ijËáÐÔÈÜÒºÖÐÖ»º¬ÓÐNH4+¡¢Cl-¡¢H+¡¢OH-4ÖÖÀë×Ó£¬ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ         

A£®¿ÉÓÉpH=3µÄHClÓëpH=11µÄNH3¡¤H2OÈÜÒºµÈÌå»ý»ìºÏ¶ø³É

B£®¸ÃÈÜÒºÖÐÀë×Ó¼äÒ»¶¨Âú×㣺c£¨NH4+£©+c£¨H+£©= c£¨OH-£©+c£¨Cl-£©

C£®¼ÓÈëÊÊÁ¿NH3¡¤H2O£¬ÈÜÒºÖÐÀë×ÓŨ¶È¿ÉÄÜΪ£ºc£¨NH4+£©>c£¨C1-£©>c£¨OH-£©>c£¨H+£©

D£®¸ÃÈÜÒº¿ÉÄÜÓɵÈÎïÖʵÄÁ¿Å¨¶È¡¢µÈÌå»ýµÄHClÈÜÒººÍNH3¡¤H2OÈÜÒº»ìºÏ¶ø³É

 

¡¾´ð°¸¡¿

A

¡¾½âÎö¡¿ÂÔ

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Óлú»¯ºÏÎïFÊǺϳɵç×Ó±¡Ä¤²ÄÁϸ߾ÛÎïZ ºÍÔöËܼÁPµÄÖØÒªÔ­ÁÏ¡£

£¨1£©Ä³Í¬Ñ§Éè¼ÆÁËÓÉÒÒÏ©ºÏ³É¸ß¾ÛÎïZµÄ3Ìõ·Ïߣ¨I¡¢II¡¢III£©ÈçÏÂͼËùʾ¡£

¢Ù 3ÌõºÏ³É·ÏßÖУ¬ÄãÈÏΪ·ûºÏ¡°Ô­×Ó¾­¼Ã¡±ÒªÇóµÄºÏ³É·ÏßÊÇ£¨ÌîÐòºÅ¡°I¡±¡¢¡°II¡±»ò¡°III¡±£©     ¡£

¢Ú XµÄ½á¹¹¼òʽÊÇ      ¡£

¢Û 1 mol FÔÚO2Öгä·ÖȼÉÕ£¬ÏûºÄ7.5 mol O2£¬Éú³É8molCO2 ºÍ3molH2O£¬1molFÓë×ãÁ¿NaHCO3ÈÜÒº·´Ó¦Éú³É2 mol CO2¡£Æä·Ö×ÓÄÚµÄÇâÔ­×Ó´¦ÓÚ3ÖÖ²»Í¬µÄ»¯Ñ§»·¾³¡£

F·Ö×ÓÖк¬Óеĺ¬Ñõ¹ÙÄÜÍŵÄÃû³ÆÊÇ           ¡£

Y£«F¡úZ·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ                                         ¡£

£¨2£©ÒÑÖª£º£¨R¡¢R£§´ú±íÌþ»ù»òÇâÔ­×Ó£©¡£ºÏ³ÉPµÄ·ÏßÈçÏÂͼËùʾ¡£D·Ö×ÓÖÐÓÐ8¸ö̼ԭ×Ó£¬ÆäÖ÷Á´ÉÏÓÐ6¸ö̼ԭ×Ó£¬ÇÒ·Ö×ÓÄÚÖ»º¬ÓÐÁ½¸ö¡ªCH3¡£

     

¢Ù A¡úB·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ                                           ¡£  

¢Ú B¡úCµÄ·´Ó¦ÖУ¬B·Ö×ÓÔÚ¼ÓÈÈÌõ¼þÏÂÍÑÈ¥Ò»¸öË®·Ö×Ó£¬Éú³ÉC£»C·Ö×ÓÖÐÖ»ÓÐ1¸ö̼ԭ×ÓÉÏÎÞÇâÔ­×Ó¡£CµÄ½á¹¹¼òʽÊÇ                              ¡£

¢Û PµÄ½á¹¹¼òʽÊÇ                      ¡£

¢Ü ·ûºÏÏÂÁÐÌõ¼þµÄBµÄͬ·ÖÒì¹¹Ìå¹²ÓУ¨ÌîÊý×Ö£©       ÖÖ¡£

a£®ÔÚËáÐÔÌõ¼þÏÂË®½âΪMºÍN    b£®Ò»¶¨Ìõ¼þÏÂM¿ÉÒÔת»¯ÎªN

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º»Æ¸ÔÖÐѧ2010½ì¸ßÈý6ÔÂÊÊÓ¦ÐÔ¿¼ÊÔÀí×ÛÄÜÁ¦²âÊÔ»¯Ñ§A ¾í ÌâÐÍ£ºÌî¿ÕÌâ

Óлú»¯ºÏÎïFÊǺϳɵç×Ó±¡Ä¤²ÄÁϸ߾ÛÎïZ ºÍÔöËܼÁPµÄÖØÒªÔ­ÁÏ¡£
£¨1£©Ä³Í¬Ñ§Éè¼ÆÁËÓÉÒÒÏ©ºÏ³É¸ß¾ÛÎïZµÄ3Ìõ·Ïߣ¨I¡¢II¡¢III£©ÈçÏÂͼËùʾ¡£

¢Ù 3ÌõºÏ³É·ÏßÖУ¬ÄãÈÏΪ·ûºÏ¡°Ô­×Ó¾­¼Ã¡±ÒªÇóµÄºÏ³É·ÏßÊÇ£¨ÌîÐòºÅ¡°I¡±¡¢¡°II¡±»ò¡°III¡±£©     ¡£
¢Ú XµÄ½á¹¹¼òʽÊÇ     ¡£
¢Û 1 mol FÔÚO2Öгä·ÖȼÉÕ£¬ÏûºÄ7.5 mol O2£¬Éú³É8molCO2ºÍ3molH2O£¬1mol FÓë×ãÁ¿NaHCO3ÈÜÒº·´Ó¦Éú³É2 mol CO2¡£Æä·Ö×ÓÄÚµÄÇâÔ­×Ó´¦ÓÚ3ÖÖ²»Í¬µÄ»¯Ñ§»·¾³¡£
F·Ö×ÓÖк¬Óеĺ¬Ñõ¹ÙÄÜÍŵÄÃû³ÆÊÇ          ¡£
Y£«F¡úZ·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ                                         ¡£
£¨2£©ÒÑÖª£º£¨R¡¢R£§´ú±íÌþ»ù»òÇâÔ­×Ó£©¡£ºÏ³ÉPµÄ·ÏßÈçÏÂͼËùʾ¡£D·Ö×ÓÖÐÓÐ8¸ö̼ԭ×Ó£¬ÆäÖ÷Á´ÉÏÓÐ6¸ö̼ԭ×Ó£¬ÇÒ·Ö×ÓÄÚÖ»º¬ÓÐÁ½¸ö¡ªCH3¡£

¢Ù A¡úB·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ                                           ¡£  
¢Ú B¡úCµÄ·´Ó¦ÖУ¬B·Ö×ÓÔÚ¼ÓÈÈÌõ¼þÏÂÍÑÈ¥Ò»¸öË®·Ö×Ó£¬Éú³ÉC£»C·Ö×ÓÖÐÖ»ÓÐ1¸ö̼ԭ×ÓÉÏÎÞÇâÔ­×Ó¡£CµÄ½á¹¹¼òʽÊÇ                              ¡£
¢Û PµÄ½á¹¹¼òʽÊÇ                      ¡£
¢Ü ·ûºÏÏÂÁÐÌõ¼þµÄBµÄͬ·ÖÒì¹¹Ìå¹²ÓУ¨ÌîÊý×Ö£©      ÖÖ¡£
a£®ÔÚËáÐÔÌõ¼þÏÂË®½âΪMºÍN     b£®Ò»¶¨Ìõ¼þÏÂM¿ÉÒÔת»¯ÎªN

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º»Æ¸ÔÖÐѧ2010½ì¸ßÈý6ÔÂÊÊÓ¦ÐÔ¿¼ÊÔÀí×ÛÄÜÁ¦²âÊÔ»¯Ñ§A¾í ÌâÐÍ£ºÌî¿ÕÌâ

Óлú»¯ºÏÎïFÊǺϳɵç×Ó±¡Ä¤²ÄÁϸ߾ÛÎïZ ºÍÔöËܼÁPµÄÖØÒªÔ­ÁÏ¡£

£¨1£©Ä³Í¬Ñ§Éè¼ÆÁËÓÉÒÒÏ©ºÏ³É¸ß¾ÛÎïZµÄ3Ìõ·Ïߣ¨I¡¢II¡¢III£©ÈçÏÂͼËùʾ¡£

¢Ù 3ÌõºÏ³É·ÏßÖУ¬ÄãÈÏΪ·ûºÏ¡°Ô­×Ó¾­¼Ã¡±ÒªÇóµÄºÏ³É·ÏßÊÇ£¨ÌîÐòºÅ¡°I¡±¡¢¡°II¡±»ò¡°III¡±£©      ¡£

¢Ú XµÄ½á¹¹¼òʽÊÇ      ¡£

¢Û 1 mol FÔÚO2Öгä·ÖȼÉÕ£¬ÏûºÄ7.5 mol O2£¬Éú³É8molCO2 ºÍ3molH2O£¬1mol FÓë×ãÁ¿NaHCO3ÈÜÒº·´Ó¦Éú³É2 mol CO2¡£Æä·Ö×ÓÄÚµÄÇâÔ­×Ó´¦ÓÚ3ÖÖ²»Í¬µÄ»¯Ñ§»·¾³¡£

F·Ö×ÓÖк¬Óеĺ¬Ñõ¹ÙÄÜÍŵÄÃû³ÆÊÇ           ¡£

Y£«F¡úZ·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ                                          ¡£

£¨2£©ÒÑÖª£º£¨R¡¢R£§´ú±íÌþ»ù»òÇâÔ­×Ó£©¡£ºÏ³ÉPµÄ·ÏßÈçÏÂͼËùʾ¡£D·Ö×ÓÖÐÓÐ8¸ö̼ԭ×Ó£¬ÆäÖ÷Á´ÉÏÓÐ6¸ö̼ԭ×Ó£¬ÇÒ·Ö×ÓÄÚÖ»º¬ÓÐÁ½¸ö¡ªCH3¡£

     

¢Ù A¡úB·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ                                            ¡£  

¢Ú B¡úCµÄ·´Ó¦ÖУ¬B·Ö×ÓÔÚ¼ÓÈÈÌõ¼þÏÂÍÑÈ¥Ò»¸öË®·Ö×Ó£¬Éú³ÉC£»C·Ö×ÓÖÐÖ»ÓÐ1¸ö̼ԭ×ÓÉÏÎÞÇâÔ­×Ó¡£CµÄ½á¹¹¼òʽÊÇ                               ¡£

¢Û PµÄ½á¹¹¼òʽÊÇ                       ¡£

¢Ü ·ûºÏÏÂÁÐÌõ¼þµÄBµÄͬ·ÖÒì¹¹Ìå¹²ÓУ¨ÌîÊý×Ö£©       ÖÖ¡£

a£®ÔÚËáÐÔÌõ¼þÏÂË®½âΪMºÍN     b£®Ò»¶¨Ìõ¼þÏÂM¿ÉÒÔת»¯ÎªN

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010-2011ѧÄêºÓÄÏÊ¡Ðí²ýÐÂÏçƽ¶¥É½¸ßÈýµÚÈý´Îµ÷Ñп¼ÊÔ£¨Àí×Û£©»¯Ñ§²¿·Ö ÌâÐÍ£ºÌî¿ÕÌâ

¸ß¾ÛÎïZÊÇÒ»ÖÖÖØÒªµÄµç×Ó±¡Ä¤²ÄÁÏ£¬PÊÇÒ»Öֹ㷺ӦÓõÄÔöËܼÁ¡£ÒÔÏÂÊÇijÑо¿Ð¡×éÉè¼ÆµÄËûÃǵĺϳÉ·Ïß¡£

 

       ÒÑÖªÒÔÏÂÐÅÏ¢£º

       ¢Ùl molFÔÚO2Öгä·ÖȼÉÕ£¬ÏûºÄ7£®5 rriol O2£¬Éú³É8 nl01 C02ºÍ3 mol H20£¬l mol FÓë×ãÁ¿NaHCO3ÈÜÒº·´Ó¦Éú³É2 rriol C02¡£ÆäºË´Å¹²ÕñÇâÆ×ÖÐÓÐ3¸öÎüÊÕ·å¡£

       ¢ÚÒÑÖª£º£¨R¡¢R£¬´ú±íÌþ»ù»òÇâÔ­×Ó£©¡£

 

¢ÛAÊÇÒ»ÖÖÈ©£¬B¡úCµÄ·´Ó¦ÖУ¬B·Ö×ÓÔÚ¼ÓÈÈÌõ¼þÏÂÍÑÈ¥Ò»¸öË®·Ö×Ó£¬Éú³ÉC£»C·Ö×ÓÖÐÖ»ÓÐ1¸ö̼ԭ×ÓÉÏÎÞÇâÔ­×Ó¡£

¢ÜD·Ö×ÓÖÐÓÐ8¸ö̼ԭ×Ó£¬ÆäÖ÷Á´ÉÏÓÐ6¸ö̼ԭ×Ó£¬ÇÒ·Ö×ÓÄÚÖ»º¬ÓÐÒ»¸ö¡ªCH3¡£

Çë»Ø´ðÒÔÏÂÎÊÌ⣺

£¨1£©XµÄ½á¹¹¼òʽÊÇ               ¡£F·Ö×ÓÖк¬Óеĺ¬Ñõ¹ÙÄÜÍŵÄÃû³ÆÊÇ            ¡£

£¨2£©Ð´³ö·´Ó¦¢ÙµÄ»¯Ñ§·½³Ìʽ            ¡£·´Ó¦¢ÚµÄ»¯Ñ§·½³ÌʽÊÇ              ¡£

£¨3£©·´Ó¦¢ÛµÄ»¯Ñ§·½³ÌʽÊÇ             ¡£

£¨4£©CµÄ½á¹¹¼òʽÊÇ            _£»PµÄ½á¹¹¼òʽÊÇ-¡£

£¨5£©·ûºÏÏÂÁÐÌõ¼þµÄBµÄͬ·ÖÒì¹¹Ìå¹²ÓУ¨ÌîÊý×Ö£©             ÖÖ¡£

a£®ÔÚËáÐÔÌõ¼þÏÂË®½âΪMºÍN    b£®-¶¨Ìõ¼þÏÂM¿ÉÒÔת»¯ÎªN

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A£®Ä³ËáÐÔÈÜÒºÖÐÖ»º¬ÓÐNa+.CH3COO-.H+.OH-,Ïò¸ÃÈÜÒºÖмÓÈëÊÊÁ¿°±Ë®£¬c(CH3COO-)Ò»¶¨´óÓÚc(Na+)Óë c(NH4+)Ö®ºÍ¡£

B£®³£ÎÂϽ«0.01molCH3COONaºÍ0.004molHClÈÜÓÚË®£¬ÅäÖƳÉ0.5L»ìºÏÈÜÒº¡£ÈÜÒºÖÐn(CH3COO-)+ n(OH-)-n(H+)=0.006mol

C£®ÏòÏõËáÄÆÈÜÒºÖеμÓÏ¡ÑÎËáµÃµ½µÄpH£½5µÄ»ìºÏÈÜÒº£ºc(Na£«)<c(NO3£­)

D£®ÎïÖʵÄÁ¿Å¨¶ÈÏàµÈµÄNaFºÍNaCNÈÜÒºÖÐÒõÀë×Ó×ÜÊýÏàµÈ

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸