ÌâÄ¿ÄÚÈÝ

ÄÜÔ´ºÍ»·¾³Êǵ±½ñÈËÀàÃæÁÙµÄÁ½´óÎÊÌ⣮
£¨1£©½üÄêÀ´£¬ÎÒÊв»ÉÙ¹«½»³µºÍ³ö×â³µÉÏÓ¡ÓС°CNG¡±µÄ±êÖ¾£¬´ú±íËüÃÇÊÇÒÔѹËõÌìÈ»ÆøΪȼÁϵÄÆû³µ£®
¢ÙÌìÈ»ÆøµÄÖ÷Òª³É·ÖÊÇ______£¬ÆäÖÐ̼ԪËصĻ¯ºÏ¼ÛΪ______£»
¢ÚÓÐÈËÈÏΪ£¬ÌìÈ»Æø×÷ȼÁϿɱÜÃâÎÂÊÒЧӦµÄ·¢Éú£®ÄãµÄ¿´·¨ÊÇ______£¨Ìî¡°ÊÇ¡±»ò¡°·ñ¡±£©£¬ÀíÓÉÊÇ£º______£®
£¨2£©È¼Æø°²È«ÊǼÒÍ¥Éú»îµÄÍ·µÈ´óÊ£®ÎªÁË·ÀֹȼÆøÒÝÉ¢£¬³£ÔÚȼÆøÖмÓÈëÉÙÁ¿µÄÓÐÌØÊâÆøζµÄÒÒÁò´¼£¨C2H5SH£©£®ÒÒÁò´¼³ä·ÖȼÉÕʱ²úÉú¶þÑõ»¯Ì¼¡¢¶þÑõ»¯ÁòºÍË®£®Çëд³öÒÒÁò´¼³ä·ÖȼÉյĻ¯Ñ§·´Ó¦·½³Ìʽ£º______£®
£¨3£©SO2ºÍNO2¶¼ÄÜÈÜÓÚÓêË®ÐγÉËáÓê¶øÆÆ»µ»·¾³£®ËáÓê¿ÉÄܶԻ·¾³Ôì³ÉµÄΣº¦ÓÐ______£¨¾ÙÒ»Àý£©£»Îª¼õÉÙËáÓêµÄ²úÉú£¬ÓÐÏÂÁдëÊ©£º¢ÙÉÙÓÃú×÷ȼÁÏ£»   ¢Ú°Ñ¹¤³§µÄÑ̴ѽ¨¸ß£»  ¢ÛȼÁÏÍÑÁò£»  ¢Ü¿ª·¢ÐÂÄÜÔ´£®ÆäÖÐÓÐЧµÄ´ëÊ©ÊÇ______£¨ÌîÐòºÅ£©£®
£¨1£©¢ÙÌìÈ»ÆøµÄÖ÷Òª³É·ÖÊǼ×Í飻¼×ÍéµÄ»¯Ñ§Ê½ÎªCH4£¬Éè̼µÄ»¯ºÏ¼ÛΪX£¬ÔòX=0-£¨+1£©¡Á4=-4£®
¹Ê´ð°¸Îª£º¼×Í飻-4£®
¢ÚÌìÈ»ÆøȼÉÕÉú³É¶þÑõ»¯Ì¼ºÍË®£¬ÆäÖжþÑõ»¯Ì¼ÊÇÎÂÊÒЧӦµÄÖ÷ÒªÆøÌ壬ËùÒÔÓÃÌìÈ»Æø×÷ȼÁϿɱÜÃâÎÂÊÒЧӦµÄ·¢ÉúµÄ˵·¨ÊÇ´íÎóµÄ£®
¹Ê´ð°¸Îª£º·ñ£»¼×ÍéȼÉÕÉú³ÉÎÂÊÒÆøÌå¶þÑõ»¯Ì¼£®
£¨2£©ÒÒÁò´¼È¼ÉÕʱ²úÉú¶þÑõ»¯Ì¼¡¢¶þÑõ»¯ÁòºÍË®£¬»¯Ñ§·½³ÌʽΪ£º2C2H5SH+9O2
 µãȼ 
.
 
2SO2+4CO2+6H2O£®
¹Ê´ð°¸Îª£º2C2H5SH+9O2
 µãȼ 
.
 
2SO2+4CO2+6H2O£®
£¨3£©ËáÓêÄܹ»¸¯Ê´½¨ÖþÎʹɭÁÖ¡¢×¯¼ÚµÈËÀÍö£»¼õÉÙÓÃú×öȼÁÏ¡¢È¼ÁÏÍÑÁò¡¢¿ª·¢ÐÂÄÜÔ´µÈ¿ÉÒÔ¼õÉÙËáÓêµÄ²úÉú£»°Ñ¹¤³§µÄÑ̴ѼӸ߲»»á¼õÉÙËáÓêµÄ²úÉú£®
¹Ê´ð°¸Îª£º¸¯Ê´½¨ÖþÎ¢Ù¢Û¢Ü£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨2006?Õò½­£©ÄÜÔ´ºÍ»·¾³Êǵ±½ñÈËÀàÃæÁÙµÄÁ½´óÎÊÌ⣮Ŀǰ£¬»¯Ê¯È¼ÁÏÊÇÈËÀàÉú²ú¡¢Éú»îµÄÖ÷ÒªÄÜÔ´£®Ëæ×ÅÈ«ÇòÄÜԴʹÓÃÁ¿µÄÔö³¤¼°²»¿ÆѧʹÓ㬻¯Ê¯È¼ÁϵȲ»¿ÉÔÙÉúÄÜÔ´½«ÈÕÒæ¿Ý½ß£¬²¢¶Ô»·¾³²úÉúÑÏÖØÓ°Ï죮Õâ¾ÍÆÈÇÐÒªÇóÈËÃÇ¿ª·¢ÇâÄÜ¡¢Ì«ÑôÄܵÈÐÂÄÜÔ´£®
£¨1£©ÇâÄÜÀ´Ô´¹ã·º£¬ÊǸßÄÜÇå½àÄÜÔ´£®ÇâÆø³ýÁË¿ÉÖ±½ÓȼÉÕ»ñµÃÈÈÁ¿Í⣬»¹¿ÉÉè¼Æ³ÉȼÁϵç³Ø£®Ä¿Ç°Ê¹ÓÃÇâÄÜ»¹ÃæÁÙ×ÅÇâÆøµÄÖÆÈ¡ºÍÇâÆøµÄ´¢´æÁ½¸öÎÊÌ⣮ÈçͼΪģÄâÇâ¡¢ÑõȼÁϵç³ØµÄʵÑé×°Öã®X¡¢Y¾ùΪɴ²¼´ü×ö³ÉµÄÈÝÆ÷£¬×°ÓлîÐÔÌ¿£¬²¢ÓëÄÚ²¿µÄʯī°ô½ôÃܽӴ¥£¬µ¼ÏßÓëʯīÁ¬½Ó£¨Ê¯Ä«ºÍ»îÐÔÌ¿¶¼Äܵ¼µç£©£®°ÑX¡¢Y½þÔÚË®ÖУ¬²¢ÔÚË®ÖмÓһЩϡÁòËᣬÒÔÔöÇ¿Ë®µÄµ¼µçÄÜÁ¦£®µ±µç¼üKÓëb½Óͨ£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
2H2O
 Í¨µç 
.
 
2H2¡ü+O2¡ü
2H2O
 Í¨µç 
.
 
2H2¡ü+O2¡ü
£®Í¨µçÒ»¶Îʱ¼äÄÚûÓп´µ½ÆøÅÝÒݳö£¬ÊÇÒòΪ»îÐÔÌ¿¾ßÓÐ
Îü¸½
Îü¸½
ÐÔ£¬´Ó¶ø½â¾öÁËÆøÌå´¢´æµÄÎÊÌ⣮´Ëʱ½«µç¼üKÓëa½Óͨ£¬µÆÅÝ·¢¹â£¬ËµÃ÷ÄÜÁ¿·¢ÉúÁËÈçÏÂת»¯£ºÇâÄÜ¡ú
µç
µç
ÄÜ¡ú¹âÄܺÍÈÈÄÜ£®
£¨2£©Ò»Ð©³£ÓÃȼÁÏȼÉÕʱµÄÈÈÖµÈçÏÂ±í£º
ȼÁÏ ÌìÈ»Æø£¨ÒÔCH4¼Æ£© Òº»¯Æø£¨ÒÔC4H10¼Æ£© ԭú£¨ÒÔC¼Æ£©
ÈÈÖµ£¨kJ/g£© Ô¼56 Ô¼50 Ô¼33
¢ÙÄ¿Ç°²»Ìᳫֱ½ÓÓÃԭú×÷ȼÁϵÄÖ÷ÒªÔ­ÒòÊÇ£ºa
ԭúº¬ÓÐÁòµÈÓк¦ÔªËØ£¬È¼ÉÕ²úÉúSO2ÎÛȾ¿ÕÆø
ԭúº¬ÓÐÁòµÈÓк¦ÔªËØ£¬È¼ÉÕ²úÉúSO2ÎÛȾ¿ÕÆø
£¬b
ȼÉÕ²úÉú´óÁ¿·Û³¾ÎÛȾ¿ÕÆøÈÈÖµµÍµÈ
ȼÉÕ²úÉú´óÁ¿·Û³¾ÎÛȾ¿ÕÆøÈÈÖµµÍµÈ
£®
¢Ú»¯Ê¯È¼ÁÏȼÉշųöµÄCO2ÊÇÒ»ÖÖÎÂÊÒÆøÌ壬´óÁ¿Ïò¿ÕÆøÖÐÅÅ·ÅCO2½«¶ÔÆøºò²úÉú¸ºÃæÓ°Ï죮ÊÔͨ¹ý¼ÆËã˵Ã÷£¬µ±ÉϱíÖÐÌìÈ»ÆøºÍÒº»¯Æø³ä·ÖȼÉÕ»ñµÃÏàµÈµÄÈÈÁ¿Ê±£¬ÄÄÒ»ÖÖȼÁϷųöµÄCO2½ÏÉÙ£¿
ÄÜÔ´ºÍ»·¾³Êǵ±½ñÈËÀàÃæÁÙµÄÁ½´óÎÊÌ⣮Ŀǰ£¬»¯Ê¯È¼ÁÏÊÇÈËÀàÉú²ú¡¢Éú»îµÄÖ÷ÒªÄÜÔ´£®ÎÒ¹úÌìÈ»ÆøµÄ´¢Á¿·á¸»£¬ºÏÀí¿ª·¢ºÍʹÓÃÊǹúÈËÃæÁٵĹ²Í¬¿ÎÌ⣮
£¨1£©½üÄêÀ´£¬ÎÒÃÇ»áÔÚÐí¶à³ÇÊеĽÖÍ··¢ÏÖ£¬²»ÉÙ¹«½»³µºÍ³ö×â³µÉÏÓ¡ÓС°CNG¡±µÄ±êÖ¾£¬´ú±íËüÃÇÊÇÒÔѹËõÌìÈ»ÆøΪȼÁϵÄÆû³µ£®
¢ÙÌìÈ»ÆøµÄÖ÷Òª³É·ÖÊǼ×Íé--CH4£¬ÆäÖÐ̼ԪËصĻ¯ºÏ¼ÛΪ
-4
-4
£»
¢ÚʹÓÃÌìÈ»Æø×÷ȼÁÏÄÜ·ñ±ÜÃâÎÂÊÒЧӦµÄ·¢Éú£¿
²»ÄÜ
²»ÄÜ
£¨Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±£©£¬ÀíÓÉÊÇ£º
¼×Íé±¾Éí¾ÍÊÇÎÂÊÒÆøÌ壬ËüȼÉյIJúÎï¶þÑõ»¯Ì¼Ò²ÊÇÎÂÊÒÆøÌå
¼×Íé±¾Éí¾ÍÊÇÎÂÊÒÆøÌ壬ËüȼÉյIJúÎï¶þÑõ»¯Ì¼Ò²ÊÇÎÂÊÒÆøÌå
£®
£¨2£©¡°Î÷Æø¶«Ê䡱¹¤³ÌÒѻݼ°Ç§¼ÒÍò»§£¬Ô­±¾ÒÔÒº»¯Ê¯ÓÍÆø»òúÆøΪȼÁϵļÒÍ¥Õý×¼±¸¸ÄÓÃÌìÈ»Æø×÷ȼÁÏ£®ÌìÈ»Æø¡¢Òº»¯Ê¯ÓÍÆøȼÉյķ´Ó¦¿É±íʾΪÊÇ£ºCH4+2O2
 µãȼ 
.
 
2CO2+H2O¡¢C3H8+5O2 
 µãȼ 
.
 
3CO2+4H2O
¢ÙСÃ÷¼ÒÏÖÓÐÒ»Ì×ÒÔÒº»¯Ê¯ÓÍÆøΪȼÁϵÄÔî¾ß£¬ÈôÒª¸ÄÓÃÌìÈ»Æø×÷ȼÁÏ£¬Ó¦²ÉÓõĴëÊ©ÊÇ£¨×¢£ºÍ¬Î¡¢Í¬Ñ¹Ï£¬ÏàͬÌå»ýµÄ²»Í¬ÆøÌå¾ßÓÐÏàͬµÄ·Ö×ÓÊý£©
A
A

A£®¼õÉÙ¿ÕÆøµÄ½øÈëÁ¿£¬Ôö´óÒº»¯ÆøµÄ½øÈëÁ¿          B£®Ôö´óÁ½ÕߵĽøÈëÁ¿
C£®Ôö´ó¿ÕÆø½øÈëÁ¿»ò¼õÉÙÒº»¯ÆøµÄ½øÈëÁ¿            D£®¼õÉÙÁ½ÕߵĽøÈëÁ¿
¢ÚȼÆøй©»áÔì³ÉΣÏÕ£¬È¼Æø°²È«ÊǼÒÍ¥Éú»îµÄÍ·µÈ´óÊ£®
a£®ÎªÁË·ÀֹȼÆøÒÝÉ¢£¬³£ÔÚȼÆøÖмÓÈëÉÙÁ¿µÄÓÐÌØÊâÆøζµÄÒÒÁò´¼£¨C2H5SH£©£®ÒÒÁò´¼³ä·ÖȼÉÕʱ²úÉú¶þÑõ»¯Ì¼¡¢¶þÑõ»¯ÁòºÍË®£®ÇëÊéдÒÒÁò´¼³ä·ÖȼÉյĻ¯Ñ§·´Ó¦·½³Ìʽ£º
2C2H5SH+9O2
 µãȼ 
.
 
4CO2+2X+6H2O
2C2H5SH+9O2
 µãȼ 
.
 
4CO2+2X+6H2O
£»
b£®ÔÚ¼ÒÖа²×°±¨¾¯Æ÷ÊÇ·ÀֹȼÆøÒÝÉ¢µÄÁíÒ»´ëÊ©£¬µ±±¨¾¯Æ÷½Ó´¥µ½Ò»¶¨Á¿µÄй©ÆøÌåʱ£¬»á·¢³öÏìÉù£®ÓÐλͬѧ¼ÒÖÐËùʹÓõÄȼÁÏÊÇÌìÈ»Æø£¬ÊÔÅжϱ¨¾¯Æ÷°²×°µÄλÖÃÓ¦ÈçͼÖÐ
A
A
£¨ÌîA»òB£©Ëùʾ£¬ÄãÅжϵÄÀíÓÉÊÇ
¼×ÍéµÄÃܶÈСÓÚ¿ÕÆø
¼×ÍéµÄÃܶÈСÓÚ¿ÕÆø
£®
£¨3£©ÌìÈ»ÆøµÄÖ÷Òª³É·ÖÈôÔÚú¿óµÄ¿ó¾®Àï´ïµ½Ò»¶¨µÄŨ¶È£¬ÓöÃ÷»ð¾Í»á·¢Éú±¬Õ¨£¬³£³ÆΪÍß˹±¬Õ¨£®ÎªÁË·ÀÖ¹Íß˹±¬Õ¨£¬Ãº¿óµÄ¿ó¾®ÄÚ±ØÐë²ÉÈ¡µÄ°²È«´ëÊ©ÊÇ
±£³Öͨ·ç£¨»ò±ÜÃâ²úÉú»ð»¨»ò°²Öñ¨¾¯×°Öõȣ©
±£³Öͨ·ç£¨»ò±ÜÃâ²úÉú»ð»¨»ò°²Öñ¨¾¯×°Öõȣ©
£®
ÄÜÔ´ºÍ»·¾³Êǵ±½ñÈËÀàÃæÁÙµÄÁ½´óÎÊÌ⣮Ŀǰ£¬»¯Ê¯È¼ÁÏÊÇÈËÀàÉú²ú¡¢Éú»îµÄÖ÷ÒªÄÜÔ´£®Ëæ×ÅÈ«ÇòÄÜԴʹÓÃÁ¿µÄÔö³¤¼°²»¿ÆѧʹÓ㬻¯Ê¯È¼ÁϵȲ»¿ÉÔÙÉúÄÜÔ´½«ÈÕÒæ¿Ý½ß£¬²¢¶Ô»·¾³²úÉúÑÏÖØÓ°Ï죮Õâ¾ÍÆÈÇÐÒªÇóÈËÃÇ¿ª·¢ÇâÄÜ¡¢Ì«ÑôÄܵÈÐÂÄÜÔ´£®
£¨1£©ÇâÄÜÀ´Ô´¹ã·º£¬ÊǸßÄÜÇå½àÄÜÔ´£®ÇâÆø³ýÁË¿ÉÖ±½ÓȼÉÕ»ñµÃÈÈÁ¿Í⣬»¹¿ÉÉè¼Æ³ÉȼÁϵç³Ø£®Ä¿Ç°Ê¹ÓÃÇâÄÜ»¹ÃæÁÙ×ÅÇâÆøµÄÖÆÈ¡ºÍÇâÆøµÄ´¢´æÁ½¸öÎÊÌ⣮ÈçͼΪģÄâÇâ¡¢ÑõȼÁϵç³ØµÄʵÑé×°Öã®X¡¢Y¾ùΪɴ²¼´ü×ö³ÉµÄÈÝÆ÷£¬×°ÓлîÐÔÌ¿£¬²¢ÓëÄÚ²¿µÄʯī°ô½ôÃܽӴ¥£¬µ¼ÏßÓëʯīÁ¬½Ó£¨Ê¯Ä«ºÍ»îÐÔÌ¿¶¼Äܵ¼µç£©£®°ÑX¡¢Y½þÔÚË®ÖУ¬²¢ÔÚË®ÖмÓһЩϡÁòËᣬÒÔÔöÇ¿Ë®µÄµ¼µçÄÜÁ¦£®µ±µç¼üKÓëb½Óͨ£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______£®Í¨µçÒ»¶Îʱ¼äÄÚûÓп´µ½ÆøÅÝÒݳö£¬ÊÇÒòΪ»îÐÔÌ¿¾ßÓÐ______ÐÔ£¬´Ó¶ø½â¾öÁËÆøÌå´¢´æµÄÎÊÌ⣮´Ëʱ½«µç¼üKÓëa½Óͨ£¬µÆÅÝ·¢¹â£¬ËµÃ÷ÄÜÁ¿·¢ÉúÁËÈçÏÂת»¯£ºÇâÄÜ¡ú______ÄÜ¡ú¹âÄܺÍÈÈÄÜ£®
£¨2£©Ò»Ð©³£ÓÃȼÁÏȼÉÕʱµÄÈÈÖµÈçÏÂ±í£º
ȼÁÏ ÌìÈ»Æø£¨ÒÔCH4¼Æ£© Òº»¯Æø£¨ÒÔC4H10¼Æ£© ԭú£¨ÒÔC¼Æ£©
ÈÈÖµ£¨kJ/g£© Ô¼56 Ô¼50 Ô¼33
¢ÙÄ¿Ç°²»Ìᳫֱ½ÓÓÃԭú×÷ȼÁϵÄÖ÷ÒªÔ­ÒòÊÇ£ºa______£¬b______£®
¢Ú»¯Ê¯È¼ÁÏȼÉշųöµÄCO2ÊÇÒ»ÖÖÎÂÊÒÆøÌ壬´óÁ¿Ïò¿ÕÆøÖÐÅÅ·ÅCO2½«¶ÔÆøºò²úÉú¸ºÃæÓ°Ï죮ÊÔͨ¹ý¼ÆËã˵Ã÷£¬µ±ÉϱíÖÐÌìÈ»ÆøºÍÒº»¯Æø³ä·ÖȼÉÕ»ñµÃÏàµÈµÄÈÈÁ¿Ê±£¬ÄÄÒ»ÖÖȼÁϷųöµÄCO2½ÏÉÙ£¿
¾«Ó¢¼Ò½ÌÍø

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

¾«Ó¢¼Ò½ÌÍø