ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿½ð¾§ÌåµÄ×îСÖظ´µ¥Ôª(Ò²³Æ¾§°û)ÊÇÃæÐÄÁ¢·½Ì壬ÈçͼËùʾ£¬¼´ÔÚÁ¢·½ÌåµÄ8¸ö¶¥µã¸÷ÓÐÒ»¸ö½ðÔ­×Ó£¬¸÷¸öÃæµÄÖÐÐÄÓÐÒ»¸ö½ðÔ­×Ó£¬Ã¿¸ö½ðÔ­×Ó±»ÏàÁڵľ§°ûËù¹²ÓС£½ðÔ­×ÓµÄÖ±¾¶Îªd£¬ÓÃNA±íʾ°¢·ü¼ÓµÂÂÞ³£Êý£¬M±íʾ½ðµÄĦ¶ûÖÊÁ¿¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)½ð¾§Ìåÿ¸ö¾§°ûÖк¬ÓÐ________¸ö½ðÔ­×Ó¡£

(2)Óû¼ÆËãÒ»¸ö¾§°ûµÄÌå»ý£¬³ý¼Ù¶¨½ðÔ­×ÓÊǸÕÐÔСÇòÍ⣬»¹Ó¦¼Ù¶¨_______________¡£

(3)Ò»¸ö¾§°ûµÄÌå»ýÊÇ____________¡£

(4)½ð¾§ÌåµÄÃܶÈÊÇ____________¡£

¡¾´ð°¸¡¿£¨1£©4

£¨2£©¾àÀë×î½üµÄÁ½½ðÔ­×Ó¼äÏà½Ó´¥£¬¼´ÏàÇÐ

£¨3£©2d3

£¨4£©

¡¾½âÎö¡¿

ÀûÓþù̯·¨½âÌ⣬8¸ö¶¥½ÇÉϽðÔ­×ÓÓÐÊôÓڸþ§°û£¬Ã¿¸öÃæÉϽðÔ­×ÓÓÐÊôÓڸþ§°û£¬¹²ÓÐ6¸ö£¬¹Êÿ¸ö¾§°ûÖнðÔ­×Ó¸öÊý£½8¡Á£«6¡Á£½4¡£¼ÙÉè¾àÀë×î½üµÄÁ½½ðÔ­×Ó¼äÏàÇУ¬ÔòÓÐÕý·½ÐεĶԽÇÏßΪ2d¡£Õý·½Ðα߳¤Îªd¡£ËùÒÔV¾§£½£¨d£©3£½2d3£¬Vm£½NA£½d3NA£¬ËùÒԦѣ½£½¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿ÇâÆøÊÇÒ»ÖÖÀíÏëµÄÂÌÉ«ÄÜÔ´¡£ÀûÓÃÉúÎïÖÊ·¢½ÍµÃµ½µÄÒÒ´¼ÖÆÈ¡ÇâÆø£¬¾ßÓÐÁ¼ºÃµÄÓ¦ÓÃÇ°¾°¡£ÒÒ´¼Ë®ÕôÆøÖØÕûÖÆÇâµÄ²¿·Ö·´Ó¦¹ý³ÌÈçͼ1Ëùʾ£º

ÒÑÖª£º·´Ó¦IºÍ·´Ó¦¢òµÄƽºâ³£ÊýËæζȱ仯ÇúÏßÈçͼ2Ëùʾ£º

£¨1£©¢Ùд³ö·´Ó¦IÖеĻ¯Ñ§·½³ÌʽÊÇ________________¡£

¢ÚÈô·´Ó¦IÔÚºãκãѹÌõ¼þÏ£¬ÏòÌåϵÖгäÈëN2£¬ÒÒ´¼µÄƽºâת»¯ÂÊ___(Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±)£¬ÀíÓÉÊÇ___________¡£

£¨2£©·´Ó¦¢òÔÚ½øÆø±È[n(CO)£ºn(H2O)]²»Í¬£¬²âµÃÏàÓ¦µÄCOµÄƽºâת»¯ÂʼûÏÂͼ3(¸÷µã¶ÔÓ¦µÄ·´Ó¦Î¶ȿÉÄÜÏàͬ£¬Ò²¿ÉÄܲ»Í¬)¡£

¢ÙͼÖÐD¡¢EÁ½µã¶ÔÓ¦µÄ·´Ó¦Î¶ȷֱðΪTDºÍTE¡£ÅжϣºTD______________TE(Ìî¡°<¡±¡°=¡±»ò¡°>¡±)£»

¢Úµ±²»Í¬µÄ½øÆø±È´ïµ½ÏàͬµÄCOƽºâת»¯ÂÊʱ£¬¶ÔÓ¦µÄ·´Ó¦Î¶ȺͽøÆø±ÈµÄ¹ØϵÊÇ___(ÓüòÒªµÄÎÄ×ÖÐðÊö)¡£

£¨3£©ÒÑÖª£º

a¡¢2CH3OH(g) + CO2(g) CO(OCH3)2(g) + H2O(g) ¡÷H = -15.5 kJ/mol£¬¸Ã·´Ó¦ÔÚ0 ¡æʱK¡Ö10-4.5£»

b¡¢2CH3OH(g) + CO2(g) + (g) CO(OCH3)2(g) + (g)¡÷H = -110.7 kJ/mol¡£ÒÀ¾ÝÒÔÉÏÊý¾Ý£¬ÎÄÏ×ÈÏΪ·´Ó¦aûÓй¤Òµ¼ÛÖµ£¬ÄãÈÏΪÆäÀíÓÉÊÇ______________________£»µ«·´Ó¦bÒýÈë»·Ñõ±ûÍé()¿ÉÓÐЧ´Ù½øCO2ÓëCH3OH·´Ó¦Éú³ÉCO(OCH3)2£¬ÆäÔ­ÒòÊÇ_________________________________¡£

¡¾ÌâÄ¿¡¿ÎÒ¹ú»¯Ñ§¼ÒÊ×´ÎʵÏÖÁË좴߻¯µÄ»·¼Ó³É·´Ó¦£¬²¢ÒÀ¾Ý¸Ã·´Ó¦£¬·¢Õ¹ÁËÒ»ÌõºÏ³ÉÖвÝÒ©»îÐԳɷÖé²ÔÊõ´¼µÄÓÐЧ·Ïß¡£

ÒÑÖª»·¼Ó³É·´Ó¦£º

£¨¡¢¿ÉÒÔÊÇ»ò£©

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Ã©²ÔÊõ´¼µÄ·Ö×ÓʽΪ____________£¬Ëùº¬¹ÙÄÜÍÅÃû³ÆΪ____________£¬·Ö×ÓÖÐÊÖÐÔ̼ԭ×Ó£¨Á¬ÓÐËĸö²»Í¬µÄÔ­×Ó»òÔ­×ÓÍÅ£©µÄÊýĿΪ____________¡£

£¨2£©»¯ºÏÎïBµÄºË´Å¹²ÕñÇâÆ×ÖÐÓÐ______¸öÎüÊշ壻ÆäÂú×ãÒÔÏÂÌõ¼þµÄͬ·ÖÒì¹¹Ì壨²»¿¼ÂÇÊÖÐÔÒì¹¹£©ÊýĿΪ______¡£

¢Ù·Ö×ÓÖк¬ÓÐ̼̼Èþ¼üºÍÒÒõ¥»ù

¢Ú·Ö×ÓÖÐÓÐÁ¬ÐøËĸö̼ԭ×ÓÔÚÒ»ÌõÖ±ÏßÉÏ

д³öÆäÖÐ̼̼Èþ¼üºÍÒÒõ¥»ùÖ±½ÓÏàÁ¬µÄͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ____________¡£

£¨3£©µÄ·´Ó¦ÀàÐÍΪ____________¡£

£¨4£©µÄ»¯Ñ§·½³ÌʽΪ__________________£¬³ýÍâ¸Ã·´Ó¦ÁíÒ»²úÎïµÄϵͳÃüÃûΪ____________¡£

£¨5£©ÏÂÁÐÊÔ¼Á·Ö±ðÓëºÍ·´Ó¦£¬¿ÉÉú³ÉÏàͬ»·×´²úÎïµÄÊÇ______£¨ÌîÐòºÅ£©¡£

a£® b£® c£®ÈÜÒº

£¨6£©²Î¿¼ÒÔÉϺϳÉ·Ïß¼°Ìõ¼þ£¬Ñ¡ÔñÁ½ÖÖÁ´×´²»±¥ºÍõ¥£¬Í¨¹ýÁ½²½·´Ó¦ºÏ³É»¯ºÏÎÔÚ·½¿òÖÐд³ö·ÏßÁ÷³Ìͼ£¨ÆäËûÊÔ¼ÁÈÎÑ¡£©¡£_______

¡¾ÌâÄ¿¡¿ÏÖÓÐËÄÖÖ¶ÌÖÜÆÚÔªËØA£¬B£¬C£¬D£¬ÒÑÖª£º¢ÙC£¬DÔÚͬһÖÜÆÚ£¬A£¬BÔÚͬһÖ÷×壻¢ÚËüÃÇ¿ÉÒÔ×é³É»¯ºÏÎïA2C¡¢B2C2¡¢DC2¡¢D2A4µÈ£»¢ÛBµÄÑôÀë×ÓÓëCµÄÒõÀë×ӵĺËÍâµç×ÓÅŲ¼Ïàͬ£»¢ÜB2C2ͬA2C»òDC2·´Ó¦¶¼Éú³ÉÆøÌåC2£¬BÓëA2C·´Ó¦Éú³ÉÆøÌåA2£¬A2ÓëÆøÌåC2°´Ìå»ý±È2¡Ã1»ìºÏºóµãȼ·¢Éú±¬Õ¨£¬Æä²úÎïÊÇÒ»ÖÖ³£ÎÂϳ£¼ûµÄÎÞÉ«ÎÞζµÄÒºÌå¡£Çë»Ø´ð£º

(1)д³öÔªËØ·ûºÅ£ºA________¡¢B________¡¢C________¡¢D________¡£

(2)ÔÚA2C¡¢B2C2¡¢DC2ºÍD2A4ÖУ¬Í¬Ê±º¬ÓÐÀë×Ó¼üºÍ·Ç¼«ÐÔ¹²¼Û¼üµÄ»¯ºÏÎïµÄµç×ÓʽΪ__________£»°´Ô­×Ó¹ìµÀÖصþ·½Ê½£¬Æä·Ç¼«ÐÔ¼üµÄÀàÐÍÊÇ________¡£»¯ºÏÎïDC2µÄ½á¹¹Ê½__________¡£

(3)A2C·Ö×ӵĵç×Óʽ____________£¬°´Ô­×Ó¹ìµÀÖصþ·½Ê½£¬Æä¹²¼Û¼üµÄÀàÐÍÊÇ____________¡£D2A4ÊÇƽÃæÐηÖ×Ó£¬Æä·Ö×ÓÖк¬ÓÐ______¸ö¦Ò¼ü£¬______¸ö¦Ð¼ü¡£

(4)д³ö»¯Ñ§·½³Ìʽ»òÀë×Ó·½³Ìʽ£º

B2C2ÓëA2C·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º___________________________________________£»

B2C2ÓëDC2·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º___________________________________________£»

BÓëA2C·´Ó¦µÄÀë×Ó·½³Ìʽ£º______________________________________________¡£

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

¾«Ó¢¼Ò½ÌÍø