO4 + 2H2O

?³äµçʱ£ºÑô¼« PbSO4 + 2H2O == PbO2 + 4H+ + SO24 + 2e

?????? Òõ¼« PbSO4 + 2e== PbSO4 + 2e

16£® ǦµÄÑõ»¯ÎïÓм¸ÖÖ£¬·Ö±ð¼òÊöÆäÐÔÖÊ£¿ÈçºÎÓÃʵÑé·½·¨Ö¤ÊµPb3O4ÖÐǦÓв»Í¬¼Û̬£¿

´ð£ºÇ¦µÄÑõ»¯ÎïÓУºPbO2£¬Pb2O3£¬Pb3O4£¬PbO

Ó¦ÓÃÒÔÏ·´Ó¦£ºPb3O4?4HNO3?2Pb£¨NO3£©2?PbO2??2H2O

17£® ÔÚ298Kʱ£¬½«º¬ÓÐSn(ClO4)2ºÎPb(ClO4)2µÄijÈÜÒºÓë¹ýÁ¿·Ûĩ״Sn-PbºÏ½ðÒ»ÆðÕñµ´£¬²âµÃÈÜÒºÖÐÆ½ºâŨ¶È[Sn2+]/[Pb2+]Ϊ0.46¡£

ÒÑÖª ?Pb2?/Pb=-0.126V£¬¼ÆËã?Sn2?/SnÖµ¡£ ½â£ºPb???Sn2??Sn?Pb2?

2?0.05916[Sn]?lg?0; ´ïµ½Æ½ºâʱ,E=0,¼´E?E?2?2[Pb]ËùÒÔ,?Sn?2?/Sn?(?0.126)?0.059160.46lg?0 21??Sn2?/Sn??0.136ev

18£® Íê³É²¢Å䯽ÏÂÁл¯Ñ§·´Ó¦·½³Ìʽ£º

(1) Sn+HCl¡ú (2) Sn+Cl2¡ú (3) SnCl2+FeCl3¡ú (4) SnCl4+H2O¡ú

(5) SnS+Na2S2¡ú (6) SnS3+H+¡ú (7) Sn+SnCl4¡ú (8) PbS+HNO3¡ú

18

2?

(9) Pb3O4+HI(¹ýÁ¿)¡ú (10)Pb2++OH-(¹ýÁ¿)¡ú ´ð£º

1)Sn?2HCl?SnCl2?H2?2)Sn?2Cl2?SnCl43)SnCl2?2FeCl3?SnCl4?2FeCl24)3SnCl4?3H2O?SnO2?H2O?2H2SnCl65)SnS?Na2S2?Na2SnS32?6)SnS3?2H??SnS2?H2S

7)Sn?SnCl4?2SnCl28)3PbS?8HNO3?3Pb(NO3)2?3S??2NO??4H2O??9)Pb3O4?15I??8H??3PbI24?I3?4H2O10)Pb2??4OH?(¹ýÁ¿£©?Pb£¨OH£©4£»Pb£¨OH£©42?2??ClO??PbO2??Cl??2OH??H2O19£® ÏÖÓÐÒ»°×É«¹ÌÌåA£¬ÈÜÓÚË®²úÉú°×É«³ÁµíB£¬B¿ÉÈÜÓÚŨÑÎËá¡£Èô½«¹ÌÌåAÈÜÓÚÏ¡ÏõËáÖУ¬µÃÎÞÉ«ÈÜÒºC¡£½«AgNO3ÈÜÒº¼ÓÈëCÖУ¬Îö³ö°×É«³ÁµíD¡£DÈÜÓÚ°±Ë®Öеõ½ÈÜÒºE£¬ EËữºóÓÖ²úÉú°×É«³ÁµíD¡£

½«H2SÆøÌåͨÈëÈÜÒºCÖУ¬²úÉúרɫ³ÁµíF£¬FÈÜÓÚ(NH4)2SxÐγÉÈÜÒºG¡£ËữÈÜÒºG£¬µÃµ½»ÆÉ«³ÁµíH¡£

ÉÙÁ¿ÈÜÒºC¼ÓÈëHgCl2ÈÜÒºµÃ°×É«³ÁµíI£¬¼ÌÐø¼ÓÈëÈÜÒºC£¬³ÁµíIÖð½¥±ä»Ò£¬×îºó±äΪºÚÉ«³ÁµíJ¡£

ÊÔÈ·¶¨×ÖĸA£¬B£¬C£¬D£¬E£¬F£¬G£¬H£¬I£¬J¸÷±íʾʲôÎïÖÊ¡£

´ð£ºA:SnCl2 B:Sn(OH)Cl; C:Sn(NO3)2; D:AgCl; E:Ag(NH3)2Cl; F:SnS ; G:(NH4)2SnS3 H:SnS2;

I:Hg2Cl2; J:Hg µÚÊ®ÁùÕÂ

1£® ϱíÖиø³öµÚ¶þ¡¢ÈýÖÜÆÚÔªËØµÄµÚÒ»µçÀëÄÜÊý¾Ý£¨µ¥Î»kJ¡¤mol

Ϊʲô±È×óÓÒÁ½ÔªËصͼµÍ£¿ µÚ¶þÖÜÆÚ µçÀëÄÜ µÚÈýÖÜÆÚ µçÀëÄÜ Li 520 Na 496 Be 900 Mg 738 B 801 Al 578 C 1086 Si 787 N 1402 P 1012 O 1314 S 1000 F 1681 Cl 1251 Ne 2081 Ar 1521 ´ð£ºÒòΪB£¬AlÔªËØ»ù̬ԭ×ӵļÛ

?1£©ÊÔ˵Ã÷B,AlµÄµÚÒ»µçÀëÄÜ

µç×Ó²ã½á¹¹Îª£ºns2np1£¬×îÍâ²ãµÄÒ»¸öµç×ÓÈÝÒ×ʧȥ¶øÐγÉÈ«³äÂúµÄÎȶ¨½á¹¹¡£ 2£® ÔÚʵÑéÊÒÖÐÈçºÎÖÆ±¸ÒÒÅðÍ飬ÒÒÅðÍéµÄ½á¹¹ÈçºÎ£¿ ´ð£ºÇ⸺Àë×ÓÖû»·¨£º

ÒÒÃÑ 3LiAlH4+4BF3????2B2H6+3LiF+3AlF3

3NaBH4+4BF3????2B2H6+3NaBF4

B2H6 µÄ½á¹¹¼û¿Î±¾P778

3£® ˵Ã÷Èý±»¯ÅðºÍÈý±»¯ÂÁµÄ·Ðµã¸ßµÍ˳Ðò£¬²¢Ö¸³öÕôÆû·Ö×ӵĽṹ¡£

19

ÒÒÃÑ

´ð£ºÈý±»¯ÎïµÄÈ۷еã˳Ðò¼û¿Î±¾P780

Èý±»¯ÅðµÄÕôÆø·Ö×Ó¾ùΪµ¥·Ö×Ó£¬AlF3µÄÕôÆøÎªµ¥·Ö×Ó¡£¶øAlCl3µÄÕôÆøÎª¶þ¾Û·Ö×Ó£¬Æä½á¹¹¼û¿Î±¾P781

4£® »­³öB3N3H6£¨ÎÞ»ú±½£©µÄ½á¹¹¡£ ´ð£º

5£® B10H14µÄ½á¹¹ÖÐÓжàÉÙÖÖÐÎʽµÄ»¯Ñ§¼ü£¿¸÷ÓжàÉÙ¸ö£¿ ´ð£ºB10H14µÄ½á¹¹

6£® ΪʲôÅðËáÊÇÒ»ÖÖ·Ò×˹Ë᣿ÅðɰµÄ½á¹¹Ó¦ÔõÑùд·¨£¿ÅðɰˮÈÜÒºµÄËá¼îÐÔÈçºÎ£¿

´ð£ºÅðËáΪȱµç×Ó»¯ºÏÎÖÐÐÄÔ­×ÓBÉÏ»¹ÓÐÒ»¸ö¿ÕµÄp¹ìµÀ£¬ÄܽÓÊܵç×Ó¶Ô£¬Òò¶øÎªÂ·Ò×˹Ë᣻ ÅðɰµÄ½á¹¹Ê½¼û¿Î±¾P787£» ÅðɰˮÈÜÒºÏÔÇ¿¼îÐÔ¡£

7£® ÊÔÓû¯Ñ§·´Ó¦·½³Ìʽ±íʾ´ÓÅðɰ֯±¸ÏÂÁи÷»¯ºÏÎïµÄ¹ý³Ì£º

£¨1£©H3BO3 (2) BF3 (3) NaBH4

´ð£º½«ÅðɰŨÈÜÒºÓëŨÁòËá×÷ÓúóÀäÈ´µÃH3BO3

Na2[B4O5(OH)4]?H2SO4?3H2O?4H3BO3?Na2SO4

8£® ÔõÑù´ÓÃ÷·¯ÖƱ¸(1) ÇâÑõ»¯ÂÁ£¬£¨2£©ÁòËá¼Ø£¬£¨3£©ÂÁËá¼Ø£¿Ð´³ö·´Ó¦Ê½¡£

´ð£º½«Ã÷·¯KAl£¨SO4£©2 12H2OÈÜÓÚË®£¬¼ÓÈëÊÊÁ¿KOHÈÜÒºµÃµ½Al£¨OH£©3³Áµí£»

Al3++3OH-=Al£¨OH£©3?

¹ýÂËÕô·¢Å¨Ëõ¼´µÃµ½K2SO4

½«Al£¨OH£©3ÓëKOH×÷ÓõÃÎÞÉ«ÈÜÒº£¬Õô·¢Å¨Ëõºó¼´µÃµ½KAlO2 9£® д³öÏÂÁз´Ó¦·½³Ìʽ£º

£¨1£© ¹ÌÌå̼ËáÄÆÍ¬Ñõ»¯ÂÁÒ»ÆðÈÛÉÕ£¬½«ÈÛ¿é´òËéºóͶÈëË®ÖУ¬²úÉú°×É«Èé×´³Áµí£¿ £¨2£© ÂÁºÍÈÈŨNaOHÈÜÒº×÷Ó㬷ųöÆøÌ壻

£¨3£© ÂÁËáÄÆÈÜÒºÖмÓÈëÂÈ»¯ï§£¬Óа±Æø²úÉú£¬ÇÒÈÜÒºÖÐÓÐÈé°×É«Äý½º×´³Áµí£» £¨4£© Èý·ú»¯ÅðͨÈë̼ËáÄÆÈÜÒºÖС£ ´ð£º£¨1£©Na2CO3+Al2O3=2NaAlO2+CO2?

NaAlO2+2H2O=Al(OH)3?+NaOH

£¨2£©2Al+2NaOH+2H2O=2NaAlO2+3H2

£¨3£©Na[Al(OH)4]+NH4Cl=Al(OH)3?+NH3?+NaCl+H2O

£¨4£©8BF3+3Na2CO3+3H2O=2Na Al(OH)4+3H2?

10£® ÔÚ½ðÊô»î¶¯Ë³Ðò±íÖÐAlÔÚFe֮ǰ£¬¸üÔÚCu֮ǰ£¬µ«Al±ÈFe¿¹¸¯Ê´ÐÔÇ¿£¬ÕâÊÇΪʲô£¿

Cu¿ÉÒÔºÍÀäµÄŨÏõËá·´Ó¦£¬¶øAlÈ´²»ÄÜ£¬ÕâÊÇΪʲô£¿

´ð£ºAlËäÈ»ÊÇ»îÆÃ½ðÊô£¬µ«ÓÉÓÚ±íÃæÉϸ²¸ÇÁËÒ»²ãÖÂÃܵÄÑõ»¯ÎïĤ£¬Ê¹ÂÁ²»ÄܽøÒ»²½Í¬ÑõºÍË®×÷Óöø¾ßÓкܸߵÄÎȶ¨ÐÔ¡£

ÔÚÀäµÄŨÁòËáºÍŨÏõËáÖУ¬ÂÁµÄ±íÃæ±»¶Û»¯¶ø²»·¢Éú×÷Óá£

11£® ÁòͨÂÁÔÚ¸ßÎÂÏ·´Ó¦¿ÉµÃµ½Al2S3£¬µ«ÔÚË®ÈÜÒºÖÐNa2SºÍÂÁÑÎ×÷Óã¬È´²»ÄÜÉú³ÉAl2S3£¬Îª

ʲô£¿ÊÔÓû¯Ñ§·´Ó¦·½³Ìʽ±íʾ¡£

20