ÌâÄ¿ÄÚÈÝ

ijËÜ»¯¼Á£¨DEHP£©µÄ½á¹¹¼òʽΪ£º
ÆäºÏ³É·ÏßÈçÏ£º

£¨1£©DEHPµÄ·Ö×Óʽ              £»
£¨2£©·´Ó¦IµÄÀàÐÍ              £¬·´Ó¦IVµÄ²úÎïÖк¬Ñõ¹ÙÄÜÍŵÄÃû³Æ             ¡£
£¨3£©M£¬NµÄ½á¹¹¼òʽ·Ö±ðΪM            ¡¢N                     ¡£
£¨4£©·´Ó¦IVµÄ»¯Ñ§·½³Ìʽ                                                 £»
£¨5£©Ò»¶¨Ìõ¼þÏ£¬1molºÍ×ãÁ¿µÄH2·´Ó¦£¬×î¶à¿ÉÏûºÄH2      mol¡£
£¨6£©·Ö×ÓʽΪC8H6Cl4µÄÒ»ÖÖͬ·ÖÒì¹¹Ìå¿ÉÒÔͨ¹ýÀàËÆ£ºµÄ·´Ó¦µÃµ½£¬Çëд³ö¸Ãͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ£º            ¡£
£¨16·Ö£©
£¨1£©C14H18O4£¨2·Ö£©
£¨2£©È¡´ú·´Ó¦£¨1·Ö£©  ôÈ»ù£¨1·Ö£©
£¨3£© £¨2·Ö£©      CH3CH2CH2OH£¨2·Ö£©
£¨4£©+ 4 Cu(OH)2     + 2 Cu2O¡ý+4H2O£¨3·Ö£©
£¨5£© 5 £¨2·Ö£©   
£¨6£©   £¨3·Ö£©

ÊÔÌâ·ÖÎö£º
£¨1£©ÓɼüÏßʽת»¯Îª·Ö×ÓʽµÄÊéд£¬Òªµã£º½»µã¡¢×ªÕÛµã̼ԭ×Ó²»ÒªÊýÊý´í£¬Ì¼¡¢Çâ¡¢ÑõÔ­×ÓµÄÊýĿһ¶¨ÊýÇ壬Ò׵ãºC14H18O4£»
£¨2£©·´Ó¦ÀàÐ͵ÄÅжϣºI¡¢IIΪȡ´ú·´Ó¦£¬III¡¢IVΪÑõ»¯·´Ó¦£¬IVµÄ²úÎïÊÇÁÚ±½¶þ¼×Ëᣬ¹ÙÄÜÍÅÊÇôÈ»ù£»
£¨3£©½á¹¹ÍƶϣºMΪÁÚ±½¶þ¼×´¼£¬NÊÇ1-±û´¼£¬Ð´³öÆä½á¹¹¼òʽ¼´¿É£»
£¨4£©Ð´³öÁÚ±½¶þ¼×È©±»ÐÂÖÆCu(OH)2Ñõ»¯ÎªÁÚ±½¶þ¼×ËáµÄ»¯Ñ§·½³Ìʽ£¬×¢ÒâÅ䯽ºÍ·´Ó¦Ìõ¼þ²»ÒªÅª´í£»
£¨5£©1mol±½»·¿É×î¶àÓë3molH2¼Ó³É£¬1molÈ©»ùÏûºÄ1molH2£¬ºÏ¼Æ¹²ÏûºÄ5molH2£»
£¨6£©½áºÏÌâ¸øÐÅÏ¢£¬·Ö×ÓʽΪ£ºC8H6Cl4ÄÜË®½âµÃµ½ÁÚ±½¶þ¼×È©µÄ½á¹¹Îª£º¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø