ÌâÄ¿ÄÚÈÝ

ÏÂÁйØÓÚÈÜÒºÖÐÀë×ÓµÄ˵·¨ÕýÈ·µÄÊÇ                         
A£®0.1mol¡¤L-1µÄNa2CO3ÈÜÒºÖÐÀë×ÓŨ¶È¹Øϵ£ºc(Na+)=2 c(CO)+ c(HCO)+ c(H2CO3)
B£®0.1mol¡¤L-1µÄNH4ClºÍ0.1mol¡¤L-1µÄNH3¡¤H2OµÈÌå»ý»ìºÏºóÈÜÒºÖеÄÀë×ÓŨ¶È¹Øϵ£º
c(Cl£­)> c(NH)> c(OH£­)> c(H+)
C£®³£ÎÂÏ£¬´×ËáÄÆÈÜÒºÖеμÓÉÙÁ¿´×ËáʹÈÜÒºµÄpH=7£¬Ôò»ìºÏÈÜÒºÖУ¬Àë×ÓŨ¶È¹Øϵ£º
c(Na+)< c(CH3COO£­)
D£®µÈÎïÖʵÄÁ¿Å¨¶È¡¢µÈÌå»ýµÄNH4NO3ºÍKNO3ÈÜÒºÖУ¬Ç°ÕßµÄÑôÀë×ÓŨ¶ÈСÓÚºóÕßµÄÑôÀë×ÓŨ¶È
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
·Ï¾ÉÓ¡Ë¢µç·°åµÄ»ØÊÕÀûÓÿÉʵÏÖ×ÊÔ´ÔÙÉú£¬²¢¼õÉÙÎÛȾ¡£·Ï¾ÉÓ¡Ë¢µç·°å¾­·ÛËé·ÖÀ룬Äܵõ½·Ç½ðÊô·ÛÄ©ºÍ½ðÊô·ÛÄ© 
£¨1£©·Ï¾ÉÓ¡Ë¢µç·°åÖлØÊÕµÄÌúÊÇÐÂÐ͵ç³ØµÄʹÓòÄÁÏ£¬ÈçÖƳÉLiFePO4µç³Ø£¬Ëü¿ÉÓÃÓڵ綯Æû³µ¡£µç³Ø·´Ó¦Îª£ºFePO4+Li LiFePO4£¬µç³ØµÄÕý¼«²ÄÁÏÊÇLiFePO4£¬¸º¼«²ÄÁÏÊÇʯī£¬º¬Li+µ¼µç¹ÌÌåΪµç½âÖÊ¡£·ÅµçʱÆäÕý¼«·´Ó¦·½³ÌʽΪ£º                                        
£¨2£©ÓÃH2O2ºÍH2SO4µÄ»ìºÏÈÜÒº¿ÉÈܳöÓ¡Ë¢µç·°å½ðÊô·ÛÄ©ÖеÄÍ­¡£ÒÑÖª£º
Cu(s)£«2H£«(aq)£½Cu2£«(aq)£«H2(g) ¡÷H£½64.39kJ¡¤mol£­1
      2H2O2(l)£½2H2O(l)£«O2(g) ¡÷H£½£­196.46kJ¡¤mol£­1
      H2(g)£«1/2O2(g)£½H2O(l)  ¡÷H£½£­285.84kJ¡¤mol£­1
       ÔòÔÚH2SO4ÈÜÒºÖÐCu ÓëH2O2·´Ó¦Éú³ÉCu2£«ºÍH2OµÄÈÈ»¯Ñ§·½³ÌʽΪ£º                 ¡£
£¨3£©²¿·Ö½ðÊôµÄ»ØÊÕÐèÒªÑõ»¯ÐÔºÜÇ¿µÄÈÜÒº£¬Èç½ð³£Óà   ºÍ    µÄ»ìºÏÈÜÒºÈܽâ
£¨4£©ÎªÁË·ÖÀë½ðÊô·ÛÄ©³£Óõ½ÇèËᣨHCN£©ÈÜÒº£¬HCNÊÇÒ»ÖÖÓж¾ÇÒ½ÏÈõµÄËᣬÒÑÖª£º³£ÎÂÏÂHCNµÄµçÀë³Ì¶È·Ç³£Ð¡,ÆäKa=6.2¡Á10-10,0.1mol/LµÄNaCNµÄpH=11.1,0.1mol/LµÄNH4CNµÄpH=9.2,ÔòŨ¶È¶¼ÊÇ0.1mol/LµÄNaCNºÍNH4CNÈÜÒºÖÐ,CN-Ë®½â³Ì¶È´óСΪ£ºNaCN     NH4CN£¨Ì£¾  =¡¡£¼¡¡£©£¬ÀíÓÉÊÇ£º¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡

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

¾«Ó¢¼Ò½ÌÍø