ijÓÅÖÊÌðÓ£ÌÒÖк¬ÓÐÒ»ÖÖôÇ»ùËᣨÓÃM±íʾ£©£¬MµÄ̼Á´½á¹¹ÎÞÖ§Á´£¬·Ö×ÓʽΪ
C4H6O5£»1.34gMÓë×ãÁ¿µÄ̼ËáÇâÄÆÈÜÒº·´Ó¦£¬Éú³É±ê×¼×´¿öϵÄÆøÌå0.448L¡£MÔÚÒ»¶¨Ìõ¼þÏ¿ɷ¢ÉúÈçÏÂת»¯£ºMABC£¨M¡¢A¡¢B¡¢C·Ö×ÓÖÐ̼ԭ×ÓÊýÄ¿Ïàͬ£©¡£ÏÂÁÐÓйØ˵·¨Öв»ÕýÈ·µÄÊÇ              £¨    £©
A£®MµÄ½á¹¹¼òʽΪHOOC¡ªCHOH¡ªCH2¡ªCOOH
B£®BµÄ·Ö×ÓʽΪC4H4O4Br2
C£®ÓëMµÄ¹ÙÄÜÍÅÖÖÀà¡¢ÊýÁ¿ÍêÈ«ÏàͬµÄͬ·ÖÒì¹¹Ì廹ÓÐ2ÖÖ
D£®CÎïÖʲ»¿ÉÄÜÈÜÓÚË®

D

½âÎöÊÔÌâ·ÖÎö£º1.34gAµÄÎïÖʵÄÁ¿Îª0.01mol£¬MÓë×ãÁ¿µÄ̼ËáÇâÄÆÈÜÒº·´Ó¦£¬Éú³É±ê×¼×´¿öϵÄÆøÌå0.448L¼´0.02mol£¬¿ÉÒÔ»ñµÃôÈ»ùµÄÊýĿΪ2£¬È·¶¨ÓлúÎïMµÄ½á¹¹¼òʽΪHOOC¡ªCHOH¡ªCH2¡ªCOOH£¬¹ÊAÕýÈ·£»ÓëMµÄ¹ÙÄÜÍÅÖÖÀà¡¢ÊýÁ¿ÍêÈ«ÏàͬµÄͬ·ÖÒì¹¹Ì廹ÓÐ2ÖÖ£¬¹ÊBÕýÈ·£»HOOC¡ªCHOH¡ªCH2¡ªCOOHÔÚ´ß»¯¼Á¼ÓÈȵÄÌõ¼þÏ·¢ÉúÏûÈ¥·´Ó¦Éú³Éº¬ÓÐË«¼üµÄÎïÖÊA£¬AΪHOOCCH=CHCOOH£¬¶øAÓëäå·¢ÉúÁ˼ӳɷ´Ó¦£¬Éú³ÉÁËB£¬ËùÒÔBµÄ·Ö×ÓʽΪC4H4O4Br2£¬ËùÒÔBÕýÈ·£»HOOCCH=CHCOOH¿ÉÒÔÓë×ãÁ¿NaOHÈÜÒº·¢Éú¼Ó³É·´Ó¦ÉúC,CΪHOOCHOHCHOHCOOH£¬¹ÊHOOCHOHCHOHCOOH¿ÉÒÔÈÜÓÚË®£¬¹ÊD´íÎó£¬Îª±¾ÌâµÄ´ð°¸¡£
¿¼µã£ºÓлú·´Ó¦¡¢·Ö×ÓʽµÄÈ·¶¨¡¢ÓлúÎïµÄÐÔÖÊ
µãÆÀ£º±¾Ì⿼²éÁËÓлú·´Ó¦¡¢·Ö×ÓʽµÄÈ·¶¨¡¢ÓлúÎïµÄÐÔÖÊ£¬¸ÃÌâ×ÛºÏÐԽϴó£¬ÐÅÏ¢½Ï¶à£¬±¾ÌâÄѶȽϴó¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ijÓÅÖÊÌðÓ£ÌÒÖк¬ÓÐÒ»ÖÖôÇ»ùËᣨÓÃM±íʾ£©£¬MµÄ̼Á´½á¹¹ÎÞÖ§Á´£¬·Ö×ÓʽΪC4H6O5£»1.34gMÓë×ãÁ¿µÄ̼ËáÇâÄÆÈÜÒº·´Ó¦£¬Éú³É±ê×¼×´¿öϵÄÆøÌå0.448L£®MÔÚÒ»¶¨Ìõ¼þÏ¿ɷ¢ÉúÈçÏÂת»¯M
ŨÁòËá
¡÷
A
Br2
B
×ãÁ¿NaOHÈÜÒº
C£¨M¡¢A¡¢B¡¢C·Ö×ÓÖÐ̼ԭ×ÓÊýÄ¿Ïàͬ£©£®ÏÂÁÐÓйØ˵·¨Öв»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ijÓÅÖÊÌðÓ£ÌÒÖк¬ÓÐÒ»ÖÖôÇ»ùËᣨÓÃM±íʾ£©£¬MµÄ̼Á´½á¹¹ÎÞÖ§Á´£¬·Ö×ÓʽΪC4H6O5£»1.34g MÓë×ãÁ¿µÄ̼ËáÇâÄÆÈÜÒº·´Ó¦£¬Éú³É±ê×¼×´¿öϵÄÆøÌå0.448L£®MÔÚÒ»¶¨Ìõ¼þÏ¿ɷ¢ÉúÈçÏÂת»¯£ºM  
 Å¨ÁòËá 
.
¡÷
A
Br2
B
×ãÁ¿NaOHÈÜÒº
  C £¨M¡¢A¡¢B¡¢C·Ö×ÓÖÐ̼ԭ×ÓÊýÄ¿Ïàͬ£©£®ÏÂÁÐÓйØ˵·¨Öв»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014½ìºþÄÏÊ¡¸ß¶þ4Ô¶ο¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

ijÓÅÖÊÌðÓ£ÌÒÖк¬ÓÐÒ»ÖÖôÇ»ùËᣨÓÃM±íʾ£©£¬MµÄ̼Á´½á¹¹ÎÞÖ§Á´£¬·Ö×ÓʽΪ

C4H6O5£»1.34gMÓë×ãÁ¿µÄ̼ËáÇâÄÆÈÜÒº·´Ó¦£¬Éú³É±ê×¼×´¿öϵÄÆøÌå0.448L¡£MÔÚÒ»¶¨Ìõ¼þÏ¿ɷ¢ÉúÈçÏÂת»¯£ºMABC£¨M¡¢A¡¢B¡¢C·Ö×ÓÖÐ̼ԭ×ÓÊýÄ¿Ïàͬ£©¡£ÏÂÁÐÓйØ˵·¨Öв»ÕýÈ·µÄÊÇ              £¨    £©

A£®MµÄ½á¹¹¼òʽΪHOOC¡ªCHOH¡ªCH2¡ªCOOH

B£®BµÄ·Ö×ÓʽΪC4H4O4Br2

C£®ÓëMµÄ¹ÙÄÜÍÅÖÖÀà¡¢ÊýÁ¿ÍêÈ«ÏàͬµÄͬ·ÖÒì¹¹Ì廹ÓÐ2ÖÖ

D£®CÎïÖʲ»¿ÉÄÜÈÜÓÚË®

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013½ì½­Î÷Ê¡¸ß¶þÏÂѧÆÚµÚÒ»´ÎÔ¿¼»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÑ¡ÔñÌâ

ijÓÅÖÊÌðÓ£ÌÒÖк¬ÓÐÒ»ÖÖôÇ»ùËᣨÓÃM±íʾ£©£¬MµÄ̼Á´½á¹¹ÎÞÖ§Á´£¬·Ö×ÓʽΪC4H6O5£»

1.34gMÓë×ãÁ¿µÄ̼ËáÇâÄÆÈÜÒº·´Ó¦£¬Éú³É±ê×¼×´¿öϵÄÆøÌå0.448L¡£MÔÚÒ»¶¨Ìõ¼þÏ¿ɷ¢ÉúÈçÏÂת»¯£º£¨M¡¢A¡¢B¡¢C·Ö×ÓÖÐ̼ԭ×ÓÊýÄ¿Ïàͬ£©¡£ÏÂÁÐÓйØ˵·¨Öв»ÕýÈ·µÄÊÇ               £¨     £©

A£®MµÄ½á¹¹¼òʽΪHOOC¡ªCHOH¡ªCH2¡ªCOOH

B£®BµÄ·Ö×ÓʽΪC4H4O4Br2

C£®ÓëMµÄ¹ÙÄÜÍÅÖÖÀà¡¢ÊýÁ¿ÍêÈ«ÏàͬµÄͬ·ÖÒì¹¹Ì廹ÓÐ1ÖÖ

D£®CÎïÖÊ¿ÉÄÜÈÜÓÚË®

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸