ÌâÄ¿ÄÚÈÝ

¶à²ÊµÄ¡°Ì¼¡±£¬¶à×˵ÄÉú»î£¬ÈÃÎÒÃÇÒ»Æð×ß½ø¡°Ì¼¡±µÄÊÀ½ç£®
£¨1£©ÌîдÓйغ¬Ì¼ÎïÖʵĶÔÓ¦ÌØÐÔ£®
ÎïÖÊÓÃ;½ð¸ÕʯÇиÁ§Ê¯Ä«×÷µç¼«»îÐÔÌ¿¾»Ë®
¶ÔÓ¦ÌØÕ÷¢Ù
 
¢Ú
 
¢Û
 
£¨2£©ÒºÌ¬¶þÑõ»¯Ì¼Ãð»ðÆ÷¿ÉÓÃÓÚÆË¾Èµµ°¸×ÊÁÏÊÒ·¢ÉúµÄ»ðÔÖ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÓÐ
 
£¨Ìî±êºÅ£©£®
A£®ÒºÌ¬¶þÑõ»¯Ì¼Æø»¯ºó²»»áÎÛȾµµ°¸×ÊÁÏ
B£®¶þÑõ»¯Ì¼¿É¸²¸ÇÔÚȼÉÕÎï±íÃæ£¬¸ô¾ø¿ÕÆø
C£®ÒºÌ¬¶þÑõ»¯Ì¼Æø»¯Ê±ÎüÈÈ£¬½µµÍÁË¿ÉȼÎïµÄ×Å»ðµã
£¨3£©¹ý¶àµÄ¶þÑõ»¯Ì¼¼Ó¾çÁË¡°ÎÂÊÒЧӦ¡±£¬Ð´³öÒ»Ìõ¼õÉÙ¶þÑõ»¯Ì¼ÅŷŵĽ¨Ò飺
 
£®
£¨4£©CO2ºÍNH3¿ÉÒԺϳÉÄòËØ[CO£¨NH2£©2]£¬Í¬Ê±Éú³ÉË®£®¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
 
£®
£¨5£©Ä¿Ç°£¬ÈËÀàÒÔ»¯Ê¯È¼ÁÏΪÖ÷ÒªÄÜÔ´£®»¯Ê¯È¼ÁÏÓÐú¡¢
 
ºÍÌìÈ»Æø£®
£¨6£©ÌìÈ»ÆøÖм×ÍéÍêȫȼÉյĻ¯Ñ§·½³ÌʽΪ
 
£®
£¨7£©´Ó±í2Êý¾Ý·ÖÎö£¬ÓëúÏà±È£¬ÓÃÌìÈ»Æø×÷ȼÁϵÄÓŵãÓÐ
 
£®
1gÎïÖÊÍêȫȼÉÕ²úÉúCO2µÄÖÊÁ¿/g·Å³öµÄÈÈÁ¿/kJ
¼×Íé2.7556
̼3.6732
£¨8£©2.3gijÎïÖÊÔÚ¿ÕÆøÖÐÍêȫȼÉÕ£¬Éú³É4.4g¶þÑõ»¯Ì¼ºÍ2.7gË®£¬Ôò¸ÃÎïÖÊÖÐÒ»¶¨º¬ÓУ¨ÌîÔªËØ·ûºÅ£©
 
ÔªËØ£¬ÆäÔªËØÖÊÁ¿±ÈΪ
 
£®
¿¼µã£ºÌ¼µ¥ÖʵÄÎïÀíÐÔÖʼ°ÓÃ;,¶þÑõ»¯Ì¼¶Ô»·¾³µÄÓ°Ïì,ÖÊÁ¿Êغ㶨Âɼ°ÆäÓ¦ÓÃ,³£ÓÃȼÁϵÄʹÓÃÓëÆä¶Ô»·¾³µÄÓ°Ïì,»¯Ê¯È¼Áϼ°Æä×ÛºÏÀûÓÃ,¼¸ÖÖ³£ÓõÄÃð»ðÆ÷
רÌ⣺»¯Ñ§ÓëÄÜÔ´,̼µ¥ÖÊÓ뺬̼»¯ºÏÎïµÄÐÔÖÊÓëÓÃ;
·ÖÎö£º£¨1£©¸ù¾Ý̼µ¥ÖʵÄÐÔÖʺÍÓÃ;·ÖÎö»Ø´ð£»
£¨2£©¸ù¾Ý¶þÑõ»¯Ì¼²»È¼ÉÕÒ²²»Ö§³ÖȼÉÕ½øÐзÖÎöÅжϣ»
£¨3£©¸ù¾Ý¼õÉÙ¶þÑõ»¯Ì¼ÅŷŵĴëÊ©½øÐнâ´ð£»
£¨4£©¸ù¾Ý·´Ó¦ÎïºÍÉú³ÉÎïÊéд·´Ó¦µÄ»¯Ñ§·½³Ìʽ¼´¿É£»
£¨5£©¸ù¾Ý»¯Ê¯È¼ÁϵÄÖÖÀà·ÖÎö»Ø´ð£»
£¨6£©¸ù¾Ý¼×ÍéȼÉյķ´Ó¦Ð´³ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ£»
£¨7£©¸ù¾Ý±íÖеÄÊý¾Ý·ÖÎöµÈÖÊÁ¿µÄúºÍÌìÈ»ÆøÉú³ÉµÄ¶þÑõ»¯Ì¼µÄÖÊÁ¿ºÍ·Å³öµÄÈÈÁ¿·ÖÎö£»
£¨8£©ÔÚ»¯Ñ§·´Ó¦ÖУ¬·´Ó¦Ç°ºóÔ­×ÓµÄÖÖÀàûÓиı䣬ÊýĿûÓÐÔö¼õ£¬Ô­×ÓµÄÖÊÁ¿Ò²Ã»Óиı䣬ËùÒÔ·´Ó¦Ç°ºóÎïÖʵÄÖÊÁ¿×ܺÍÏàµÈ£®¸ù¾ÝÉú³ÉÎïCO2ºÍH2OÖÐC¡¢HÔªËØÖÊÁ¿£¬¿ÉÒÔÈ·¶¨¿ÉȼÎïÖÐC¡¢HÔªËØ±ÈºÍÔ­×Ó¸öÊý±È£®
½â´ð£º½â£º£¨1£©½ð¸ÕʯµÄÓ²¶È×î´ó£¬¿ÉÒÔÓÃÓÚÇиÁ§£¬Ê¯Ä«¾ßÓÐÓÅÁ¼µÄµ¼µçÐÔ£¬¿ÉÒÔÓÃ×÷µç¼«£¬»îÐÔÌ¿¾ßÓÐÎü¸½ÐÔ£¬¿ÉÒÔÓÃÓÚ¾»Ë®£¬¹ÊÌ
ÎïÖÊÓÃ;½ð¸ÕʯÇиÁ§Ê¯Ä«×÷µç¼«»îÐÔÌ¿¾»Ë®
¶ÔÓ¦ÌØÕ÷¢ÙÓ²¶È´ó¢Úµ¼µçÐÔ¢ÛÎü¸½ÐÔ
£¨2£©A£®ÒºÌ¬¶þÑõ»¯Ì¼Æø»¯ºó²»»áÎÛȾµµ°¸×ÊÁÏ£¬ÕýÈ·£»
B£®¶þÑõ»¯Ì¼¿É¸²¸ÇÔÚȼÉÕÎï±íÃæ£¬¸ô¾ø¿ÕÆø£¬ÕýÈ·£»
C£®ÒºÌ¬¶þÑõ»¯Ì¼Æø»¯Ê±ÎüÈÈ£¬²»ÄܽµµÍÁË¿ÉȼÎïµÄ×Å»ðµã£¬´íÎó
£¨3£©¼õÉÙ¶þÑõ»¯Ì¼ÅŷŵĴëÊ©ÓнÚÔ¼ÓÃÖ½¡¢¿ª·¢ÐÂÄÜÔ´¡¢Ê¹ÓÃÇå½àÄÜÔ´µÈ£»
£¨4£©ÔÚ¸ßθßѹÏ£¬CO2ºÍNH3¿ÉÒԺϳÉÄòËØ[CO£¨NH2£©2]£¬Í¬Ê±Éú³ÉË®£®·´Ó¦µÄ·½³ÌʽÊÇ£ºCO2+2NH3
 ¸ßθßѹ 
.
 
CO£¨NH2£©2+H2O£®
£¨5£©Ä¿Ç°£¬ÈËÀàÒÔ»¯Ê¯È¼ÁÏΪÖ÷ÒªÄÜÔ´£®»¯Ê¯È¼ÁÏÓÐú¡¢Ê¯ÓͺÍÌìÈ»Æø£®
£¨6£©ÌìÈ»ÆøÖм×ÍéÍêȫȼÉÕÉú³ÉÁ˶þÑõ»¯Ì¼ºÍË®£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºCH4+2O2
 µãȼ 
.
 
CO2+2H2O£®
£¨7£©ÓɱíÖÐÖÐÊý¾Ý¿ÉÖª£¬ÓëúÏà±È£¬ÓÃÌìÈ»Æø×÷ȼÁϵÄÓŵãÓУºµÈÖÊÁ¿µÄÌìÈ»ÆøºÍúÍêȫȼÉÕ£¬ÌìÈ»ÆøÈ¼ÉÕ²úÉú¶þÑõ»¯Ì¼µÄÖÊÁ¿Ð¡ÓÚú£¬·Å³öµÄÈÈÁ¿¸ßÓÚú£»
£¨8£©Éú³ÉÎïÖк¬ÓÐC¡¢H¡¢OÈýÖÖÔªËØ£¬ËùÒÔ¿ÉȼÎïÖÐÒ»¶¨º¬C¡¢HÔªËØ£»B¡¢4.4gCO2ÖÐCÔªËØÖÊÁ¿=4.4g¡Á
12
44
¡Á100%=1.2g£» 2.7gH2OÖÐHÔªËØÖÊÁ¿=2.7g¡Á
2
18
¡Á100%=0.3g¿ÉȼÎïÖÐC¡¢HÔªËØÖÊÁ¿±È=1.2g£º0.3g=4£º1£®
¹Ê´ðΪ£º£¨1£©¼ûÉÏ±í£»£¨2£©AB£»£¨3£©¿ª·¢ÐÂÄÜÔ´£»£¨4£©CO2+2NH3
 ¸ßθßѹ 
.
 
CO£¨NH2£©2+H2O£»£¨5£©Ê¯ÓÍ£»£¨6£©CH4+2O2
 µãȼ 
.
 
CO2+2H2O£»£¨7£©µÈÖÊÁ¿µÄÌìÈ»ÆøºÍúÍêȫȼÉÕ£¬ÌìÈ»ÆøÈ¼ÉÕ²úÉú¶þÑõ»¯Ì¼µÄÖÊÁ¿Ð¡ÓÚú£¬·Å³öµÄÈÈÁ¿¸ßÓÚú£»£¨8£©C¡¢H£¬4£º1£®
µãÆÀ£º±¾ÌâÄѶȲ»´ó£¬ÎïÖʵÄÐÔÖʾö¶¨ÎïÖʵÄÓÃ;£¬ÕÆÎÕ̼ºÍ̼µÄ»¯ºÏÎïµÄÐÔÖʺÍÓÃ;ÊÇÕýÈ·½â´ð´ËÀàÌâµÄ¹Ø¼ü£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø