ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿Åð¼°Æ仯ºÏÎïÔÚÄ͸ßκϽð¹¤Òµ¡¢´ß»¯¼ÁÖÆÔì¡¢¸ßÄÜȼÁϵȷ½ÃæÓй㷺ӦÓá£

£¨1£©Åð»ù̬ԭ×ӵĵç×ÓÅŲ¼Ê½Îª_____¡£ÅðºÍÂÁ¿É·Ö±ðÐγÉ[BF4]-ºÍ[AlF6]3-£¬[BF4]-µÄ¿Õ¼ä¹¹ÐÍΪ___£¬ÅðÔªËز»¿ÉÄÜÐγÉ[BF6]3-µÄÔ­ÒòÊÇ________________¡£

£¨2£©B2H6ÊÇÒ»ÖÖ¸ßÄÜȼÁÏ£¬ËüÓëCl2·´Ó¦Éú³ÉµÄBCl3¿ÉÓÃÓÚ°ëµ¼Ìå²ôÔÓ¹¤ÒÕ¼°¸ß´¿¹èÖÆÔì¡£ÓɵڶþÖÜÆÚÔªËØ×é³ÉµÄÓëBCl3»¥ÎªµÈµç×ÓÌåµÄÒõÀë×ÓΪ________£¬¸ÃÒõÀë×ÓµÄÖÐÐÄÔ­×ÓÔÓ»¯·½Ê½Îª_________¡£

£¨3£©°±ÅðÍé(H3N¡úBH3)ºÍTi(BH4)3¾ùΪ¹ãÊܹØ×¢µÄÐÂÐÍ´¢Çâ²ÄÁÏ¡£

¢ÙBÓëNµÄµç¸ºÐÔ:B______N(Ìî¡°£¾¡±¡°£¼¡±»ò¡°£½¡±£¬ÏÂͬ)¡£

¢ÚTi(BH4)3ÓÉTiCl3ºÍLiBH4·´Ó¦ÖƵá£Ð´³öÖƱ¸·´Ó¦µÄ»¯Ñ§·½³Ìʽ:_________¡£

£¨4£©Á×»¯Åð(BP)ÊÇÊܵ½¸ß¶È¹Ø×¢µÄÄÍÄ¥²ÄÁÏ£¬Ëü¿ÉÓÃ×÷½ðÊô±íÃæµÄ±£»¤²ã¡£ÈçͼΪÁ×»¯Å𾧰û¡£

¢ÙÁ×»¯Åð¾§ÌåÊôÓÚ______¾§Ìå(ÌÌåÀàÐÍ)£¬ÊÇ·ñº¬ÓÐÅäλ¼ü? _____ (Ìî¡°ÊÇ¡±»ò¡°·ñ¡±)

¢Ú¾§ÌåÖÐPÔ­×ÓµÄÅäλÊýΪ_____¡£

¢ÛÒÑÖªBPµÄ¾§°û±ß³¤Îªanm£¬NAΪ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÊýÖµ£¬ÔòÁ×»¯Åð¾§ÌåµÄÃܶÈΪ_____g¡¤cm-3(Óú¬a¡¢NAµÄʽ×Ó±íʾ)¡£

¡¾´ð°¸¡¿ 1s22s22p1 ÕýËÄÃæÌå ÅðÔ­×ӵļÛÉê×Ó²ãÖ»ÓÐ2sºÍ2pËĸöÔ­×Ó¹ìµÀ£¬×î¶à¿ÉÒÔÐγÉËĸö¹²¼Û¼ü CO32-»òNO3- sp2 £¼ TiCl3+3LiBH4==Ti(BH4)3+3LiCl Ô­×Ó ÊÇ 4

¡¾½âÎö¡¿·ÖÎö£ºBΪ5ºÅÔªËØ¡£¿ÉÒÔ¸ù¾ÝÖÐÐÄÔ­×Ó¼Û²ãµç×Ó¶ÔÊýÅжÏÆä¿Õ¼ä¹¹ÐÍ¡£¸ù¾ÝµÈµç×ÓÔ­ÀíÑ°ÕÒºÏÊʵĵȵç×ÓÌå¡£¿ÉÒÔ¸ù¾ÝÖÐÐÄÔ­×ӵļ۲ãµç×Ó¶ÔÊýÅжÏÆäÔÓ»¯·½Ê½¡£ÔªËصķǽðÊôÐÔԽǿ£¬Æäµç¸ºÐÔԽǿ¡£¸ù¾Ý¾§ÌåµÄÓ²¶È¿ÉÒÔÅжÏÆä¿ÉÄܵľ§ÌåÀàÐÍ¡£

Ïê½â£º£¨1£©Åð»ù̬ԭ×ӵĵç×ÓÅŲ¼Ê½Îª1s22s22p1¡£ÅðºÍÂÁ¿É·Ö±ðÐγÉ[BF4]-ºÍ[AlF6]3-£¬[BF4]-µÄ¿Õ¼ä¹¹ÐÍΪÕýËÄÃæÌ壬ÅðÔªËز»¿ÉÄÜÐγÉ[BF6]3-µÄÔ­ÒòÊÇÅðÔ­×ӵļÛÉê×Ó²ãÖ»ÓÐ2sºÍ2pËĸöÔ­×Ó¹ìµÀ£¬Ã»ÓÐd¹ìµÀ£¬¹Ê×î¶à¿ÉÒÔÐγÉËĸö¹²¼Û¼ü¡£

£¨2£©ÓɵڶþÖÜÆÚÔªËØ×é³ÉµÄÓëBCl3»¥ÎªµÈµç×ÓÌåµÄÒõÀë×ÓΪCO32-»òNO3-£¬¸ÃÒõÀë×ÓµÄÖÐÐÄÔ­×ӵļÛÄܲãµç×Ó¶ÔÊýΪ4£¬¹ÊÆäÔÓ»¯·½Ê½Îªsp2¡£

£¨3£©¢ÙBµÄ·Ç½ðÊôÐÔ±ÈNÈõ£¬¹Êµç¸ºÐÔ:B£¼N¡£

¢ÚTi(BH4)3ÓÉTiCl3ºÍLiBH4·´Ó¦ÖƵ㬸÷´Ó¦µÄ»¯Ñ§·½³ÌʽΪTiCl3+3LiBH4==Ti(BH4)3+3LiCl¡£

£¨4£©¢ÙÁ×»¯Åð(BP)ÊÇÄÍÄ¥²ÄÁÏ£¬ÆäÓ²¶È±ØÈ»ºÜ´ó£¬¹ÊÁ×»¯Åð¾§ÌåÊôÓÚÔ­×Ó¾§Ìå¡£Óɾ§°û½á¹¹¿ÉÖª£¬BºÍPµÄÅäλÊý¾ùΪ4£¬¶øBÔ­×Ó×îÍâ²ãÖ»ÓÐ3¸öµç×Ó¡¢¶øPÔ­×ÓÓйµç×Ó¶Ô£¬¹ÊBÔ­×ÓÓëPÔ­×ÓÖ®¼ä¿ÉÒÔÐγÉÅäλ¼ü£¬ÆäÖбض¨º¬ÓÐÅäλ¼ü¡£

¢Ú¾§ÌåÖÐPÔ­×ÓµÄÅäλÊýΪ4¡£

¢ÛBPµÄ¾§°û±ß³¤Îªanm£¬Ôò¾§°ûµÄÌå»ýΪ¡£Óɾ§°û½á¹¹Ê¾Òâͼ¿ÉÖª£¬Ã¿¸ö¾§°ûÖк¬ÓÐ4¸öPÔ­×ÓºÍ4¸öBÔ­×Ó£¬ÔòNA¸ö¾§°ûµÄÖÊÁ¿Îª168g£¬ÔòÁ×»¯Åð¾§ÌåµÄÃܶÈΪg¡¤cm-3¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿ÁòËáǦ¿ÉÓÃÓÚǦÐîµç³Ø¡¢ÏËάÔöÖؼÁ¡¢Í¿ÁÏ·ÖÎöÊÔ¼Á¡£¹¤ÒµÉÏͨ³£Ó÷½Ç¦¿ó(Ö÷Òª³É·ÖΪPbS)Éú²úÁòËáǦ¡£¹¤ÒÕÁ÷³ÌÈçÏ£º

ÒÑÖª£º

¢Ù25¡æ£¬Ksp(PbS)=1.0¡Á10-28£¬Ksp(PbCl2)=1.6¡Á10-5

¢ÚPbCl2(s)+2Cl-(aq)PbCl42-(aq) ¡÷H>0

¢ÛFe3+ÒÔÇâÑõ»¯ÎïÐÎʽ¿ªÊ¼³ÁµíʱµÄpHֵΪ1.9

(1)ÓÉÓÚKsp(PbS)©‚Ksp(PbCl2)£¬PbS+2HClPbCl2+H2SµÄ·´Ó¦³Ì¶ÈºÜС£¬¼ÓÈëFeCl3ÄÜÔö´ó·´Ó¦³Ì¶È£¬Ô­ÒòÊÇ____________________£»²½Öè¢Ù·´Ó¦¹ý³ÌÖпɹ۲쵽Óе­»ÆÉ«¹ÌÌåÉú³É£¬×Ü·´Ó¦µÄÀë×Ó·½³ÌʽΪ_____________________£»¸Ã²½ÖèÐè¿ØÖÆÈÜÒºµÄpH<1.9£¬Ö÷ҪĿµÄÊÇ_______________________¡£

(2)²½Öè¢ÚÖб¥ºÍʳÑÎË®µÄ×÷ÓÃÊÇ_________________________¡£

(3)²½Öè¢ÛÖÐÂËÒºA¾­¹ýÕô·¢Å¨Ëõ¡¢ÓñùˮԡÀäÈ´½á¾§ºó£¬»¹Ðè½øÐеIJÙ×÷ÊÇ__________(Ìî²Ù×÷Ãû³Æ)¡£

(4)²½Öè¢ÜÖУ¬Èô¼ÓÈëÏ¡ÁòËá³ä·Ö·´Ó¦ºóËùµÃÐü×ÇÒºÖÐc(Cl-)=1.0mol¡¤L-1£¬Ôòc(SO42-)=________[Ksp(PbSO4)=1.6¡Á10-8]¡£²½Öè¢ÝÖÐÂËÒºBÓÃÊÊÁ¿ÊÔ¼ÁX´¦Àíºó¿ÉÑ­»·ÀûÓã¬ÊÔ¼ÁXӦѡÓÃÏÂÁÐÖеÄ_____(Ìî±êºÅ)¡£

a.HNO3 b.Cl2 c.H2O2 d.Ìú·Û

(5)Á¶Ç¦ºÍÓÃǦ¶¼»áʹˮÌåÒòÖؽðÊôǦµÄº¬Á¿Ôö´ó¶øÔì³ÉÑÏÖØÎÛȾ¡£Ë®ÈÜÒºÖÐǦµÄ´æÔÚÐÎ̬Ö÷ÒªÓÐPb2+¡¢Pb(OH)+¡¢Pb(OH)2¡¢Pb(OH)3-¡¢Pb(OH)42-¡£¸÷ÐÎ̬µÄǦŨ¶È·ÖÊýxÓëÈÜÒºpH±ä»¯µÄ¹ØϵÈçͼËùʾ£º

¢Ù̽¾¿Pb2+µÄÐÔÖÊ£ºÏòº¬Pb2+µÄÈÜÒºÖÐÖðµÎµÎ¼ÓNaOHÈÜÒººóÏȱä»ë×Ç£¬¼ÌÐøµÎ¼ÓNaOHÈÜÒºÓÖ±ä³ÎÇ壬pHΪ13¡«14ʱ£¬ÈÜÒºÖз¢ÉúµÄÖ÷Òª·´Ó¦µÄÀë×Ó·½³ÌʽΪ__________________¡£

¢Ú³ýÈ¥ÈÜÒºÖеÄPb2+£º¿ÆÑÐС×éÓÃÒ»ÖÖÐÂÐÍÊÔ¼Á(DH)¡°²¶×½¡±½ðÊôÀë×Ó£¬´Ó¶øÈ¥³ýË®ÖеĺÛÁ¿Ç¦ºÍÆäËûÔÓÖÊÀë×Ó£¬ÊµÑé½á¹û¼Ç¼ÈçÏ£º

Àë×Ó

Pb2+

Ca2+

Fe3+

Mn2+

´¦ÀíǰŨ¶È/(mg¡¤L-l)

0.100

29.8

0.12

0.087

´¦ÀíºóŨ¶È/(mg¡¤ L-1)

0.004

22.6

0.04

0.053

ÈôÐÂÐÍÊÔ¼Á(DH)ÔÚÍÑǦ¹ý³ÌÖÐÖ÷Òª·¢ÉúµÄ·´Ó¦Îª£º2DH(s)+Pb2+(aq)D2Pb(s)+2H+(aq)£¬ÔòÍÑǦʱ×îºÏÊʵÄpHԼΪ_____________¡£¸ÃʵÑéÖÐǦµÄÍѳýÂÊΪ_________________¡£

¡¾ÌâÄ¿¡¿ÄÜÔ´ÊÇÈËÀàÉú´æ¡¢Éç»á·¢Õ¹²»¿É»òȱµÄÎïÖÊ£¬CO¡¢H2¡¢CH3OH¾ùÊÇÖØÒªµÄÄÜÔ´ÎïÖÊ¡£

£¨1£©ÒÑÖª:ÑõÆøÖл¯Ñ§¼üµÄ¼üÄÜΪ497kJ/mol£¬¶þÑõ»¯Ì¼ÖÐC=O¼üµÄ¼üÄÜΪ745kJ/mol¡£

2CO(g)+O2(g)=2CO2(g) ¡÷H1=-566kJ/mol

H2O(g)+CO(g)=H2(g)+CO2(g) ¡÷H2=-41kJ/mol

CH3OH(g)+O2(g)==CO2(g)+2H2O(g) ¡÷H3=-660kJ/mol

CO(g)+2H2(g)CH3OH(g) ¡÷H4

Ôòʹ1molCO(g)ÍêÈ«·Ö½â³ÉÔ­×ÓËùÐèÒªµÄÄÜÁ¿ÖÁÉÙΪ___£¬¡÷H4=___¡£

£¨2£©Ä³ÃܱÕÈÝÆ÷ÖдæÔÚ·´Ó¦:CO(g)+2H2(g)CH3OH(g)£¬ÆðʼʱÈÝÆ÷ÖÐÖ»ÓÐamol/LCOºÍbmol/LH2£¬Æ½ºâʱ²âµÃ»ìºÏÆøÌåÖÐCH3OHµÄÎïÖʵÄÁ¿·ÖÊý[¦Õ(CH3OH)]ÓëζÈ(T)¡¢Ñ¹Ç¿(p)Ö®¼äµÄ¹ØϵÈçͼËùʾ¡£

¢ÙζÈT1ºÍT2ʱ¶ÔÓ¦µÄƽºâ³£Êý·Ö±ðΪK1¡¢K2£¬ÔòK1____K2(Ìî¡°>¡±¡°<¡±¡°=¡±)£»ÈôºãÎÂ(T1)ºãÈÝÌõ¼þÏ£¬Æðʼʱa=1¡¢b=2£¬²âµÃƽºâʱ»ìºÏÆøÌåµÄѹǿΪp1kPa£¬ÔòT1ʱ¸Ã·´Ó¦µÄѹǿƽºâ³£ÊýKp=___¡£(ÓÃƽºâ·Öѹ´úÌæƽºâŨ¶È¼ÆË㣬·Öѹ=×Üѹ¡ÁÎïÖʵÄÁ¿·ÖÊý£¬Óú¬p1µÄ´úÊýʽ±íʾ)

¢ÚÈôºãκãÈÝÌõ¼þÏ£¬Æðʼʱ³äÈë1molCOºÍ2molH2£¬´ïƽºâºó£¬COµÄת»¯ÂÊΪ¦Á1£¬´Ëʱ£¬ÈôÔÙ³äÈë1molCOºÍ2molH2£¬Ôٴδïƽºâºó£¬COµÄת»¯ÂÊΪ¦Á2£¬Ôò¦Á1_____¦Á2 (Ìî¡°£¾¡±¡°£¼¡±¡°=¡±¡°ÎÞ·¨È·¶¨¡±)

¢ÛÈôºãκãÈÝÌõ¼þÏ£¬Æðʼʱa=1¡¢b=2£¬ÔòÏÂÁÐÐðÊöÄÜ˵Ã÷·´Ó¦´ïµ½»¯Ñ§Æ½ºâ״̬µÄÊÇ____¡£

A.CO¡¢H2µÄÎïÖʵÄÁ¿Å¨¶ÈÖ®±ÈΪ1:2£¬ÇÒ²»ÔÙËæʱ¼äµÄ±ä»¯¶ø±ä»¯

B.»ìºÏÆøÌåµÄÃܶȲ»ÔÙËæʱ¼äµÄ±ä»¯¶ø±ä»¯

C.»ìºÏÆøÌåµÄƽ¾ùĦ¶ûÖÊÁ¿²»ÔÙËæʱ¼äµÄ±ä»¯¶ø±ä»¯

D.Èô½«ÈÝÆ÷¸ÄΪ¾øÈȺãÈÝÈÝÆ÷ʱ£¬Æ½ºâ³£ÊýK²»Ëæʱ¼ä±ä»¯¶ø±ä»¯

£¨3£©¹¤ÒµÉϳ£Óð±Ë®ÎüÊÕº¬Ì¼È¼ÁÏȼÉÕÖвúÉúµÄÎÂÊÒÆøÌåCO2£¬Æä²úÎïÖ®Ò»ÊÇNH4HCO3¡£ÒÑÖª³£ÎÂÏÂ̼ËáµÄµçÀë³£ÊýK1=4.4¡Á10-7¡¢K2=4.7¡Á10-11£¬NH3¡¤H2OµÄµçÀë³£ÊýK=1.8¡Á10-5£¬ÔòËùµÃµ½µÄNH4HCO3ÈÜÒºÖÐc(NH4+)_____c(HCO3-)(Ìî¡°£¾¡±¡°£¼¡±¡°=¡±)¡£

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø