ÌâÄ¿ÄÚÈÝ

³£Î³£Ñ¹ÏÂ,½«a mol CO2ÆøÌåͨÈë1 L b mol/LµÄNaOHÈÜÒºÖÐ,ÏÂÁжÔËùµÃÈÜÒºµÄÃèÊö²»ÕýÈ·µÄÊÇ£¨  £©
A£®µ±a=2bʱ,Ëæ×ÅCO2ÆøÌåµÄͨÈë,ÈÜÒºÖÐÓÉË®µçÀë³öµÄc(H+)ÓÐÈçͼ±ä»¯¹Øϵ
B£®µ±a=bʱ,ËùµÃÈÜÒºÖдæÔÚ:c(OH-)+c(C)=c(H+)+c(H2CO3)
C£®µ±2a=bʱ,ËùµÃÈÜÒºÖдæÔÚ:c(Na+)£¾c(C)£¾c(OH-)£¾c(HC)£¾c(H+)
D£®µ±1/2£¼a/b£¼1ʱ,ËùµÃÈÜÒºÖÐÒ»¶¨´æÔÚ:c(Na+)=c(C)+c(HC)+c(H2CO3)
D
µ±a=2bʱ,ËæCO2µÄͨÈëCO2ºÍNaOH·´Ó¦ÏÈÉú³ÉNa2CO3,¼Ì¶øÉú³ÉNaHCO3,×îºó¹ýÁ¿µÄCO2ÒÔH2CO3ÐÎʽ´æÔÚÓÚÈÜÒºÖÐ,ÓÉÓÚÑεÄË®½â´Ù½øË®µÄµçÀë,¶øËá»ò¼îÒÖÖÆË®µÄµçÀë,A¶Ô;µ±a=bʱ,CO2ºÍNaOH·´Ó¦Éú³ÉNaHCO3,¸ù¾ÝÖÊ×ÓÊغã¿ÉÖªc(OH-)=c(H2CO3)+[c(H+)-c(C)],B¶Ô;µ±2a=bʱ,CO2ºÍNaOH·´Ó¦Éú³ÉNa2CO3,c(OH-)ÓÉCË®½âºÍË®µÄµçÀëÁ½²¿·ÖÌṩ,¹Êc(OH-)£¾c(HC),C¶Ô;µ±1/2£¼a/b£¼1ʱ,CO2ºÍNaOH·´Ó¦Éú³ÉNa2CO3¡¢NaHCO3,¸ù¾ÝÎïÁÏÊغã,c(Na+)£¾c(C)+c(HC)+c(H2CO3)£¾c(Na+),D´í¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ΪÁË̽¾¿¹ýÑõ»¯ÄƵÄÇ¿Ñõ»¯ÐÔ£¬Ä³Ñо¿ÐÔѧϰС×éÉè¼ÆÁËÈçͼËùʾµÄʵÑé×°Öá£

ʵÑé²½Öè¼°ÏÖÏóÈçÏ£º
¢Ù¼ì²é×°ÖÃÆøÃÜÐÔºó£¬×°ÈëÒ©Æ·²¢Á¬½ÓÒÇÆ÷¡£
¢Ú»ºÂýͨÈëÒ»¶¨Á¿µÄN2ºó£¬½«×°ÖÃDÁ¬½ÓºÃ(µ¼¹ÜÄ©¶ËδÉìÈ뼯ÆøÆ¿ÖÐ)£¬ÔÙÏòÔ²µ×ÉÕÆ¿ÖлºÂýµÎ¼ÓŨÑÎËᣬ¾çÁÒ·´Ó¦£¬ÓÐÆøÌå²úÉú¡£
¢ÛÒ»¶Îʱ¼äºó£¬½«µ¼¹ÜÄ©¶ËÉìÈ뼯ÆøÆ¿ÖÐÊÕ¼¯ÆøÌå¡£×°ÖÃDÖÐÊÕ¼¯µ½ÄÜʹ´ø»ðÐǵÄľÌõ¸´È¼µÄÎÞÉ«ÆøÌå¡£
¢Ü·´Ó¦½áÊøºó£¬¹Ø±Õ·ÖҺ©¶·µÄ»îÈû£¬ÔÙͨÈëÒ»¶¨Á¿µÄN2£¬ÖÁ×°ÖÃÖÐÆøÌåÎÞÉ«¡£
»Ø´ðÏÂÁÐÎÊÌ⣺
(1)×°ÖÃBÖÐʪÈóµÄºìÉ«Ö½ÌõÍËÉ«£¬Ö¤Ã÷AÖз´Ó¦ÓР      (Ìѧʽ)Éú³É¡£ÈôBÖиķÅʪÈóµÄµí·ÛKIÊÔÖ½£¬½öƾÊÔÖ½±äÀ¶µÄÏÖÏó²»ÄÜÖ¤Ã÷ÉÏÊö½áÂÛ£¬ÇëÓÃÀë×Ó·½³Ìʽ˵Ã÷Ô­Òò                               ¡£
(2)×°ÖÃCµÄ×÷ÓÃÊÇ                                                        ¡£
(3)¼×ͬѧÈÏΪO2ÊÇNa2O2±»ÑÎËáÖеÄHCl»¹Ô­ËùµÃ¡£ÒÒͬѧÈÏΪ´Ë½áÂÛ²»ÕýÈ·£¬Ëû¿ÉÄܵÄÀíÓÉΪ¢Ù                       £»¢Ú                       ¡£
(4)ʵÑéÖ¤Ã÷£¬Na2O2Óë¸ÉÔïµÄHClÄÜ·´Ó¦£¬Íê³É²¢Åäƽ¸Ã»¯Ñ§·½³Ìʽ¡£
Na2O2£«HCl=Cl2£«NaCl£«   ¸Ã·´Ó¦       (Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±)ÓÃÓÚʵÑéÊÒ¿ìËÙÖÆÈ¡´¿¾»µÄCl2£¬ÀíÓÉÊÇ                                 (ÒªÇó´ð³öÁ½µã)¡£

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

¾«Ó¢¼Ò½ÌÍø