ÌâÄ¿ÄÚÈÝ

Ï©Ìþͨ¹ý³ôÑõÑõ»¯²¢¾­Ð¿ºÍË®´¦ÀíµÃµ½È©»òͪ¡£ÀýÈ磺

£¨1£©ÒÑÖª±ûÈ©µÄȼÉÕÈÈΪ1 815 kJ¡¤mol-1£¬±ûͪµÄȼÉÕÈÈΪ1789 kJ¡¤mol-1£¬ÊÔд³ö±ûȩȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ_________________¡£

£¨2£©ÉÏÊö·´Ó¦¿ÉÓÃÀ´ÍƶÏÏ©ÌþµÄ½á¹¹¡£Ò»ÖÖÁ´×´µ¥Ï©ÌþAͨ¹ý³ôÑõÑõ»¯²¢¾­Ð¿ºÍË®´¦ÀíµÃµ½BºÍC¡£»¯ºÏÎïBº¬Ì¼69.8%£¬º¬Çâ11.6%£¬BÎÞÒø¾µ·´Ó¦£¬´ß»¯¼ÓÇâÉú³ÉD£¬DÔÚŨÁòËá´æÔÚϼÓÈÈ£¬¿ÉµÃµ½ÄÜʹäåË®ÍÊÉ«ÇÒÖ»ÓÐÒ»ÖֽṹµÄÎïÖÊE¡£·´Ó¦Í¼Ê¾ÈçÏ£º

Íê³ÉÏÂÁÐÎÊÌ⣺

¢ÙBµÄÏà¶Ô·Ö×ÓÖÊÁ¿ÊÇ__________£»

CFµÄ·´Ó¦ÀàÐÍΪ__________£»

DÖк¬ÓйÙÄÜÍŵÄÃû³Æ__________¡£

¢ÚD+FGµÄ»¯Ñ§·½³ÌʽÊÇ£º_________________¡£

¢ÛAµÄ½á¹¹¼òʽΪ___________________________________¡£

¢Ü»¯ºÏÎïAµÄijÖÖͬ·ÖÒì¹¹Ìåͨ¹ý³ôÑõÑõ»¯²¢¾­Ð¿ºÍË®´¦ÀíÖ»µÃµ½Ò»ÖÖ²úÎ·ûºÏ¸ÃÌõ¼þµÄÒì¹¹ÌåµÄ½á¹¹¼òʽÓÐ__________ÖÖ¡£

½âÎö£º£¨1£©ÊéдÈÈ»¯Ñ§·½³Ìʽʱ£¬Ê×ÏÈÓ¦Çå³þÈÈ»¯Ñ§·½³ÌʽµÄ¶¨Ò壺1 molÎïÖÊÍêȫȼÉÕÉú³ÉÎȶ¨µÄÑõ»¯ÎïʱËù·Å³öµÄÈÈÁ¿£¬´Ó¶øÐ´³ö·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ¡£

£¨2£©¢Ù¸ù¾ÝC¡¢HµÄÖÊÁ¿·ÖÊý£¬¿ÉÒÔÇó³öÑõµÄÖÊÁ¿·ÖÊý£¬¾ÝÌâÒâ·ÖÎö£¬BÖк¬ÓÐÒ»¸öÑõÔ­×Ó£¬´Ó¶øÈ·¶¨BµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª86£¬·Ö×ÓʽΪC5H10O¡£BÎÞÒø¾µ·´Ó¦£¬ËµÃ÷BΪͪ£¬B»¹Ô­ºóµÃµ½´¼£¨D£©£¬DÍÑË®ºóÉú³ÉÏ©£¬Ö»ÓÐÒ»Öֽṹ£¬Ôò½á¹¹¼òʽΪCH3CH£¬DΪ3-Îì´¼¡£ÓÉF·Ö×ÓÄÚº¬ÓÐ8¸ö̼ԭ×Ó¿ÉÖª£¬C¡¢FÖж¼º¬ÓÐ3¸ö̼ԭ×Ó£¬CΪ±ûÈ©£¬FΪ±ûËᣬ´Ó¶øµÃ³ö¡£

¢ÚCH3CH2COOH+£¨C2H5£©2CHOHCH3CH2COOCH£¨C2H5£©2+H2O

¢Û¾­¹ýÒÔÉÏ·ÖÎö£¬ºÜÈÝÒ׵óöAµÄ½á¹¹¼òʽΪ£¨CH3CH2£©2C=CHCH2CH3¡£

¢ÜAµÄ·Ö×ÓʽΪC8H16£¬ÓÉÌâÒâÖªAµÄ·Ö×ӽṹÊǶԳƵ쬴ӶøµÃ³ö¿ÉÄܽṹΪ3ÖÖ¡£

´ð°¸£º£¨1£©CH3CH2CHO(l)+4O2(g)3CO2(g)+3H2O(l);¦¤H=-1 815 kJ¡¤mol-1

£¨2£©¢Ù86  Ñõ»¯·´Ó¦  ôÇ»ù

¢ÚCH3CH2COOH+£¨C2H5£©2CHOHCH3CH2COOCH(C2H5)2+H2O

¢Û(CH3CH2)2C=CHCH2CH3

¢Ü3


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø