ÌâÄ¿ÄÚÈÝ
£¨¹²12·Ö£©
µç»¯Ñ§ÖªÊ¶ÊÇ»¯Ñ§·´Ó¦ÔÀíµÄÖØÒª²¿·Ö£¬ÒÔÏÂÊdz£¼ûµÄµç»¯Ñ§×°Öãº
£¨1£©Ä³ÐËȤС×éͬѧģÄ⹤ҵÉÏÓÃÀë×Ó½»»»Ä¤·¨ÖÆÉÕ¼îµÄ·½·¨£¬ÄÇô¿ÉÒÔÉèÏëÓÃÈçͼװÖõç½âÁòËá¼ØÈÜÒºÀ´ÖÆÈ¡ÇâÆø¡¢ÑõÆø¡¢ÁòËáºÍÇâÑõ»¯¼Ø¡£
¢Ù¸Ãµç½â²ÛµÄÑô¼«·´Ó¦Ê½Îª_______________________¡£
![]()
¢ÚµÃµ½µÄÇâÑõ»¯¼Ø´Ó £¨ÌîA,»òD£©Åŷųö¡£³£ÎÂÏ£¬Èô½«µÃµ½µÄ2LŨ¶ÈΪ0.25mol/LµÄKOHÈÜÒºÓë2L0.025mol/LµÄÁòËáÈÜÒº»ìºÏ£¨¼ÙÉèÈÜÒº»ìºÏºóºöÂÔÌå»ý±ä»¯£©£¬pHΪ
¢Û´Ëʱͨ¹ýÒõÀë×Ó½»»»Ä¤µÄÀë×ÓÊý £¨Ìî´óÓÚ£¬Ð¡ÓÚ»òµÈÓÚ£©Í¨¹ýÑôÀë×Ó½»»»Ä¤µÄÀë×ÓÊý¡£
¢ÜÈô½«ÖƵõÄÇâÆø¡¢ÑõÆøºÍÁòËáÈÜÒº×éºÏΪÇâÑõȼÁÏµç³Ø£¬Ôòµç³ØÕý¼«µÄµç¼«·´Ó¦Ê½Îª____________________________________________
£¨2£©ÈçͼËùʾÊÇÒ»Ì׵绯ѧʵÑé×°Öã¬Í¼ÖÐC¡¢D¾ùΪ²¬µç¼«£¬UΪÑÎÇÅ£¬GÊÇÁéÃôµçÁ÷¼Æ£¬ÆäÖ¸Õë×ÜÊÇÆ«ÏòµçÔ´Õý¼«¡£
![]()
¢ÙÏòB±ÖмÓÈëÊÊÁ¿½ÏŨµÄÁòËᣬ·¢ÏÖGµÄÖ¸ÕëÏòÓÒÆ«ÒÆ¡£´ËʱA±ÖеÄÖ÷ҪʵÑéÏÖÏóÊÇ______________________£¬Dµç¼«Éϵĵ缫·´Ó¦Ê½Îª______________________¡£
¢ÚÒ»¶Îʱ¼äºó£¬ÔÙÏòB±ÖмÓÈëÊÊÁ¿µÄÖÊÁ¿·ÖÊýΪ40%µÄÇâÑõ»¯ÄÆÈÜÒº£¬·¢ÏÖGµÄÖ¸ÕëÏò×óÆ«ÒÆ¡£´ËʱÕûÌ×ʵÑé×°ÖõÄ×ܵÄÀë×Ó·½³ÌʽΪ__________________________¡£
£¨1£©1·Ö ¢Ù 2H2O-4e-=== O2 +4H+»ò 4OH- -4e-===2H2O + O2
¢ÚD£¨1·Ö£©£¬13£¨1·Ö£© ¢ÛСÓÚ £¨1·Ö£©
¢Ü4H+ +O2+4e- ===2H2O £¨2·Ö£©
£¨2£©¢ÙÎÞÉ«ÈÜÒº±ä³ÉÀ¶É«£¨2·Ö£©
£«2H£«£«2e£===
£«H2O£¨2·Ö£©
¢ÚI2£«
£«H2O===2H£«£«2I££«
(»òI2£«
£«2OH£===H2O£«2I££«
)£¨2·Ö£©
¡¾½âÎö¡¿
ÊÔÌâ·ÖÎö£º£¨1£©¢Ùµç½â³ØµÄÑô¼«·¢ÉúÑõ»¯·´Ó¦£¬ÈÜÒºÖеÄÇâÑõ¸ùÀë×ӵķŵçÄÜÁ¦´óÓÚÁòËá¸ùÀë×ӵķŵçÄÜÁ¦£¬ËùÒÔÑô¼«ÉÏÇâÑõ¸ùÀë×Óʧµç×ÓÉú³ÉË®ºÍÑõÆø4OH--4e-=2H2O+O2¡ü£»
¢Ú¼ØÀë×ÓÏòÒõ¼«Òƶ¯£¬ËùÒÔKOHÔÚÒõ¼«µÃµ½£¬ÔòKOH´ÓD¿Ú·Å³ö£»2LŨ¶ÈΪ0.25mol/LµÄKOHÈÜÒºÖÐKOHµÄÎïÖʵÄÁ¿ÊÇ0.5mol£¬2L0.025mol/LµÄÁòËáÈÜÒºÖÐÁòËáµÄÎïÖʵÄÁ¿ÊÇ0.05mol£¬ËùÒÔKOH¹ýÁ¿£¬»ìºÏºóµÄÇâÑõ¸ùÀë×ÓŨ¶ÈÊÇ(0.5mol-0.05mol¡Á2)/4L=0.1mol/L£¬ËùÒÔÈÜÒºµÄpH=-lgc(H+)=13£»
¢ÛÑô¼«ÇâÑõ¸ùÀë×ӷŵ磬Òò´ËÁòËá¸ùÀë×ÓÏòÑô¼«Òƶ¯£¬Òõ¼«ÇâÀë×ӷŵ磬Òò´Ë¼ØÀë×ÓÏòÒõ¼«Òƶ¯£¬ËùÒÔͨ¹ýÏàͬµçÁ¿Ê±£¬Í¨¹ýÒõÀë×Ó½»»»Ä¤µÄÀë×ÓÊýСÓÚͨ¹ýÑôÀë×Ó½»»»Ä¤µÄÀë×ÓÊý£»
¢ÜȼÁÏÔµç³ØÖУ¬È¼ÁÏÔÚ¸º¼«ÉÏʧµç×Ó·¢ÉúÑõ»¯·´Ó¦£¬Ñõ»¯¼ÁÔÚÕý¼«Éϵõç×Ó·¢Éú»¹Ô·´Ó¦£¬¸ÃȼÁÏÔµç³ØÖУ¬ÑõÆøÊÇÑõ»¯¼Á£¬ËùÒÔÑõÆøÔÚÕý¼«Éϵõç×ÓºÍÇâÀë×Ó½áºÏÉú³ÉË®£¬µç¼«·´Ó¦Ê½Îª4H+ +O2+4e- ===2H2O£»
£¨2£©¢ÙÏòB±ÖмÓÈëÊÊÁ¿½ÏŨµÄÁòËᣬ·¢ÏÖGµÄÖ¸ÕëÏòÓÒÆ«ÒÆ£¬¡°GÊÇÁéÃôµçÁ÷¼Æ£¬ÆäÖ¸Õë×ÜÊÇÆ«ÏòµçÔ´Õý¼«¡±£¬ËùÒÔ˵µç×Ó´ÓCÅܵ½D£¬ÔòDÊÇÕý¼«£¬CÊǸº¼«£¬AÖеâÀë×Ó·¢ÉúÑõ»¯·´Ó¦£¬Ê§È¥µç×Ó£¬Éú³Éµâµ¥ÖÊÓöµ½µí·Û±äÀ¶£¬DÖÐŨÁòËὫµÍ¼ÛµÄÑÇÉéËá¸ùÀë×ÓÑõ»¯£¬¸ÃÀë×Ó·¢Éú»¹Ô·´Ó¦£¬µç¼«·´Ó¦Ê½Îª£ºAsO43-+2H++2e-=AsO32-+H2O£»
¢ÚÔÙÏòB±ÖмÓÈëÊÊÁ¿µÄÖÊÁ¿·ÖÊýΪ40%µÄÇâÑõ»¯ÄÆÈÜÒº£¬·¢ÏÖGµÄÖ¸ÕëÏò×óÆ«ÒÆ£¬ËµÃ÷CÊÇÕý¼«£¬DÊǸº¼«£¬AÖз¢Éúµâµ¥ÖÊʧµç×ӵķ´Ó¦£¬BÖз¢ÉúÑÇÉéËáʧµç×ӵķ´Ó¦£¬×Ü·´Ó¦Îª£ºAsO32-+I2+2H2O=2H++2I-+AsO43-¡£
¿¼µã£º¿¼²éµç»¯Ñ§·´Ó¦ÔÀíµÄÓ¦ÓÃ
¼×¡¢ÒÒ¡¢±ûÊÇÈýÖÖ²»º¬ÏàͬÀë×ӵĿÉÈÜÐÔÇ¿µç½âÖÊ¡£ËüÃÇËùº¬Àë×ÓÈçϱíËùʾ£¬È¡µÈÖÊÁ¿µÄÈýÖÖ»¯ºÏÎïÅäÖÆÏàͬÌå»ýµÄÈÜÒº£¬ÆäÈÜÖÊÎïÖʵÄÁ¿Å¨¶È£ºc(¼×)>c(ÒÒ)>c(±û)£¬ÔòÒÒÎïÖÊ¿ÉÄÜÊÇ
ÒõÀë×Ó | OH£¡¢NO | ÑôÀë×Ó | NH |
¢ÙMgSO4 ¢ÚNaOH ¢Û(NH4)2SO4 ¢ÜMg(NO3)2 ¢ÝNH4NO3
A£®¢Ù¢Ú B£®¢Û¢Ü C£®¢Û¢Ý D£®¢Ù¢Ý
ÏàͬζÈÏ£¬Ìå»ý¾ùΪ0.25LµÄÁ½¸öºãÈÝÃܱÕÈÝÆ÷Öз¢Éú¿ÉÄæ·´Ó¦£º
N2(g)£«3H2(g)
2NH3(g) ¡÷H£½Ò»92.6kJ/mol¡£ÊµÑé²âµÃÆðʼ¡¢Æ½ºâʱµÄÓйØÊý¾ÝÈçÏÂ±í£º
ÈÝÆ÷
| Æðʼ¸÷ÎïÖʵÄÎïÖʵÄÁ¿/mol | ´ïƽºâʱÌåϵÄÜÁ¿µÄ±ä»¯ | ||
N2 | H2 | NH3 | ||
¢Ù | 1 | 3 | 0 | ·Å³öÈÈÁ¿£º23.15kJ |
¢Ú | 0.9 | 2.7 | 0.2 | ·Å³öÈÈÁ¿£ºQ |
ÏÂÁÐÐðÊö²»ÕýÈ·µÄÊÇ
A£®ÈÝÆ÷¢Ù¡¢¢ÚÖз´Ó¦µÄƽºâ³£ÊýÏàµÈ
B£®Æ½ºâʱ£¬Á½¸öÈÝÆ÷ÖÐNH3µÄÌå»ý·ÖÊý¾ùΪ1/7
C£®ÈÝÆ÷¢ÚÖÐ´ïÆ½ºâʱ·Å³öµÄÈÈÁ¿Q£½23.15kJ
D£®ÈôÈÝÆ÷¢ÙÌå»ýΪ0.5L£¬Ôòƽºâʱ·Å³öµÄÈÈÁ¿Ð¡ÓÚ23.15kJ