ÌâÄ¿ÄÚÈÝ

£¨6·Ö£©»ðÁ¦·¢µçÔÚÄÜÔ´ÀûÓÃÖÐÕ¼½Ï´ó±ÈÖØ£¬µ«ÊÇÅŷųöµÄSO2»áÔì³ÉһϵÁл·¾³ºÍÉú̬ÎÊÌâ¡£ÀûÓú£Ë®ÍÑÁòÊÇÒ»ÖÖÓÐЧµÄ·½·¨£¬Æ乤ÒÕÁ÷³ÌÈçÏÂͼËùʾ£º

£¨1£©ÌìÈ»º£Ë®³Ê¼îÐÔ£¬Ð´³öSO2ÓëOH-·´Ó¦µÄÀë×Ó·½³Ìʽ£º                     ¡£
£¨2£©ÌìÈ»º£Ë®ÎüÊÕÁ˺¬ÁòÑÌÆøºó»áÈÜÓÐH2SO3·Ö×Ó£¬Ê¹ÓÃÑõÆø½«ÆäÑõ»¯µÄ»¯Ñ§Ô­ÀíÊÇ
                            £¨Óû¯Ñ§·½³Ìʽ±íʾ£©¡£Ñõ»¯ºóµÄ¡°º£Ë®¡±ÐèÒªÒýÈë´óÁ¿µÄÌìÈ»º£Ë®ÓëÖ®»ìºÏºó²ÅÄÜÅÅ·Å£¬¸Ã²Ù×÷µÄÖ÷ҪĿµÄÊÇ                     ¡£
£¨1£©SO2+2OH-==SO32-+H2O  £¨2£©2H2SO3+O2==2H2SO4Öк͡¢Ï¡Ê;­ÑõÆøÑõ»¯ºóº£Ë®ÖÐÉú³ÉµÄËᣨH+£©¡£
±¾Ì⿼²éÁ˹¤ÒµÓ뻯ѧµÄÃÜÇйØϵ£»£¨1£©SO2¿ÉÓë¼î·´Ó¦Éú³ÉÑκÍË®£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£ºSO2+2OH-==SO32-+H2O£»£¨2£©º¬ÁòµÄÑÌÆøÖ÷Òª³É·ÖΪSO2,SO2ÄÜÈÜÓÚË®£¬Éú³ÉH2SO3£¬¹ÊÓÉÓÚH2SO3ÖÐÁòµÄ»¯ºÏ¼ÛΪ+4¼Û£¬¹ÊH2SO3¾ßÓл¹Ô­ÐÔ£¬¿É±»ÑõÆøÑõ»¯£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ2H2SO3+O2==2H2SO4£»ÓÉÓÚÌìÈ»º£Ë®³Ê¼îÐÔ£¬Ñõ»¯ºóµÄ¡°º£Ë®¡±ÓëÖ®»ìºÏ¿É·¢ÉúÖкͷ´Ó¦£¬Ï¡Ê;­Ñõ»¯ºóº£Ë®ÖÐÉú³ÉµÄËá¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø