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°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐÈÜÒºÖÐÓйØÎïÖʵÄÁ¿Å¨¶È¹Øϵ²»ÕýÈ·µÄÊÇ
A£®10¡æʱpH=12µÄNaOHÈÜÒºÓë40¡æʱpH=12µÄNaOHÈÜÒºÖУºc(H+)ÏàµÈ
B£®25¡æʱpH=10µÄNaOHÈÜÒºÓëpH=10µÄ°±Ë®ÖУº c(Na+)=c(NH4+)
C£®ÎïÖʵÄÁ¿Å¨¶ÈÏàµÈµÄCH3COOHºÍCH3COONaÈÜÒºµÈÌå»ý»ìºÏ£º c(CH3COO-)+2c(OH-)=2c(H+)+c(CH3COOH)
D£®0.1mol¡¤L£­1(NH4)2Fe(SO4)2ÈÜÒºÖУºc(NH4+)+ c(NH3¡¤H2O) + c(Fe2+)="0.3" mol¡¤L£­1

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

·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¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

ÏÖÓеç½âÖÊÈÜÒº£º¢ÙNa2CO3¢ÚNaHCO3¢ÛNaAlO2¢ÜCH3COONa¢ÝNaOH,ÇÒÒÑÖª£ºCO2+3H2O+2====2Al(OH)3¡ý+¡£
£¨1£©µ±ÎåÖÖÈÜÒºµÄpHÏàͬʱ£¬ÆäÎïÖʵÄÁ¿Å¨¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ________£¨Ìî±àºÅ£©¡£
£¨2£©µ±ÉÏÊöÎïÖʵÄÁ¿Å¨¶È¾ùΪ0.1mol¡¤L-1µÄÎåÖÖÈÜÒº£¬Ï¡ÊÍÏàͬ±¶Êýʱ£¬ÆäpH±ä»¯×î´óµÄÊÇ________£¨Ìî±àºÅ£©¡£
£¨3£©ÔÚÉÏÊöÎåÖÖµç½âÖÊÈÜÒºÖУ¬·Ö±ð¼ÓÈëAlCl3ÈÜÒº£¬ÎÞÆøÌå²úÉúµÄÊÇ________£¨Ìî±àºÅ£©¡£
£¨4£©½«ÉÏÊö¢Ù¢Ú¢Û¢ÜÕâËÄÖÖµç½âÖÊÈÜÒº»ìºÏ£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ______________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨1£©pH£½3µÄ´×ËáºÍpH£½11µÄÇâÑõ»¯ÄÆÈÜÒºµÈÌå»ý»ìºÏºóÈÜÒº³Ê       ÐÔ(Ìî¡°Ëᡱ¡°ÖС±¡°¼î¡±)£¬ÈÜÒºÖÐc(Na+)         c(CH3COO-)(Ìî  >  <   =)£»
£¨2£©½«m mol/LµÄ´×ËáºÍn mol/LµÄÇâÑõ»¯ÄÆÈÜÒºµÈÌå»ý»ìºÏºóÈÜÒºµÄpH£½7£¬mÓënµÄ´óС¹ØϵÊÇm    n (Ìî  >  <   = )¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÒÑÖªH2C2O4ÊǶþÔªÈõËᣬËáʽÑÎNaHC2O4µÄÈÜÒº³ÊËáÐÔ¡£30¡æʱ£¬Å¨¶È¾ùΪ0.1mol?L£­1 NaHC2O4ÈÜÒººÍNa2C2O4ÈÜÒºÖоù´æÔڵĹØϵÊÇ   
A£®c(H+)?c(OH£­) =1¡Á10£­14
B£®c(H+) + c(H2C2O4) =c(C2O42-)+ c(OH£­)
C£®c(Na+)+ c(H+)= c(OH£­)+ c(HC2O4-£­)+ 2c(C2O42£­)
D£®c(OH£­) = c(H+) +c(HC2O4£­)+ 2c(H2C2O4)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨3·Ö£©½«µÈÌå»ýµÄ°±Ë®ÓëÑÎËáÈÜÒº»ìºÍºó£¬Èô»ìºÍÈÜÒºÖÐ[NH]=[Cl-]£¬ÔòÈÜÒºÖеÄpHֵΪ______7£¬»ìºÍÇ°[NH3¡¤H2O]______[HCl]£¬°±Ë®ÖÐ[OH-]______ÑÎËáÖÐ[H+]¡££¨Ìî>¡¢<»ò=£¬ÏÂͬ£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨8·Ö£©Ð´³öÏÂÁÐÎïÖʵçÀë»òÑÎË®½âµÄÀë×Ó·½³Ìʽ£º
(1)NaHSO4ÈÜÓÚË®_______________________________________________________
(2) H2SµÄµçÀë________________________________________________________ 
(3)ÂÈ»¯ÌúÈÜÒº£º                                                      £»
(4)̼ËáÄÆÈÜÒº£º                                                        £»
(5)AlCl3ÓëNaHCO3»ìºÏ____________________________________________________

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐÈÜÒºÖÐÀë×ÓŨ¶ÈµÄ¹Øϵһ¶¨ÕýÈ·µÄÊÇ
A£®Na2CO3ºÍNaHCO3ÈÜÒºÖУºc(Na£«)£«c(H£«)£½c(OH£­)£«c(HCO3£­)£«c(CO32£­)
B£®ÏàͬÌõ¼þÏ£¬pH=5µÄ¢ÙNH4ClÈÜÒº¡¢¢ÚCH3COOHÈÜÒº¡¢¢ÛÏ¡ÑÎËáÈÜÒºÖÐÓÉË®µçÀë³öµÄC(H+)£º
¢Ù£¾¢Ú£¾¢Û
C£®µÈÎïÖʵÄÁ¿µÄÒ»ÔªÈõËáHXÓëÆä¼ØÑÎKXµÄ»ìºÏÈÜÒºÖУº2c(K£«)£½c(X£­)£«c(HX)
D£®pH£½3µÄÒ»ÔªËáHXºÍpH£½11µÄÒ»Ôª¼îMOHµÈÌå»ý»ìºÏ£ºc(M£«)£½c(X£­)£¾c(H£«)£½c(OH£­)

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸