ÒÑÖª£ºCH3CH2CH2CH3(g)£«6.5O2(g)¡ú 4CO2(g)£«5H2O(l)£»DH =£­2878 kJ¡¤mol£­1

(CH3)2CHCH3(g)£«6.5O2(g)¡ú 4CO2(g)£«5H2O(l)£»DH =£­2869 kJ¡¤mol£­1

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A£®Õý¶¡Íé·Ö×Ó´¢´æµÄÄÜÁ¿´óÓÚÒ춡Íé·Ö×Ó

B£®Õý¶¡ÍéµÄÎȶ¨ÐÔ´óÓÚÒ춡Íé

C£®Ò춡Íéת»¯ÎªÕý¶¡ÍéµÄ¹ý³ÌÊÇÒ»¸ö·ÅÈȹý³Ì

D£®Ò춡Íé·Ö×ÓÖеÄ̼Çâ¼ü±ÈÕý¶¡ÍéµÄ¶à

 

¡¾´ð°¸¡¿

A

¡¾½âÎö¡¿

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2009?¹ãÖݶþÄ££©ÏûÈ¥·´Ó¦ÊÇÓлúºÏ³ÉÖÐÒýÈë²»±¥ºÍ¼üµÄ³£¼û;¾¶£¬´ó²¿·Ö±´úÌþºÍ´¼¶¼ÓÐÕâÒ»ÐÔÖÊ£®
£¨1£©Ò»¶¨Ìõ¼þϲ¿·Ö±´úÌþÍѱ»¯Çâ»òÍÑË®ºó£¬ÏàÓ¦²úÎï¼°Æä²úÂÊÈçÏÂ±í£º
±´úÌþ»ò´¼ ÏàÓ¦²úÎï¼°Æä²úÂÊ
  
81%            19%
     
80%            20%
  
80%            20%
      
90%             10%
·ÖÎö±íÖÐÊý¾Ý£¬µÃµ½Â±´úÌþºÍ´¼·¢ÉúÏûÈ¥·´Ó¦Ê±ÍÑÇâλÖÃÇãÏòµÄÖ÷Òª½áÂÛÊÇ
ÇâÔ­×ÓÖ÷Òª´Óº¬Çâ½ÏÉÙµÄÏàÁÚ̼ԭ×ÓÉÏÍÑÈ¥£¨»ò£ºÖ÷Òª²úÎïÊÇË«¼ü̼ԭ×ÓÉÏÁ¬ÓÐÍé»ù½Ï¶àµÄÏ©Ìþ£©
ÇâÔ­×ÓÖ÷Òª´Óº¬Çâ½ÏÉÙµÄÏàÁÚ̼ԭ×ÓÉÏÍÑÈ¥£¨»ò£ºÖ÷Òª²úÎïÊÇË«¼ü̼ԭ×ÓÉÏÁ¬ÓÐÍé»ù½Ï¶àµÄÏ©Ìþ£©

£®
£¨2£©ÁоÙÒ»¸ö²»ÄÜ·¢ÉúÏûÈ¥·´Ó¦µÄ´¼£¬Ð´³ö½á¹¹¼òʽ£º
£®
£¨3£©ÒÑÖªÔÚÁòËáµÄ×÷ÓÃÏÂÍÑË®£¬Éú³ÉÎïÓÐ˳-2-ÎìÏ©£¨Õ¼25%£©ºÍ·´-2-ÎìÏ©£¨Õ¼75%£©Á½ÖÖ£®Ð´³ö2-ÎìÏ©µÄ˳ʽ½á¹¹£º

£¨4£©ÒÔÏÂÊÇÓÉ2-äåÎìÍéºÏ³ÉȲÌþCµÄ·´Ó¦Á÷³Ì£º

AµÄ½á¹¹¼òʽΪ
CH3CH2CH=CHCH3
CH3CH2CH=CHCH3
CµÄ½á¹¹¼òʽΪ
CH3CH2C¡ÔCCH3
CH3CH2C¡ÔCCH3
 Á÷³ÌµÄ·´Ó¦ÊôÓÚÏûÈ¥·´Ó¦µÄÊÇ
¢Ù¢Û
¢Ù¢Û
£¨Ìî±àºÅ£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2008?·ðɽһģ£©·Ö×ÓʽΪC13H16O2µÄÌè¸÷Ëá±½ÒÒõ¥¹ã·ºÓÃÓÚÏ㾫µÄµ÷Ïã¼Á£®ÒÑÖª£º
¢ÙR-CH2Cl+NaOH
¡÷
R-CH2OH+NaCl
¢Ú
ΪÁ˺ϳɸÃÎijʵÑéÊҵĿƼ¼ÈËÔ±Éè¼ÆÁËÏÂÁкϳÉ·Ïߣº

ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©ÊÔд³ö£ºAµÄÃû³Æ
2-¼×»ù-1-¶¡Ï©
2-¼×»ù-1-¶¡Ï©
£»DµÄ½á¹¹¼òʽ
£®
£¨2£©·´Ó¦¢ÜµÄÒ»°ãÌõ¼þÊÇ
ŨH2SO4/¡÷
ŨH2SO4/¡÷
£®
£¨3£©ÉÏÊöºÏ³É·ÏßÖÐÊôÓÚÈ¡´ú·´Ó¦µÄÊÇ
¢Ú¢Ý¢Þ
¢Ú¢Ý¢Þ
£¨Ìî±àºÅ£©£®
£¨4£©·´Ó¦¢ÙµÄ»¯Ñ§·½³ÌʽΪ
CH3CH2C£¨CH3£©=CH2+Br2¡úCH3CH2C£¨CH3£©BrCH2Br
CH3CH2C£¨CH3£©=CH2+Br2¡úCH3CH2C£¨CH3£©BrCH2Br
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º0110 Ä£ÄâÌâ ÌâÐÍ£ºÌî¿ÕÌâ

£¨Ò»£©ÏÂÁÐÊÇA¡¢B¡¢C¡¢D¡¢EÎåÖÖ¶ÌÖÜÆÚÔªËصÄijЩЩÐÔÖÊ
(1)ÔªËØAÊÇÐγÉÓлúÎïµÄÖ÷ÒªÔªËØ£¬ÏÂÁзÖ×ÓÖк¬ÓÐspºÍsp3ÔÓ»¯·½Ê½µÄÊÇ__________
A£® B£®CH4 C£®CH2£½CHCH3 D£®CH3CH2C¡ÔCH E£®CH3CH3
(2)ÓëAD2»¥ÎªµÈµç×ÓÌåµÄ·Ö×Ó¡¢Àë×ӵĻ¯Ñ§Ê½ÒÀ´ÎΪ________¡¢________£¨¸÷д1ÖÖ£©
(3)ÏàͬÌõ¼þÏ£¬AD2ÓëBD2·Ö×ÓÁ½ÕßÔÚË®ÖеÄÈܽâ¶È½Ï´óµÄÊÇ________£¨Ð´·Ö×Óʽ£©£¬ÀíÓÉÊÇ________________________£»
£¨¶þ£©ÏÂÁÐÊDz¿·Ö½ðÊôÔªËصĵçÀëÄÜ
£¨4£©ÒÑÖªX¡¢Y¡¢ZµÄ¼Û²ãµç×Ó¹¹ÐÍΪns1£¬ÔòÈýÖÖ½ðÊôµÄÂÈ»¯ÎRCl£©µÄÈÛµãÓɵ͵½¸ßµÄ˳ÐòΪ________________¡£
£¨5£©CuClÓÃ×÷Óлú»úºÏ³É´ß»¯¼Á£¬²¢ÓÃÓÚÑÕÁÏ£¬·À¸¯µÈ¹¤Òµ¡£CuClµÄ¾§Ìå½á¹¹ÈçͼËùʾ¡£ÔªËØCu»ù̬ԭ×ӵĵç×ÓÅŲ¼Ê½________£¬Óëͬһ¸öCl£­ÏàÁ¬µÄCu+ÓÐ________¸ö¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£ºÍ¬²½Ìâ ÌâÐÍ£ºÍƶÏÌâ

ËÜÁÏÊÇÈÕ³£Éú»îÖг£Óõĺϳɸ߷Ö×Ó²ÄÁÏ£¬ÒÔ¼×ÍéΪÖ÷ÒªÔ­ÁϺϳɳ£ÓÃËÜÁÏHµÄÏß·ÈçÏ£º
ÒÑÖª£ºCH3CH2X CH3CH2C¡ÔCCH3 »Ø´ðÏÂÁÐÎÊÌ⣺
(1)·´Ó¦¢ÙÓ¦¿ØÖƵÄÌõ¼þÊÇ___________ £¬·´Ó¦¢ÜµÄÀàÐÍÊÇ______________£»
(2)д³öÏÂÁз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º·´Ó¦¢Ú ______________£¬·´Ó¦¢Þ_______________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨Ò»£©ÏÂÁÐÊÇA¡¢B¡¢C¡¢D¡¢EÎåÖÖ¶ÌÖÜÆÚÔªËصÄijЩЩÐÔÖÊ

 

A

B

C

D

E

»¯ºÏ¼Û

£­4

£­2

£­1

£­2

£­1

µç¸ºÐÔ

2.5

2.5

3.0

3.5

4.0

(1)ÔªËØAÊÇÐγÉÓлúÎïµÄÖ÷ÒªÔªËØ£¬ÏÂÁзÖ×ÓÖк¬ÓÐspºÍsp3 ÔÓ»¯·½Ê½µÄÊÇ        

A£® B£®CH4   C£®CH2£½CHCH3    D£®CH3CH2C¡ÔCH   E£®CH3CH3

(2)ÓëAD2»¥ÎªµÈµç×ÓÌåµÄ·Ö×Ó¡¢Àë×ӵĻ¯Ñ§Ê½ÒÀ´ÎΪ          ¡¢         £¨¸÷д1ÖÖ£©

(3)ÏàͬÌõ¼þÏ£¬AD2ÓëBD2·Ö×ÓÁ½ÕßÔÚË®ÖеÄÈܽâ¶È½Ï´óµÄÊÇ           £¨Ð´·Ö×Óʽ£©£¬ÀíÓÉÊÇ                                                 £»

£¨¶þ£©ÏÂÁÐÊDz¿·Ö½ðÊôÔªËصĵçÀëÄÜ

X

Y

Z

µÚÒ»µçÀëÄÜ(kJ/mol)

520.2

495.8

418.8

£¨4£©ÒÑÖªX¡¢Y¡¢ZµÄ¼Û²ãµç×Ó¹¹ÐÍΪns1£¬ÔòÈýÖÖ½ðÊôµÄÂÈ»¯ÎRCl£©µÄÈÛµãÓɵ͵½¸ßµÄ˳ÐòΪ                                     ¡£

£¨5£©CuClÓÃ×÷ÓлúºÏ³É´ß»¯¼Á, ²¢ÓÃÓÚÑÕÁÏ, ·À¸¯µÈ¹¤Òµ¡£CuClµÄ¾§Ìå½á¹¹ÈçͼËùʾ¡£ÔªËØCu»ù̬ԭ×ӵĵç×ÓÅŲ¼Ê½               £¬Óëͬһ¸öCl£­ÏàÁ¬µÄ Cu+ÓР        ¸ö¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸