ÌâÄ¿ÄÚÈÝ

ÓлúÎï±û(C13H18O2)ÊÇÒ»ÖÖÏãÁÏ£¬ÆäºÏ³É·ÏßÈçͼËùʾ¡£ÆäÖÐAµÄÏà¶Ô·Ö×ÓÖÊÁ¿Í¨¹ýÖÊÆ×·¨²âµÃΪ56£¬ËüµÄºË´Å¹²ÕñÇâÆ×ÏÔʾֻÓÐÈý×é·å£»D¿ÉÒÔ·¢ÉúÒø¾µ·´Ó¦£¬ÔÚ´ß»¯¼Á´æÔÚÌõ¼þÏÂ1 mol DÓë2 mol H2·´Ó¦¿ÉÒÔÉú³ÉÒÒ£»±ûÖк¬ÓÐÁ½¸ö-CH3

£¨1£©AµÄ½á¹¹¼òʽΪ??????????? £»ÒҵķÖ×ÓʽΪ???????????? ¡£

£¨2£©CÓëÐÂÖÆCu(OH)2Ðü×ÇÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ_________________¡£

£¨3£©DËùº¬¹ÙÄÜÍŵÄÃû³ÆÊÇ???????????????????? £»DÓжàÖÖͬ·ÖÒì¹¹Ì壬ÆäÖÐÓëÆäËùº¬¹ÙÄÜÍÅÏàͬµÄͬ·ÖÒì¹¹ÌåÓÐ????????? ÖÖ£¨²»¿¼ÂÇÁ¢ÌåÒì¹¹£©¡£

£¨4£©¼×ÓëÒÒ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ????????????????????????????????????? ¡£

£¨5£©Ð´³öÂú×ãÏÂÁÐÌõ¼þµÄÓлúÎïµÄ½á¹¹¼òʽ????????????????????????? ¢¡ÓëÒÒ»¥ÎªÍ¬·ÖÒì¹¹Ì壻¢¢ÓöFeCl3ÈÜÒºÏÔ×ÏÉ«£»¢£Æä±½»·ÉϵÄÒ»äå´úÎïÖ»ÓÐÁ½ÖÖ¡£

 

¡¾´ð°¸¡¿

£¨1£©(CH3)2C=CH2? C9H12O

£¨2£©(CH3)2CHCHO+2Cu(OH)2+NaOH (CH3)2CHCOONa+Cu2O¡ý+3H2O

£¨3£©Ì¼Ì¼Ë«¼ü¡¢È©»ù? 4

£¨4£©

£¨5£©??? ?

¡¾½âÎö¡¿

ÊÔÌâ·ÖÎö£º£¨1£©ÓÉAµÄÐÔÖʼ°ÌâÄ¿ÌṩµÄÐÅÏ¢¿ÉÖªAΪ²»±¥ºÍµÄÏ©Ìþ¡£ÉèÈ¥·Ö×ÓʽΪCnH2n ¡£14n=56¡£½âµÃn=4£®¼´AΪC4H8¡£ÒòΪËüµÄºË´Å¹²ÕñÇâÆ×ÏÔʾֻÓÐÈý×é·å£¬ËµÃ÷º¬ÓÐÈýÖÖHÔ­×Ó£¬ÔòAΪCH2=C(CH3)2£® BΪ2-¼×»ù±û´¼ (CH3)2CHCH2OH£® B´ß»¯Ñõ»¯ÎªC£º2-¼×»ù±ûÈ© (CH3)2CHCHO¡£CÓëÐÂÖÆµÄÇâÑõ»¯Í­Ðü×ÇÒº¹²ÈÈ£¬Öó·Ð¿ÉµÃ¼×£º2-¼×»ù±ûËá (CH3)2CHCOOH¡£ÒòΪ¼×ÒÒ·¢Éúõ¥»¯·´Ó¦µÃµ½õ¥C13H18O2ºÍË®£¬ËùÒÔÔÚÒҵķÖ×ÓÖк¬Óеĸ÷ÖÖÔªËØµÄÔ­×Ó¸öÊýΪC£º13-4=9£»H£º18+2-8=12£®O£º2+1-1=2£®ÒÒ·Ö×ÓʽΪC9H12O¡££¨2£©2-¼×»ù±ûÈ© (CH3)2CHCHOÓëÐÂÖÆCu(OH)2Ðü×ÇÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ(CH3)2CHCHO+2Cu(OH)2+NaOH (CH3)2CHCOONa+Cu2O¡ý+3H2O£¨3£©D¿ÉÒÔ·¢ÉúÒø¾µ·´Ó¦£¬Ö¤Ã÷DÖк¬ÓÐÈ©»ù£¨-CHO£©£»ÔÚ´ß»¯¼Á´æÔÚÌõ¼þÏÂ1 mol DÓë2 mol H2·´Ó¦¿ÉÒÔÉú³ÉÒÒ£¬ÔòÒÒÖл¹º¬ÓÐ̼̼˫¼ü¡£Òò´ËDËùº¬¹ÙÄÜÍŵÄÃû³ÆÊÇ̼̼˫¼ü¡¢È©»ù¡£DÓжàÖÖͬ·ÖÒì¹¹Ì壬ÆäÖÐÓëÆäËùº¬¹ÙÄÜÍÅÏàͬµÄͬ·ÖÒì¹¹ÌåÓÐ4ÖÖ¡£ËüÃÇ·Ö±ðÊÇ£º£»£»£»¡£

£¨4£©¼×ÓëÒÒ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ

£¨5£©Ð´³öÂú×ãÌõ¼þ£º¢¡ÓëÒÒ»¥ÎªÍ¬·ÖÒì¹¹Ì壻¢¢ÓöFeCl3ÈÜÒºÏÔ×ÏÉ«£»¢£Æä±½»·ÉϵÄÒ»äå´úÎïÖ»ÓÐÁ½ÖÖµÄÓлúÎïµÄ½á¹¹¼òʽ ºÍ¡£

¿¼µã£º¿¼²éÓлúÎïµÄÍÆ¶Ï¡£°üÀ¨ÓлúÎïµÄ·Ö×Óʽ¡¢½á¹¹Ê½¡¢½á¹¹¼òʽ¡¢»¯Ñ§·½³ÌʽµÄÊéд¡¢Í¬·ÖÒì¹¹ÌåµÄÖÖÀ༰¹ÙÄÜÍŵÄÃû³ÆµÈ֪ʶ¡£

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
[»¯Ñ§--Óлú»¯Ñ§»ù´¡]
£¨Ò»£©ËÕºÏÏã´¼¿ÉÒÔÓÃ×÷ʳÓÃÏ㾫£¬Æä½á¹¹¼òʽÈçͼËùʾ£®
£¨1£©ËÕºÏÏã´¼µÄ·Ö×ÓʽΪ
C8H10O
C8H10O
£¬Ëü²»ÄÜ·¢ÉúµÄÓлú·´Ó¦ÀàÐÍÓУ¨ÌîÊý×ÖÐòºÅ£©
¢Ü¢Þ
¢Ü¢Þ
£®
¢ÙÈ¡´ú·´Ó¦           ¢Ú¼Ó³É·´Ó¦       ¢ÛÏûÈ¥·´Ó¦
¢Ü¼Ó¾Û·´Ó¦           ¢ÝÑõ»¯·´Ó¦       ¢ÞË®½â·´Ó¦
£¨¶þ£©ÓлúÎï±ûÊÇÒ»ÖÖÏãÁÏ£¬ÆäºÏ³É·ÏßÈçͼ£®ÆäÖм׵ÄÏà¶Ô·Ö×ÓÖÊÁ¿Í¨¹ýÖÊÆ×·¨²âµÃΪ88£¬ËüµÄºË´Å¹²ÕñÇâÆ×ÏÔʾֻÓÐÈý×é·å£»ÒÒÓëËÕºÏÏã´¼»¥ÎªÍ¬ÏµÎ


£¨2£©°´ÕÕϵͳÃüÃû·¨£¬AµÄÃû³ÆÊÇ
2-¼×»ù±ûÏ©
2-¼×»ù±ûÏ©
£®
£¨3£©CÓëÐÂÖÆCu£¨OH£©2Ðü×ÇÒº·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
£¨CH3£©2CHCHO+2Cu£¨OH£©2
¡÷
£¨CH3£©2CHCOOH+Cu2O¡ý+2H2O
£¨CH3£©2CHCHO+2Cu£¨OH£©2
¡÷
£¨CH3£©2CHCOOH+Cu2O¡ý+2H2O
£®
£¨4£©±ûÖк¬ÓÐÁ½¸ö-CH3£¬ÔÚ´ß»¯¼Á´æÔÚÏÂ1mol DÓë2mol H2¿ÉÒÔ·´Ó¦Éú³ÉÒÒ£¬ÔòDµÄ½á¹¹¼òʽΪ
-CH=CHCHO
-CH=CHCHO
£®
£¨5£©¼×ÓëÒÒ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
£¨CH3£©2CHCOOH+-CH2CH2CH2OH£¨CH3£©2CHCOOCH2CH2CH2-+H2O
£¨CH3£©2CHCOOH+-CH2CH2CH2OH£¨CH3£©2CHCOOCH2CH2CH2-+H2O
£®
£¨6£©Ð´³ö·ûºÏÏÂÁÐÌõ¼þµÄÒÒµÄͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ£º

¢Ù±½»·ÉÏÓÐ3¸öÈ¡´ú»ù»ò¹ÙÄÜÍÅ
¢ÚÏÔÈõËáÐÔ
¢Û±½»·ÉÏ 3¸öÈ¡´ú»ù»ò¹ÙÄÜÍÅ»¥²»ÏàÁÚ£®

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

¾«Ó¢¼Ò½ÌÍø