ÌâÄ¿ÄÚÈÝ
1£¬3-±û¶þ´¼ÊÇÉú²úÐÂÐ͸߷Ö×Ó²ÄÁÏPTTµÄÖ÷ÒªÔÁÏ£¬Ä¿Ç°ÆäÉú²ú·ÏßÓÐÒÔϼ¸ÖÖ£º
[·Ïß1]
[·Ïß2]
[·Ïß3]
[·Ïß2]
[·Ïß3]
£¨1£©ÓлúÎïAº¬ÓеĹÙÄÜÍÅΪ____________¡£
£¨2£©´ÓºÏ³ÉÔÁÏÀ´Ô´µÄ½Ç¶È¿´£¬ÄãÈÏΪ×î¾ßÓз¢Õ¹Ç°¾°µÄ·ÏßÊÇ _______£¨Ìî1¡¢2»ò3£©£¬ÀíÓÉÊÇ ____________¡£
£¨3£©ÒÔ1£¬3-±û¶þ´¼Óë¶Ô±½¶þ¼×Ëᣨ£©ÎªÔÁÏ¿ÉÒԺϳɾÛõ¥PTT£¬Ð´³öÆ仯ѧ·½³Ìʽ ______________¡£
£¨4£©ÒÑÖª±û¶þËá¶þÒÒõ¥ÄÜ·¢ÉúÒÔÏ·´Ó¦£º
£¨2£©´ÓºÏ³ÉÔÁÏÀ´Ô´µÄ½Ç¶È¿´£¬ÄãÈÏΪ×î¾ßÓз¢Õ¹Ç°¾°µÄ·ÏßÊÇ _______£¨Ìî1¡¢2»ò3£©£¬ÀíÓÉÊÇ ____________¡£
£¨3£©ÒÔ1£¬3-±û¶þ´¼Óë¶Ô±½¶þ¼×Ëᣨ£©ÎªÔÁÏ¿ÉÒԺϳɾÛõ¥PTT£¬Ð´³öÆ仯ѧ·½³Ìʽ ______________¡£
£¨4£©ÒÑÖª±û¶þËá¶þÒÒõ¥ÄÜ·¢ÉúÒÔÏ·´Ó¦£º
ÒÔ±û¶þËá¶þÒÒõ¥¡¢1,3-±û¶þ´¼¡¢ÒÒ´¼ÎªÔÁÏ£¨ÎÞ»úÎïÈÎÑ¡£©ºÏ³É£¬ÔÙת»¯Îª ¡£µÄͬ·ÖÒì¹¹Ìå²»¿ÉÄÜÊôÓÚ__________¡£
a£®´¼ b£®·Ó c£®È© d£®õ¥
£¨5£©ÒªºÏ³É£¬±ØÐëÏȺϳÉÄÄЩÎïÖÊ£¿£¨ÓúϳɸÃÎïÖʵĻ¯Ñ§·½³Ìʽ»Ø´ð£© __________¡£
£¨6£©»¯ºÏÎïB¡«EµÄת»¯¹ØϵÈçÏÂͼËùʾ£¬ÆäÖÐ×ãÁ¿µÄÐÂÖÆCu(OH)2Ðü×ÇÒºÓë1 mol B·´Ó¦¿ÉÉú³É1 mol Cu2O ºÍ1 mol C¡£Ð´³ö·ûºÏÏÂÁÐÌõ¼þµÄDµÄͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ¡£
¢Ùº¬ÓÐÈý¸ö¼×»ù£» ¢ÚÄÜÓëÒÒ´¼·¢Éúõ¥»¯·´Ó¦£» ¢Û-OH¡¢-BrÁ¬ÔÚͬһ¸ö̼Ô×ÓÉÏ¡£
________________________________________________________________
a£®´¼ b£®·Ó c£®È© d£®õ¥
£¨5£©ÒªºÏ³É£¬±ØÐëÏȺϳÉÄÄЩÎïÖÊ£¿£¨ÓúϳɸÃÎïÖʵĻ¯Ñ§·½³Ìʽ»Ø´ð£© __________¡£
£¨6£©»¯ºÏÎïB¡«EµÄת»¯¹ØϵÈçÏÂͼËùʾ£¬ÆäÖÐ×ãÁ¿µÄÐÂÖÆCu(OH)2Ðü×ÇÒºÓë1 mol B·´Ó¦¿ÉÉú³É1 mol Cu2O ºÍ1 mol C¡£Ð´³ö·ûºÏÏÂÁÐÌõ¼þµÄDµÄͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ¡£
¢Ùº¬ÓÐÈý¸ö¼×»ù£» ¢ÚÄÜÓëÒÒ´¼·¢Éúõ¥»¯·´Ó¦£» ¢Û-OH¡¢-BrÁ¬ÔÚͬһ¸ö̼Ô×ÓÉÏ¡£
________________________________________________________________
£¨1£©ÓлúÎïAº¬ÓеĹÙÄÜÍÅΪ£OH¡¢£CHO
£¨2£©1£»ÀíÓÉÊÇ·Ïß1ÒÔ¿ÉÔÙÉú×ÊÔ´µí·ÛΪÔÁÏ£¬Â·Ïß2¡¢3µÄÔÁÏΪʯÓͲúÆ·£¬¶øʯÓÍÊDz»¿ÉÔÙÉú×ÊÔ´
£¨3£©
£¨4£©b
£¨5£©HO-CH2CH2CH2-OH + 2HBr ¡úBrCH2CH2CH2Br + 2H2O£»
2CH3CH2OH + 2Na ¡ú 2CH3CH2ONa + H2¡ü
£¨6£©2
£¨2£©1£»ÀíÓÉÊÇ·Ïß1ÒÔ¿ÉÔÙÉú×ÊÔ´µí·ÛΪÔÁÏ£¬Â·Ïß2¡¢3µÄÔÁÏΪʯÓͲúÆ·£¬¶øʯÓÍÊDz»¿ÉÔÙÉú×ÊÔ´
£¨3£©
£¨4£©b
£¨5£©HO-CH2CH2CH2-OH + 2HBr ¡úBrCH2CH2CH2Br + 2H2O£»
2CH3CH2OH + 2Na ¡ú 2CH3CH2ONa + H2¡ü
£¨6£©2
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿