ÌâÄ¿ÄÚÈÝ
A¡¢B¡¢C¡¢D¡¢E¡¢F¡¢G¡¢L¡¢I¾ÅÖÖÖ÷×åÔªËØ·Ö²¼ÔÚÈý¸ö²»Í¬µÄ¶ÌÖÜÆÚ£¬ËüÃǵÄÔ×ÓÐòÊýÒÀ´ÎÔö´ó£¬ÆäÖÐB¡¢C¡¢DΪͬһÖÜÆÚ£¬AÓëE¡¢BÓëG¡¢DÓëL·Ö±ðΪͬһÖ÷×壬C¡¢D¡¢FÈýÖÖÔªËØµÄÔ×ÓÐòÊýÖ®ºÍΪ28£¬FµÄÖÊ×ÓÊý±ÈD¶à5£¬DµÄ×îÍâ²ãµç×ÓÊýÊÇFµÄ2±¶£¬CºÍDµÄ×îÍâ²ãµç×ÓÊýÖ®ºÍΪ11¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÒÔÉϰËÖÖÔªËØÖзǽðÊôËùÐγɵÄ×î¼òµ¥ÆøÌ¬Ç⻯ÎïÎȶ¨ÐÔ×îÈõµÄÊÇ£¨Ìѧʽ£©_________£¬E¡¢F¡¢LËùÐγɵļòµ¥Àë×ӵİ뾶ÓÉ´óµ½Ð¡µÄ˳ÐòΪ£¨ÓÃÀë×Ó·ûºÅ±íʾ£©__ __ > > ¡£
£¨2£©ÓÉL¡¢IÁ½ÔªËؿɰ´Ô×Ó¸öÊý±È1:1×é³É»¯ºÏÎïX£¬»¯ºÏÎïXÖи÷Ô×Ó¾ùÂú×ã8µç×ÓµÄÎȶ¨½á¹¹£¬ÔòXµÄµç×ÓʽΪ ¡£¹ÌÌ廯ºÏÎïE2D2ͶÈëµ½»¯ºÏÎïE2LµÄË®ÈÜÒºÖУ¬Ö»¹Û²ìµ½ÓгÁµí²úÉúµÄ£¬Ð´³ö¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º ¡£
£¨3£©ÔÚ10 LµÄÃܱÕÈÝÆ÷ÖУ¬Í¨Èë2mol LD2ÆøÌåºÍ1 mol D2ÆøÌ壬һ¶¨Î¶ÈÏ·´Ó¦ºóÉú³ÉLD3ÆøÌ壬µ±·´Ó¦´ïµ½Æ½ºâʱ£¬D2µÄŨ¶ÈΪ0.01 mol¡¤L-1£¬Í¬Ê±·Å³öÔ¼177 kJµÄÈÈÁ¿£¬ÔòƽºâʱLD2µÄת»¯ÂÊΪ £»¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ £»´Ëʱ¸Ã·´Ó¦µÄƽºâ³£ÊýK= ¡£
(4) ÓÐÈËÉèÏëѰÇóºÏÊʵĴ߻¯¼ÁºÍµç¼«²ÄÁÏ£¬ÒÔC2¡¢A2Ϊµç¼«·´Ó¦ÎÒÔHClÒ»NH4ClÈÜҺΪµç½âÖÊÈÜÒºÖÆÔìÐÂÐÍȼÁÏµç³Ø£¬ÊÔд³ö¸Ãµç³ØµÄÕý¼«·´Ó¦Ê½ £»·ÅµçʱÈÜÒºÖÐH+ÒÆÏò £¨Ìî¡°Õý¡±»ò¡°¸º¡±£©¼«¡£
£¨1£©SiH4 £¨1·Ö£© S2- ©ƒ Na+ ©ƒ Al3+ £¨¹²1·Ö£©
(2) £¨1·Ö£©
Na2O2+ S2-+2 H2O==S+2 Na++4OH- £¨2·Ö£©
(3)90% (1·Ö) 2SO2 (g)+O2 (g) 2SO2 (g) ¦¤H£½£196.7 kJ¡¤mol£1 £¨2·Ö£©
8100 £¨1·Ö£©
(4) N2 + 6e-+ 8H+==2NH4+£¨2·Ö£© Õý£¨1·Ö£©
½âÎö:ÂÔ