ijÌþA 0.2 mol ÔÚÑõÆøÖÐÍêȫȼÉÕºó£¬Éú³ÉCO2ºÍH2O¸÷1.2 mol¡£ÊԻشð£º
£¨1£©ÌþAµÄ·Ö×ÓʽΪ_____________¡£
£¨2£©ÈôÈ¡Ò»¶¨Á¿µÄÌþAÍêȫȼÉÕºó£¬Éú³ÉCO2ºÍH2O¸÷3 mol£¬ÔòÓÐ________gÌþA²Î¼ÓÁË·´Ó¦£¬È¼ÉÕʱÏûºÄ±ê×¼×´¿öϵÄÑõÆø___________L¡£
£¨3£©ÈôÌþA²»ÄÜʹäåË®ÍÊÉ«£¬µ«ÔÚÒ»¶¨Ìõ¼þÏÂÄÜÓëÂÈÆø·¢ÉúÈ¡´ú·´Ó¦£¬ÆäÒ»ÂÈÈ¡´úÎïÖ»ÓÐÒ»ÖÖ£¬ÔòÌþAµÄ½á¹¹¼òʽΪ__________________¡£
£¨4£©ÈôÌþAÄÜʹäåË®ÍÊÉ«£¬ÔÚ´ß»¯¼Á×÷ÓÃÏ£¬ÓëH2¼Ó³É£¬Æä¼Ó³É²úÎᄇⶨ·Ö×ÓÖк¬ÓÐ4¸ö¼×»ù£¬ÌþA¿ÉÄÜÓеĽṹ¼ò¼òʽ__________________________________________¡£
£¨11·Ö£©¢ÅC6H12 £¨1·Ö£© ¢Æ42 £¨1·Ö£© 100.8 £¨1·Ö£©
¢Ç £¨1·Ö£© ¢È(CH3)3CCH=CH2¡¢CH2=C(CH3)CH(CH3)2¡¢(CH3)2C=C(CH3)2£¨Ã¿¸ö2·Ö£¬¹²
½âÎöÊÔÌâ·ÖÎö£º£¨1£©ÌþA 0.2 mol ÔÚÑõÆøÖÐÍêȫȼÉÕºó£¬Éú³ÉCO2ºÍH2O¸÷1.2 mol£¬ÔòA·Ö×ÓÖÐ̼ÇâÔ×ӵĸöÊý·Ö±ðÊÇ6¸öºÍ12¸ö£¬Ôò»¯Ñ§Ê½Ó¦¸ÃÊÇC6H12¡£
£¨2£©ÌþAÍêȫȼÉÕºó£¬Éú³ÉCO2ºÍH2O¸÷3 mol£¬Ôò¸ù¾ÝAµÄ»¯Ñ§Ê½¿ÉÖª£¬²Î¼Ó·´Ó¦µÄAµÄÎïÖʵÄÁ¿ÊÇ0.5mol£¬ÔòAµÄÖÊÁ¿ÊÇ0.5mol¡Á84g/mol£½42g¡£
£¨3£©ÌþA²»ÄÜʹäåË®ÍÊÉ«£¬ËµÃ÷·Ö×ÓÖÐûÓÐ̼̼˫¼ü¡£µ«ÔÚÒ»¶¨Ìõ¼þÏÂÄÜÓëÂÈÆø·¢ÉúÈ¡´ú·´Ó¦£¬ÆäÒ»ÂÈÈ¡´úÎïÖ»ÓÐÒ»ÖÖ£¬Õâ˵Ã÷·Ö×ÓµÄÇâÔ×ÓÖ»ÓÐÒ»À࣬ËùÒÔAÊÇ»·¼ºÍ飬½á¹¹¼òʽÊÇ¡£
£¨4£©ÈôÌþAÄÜʹäåË®ÍÊÉ«£¬ËµÃ÷·Ö×ÓÊǺ¬ÓÐ̼̼˫¼üµÄ¡£ÔÚ´ß»¯¼Á×÷ÓÃÏ£¬ÓëH2¼Ó³É£¬Æä¼Ó³É²úÎᄇⶨ·Ö×ÓÖк¬ÓÐ4¸ö¼×»ù£¬Òò´ËÌþA¿ÉÄÜÓеĽṹ¼ò¼òʽ(CH3)3CCH=CH2¡¢CH2=C(CH3)CH(CH3)2¡¢(CH3)2C=C(CH3)2¡£
¿¼µã£º¿¼²éÓлúÎﻯѧʽ¡¢½á¹¹¼òʽµÄÅжÏÒÔ¼°ÌþȼÉÕµÄÓйؼÆËã
µãÆÀ£º¸ÃÌâÊǸ߿¼Öеij£¼ûÌâÐÍ£¬ÊôÓÚÖеÈÄѶȵÄÊÔÌâ¡£ÊÔÌâÄÑÒ×ÊÊÖУ¬×ÛºÏÐÔÇ¿£¬²àÖضÔѧÉúÄÜÁ¦µÄÅàÑøºÍ½âÌâ·½·¨µÄÖ¸µ¼ÓëѵÁ·£¬ÓÐÖúÓÚÅàÑøѧÉúµÄÂß¼ÍÆÀíÄÜÁ¦ºÍ·¢É¢Ë¼Î¬ÄÜÁ¦¡£¸ÃÌâµÄ¹Ø¼üÊÇÃ÷È·¸÷ÖÖ¹ÙÄÜÍŽṹºÍÐÔÖÊ£¬È»ºó½áºÏÌâÒâ¾ßÌåÎÊÌâ¡¢¾ßÌå·ÖÎö¼´¿É£¬ÓÐÖúÓÚÅàÑøѧÉúµÄÄæÏò˼άÄÜÁ¦ºÍÓ¦ÊÔÄÜÁ¦¡£
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
(1)
(2)ÈôAÄÜʹäåË®ÍÊÉ«£¬ÇÒÔÚ´ß»¯¼Á´æÔÚÏÂÓëH2¼Ó³ÉµÄ²úÎï·Ö×ÓÖк¬ÓÐ4¸ö¼×»ù£¬ÔòA¿ÉÄܵĽṹ¼òʽΪ(ÈÎдһÖÖ) __________________________________¡£
(3)ijÓлúÎïµÄ·Ö×ÓʽΪCxHyO2£¬ÈôxµÄÖµÓëA·Ö×ÓÖеÄ̼Ô×Ó¸öÊýÏàͬ£¬Ôò¸Ã·Ö×ÓÖÐyµÄ×î´óֵΪ__________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ijÌþA 0.2 molÔÚO2Öгä·ÖȼÉÕʱ£¬Éú³É»¯ºÏÎïB¡¢C¸÷1.2 mol£¬ÊԻشð£º
£¨1£©ÌþAµÄ·Ö×Óʽ________£¬B¡¢CµÄ·Ö×Óʽ·Ö±ðÊÇ________¡¢________¡£
£¨2£©ÈôÈ¡Ò»¶¨Á¿µÄÌþAȼÉÕºóÉú³ÉB¡¢C¸÷3 mol£¬ÔòÓÐ________gÌþA²Î¼ÓÁË·´Ó¦£¬È¼ÉÕʱÏûºÄ±ê×¼×´¿öÏÂO2________L¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ijÌþA 0.2 molÔÚO2Öгä·ÖȼÉÕʱ£¬Éú³É»¯ºÏÎïB¡¢C¸÷1.2 mol£¬ÊÔÍê³ÉÏÂÁÐÎÊÌ⣺
(1)ÌþAµÄ·Ö×Óʽ_________£¬B¡¢CµÄ·Ö×Óʽ·Ö±ðÊÇ_________¡¢_________¡£
(2)ÈôÈ¡Ò»¶¨Á¿µÄÌþAȼÉÕºóÉú³ÉB¡¢C¸÷3 mol£¬ÔòÓÐ_________ gÌþA²Î¼ÓÁË·´Ó¦£¬È¼ÉÕʱÏûºÄ±ê×¼×´¿öÏÂO2_________L¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄêÕã½Ê¡²ÔÄÏÏØÊ÷ÈËÖÐѧ¸ß¶þµÚ¶þ´ÎÔ¿¼»¯Ñ§£¨Àí£©ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ
£¨14·Ö£©Ä³ÌþA 0.2 mol ÔÚÑõÆøÖÐÍêȫȼÉÕºó£¬Éú³ÉCO2ºÍH2O¸÷1.2 mol¡£ÊԻشð£º
£¨1£©ÌþAµÄ·Ö×ÓʽΪ_____________¡£
£¨2£©ÈôÈ¡Ò»¶¨Á¿µÄÌþAÍêȫȼÉÕºó£¬Éú³ÉCO2ºÍH2O¸÷3 mol£¬ÔòÓÐ________gÌþA²Î¼ÓÁË·´Ó¦£¬È¼ÉÕʱÏûºÄ±ê×¼×´¿öϵÄÑõÆø___________L¡£
£¨3£©ÈôÌþA²»ÄÜʹäåË®ÍÊÉ«£¬µ«ÔÚÒ»¶¨Ìõ¼þÏÂÄÜÓëÂÈÆø·¢ÉúÈ¡´ú·´Ó¦£¬ÆäÒ»ÂÈÈ¡´úÎïÖ»ÓÐÒ»ÖÖ£¬ÔòÌþAµÄ½á¹¹¼òʽΪ__________________¡£
£¨4£©ÈôÌþAÄÜʹäåË®ÍÊÉ«£¬ÔÚ´ß»¯¼Á×÷ÓÃÏ£¬ÓëH2¼Ó³É£¬Æä¼Ó³É²úÎᄇⶨ·Ö×ÓÖк¬ÓÐ4¸ö¼×»ù£¬ÌþA¿ÉÄÜÓеĽṹ¼òʽΪ_______________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012Äêɽ¶«ÁijÇÝ·ÏØʵÑé¸ßÖи߶þµÚÈý´ÎÄ£¿é²âÊÔ»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ
£¨Ã¿¿Õ2·Ö£¬¹²10·Ö£©Ä³ÌþA 0.2 mol ÔÚÑõÆøÖÐÍêȫȼÉÕºó£¬Éú³ÉCO2ºÍH2O¸÷1.2 mol¡£ÊԻشð£º
£¨1£©ÌþAµÄ·Ö×ÓʽΪ_____________¡£
£¨2£©ÈôÈ¡Ò»¶¨Á¿µÄÌþAÍêȫȼÉÕºó£¬Éú³ÉCO2ºÍH2O¸÷3 mol£¬ÔòÓÐ________gÌþA²Î¼ÓÁË·´Ó¦£¬È¼ÉÕʱÏûºÄ±ê×¼×´¿öϵÄÑõÆø___________L¡£
£¨3£©ÈôÌþA²»ÄÜʹäåË®ÍÊÉ«£¬µ«ÔÚÒ»¶¨Ìõ¼þÏÂÄÜÓëÂÈÆø·¢ÉúÈ¡´ú·´Ó¦£¬ÆäÒ»ÂÈÈ¡´úÎïÖ»
ÓÐÒ»ÖÖ£¬ÔòÌþAµÄ½á¹¹¼òʽΪ__________________¡£
£¨4£©ÈôÌþAÄÜʹäåË®ÍÊÉ«£¬ÔÚ´ß»¯¼Á×÷ÓÃÏ£¬ÓëH2¼Ó³É£¬Æä¼Ó³É²úÎᄇⶨ·Ö×ÓÖк¬ÓÐ
4¸ö¼×»ù£¬ÌþA¿ÉÄÜÓеĽṹ¼òʽΪ______________£¨Ð´³öÒ»ÖÖ¼´¿É£©
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com