ÌâÄ¿ÄÚÈÝ

£¨10·Ö£©ÔªËØÂȼ°Æä»¯ºÏÎïÔÚÉú²ú¡¢Éú»î¡¢¿ÆÑÐÖÐÓй㷺µÄÓ¦Óá£
¢Å25¡æÊ±£¬PbCl2¹ÌÌåÔÚ²»Í¬Å¨¶ÈÑÎËá(mol¡¤L£­1)ÖеÄÈܽâ¶È(mmol¡¤L£­1)Èçͼ¡£

¢ÙÔÚÖÆ±¸PbCl2µÄʵÑéÖУ¬Ï´µÓPbCl2¹ÌÌå×îºÃÑ¡Óà        ¡£
a£®ÕôÁóË®             b£®1mol¡¤L£­1ÑÎËá
c£®5 mol¡¤L£­1ÑÎËá      d£®10mol¡¤L£­1ÑÎËá
¢Úµ±ÑÎËáµÄŨ¶ÈСÓÚ1mol¡¤L£­1ʱ£¬Ëæ×ÅÑÎËáŨ¶ÈµÄÔö´ó£¬PbCl2 µÄÈܽâ¶È¼õС£¬ÆäÔ­ÒòÊÇ                           ¡£
¢ÆTCCA¹ã·ºÓÃÓÚÆ¯°×¡¢É±¾úÏû¶¾£¬ÆäѧÃûΪÈýÂȾùÈýàº-2,4,6-Èýͪ£¬·Ö×ÓʽΪ£ºC3Cl3N3O3¡£
¢ÙTCCA·Ö×Ó¾ßÓÐÍêÈ«¶Ô³ÆµÄ½á¹¹£¬²¢º¬ÓÐÒ»¸öÁùÔª»·£¬ÔòÆä½á¹¹¼òʽΪ         ¡£
¢ÚʹÓÃTCCAʱ£¬ÐèÏȽ«¸ÃÎïÖÊÈܽâÓÚË®£¬ÆäË®½â²úÎï֮һΪC3H3N3O3£¬ÁíÒ»ÖÖ²ú
Îï¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬Äܹ»É±¾úÏû¶¾¡£Ð´³öÁíÒ»ÖÖ²úÎïµÄµç×Óʽ                ¡£
¢Ç¸ßÂÈËáï§(AP)×÷ΪһÖÖÓÅÁ¼µÄ¹ÌÌåÍÆ½ø¼Á±»ÓÃÓÚµ¼µ¯ºÍ»ð¼ý·¢É䡣Ŀǰ£¬½ÏΪÏȽøµÄÖÆ±¸·½·¨Êǵç½â¸ß´¿´ÎÂÈËáµÃµ½¸ß´¿¸ßÂÈËᣬÔÙÓë¸ß´¿°±½øÐÐÅçÎí·´Ó¦ÖÆ³É¸ßÂÈËáï§¡£Ð´³öÓÉ´ÎÂÈËáµç½âÖÆ±¸¸ßÂÈËáµÄÑô¼«·´Ó¦Ê½£º                            ¡£
¢Å¢Ùb  ¢ÚCl£­Å¨¶ÈÔö´ó£¬PbCl2µÄÈÜ½âÆ½ºâÄæÏòÒÆ¶¯
¢Æ¢Ù ¢Ú
¢Ç¢ÙHClO+3H2O£­6e£­£½ClO4¡ª+7H+
ÂÔ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨6·Ö£©ÇëÓÃѧ¹ýµÄ֪ʶ·ÖÎöÆäÖеĻ¯Ñ§Ô­Àí²¢Ð´³ö»¯Ñ§·½³Ìʽ£¬ÊôÓÚÀë×Ó·´Ó¦µÄÇëд³öÀë×Ó·½³Ìʽ¡££¨1£©½«ÂÈÆøÍ¨ÈëÊìʯ»Ò¼´¿ÉÖÆµÄƯ°×·Û¡¾Æ¯°×·ÛµÄÓÐЧ³É·ÖÊÇ´ÎÂÈËá¸ÆCa(ClO)2,ÉÌÆ·Æ¯°×·ÛÍùÍùº¬ÓÐCa(OH)2µÈÔÓÖÊ¡¿                                 ¡££¨2£©Æ¯°×·ÛÖ®ËùÒÔ¾ßÓÐÆ¯°××÷ÓÃÊÇÓÉÓÚCa(ClO)2ÔÚË®ÈÜÒºÖз¢ÉúË®½â·´Ó¦Éú³É¾ßÓÐÇ¿Ñõ»¯ÐÔµÄÎïÖÊ                                                                  ¡££¨3£©Ca(OH)2ÔÓÖʵĴæÔÚʹÈÜÒºµÄ¼îÐÔÔöÇ¿£¬Òò´ËƯ°××÷ÓýøÐлºÂý¡£ÒªÔÚ¶Ìʱ¼äÊܵ½Æ¯°×Ч¹û£¬±ØÐë³ýÈ¥Ca(OH)2£¬ËùÒÔ¹¤ÒµÉÏʹÓÃÆ¯°×·ÛÊdz£¼ÓÈëÉÙÁ¿ÈõËáÈç´×ËáµÈ£¬»ò¼ÓÈëÉÙÁ¿µÄÏ¡ÑÎËá                                                             ¡¢
                                   ¡££¨4£©¼ÒͥʹÓÃÆ¯°×·Û²»±Ø¼ÓËᣬÒòΪˮÖеÄCO2Ò²Æðµ½ÁËÈõËáµÄ×÷Óà                                                       ¡££¨5£©²»Òª½«Æ¯°×·ÛÓë½à²Þ¼ÁµÈÇ¿ËáÐÔÎïÖÊ»ìºÏʹÓã¬ÒòΪÔÚÇ¿ËáÐÔÌõ¼þ϶þÕß¿ÉÒÔ·¢ÉúÑõ»¯»¹Ô­·´Ó¦Éú³ÉÒ»ÖÖÓж¾ÆøÌ壺                                   
¹¤ÒµºÏ³É°±ÓëÖÆ±¸ÏõËáÒ»°ã¿ÉÁ¬ÐøÉú²ú£¬Á÷³ÌÈçÏ£º

£¨1£©¹¤ÒµÉú²úʱ£¬ÖÆÈ¡ÇâÆøµÄÒ»¸ö·´Ó¦Îª£ºCO+H2O(g)CO2+H2
¢Ùt¡æÊ±£¬Íù1LÃܱÕÈÝÆ÷ÖгäÈë0.2mol COºÍ0.3molË®ÕôÆø¡£·´Ó¦½¨Á¢Æ½ºâºó£¬ÌåϵÖÐc(H2)=0.12mol¡¤L£­1¡£¸ÃζÈÏ´˷´Ó¦µÄƽºâ³£ÊýK=_____£¨Ìî¼ÆËã½á¹û£©¡£
¢Ú±£³ÖζȲ»±ä£¬ÏòÉÏÊöƽºâÌåϵÖÐÔÙ¼ÓÈë0.1molCO£¬µ±·´Ó¦ÖØÐ½¨Á¢Æ½ºâʱ£¬Ë®ÕôÆøµÄת»¯ÂʦÁ(H2O)=________¡£
£¨2£©ºÏ³ÉËþÖз¢Éú·´Ó¦N2(g)+3H2(g)2NH3(g)£»¡÷H<0¡£Ï±íΪ²»Í¬Î¶Èϸ÷´Ó¦µÄƽºâ³£Êý¡£ÓÉ´Ë¿ÉÍÆÖª£¬±íÖÐ
T/K
T1
573
T2
K
1.00¡Á107
2.45¡Á105
1.88¡Á103
T1____573K£¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©¡£
£¨3£©NH3ºÍO2ÔÚ²¬Ïµ´ß»¯¼Á×÷ÓÃÏ´Ó145¡æ¾Í¿ªÊ¼·´Ó¦£º
4NH3(g)+5O2(g)4NO(g)+6H2O(g)¡¡¡÷H=£­905kJ¡¤mol£­1
²»Í¬Î¶ÈÏÂNO²úÂÊÈçͼËùʾ¡£Î¶ȸßÓÚ900¡æÊ±£¬
NO²úÂÊϽµµÄÔ­Òò                               ¡£
£¨4£©ÎüÊÕËþÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º
                                                                     ¡£
£¨5£©ÉÏÊö¹¤ÒµÁ÷³ÌÖУ¬²ÉÓÃÁËÑ­»·²Ù×÷¹¤ÒÕµÄÊÇ     £¨ÌîÐòºÅ£©
£¨6£©ÏõËá³§µÄÎ²Æøº¬ÓеªÑõ»¯ÎÈç¹û²»¾­´¦ÀíÖ±½ÓÅŷŽ«ÎÛȾ¿ÕÆø¡£Ä¿Ç°¿ÆÑ§¼Ò̽Ë÷ÀûÓÃȼÁÏÆøÌåÖеļ×ÍéµÈ½«µªÑõ»¯ÎﻹԭΪµªÆøºÍË®£¬Æä·´Ó¦»úÀíΪ£º
CH4(g)+4NO2(g)£½4NO(g)+CO2(g)+2H2O(g)£» ¡÷H=£­574kJ¡¤mol£­1
CH4(g)+4NO(g)£½2N2(g)+CO2(g)+2H2O(g)£»  ¡÷H=£­1160kJ¡¤mol£­1
Ôò¼×ÍéÖ±½Ó½«NO2»¹Ô­ÎªN2µÄÈÈ»¯Ñ§·½³ÌʽΪ£º_____________________________¡£

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

¾«Ó¢¼Ò½ÌÍø