CH3¡ªCHCH¡ªCHCH2CH3CHO£«OHC¡ªCHO£«HCHO£¬
CH3¡ªCC¡ªCH2¡ªCCHCH3COOH£«HOOC¡ªCH2¡ªCOOH£«HCOOH£¬
ijÌþ·Ö×ÓʽΪC10H10£¬ÔÚ³ôÑõ×÷ÓÃÏ·¢Éú·´Ó¦£º
C10H10CH3COOH£«3HOOC¡ªCHO£«CH3CHO£¬ÊԻشð£º
¢ÙC10H10·Ö×ÓÖк¬________¸öË«¼ü£¬________¸öÈý¼ü£»¢ÚC10H10µÄ½á¹¹¼òʽΪ________¡£
¢Ù2 2 ¢ÚCH3CC£CH=CH£CC£CH Ìáʾ£º
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º £¨1£©Èô·Ö×ÓʽΪC4H6µÄijÌþÖÐËùÓеÄ̼Ô×Ó¶¼ÔÚͬһÌõÖ±ÏßÉÏ£¬Ôò¸ÃÌþµÄ½á¹¹¼òʽΪ CH3C¡ÔCCH3 CH3C¡ÔCCH3 £®£¨2£©·Ö×ÓʽΪC9H20µÄijÌþÆäÒ»ÂÈ´úÎïÖ»ÓÐÁ½ÖÖ²úÎд³ö·ûºÏÒªÇóµÄ½á¹¹¼òʽ £¨CH3£©3CCH2C£¨CH3£©3 £¨CH3£©3CCH2C£¨CH3£©3 £®£¨3£©·Ö×ÓʽΪC5H10µÄÏ©ÌþÖв»´æÔÚ˳·´Òì¹¹µÄÎïÖÊÓÐ 4 4 ÖÖ £¨4£©·Ö×ÓʽΪC5H12O2µÄ¶þÔª´¼ÓжàÖÖͬ·ÖÒì¹¹Ì壬ÆäÖÐÄܹ»Ñõ»¯³ÉÖ÷Á´ÉÏ̼Ô×ÓÊýΪ3µÄ¶þԪȩÓÐ 2 2 ÖÖ£¨5£©ÒÑ֪ϩÌþ¡¢È²ÌþÔÚ³ôÑõ×÷ÓÃÏ·¢ÉúÒÔÏ·´Ó¦£º CH3-CH¨TCH-CH2-CH¨TCH2
CH3-C¡ÔC-CH2-C¡ÔCH
ijÌþ»¯Ñ§Ê½ÎªC10H10£¬ÔÚ³ôÑõ×÷ÓÃÏ·¢Éú·´Ó¦£º C10H10
¢ÙC10H10·Ö×ÓÖк¬ 2 2 ¸öË«¼ü£¬2 2 ¸öÈý¼ü£®¢ÚC10H10½á¹¹¼òʽΪ CH3-C¡ÔC-CH¨TCH-C¡ÔC-CH¨TCH-CH3 CH3-C¡ÔC-CH¨TCH-C¡ÔC-CH¨TCH-CH3 £®²é¿´´ð°¸ºÍ½âÎö>> ¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º ÒÑ֪ϩÌþ¡¢È²ÌþÔÚ³ôÑõµÄ×÷ÓÃÏ·¢ÉúÒÔÏ·´Ó¦£º CH3-CH=CH-CH2-CH=CH2¡úCH3CHO+OHC-CH2-CHO+HCHO CH3-C¡ÔC-CH2-CCH¡úCH3COOH+HOOCCH2COOH+HCOOH ijÌþ·Ö×ÓʽΪC10H10£¬ÔÚ³ôÑõ×÷ÓÃÏ·¢Éú·´Ó¦£ºC10H10¡úCH3COOH+3HOOC-CHO+CH3CHO £¨1£©C10H10·Ö×ÓÖк¬ÓÐ 2 2 ¸öË«¼ü£¬2 2 ¸öÈý¼ü£¨2£©C10H10·Ö×ӽṹ¼òʽΪ CH3-C¡ÔC-CH¨TCH-C¡ÔC-CH¨TCH-CH3 CH3-C¡ÔC-CH¨TCH-C¡ÔC-CH¨TCH-CH3 £®²é¿´´ð°¸ºÍ½âÎö>> ¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º ÒÑ֪ϩÌþ¡¢È²ÌþÔÚ³ôÑõ×÷ÓÃÏ·¢ÉúÒÔÏ·´Ó¦£º ijÌþ»¯Ñ§Ê½ÎªC10H10£¬ÔÚ³ôÑõ×÷ÓÃÏ·¢Éú·´Ó¦£º (1)C10H10·Ö×ÓÖк¬_______________¸öË«¼ü£¬_______________¸öÈý¼ü¡£ (2)C10H10½á¹¹¼òʽΪ_________________________________________________¡£ ²é¿´´ð°¸ºÍ½âÎö>> ¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º ÒÑ֪ϩÌþ¡¢È²ÌþÔÚ³ôÑõ×÷ÓÃÏ·¢ÉúÒÔÏ·´Ó¦£º £¨1£©·Ö×ÓʽΪC10H10µÄijÌþ£¬ÔÚ³ôÑõ×÷ÓÃÏ·¢Éú·´Ó¦Îª£º Ôò´ËC10H10µÄ½á¹¹¼òʽÊÇ£º__________________________________¡£ £¨2£©ÓÐÈ˳¢ÊÔןϳÉC10H10µÄ±¥ºÍÌþ£¬Äã¹À¼ÆÈôÄܳɹ¦µÄ»°£¬Æä½á¹¹ÔõÑù£¿ÊÔÓÃÏßÌõ±íʾ·Ö×ÓÖС°Ì¼¼Ü¡±µÄ¿Õ¼ä¹¹ÐÍ£¨²»±Øд³öC¡¢H·ûºÅ£©____________¡£ £¨3£©ÊÔÍÆÂÛ£ºÈç¹ûCnHnµÄ±¥ºÍÌþÄܹ»ºÏ³ÉµÄ»°£¬Æä¿Õ¼ä½á¹¹µÄÌصã¿ÉÄÜÊÇ_______________£»nµÄ×îСֵΪ___________¡£ ²é¿´´ð°¸ºÍ½âÎö>> ¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º ¿Æѧ¼ÒÖÂÁ¦ÓÚ¶þÑõ»¯Ì¼µÄ¡°×éºÏת»¯¡±¼¼ÊõµÄÑо¿£¬°Ñ¹ý¶à¶þÑõ»¯Ì¼×ª»¯ÎªÓÐÒæÓÚÈËÀàµÄÎïÖÊ¡£ (1)Èç¹û½«CO2ºÍH2ÒÔ1¡Ã4µÄ±ÈÀý»ìºÏ£¬Í¨Èë·´Ó¦Æ÷£¬ÔÚÊʵ±µÄÌõ¼þÏ·´Ó¦£¬¿É»ñµÃÒ»ÖÖÖØÒªµÄÄÜÔ´¡£ÇëÍê³É»¯Ñ§·½³Ìʽ£ºCO2+4H2_________+2H2O¡£ (2)Èô½«CO2ºÍH2ÒÔ1¡Ã3µÄ±ÈÀý»ìºÏ£¬Ê¹Ö®·¢Éú·´Ó¦Éú³ÉijÖÖÖØÒªµÄ»¯¹¤ÔÁϺÍË®£¬ÔòÉú³ÉµÄ¸ÃÖØÒª»¯¹¤ÔÁÏ¿ÉÄÜÊÇ A.ÍéÌþ B.Ï©Ìþ C.ȲÌþ D.·¼ÏãÌþ (3)ÒÑÖªÔÚ443 K¡ª473 Kʱ£¬ÓÃîÜ(Co)×÷´ß»¯¼Á¿ÉʹCO2ºÍH2Éú³ÉC5¡ªC8µÄÍéÌþ£¬ÕâÊÇÈ˹¤ºÏ³ÉÆûÓ͵ķ½·¨Ö®Ò»¡£Òª´ïµ½¸ÃÆûÓ͵ÄÒªÇó£¬CO2ºÍH2µÄÌå»ý±ÈµÄÈ¡Öµ·¶Î§ÊÇV(H2)£ºV(CO2)£º_________________________________________________________________¡£ (4)ÒÑÖªÔÚÒ»¶¨Ìõ¼þÏ£¬ºÏ³ÉÄòËصķ´Ó¦Îª£ºCO2(g)+2NH3(g)CO(NH2)2(s)+H2O(g);¦¤H=-127 kJ/mol£¬ÔÚ¸ÃÌõ¼þÏ£¬½«44 g CO2Óë40 g NH3³ä·Ö»ìºÏ£¬·´Ó¦·Å³öµÄÈÈÁ¿Ò»¶¨___________(Ìî¡°´óÓÚ¡±¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±)127 kJ¡£ÇëÉè¼Æ¹¤ÒµºÏ³ÉÄòËصÄÌõ¼þ___________¡£ ²é¿´´ð°¸ºÍ½âÎö>> ͬ²½Á·Ï°²á´ð°¸ °Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁÐ±í 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°æȨÉùÃ÷£º±¾Õ¾ËùÓÐÎÄÕ£¬Í¼Æ¬À´Ô´ÓÚÍøÂ磬Öø×÷Ȩ¼°°æȨ¹éÔ×÷ÕßËùÓУ¬×ªÔØÎÞÒâÇÖ·¸°æȨ£¬ÈçÓÐÇÖȨ£¬Çë×÷ÕßËÙÀ´º¯¸æÖª£¬ÎÒÃǽ«¾¡¿ì´¦Àí£¬ÁªÏµqq£º3310059649¡£ ICP±¸°¸ÐòºÅ: »¦ICP±¸07509807ºÅ-10 ¶õ¹«Íø°²±¸42018502000812ºÅ |