ÒÑ֪ϩÌþ¡¢È²Ìþ£¬¾­³ôÑõ×÷Ó÷¢Éú·´Ó¦£º

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________¡£

 

´ð°¸£º
½âÎö£º

¢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×ÓʽΪC4H6µÄijÌþÖÐËùÓеÄ̼ԭ×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´ÉÏ̼ԭ×ÓÊýΪ3µÄ¶þԪȩÓÐ
2
2
ÖÖ
£¨5£©ÒÑ֪ϩÌþ¡¢È²ÌþÔÚ³ôÑõ×÷ÓÃÏ·¢ÉúÒÔÏ·´Ó¦£º
CH3-CH¨TCH-CH2-CH¨TCH2
¢ÙO3
¢ÚH2O
CH3CHO+OHC-CH2-CHO+HCHO
CH3-C¡ÔC-CH2-C¡ÔCH 
¢ÙO3
¢ÚH2O
CH3COOH+HOOCCH2COOH+HCOOH
ijÌþ»¯Ñ§Ê½ÎªC10H10£¬ÔÚ³ôÑõ×÷ÓÃÏ·¢Éú·´Ó¦£º
C10H10
¢ÙO3
¢ÚH2O
CH3COOH+3HOOC-CHO+CH3CHO
¢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°¸