ÌâÄ¿ÄÚÈÝ

£¨1£©¾ßÓÐÖ§Á´µÄ»¯ºÏÎïAµÄ·Ö×ÓʽΪC4H6O2£¬A¿ÉÒÔʹBr2µÄËÄÂÈ»¯Ì¼ÈÜÒºÍÊÉ«£®1molAºÍ1mol NaHCO3ÄÜÍêÈ«·´Ó¦£¬ÔòAµÄ½á¹¹¼òʽÊÇ
 

£¨2£©»¯ºÏÎïBº¬ÓÐC¡¢H¡¢OÈýÖÖÔªËØ£¬·Ö×ÓÁ¿Îª60£¬ÆäÖÐ̼µÄÖÊÁ¿·ÖÊýΪ60%£¬ÇâµÄÖÊÁ¿·ÖÊýΪ13.33%£®BÔÚ´ß»¯¼ÁCuµÄ×÷ÓÃϱ»Ñõ»¯³ÉC£¬CÄÜ·¢ÉúÒø¾µ·´Ó¦£¬ÔòBµÄ½á¹¹¼òʽÊÇ
 

£¨3£©Ä³ÓлúÎïÍêȫȼÉÕ£¬Éú³É±ê×¼×´¿öÏÂCO2µÄÌå»ýΪ4.48L£¬H2OµÄÖÊÁ¿Îª5.4g£®
¢ÙÈôÓлúÎïµÄÖÊÁ¿Îª4.6g£¬´ËÓлúÎïµÄ·Ö×ÓʽΪ
 

¢ÚÈôÓлúÎïµÄÖÊÁ¿Îª6.2g£¬ÇÒ´ËÓлúÎïlmolÄܺÍ×ãÁ¿µÄ½ðÊôÄÆ·´Ó¦Éú³ÉlmolH2£¬´ËÓлúÎïµÄ½á¹¹¼òʽΪ
 
£¨Á½¸öôÇ»ù²»ÄÜÁ¬ÔÚͬһ¸ö̼ԭ×ÓÉÏ£©
¿¼µã£ºÓйØÓлúÎï·Ö×Óʽȷ¶¨µÄ¼ÆËã,ÓлúÎïʵÑéʽºÍ·Ö×ÓʽµÄÈ·¶¨
רÌ⣺Ìþ¼°ÆäÑÜÉúÎïµÄȼÉÕ¹æÂÉ
·ÖÎö£º£¨1£©¾ßÓÐÖ§Á´µÄ»¯ºÏÎïAµÄ·Ö×ÓʽΪC4H6O2£¬1molAºÍ1mol NaHCO3ÄÜÍêÈ«·´Ó¦£¬º¬ÓÐ-COOH£¬A¿ÉÒÔʹBr2µÄËÄÂÈ»¯Ì¼ÈÜÒºÍÊÉ«£¬»¹º¬ÓÐ1¸öC=CË«¼ü£¬¾Ý´ËÊéдAµÄ½á¹¹¼òʽ£»
£¨2£©¼ÆËã·Ö×ÓʽÖÐC¡¢H¡¢OÔ­×ÓÊýĿȷ¶¨B·Ö×Óʽ£¬BÔÚ´ß»¯¼ÁCuµÄ×÷ÓÃϱ»Ñõ»¯³ÉC£¬CÄÜ·¢ÉúÒø¾µ·´Ó¦£¬ÔòBÊôÓÚ´¼£¬ÇÒôÇ»ùÁ¬½ÓµÄ̼ԭ×ÓÉϺ¬ÓÐ2¸öHÔ­×Ó£»
£¨3£©¢Ù¸ù¾Ýn=
V
Vm
¼ÆËã¶þÑõ»¯Ì¼µÄÎïÖʵÄÁ¿£¬¸ù¾Ýn=
m
M
¼ÆËãË®µÄÎïÖʵÄÁ¿£¬½áºÏÖÊÁ¿ÊغãÅжÏÊÇ·ñº¬ÓÐÑõÔªËØ£¬¼ÆËãÔªËØÔ­×Ó¸öÊýÖ®±È£¬¾Ý´ËÈ·¶¨ÊµÑéʽ£¬½ø¶øÈ·¶¨·Ö×Óʽ£»
¢Ú¸ù¾Ý¢ÙÖеķÖÎöÈ·¶¨·Ö×Óʽ£¬´ËÓлúÎïlmolÄܺÍ×ãÁ¿µÄ½ðÊôÄÆ·´Ó¦Éú³ÉlmolH2£¬ÖÁÉÙº¬ÓÐ-OH¡¢-COOHÖеÄÒ»ÖÖ£¬½áºÏ·Ö×Óʽȷ¶¨º¬ÓеĹÙÄÜÍÅ£¬¾Ý´ËÅжÏBµÄ½á¹¹¼òʽ£®
½â´ð£º ½â£º£¨1£©¾ßÓÐÖ§Á´µÄ»¯ºÏÎïAµÄ·Ö×ÓʽΪC4H6O2£¬1molAºÍ1mol NaHCO3ÄÜÍêÈ«·´Ó¦£¬º¬ÓÐ-COOH£¬A¿ÉÒÔʹBr2µÄËÄÂÈ»¯Ì¼ÈÜÒºÍÊÉ«£¬»¹º¬ÓÐ1¸öC=CË«¼ü£¬¹ÊAµÄ½á¹¹¼òʽΪCH2=C£¨CH3£©COOH£¬
¹Ê´ð°¸Îª£ºCH2=C£¨CH3£©COOH£»
£¨2£©·Ö×ÓʽÖÐN£¨C£©=
60¡Á60%
12
=3¡¢N£¨H£©=
60¡Á13.33%
1
=8¡¢N£¨O£©=
60-12¡Á3-8
16
=1£¬¹ÊÓлúÎïBµÄ·Ö×ÓʽΪC3H8O£¬BÔÚ´ß»¯¼ÁCuµÄ×÷ÓÃϱ»Ñõ»¯³ÉC£¬CÄÜ·¢ÉúÒø¾µ·´Ó¦£¬ÔòBÊôÓÚ´¼£¬ÇÒôÇ»ùÁ¬½ÓµÄ̼ԭ×ÓÉϺ¬ÓÐ2¸öHÔ­×Ó£¬¹ÊBµÄ½á¹¹¼òʽΪCH3CH2CH2OH£¬
¹Ê´ð°¸Îª£ºCH3CH2CH2OH£»
£¨3£©¶þÑõ»¯Ì¼µÄÎïÖʵÄÁ¿=
4.48L
22.4L/mol
=0.2mol£¬Ë®µÄÎïÖʵÄÁ¿=
5.4g
18g/mol
=0.3mol£¬
¢ÙÈôÓлúÎïµÄÖÊÁ¿Îª4.6g£¬ÔòÑõÔªËØÖÊÁ¿=4.6g-0.2mol¡Á12g/mol-0.3mol¡Á2¡Á1g/mol=1.6g£¬ÔòÑõÔ­×ÓÎïÖʵÄÁ¿=
1.6g
16g/mol
=0.1mol£¬¹ÊÓлúÎïÖÐC¡¢H¡¢OÔ­×Ó¸öÊýÖ®±È=0.2mol£º0.6mol£º0.1mol=2£º6£º1£¬ÓлúÎïµÄʵÑéʽΪC2H6O£¬HÔ­×ÓÒѾ­±¥ºÍ£¬¹ÊʵÑéÊÒ¼´Îª·Ö×Óʽ£¬
¹Ê´ð°¸Îª£ºC2H6O£»
¢ÚÈôÓлúÎïµÄÖÊÁ¿Îª6.2g£¬ÔòÑõÔªËØÖÊÁ¿=6.2g-0.2mol¡Á12g/mol-0.3mol¡Á2¡Á1g/mol=3.2g£¬ÔòÑõÔ­×ÓÎïÖʵÄÁ¿=
3.2g
16g/mol
=0.2mol£¬¹ÊÓлúÎïÖÐC¡¢H¡¢OÔ­×Ó¸öÊýÖ®±È=0.2mol£º0.6mol£º0.2mol=2£º6£º2£¬ÓлúÎïµÄʵÑéʽΪC2H6O2£¬HÔ­×ÓÒѾ­±¥ºÍ£¬¹ÊʵÑéÊÒ¼´Îª·Ö×Óʽ£¬´ËÓлúÎïlmolÄܺÍ×ãÁ¿µÄ½ðÊôÄÆ·´Ó¦Éú³ÉlmolH2£¬Ôò·Ö×ÓÖк¬ÓÐ2¸ö-OH£¬Á½¸öôÇ»ù²»ÄÜÁ¬ÔÚͬһ¸ö̼ԭ×ÓÉÏ£¬Ôò¸ÃÓлúÎïµÄ½á¹¹¼òʽΪHO-CH2-CH2-OH£¬
¹Ê´ð°¸Îª£ºHO-CH2-CH2-OH£®
µãÆÀ£º±¾Ì⿼²éÓлúÎï·Ö×Óʽ¡¢½á¹¹Ê½µÄÈ·¶¨µÈ£¬ÄѶÈÖеȣ¬×¢ÒâÕÆÎÕ¹ÙÄÜÍŵÄÐÔÖÊ£¬£¨3£©ÖÐ×¢ÒâÀûÓÃÖÊÁ¿ÊغãÈ·¶¨ÑõÔªËØ£¬¸ù¾ÝʵÑéʽÖÐC¡¢HÊýÄ¿¹ØÏµÈ·¶¨·Ö×Óʽ£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
A¡¢B¡¢C¡¢D¡¢E¡¢FÊÇÔ­×ÓÐòÊýÒÀ´ÎÔö´óµÄ¶ÌÖÜÆÚÖ÷×åÔªËØ£¬AµÄ×îÍâ²ãµç×ÓÊýÊÇ´ÎÍâ²ãµç×ÓÊý2±¶£¬BÊǶÌÖÜÆÚÖнðÊôÐÔ×îÇ¿µÄÔªËØ£¬CÊÇͬÖÜÆÚÖÐÑôÀë×Ó°ë¾¶×îСµÄÔªËØ£¬DÔªËØµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïÓëBÔªËØµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯Îï·´Ó¦µÄ²úÎïMÊÇÖÆ±¸Ä¾²Ä·À»ð¼ÁµÄÔ­ÁÏ£¬EµÄ×îÍâ²ãµç×ÓÊýÓëÄÚ²ãµç×ÓÊýÖ®±ÈΪ3£º5£®Çë»Ø´ð£º
£¨1£©FÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃ
 
£®
£¨2£©ÔªËصķǽðÊôÐÔΪ£¨Ô­×ӵĵõç×ÓÄÜÁ¦£©£ºA
 
E£¨Ìî¡°Ç¿ÓÚ¡±»ò¡°ÈõÓÚ¡±£©£®
£¨3£©Ð´³öMµÄË®ÈÜÒº¾ÃÖÃÓÚ¿ÕÆøÖбä»ë×ǵÄÀë×Ó·½³Ìʽ
 
£®
£¨4£©Ð´³öBºÍE×é³ÉµÄ»¯ºÏÎïµÄµç×Óʽ£º
 
£®
£¨5£©Bµ¥ÖÊÓëÑõÆø·´Ó¦µÄ²úÎïÓëCµÄµ¥ÖÊͬʱ·ÅÈëË®ÖУ¬²úÉúÁ½ÖÖÎÞÉ«ÆøÌ壬Èç¹ûÕâÁ½ÖÖÆøÌåÇ¡ºÃÄÜÍêÈ«·´Ó¦£¬ÓëÑõÆø·´Ó¦µÄBµ¥ÖʺͷÅÈëË®ÖеÄCµ¥ÖʵÄÖÊÁ¿±ÈΪ
 
£®
£¨6£©¹¤ÒµÉϽ«¸ÉÔïµÄFµ¥ÖÊͨÈëÈÛÈÚµÄEµ¥ÖÊÖпÉÖÆµÃ»¯ºÏÎïE2F2£®¸ÃÎïÖÊ¿ÉÓëË®·´Ó¦Éú³ÉÒ»ÖÖÄÜʹƷºìÈÜÒºÍÊÉ«µÄÆøÌ壬0.2mol¸ÃÎïÖʲμӷ´Ó¦Ê±×ªÒÆ0.3molµç×Ó£¬ÆäÖÐÖ»ÓÐÒ»ÖÖÔªËØ»¯ºÏ¼Û·¢Éú¸Ä±ä£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
 
£®
£¨7£©ÔÚ298KÏ£¬A¡¢CµÄµ¥Öʸ÷1molÍêȫȼÉÕ£¬·Ö±ð·Å³öÈÈÁ¿a kJºÍb kJ£®ÓÖÖªÒ»¶¨Ìõ¼þÏ£¬CµÄµ¥ÖÊÄܽ«A´ÓËüµÄ×î¸ß¼ÛÑõ»¯ÎïÖÐÖû»³öÀ´£¬Èô´ËÖû»·´Ó¦Éú³É3mol AµÄµ¥ÖÊ£¬Ôò¸Ã·´Ó¦ÔÚ298Kϵġ÷H=
 
£¨×¢£ºÌâÖÐËùÉèµ¥ÖʾùΪ×îÎȶ¨µ¥ÖÊ£©£®

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø