Çë¸ù¾ÝÏÂͼ×÷´ð£º

ÒÑÖª£ºÒ»¸ö̼ԭ×ÓÉÏÁ¬ÓÐÁ½¸öôÇ»ùʱ£¬Ò×·¢ÉúÏÂÁÐת»¯£º

£¨1£©EÖк¬ÓеĹÙÄÜÍÅÊÇ                  ¡£

£¨2£©·´Ó¦¢ÛµÄ»¯Ñ§·½³ÌʽÊÇ                  ¡£

£¨3£©ÒÑÖªBµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª162£¬ÆäÍêȫȼÉյIJúÎïÖÐn£¨CO2£©£ºn£¨H2O£©=2£º1£¬ÔòBµÄ·Ö×ÓʽΪ                  ¡£

£¨4£©FÊǸ߷Ö×Ó¹â×è¼ÁÉú²úµÄÖ÷ÒªÔ­ÁÏ¡£F¾ßÓÐÈçÏÂÌص㣺

¢ÙÄܸúFeC13ÈÜÒº·¢ÉúÏÔÉ«·´Ó¦£»

¢ÚÄÜ·¢Éú¼Ó¾Û·´Ó¦£»

¢Û±½»·ÉϵÄÒ»ÂÈ´úÎïÖ»ÓÐÁ½ÖÖ¡£

FÔÚÒ»¶¨Ìõ¼þÏ·¢Éú¼Ó¾Û·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                  ¡£

£¨5£©»¯ºÏÎïGÊÇFµÄͬ·ÖÒì¹¹Ì壬ÊôÓÚ·¼Ïã×廯ºÏÎÄÜ·¢ÉúÒø¾µ·´Ó¦¡£GÓжàÖֽṹ£¬Ð´³öÆäÖÐÒ»ÖֵĽṹ¼òʽ                  ¡£

£¨1£©ôÈ»ù£»

£¨2£©CH3CHO+2Cu£¨OH£©2 CH3COOH+Cu2O¡ý+2H2O

£¨3£©C10H10O2£»

£¨4£© 

£¨5£©»ò   »ò¼äλ¡¢ÁÚλÈÎÒ»ÖÖ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Çë¸ù¾ÝÏÂͼ×÷´ð£º

ÒÑÖª£ºÒ»¸ö̼ԭ×ÓÉÏÁ¬ÓÐÁ½¸öôÇ»ùʱ£¬Ò×·¢ÉúÏÂÁÐת»¯£º
£¨1£©·´Ó¦¢ÙËùÊôµÄÓлú·´Ó¦ÀàÐÍÊÇ
È¡´ú£¨»òË®½â£©
È¡´ú£¨»òË®½â£©
·´Ó¦£®
£¨2£©·´Ó¦¢ÛµÄ»¯Ñ§·½³Ìʽ
CH3CHO+2Cu£¨OH£©2
¡÷
CH3COOH+Cu2O+2H2O
CH3CHO+2Cu£¨OH£©2
¡÷
CH3COOH+Cu2O+2H2O
£®
£¨3£©ÒÑÖªBµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª162£¬ÆäÍêȫȼÉյIJúÎïÖÐn£¨CO2£©£ºn £¨H2O£©=2£º1£¬ÔòBµÄ·Ö×ÓʽΪ
C10H10O2
C10H10O2
£®
£¨4£©FÊǸ߷Ö×Ó¹â×è¼ÁÉú²úÖеÄÖ÷ÒªÔ­ÁÏ£®F¾ßÓÐÈçÏÂÌص㣺¢ÙÄܸúFeCl3ÈÜÒº·¢ÉúÏÔÉ«·´Ó¦£»¢ÚÄÜ·¢Éú¼Ó¾Û·´Ó¦£»¢Û±½»·ÉϵÄÒ»ÂÈ´úÎïÖ»ÓÐÁ½ÖÖ£®FÔÚÒ»¶¨Ìõ¼þÏ·¢Éú¼Ó¾Û·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
£®
£¨5£©»¯ºÏÎïGÊÇFµÄͬ·ÖÒì¹¹Ì壬ÊôÓÚ·¼Ïã×廯ºÏÎÄÜ·¢ÉúÒø¾µ·´Ó¦£®GµÄ½á¹¹ÓÐ
4
4
ÖÖ£®
£¨6£©»¯ºÏÎïHÊÇBµÄͬ·ÖÒì¹¹Ì壬H·Ö×ÓÖк¬ÓеIJ¿·Ö½á¹¹Îª£¬ËüµÄË®½â²úÎï¾­¾ÛºÏ·´Ó¦ºó¿ÉµÃµ½¸ß¾ÛÎCaHbO2£©nHÓжàÖֽṹ£¬Ð´³öÆäÖÐÒ»ÖֵĽṹ¼òʽ
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2010?ÌÆɽÈýÄ££©Çë¸ù¾ÝÏÂͼ×÷´ð£º

£¨1£©EÖк¬ÓеĹÙÄÜÍŵÄÃû³ÆÊÇ
È©»ù
È©»ù
£®
£¨2£©BµÄ½á¹¹¼òʽΪ
£®
£¨3£©Ð´³öÏÂÁз´Ó¦ÀàÐÍ£ºC¡úE
Ñõ»¯·´Ó¦
Ñõ»¯·´Ó¦
£¬B¡úF+G
Ë®½â·´Ó¦
Ë®½â·´Ó¦
£®
£¨4£©Ð´³öÏÂÁз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
A¡úC+D
HOOC-COOCH2CH2Cl+3NaOH¡úNaOOC-COONa+NaCl+H2O+HOCH2CH2OH
HOOC-COOCH2CH2Cl+3NaOH¡úNaOOC-COONa+NaCl+H2O+HOCH2CH2OH
£®G¡úH
£®
£¨5£©H·Ö×ÓÖй²ÏßµÄÔ­×ÓÊý×î¶àΪ
4
4
¸ö£®
£¨6£©·ûºÏÏÂÁÐÌõ¼þµÄBµÄͬ·ÖÒì¹¹Ì壨²»º¬B£©ÓÐ
12
12
ÖÖ£®
¢ÙËüÊÇ·¼Ïã×廯ºÏÎ
¢Ú·Ö×ÓÖдæÔÚÁ½¸öÁ¬ÐøµÄ¡°-CH2-¡±»ùÍÅ£»
¢ÛËüΪ¶þÔªõ¥Àࣨº¬ÓÐÁ½¸ö¡°¡±»ùÍÅ£©»¯ºÏÎÇÒ²»·¢ÉúÒø¾µ·´Ó¦£®Ð´³öÆäÖÐÒ»¸ö1mol¸ÃÎïÖÊ×î¶àÏûºÄ3molNaOHµÄͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2012?ºìÇÅÇøһģ£©Çë¸ù¾ÝÏÂͼ×÷´ð£º

ÒÑÖª£ºÒ»¸ö̼ԭ×ÓÉÏÁ¬ÓÐÁ½¸öôÇ»ùʱ£¬Ò×·¢ÉúÏÂÁÐת»¯£º

£¨1£©AµÄºË´Å¹²ÕñÇâÆ×ͼÖй²ÓÐ
3
3
×é·å£¬·åÃæ»ýÖ®±ÈΪ
3£º3£º1
3£º3£º1
£»EÖк¬ÓеĹÙÄÜÍŵĽṹ¼òʽÊÇ
-COOH
-COOH
£®
£¨2£©·´Ó¦¢ÙµÄ»¯Ñ§·½³Ìʽ
CH3COOCH£¨Cl£©CH3+2NaOH
¡÷
CH3COONa+CH3CHO+NaCl+H2O
CH3COOCH£¨Cl£©CH3+2NaOH
¡÷
CH3COONa+CH3CHO+NaCl+H2O
£®
£¨3£©ÒÑÖªBµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª162£¬ÆäÍêȫȼÉյIJúÎïÖÐn£¨CO2£©£ºn£¨H2O£©=2£º1£¬ÔòBµÄ·Ö×ÓʽΪ
C10H10O2
C10H10O2
£¬·´Ó¦¢ÝµÄ·´Ó¦ÀàÐÍΪ
Ë®½â·´Ó¦
Ë®½â·´Ó¦
£®
£¨4£©·´Ó¦¢ÛµÄ»¯Ñ§·½³Ìʽ
CH3CHO+2Cu£¨OH£©2+NaOH
¡÷
CH3COONa+Cu2O¡ý+3H2O
CH3CHO+2Cu£¨OH£©2+NaOH
¡÷
CH3COONa+Cu2O¡ý+3H2O
£®
£¨5£©FÊǸ߷Ö×Ó¹â×è¾£Éú²úÖеÄÖ÷ÒªÔ­ÁÏ£®F¾ßÓÐÈçÏÂÌص㣺¢ÙÄܸúFeCl3ÈÜÒº·¢ÉúÏÔÉ«·´Ó¦£»¢ÚÄÜ·¢Éú¼Ó¾Û·´Ó¦£»¢Û±½»·ÉϵÄÒ»ÂÈ´úÎïÖ»ÓÐÁ½ÖÖ£®FÔÚÒ»¶¨Ìõ¼þÏ·¢Éú¼Ó¾Û·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
£®
£¨6£©»¯ºÏÎïGÊÇFµÄͬ·ÖÒì¹¹Ì壬ÊôÓÚ·¼Ïã×廯ºÏÎÄÜ·¢ÉúÒø¾µ·´Ó¦£®GÓжàÖֽṹ£¬Ð´³öÆäÖÐÁ½ÖֵĽṹ¼òʽ
¡¢¡¢¡¢ÈÎÒâ2ÖÖ
¡¢¡¢¡¢ÈÎÒâ2ÖÖ
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

(¹²15·Ö) Çë¸ù¾ÝÏÂͼ×÷´ð£º

ÒÑÖª£ºÒ»¸ö̼ԭ×ÓÉÏÁ¬ÓÐÁ½¸öôÇ»ùʱ£¬Ò×·¢ÉúÏÂÁÐת»¯£º

£¨1£©·´Ó¦¢ÙËùÊôµÄÓлú·´Ó¦ÀàÐÍÊÇ_______________·´Ó¦¡£

£¨2£©·´Ó¦¢ÛµÄ»¯Ñ§·½³Ìʽ______________________________________________________¡£

£¨3£©ÒÑÖªBµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª162£¬ÆäÍêȫȼÉյIJúÎïÖÐn(CO2)¡Ãn (H2O) =2¡Ã1£¬ÔòBµÄ·Ö×ÓʽΪ                        ¡£

£¨4£©FÊǸ߷Ö×Ó¹â×è¼ÁÉú²úÖеÄÖ÷ÒªÔ­ÁÏ¡£F¾ßÓÐÈçÏÂÌص㣺¢ÙÄܸúFeCl3ÈÜÒº·¢ÉúÏÔÉ«·´Ó¦£»¢ÚÄÜ·¢Éú¼Ó¾Û·´Ó¦£»¢Û±½»·ÉϵÄÒ»ÂÈ´úÎïÖ»ÓÐÁ½ÖÖ¡£FÔÚÒ»¶¨Ìõ¼þÏ·¢Éú¼Ó¾Û·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                                       ¡£

£¨5£©»¯ºÏÎïGÊÇFµÄͬ·ÖÒì¹¹Ì壬ÊôÓÚ·¼Ïã×廯ºÏÎÄÜ·¢ÉúÒø¾µ·´Ó¦¡£GµÄ½á¹¹ÓÐ    ÖÖ¡£

£¨6£©»¯ºÏÎïHÊÇBµÄͬ·ÖÒì¹¹Ì壬H·Ö×ÓÖк¬ÓеIJ¿·Ö½á¹¹Îª£¬ËüµÄË®½â²úÎï¾­¾ÛºÏ·´Ó¦ºó¿ÉµÃµ½¸ß¾ÛÎï(CaHbO2)n.¡£HÓжàÖֽṹ£¬Ð´³öÆäÖÐÒ»ÖֵĽṹ¼òʽ                                   ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010Ä꺣ÄÏÊ¡¸ßÈýÎåУÁª¿¼»¯Ñ§ÊÔÌâ ÌâÐÍ£ºÌî¿ÕÌâ

¡¶Óлú»¯Ñ§»ù´¡¡·
18-1£®£¨6·Ö£¬¸ÃÌâΪ¶àÏîÑ¡ÔñÌ⣬ȫѡ¶ÔµÃ6·Ö£¬Â©Ñ¡°´±ÈÀý¸ø·Ö£¬´íÑ¡»ò¶àѡΪ0·Ö£©
ÏÂÁйØÓÚÓлúÎïµÄ˵·¨ÖУ¬ÕýÈ·µÄÊÇ£¨   £©

A£®ÕáÌÇÊǸ߷Ö×Ó»¯ºÏÎÆäË®½â²úÎïÄÜ·¢ÉúÒø¾µ·´Ó¦
B£®Ï𽺺ÍÏËά²»Ò»¶¨¶¼ÊǺϳɸ߷Ö×Ó²ÄÁÏ
C£®³£ÎÂϵí·ÛÓöµâ¾Æ±äÀ¶É«£¬ÆÏÌÑÌÇÄÜÓëÐÂÖÆCu£¨OH£©2·¢Éú·´Ó¦
D£®½«Ä³ÌþÀàµÄ»ìºÏÆøÌå¸úCl2»ìºÏ¹âÕÕ£¬·¢ÏÖÓÐÓÍ×´ÒºµÎÉú³É£¬ËµÃ÷»ìºÏÆøÌå¿Ï¶¨º¬Óм×Íé
E£®·Ö×ÓÖÐËùÓÐÔ­×Ó²»¿ÉÄܹ²Æ½Ãæ
18-2£®£¨14·Ö£©ÒÑÖª£ºÒ»¸ö̼ԭ×ÓÉÏÁ¬ÓÐÁ½¸öôÇ»ùʱ£¬Ò×·¢ÉúÏÂÁÐת»¯£º

Çë¸ù¾ÝÏÂͼ×÷´ð£º

£¨1£©AµÄºË´Å¹²ÕñÇâÆ×ÖгöÏÖ      ×é·å£»EÖк¬ÓеĹÙÄÜÍŵÄÃû³ÆÊÇ        £»
£¨2£©·´Ó¦¢ÛµÄ»¯Ñ§·½³ÌʽΪ                                               £»
£¨3£©ÒÑÖªBµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª162£¬ÆäÍêȫȼÉյIJúÎïÖÐCO 2ºÍH2OµÄÎïÖʵÄÁ¿Ö®±ÈΪ n£¨CO2£©©Un £¨H2O£©£½2©U1£¬ÔòBµÄ·Ö×ÓʽΪ                   £»
£¨4£©FÊǸ߷Ö×Ó¹â×è¼ÁÉú²úÖеÄÖ÷ÒªÔ­ÁÏ¡£F¾ßÓÐÈçÏÂÌص㣺¢ÙÊôÓÚ·¼Ïã×廯ºÏÎ¢ÚÄܸúFeCl3ÈÜÒº·¢ÉúÏÔÉ«·´Ó¦£»¢ÛÄÜ·¢Éú¼Ó¾Û·´Ó¦£»¢Ü±½»·ÉϵÄÒ»ÂÈ´úÎïÖ»ÓÐÁ½ÖÖ¡£FÔÚÒ»¶¨Ìõ¼þÏ·¢Éú¼Ó¾Û·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                               £»
£¨5£©»¯ºÏÎïGÊÇFµÄͬ·ÖÒì¹¹Ì壬ÊôÓÚ·¼Ïã×廯ºÏÎÄÜ·¢ÉúÒø¾µ·´Ó¦¡£GÓжàÖÖ¿ÉÄܵĽṹ£¬Çëд³öËüÃǵĽṹ¼òʽ                                         ¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸