ÌâÄ¿ÄÚÈÝ

ij³§ÒÔ¸ÊÕáΪԭÁÏÖÆÌÇ£¬¶Ô²úÉúµÄ´óÁ¿¸ÊÕáÔü°´ÏÂͼËùʾת»¯½øÐÐ×ÛºÏÀûÓá£ÆäÖÐBÊÇAË®½âµÄ×îÖÕ²úÎCµÄ»¯Ñ§Ê½ÎªC3H6O3£¬Ò»¶¨Ìõ¼þÏÂ2¸öC·Ö×Ó¼äÍÑÈ¥2¸öË®·Ö×Ó¿ÉÉú³ÉÒ»ÖÖÁùÔª»·×´»¯ºÏÎD¿ÉʹäåË®ÍÊÉ«£»FÊǾßÓÐÏãζµÄÒºÌå¡£(ͼÖв¿·Ö·´Ó¦Ìõ¼þ¼°²úÎïûÓÐÁгö)

»Ø´ðÏÂÁÐÎÊÌ⣺
(1)AµÄÃû³ÆÊÇ________£¬FµÄ½á¹¹¼òʽ__________£¬
DµÄ½á¹¹¼òʽ________¡£
(2)C¡úDµÄ·´Ó¦ÀàÐÍΪ________¡£D¡úEµÄ·´Ó¦ÀàÐÍΪ________¡£
(3)д³öÏÂÁз´Ó¦µÄ·½³Ìʽ£º
A¡úB£º______________________________£»
G¡úH£º______________________________________¡£
(4)H·Ö×ÓËùº¬¹ÙÄÜÍŵÄÃû³ÆÊÇ________£¬ÊµÑéÊÒÖг£ÓÃÓÚ¼ìÑé¸Ã¹ÙÄÜÍŵÄÊÔ¼ÁµÄÃû³ÆÊÇ(ֻдһÖÖ)_____________________________¡£

(1)ÏËάËØ¡¡CH3COOCH2CH3
CH2===CH¡ªCOOH
(2)ÏûÈ¥·´Ó¦¡¡õ¥»¯·´Ó¦(»òÈ¡´ú·´Ó¦)
(3)(C6H10O5)n£«nH2O nC6H12O6
2CH3CH2OH£«O2 2CH3CHO£«2H2O
(4)È©»ù¡¡ÐÂÖÆÇâÑõ»¯Í­Ðü×ÇÒº(ì³ÁÖÊÔ¼Á)»òÒø°±ÈÜÒº(ÏõËáÒøµÄ°±ÈÜÒº¡¢ÍÐÂ×ÊÔ¼Á)

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø