ÌâÄ¿ÄÚÈÝ

12£®Á¼ºÃµÄÉú̬»·¾³¿ÉÒÔÌáÉýÉú»îÖÊÁ¿£®
£¨1£©2013Äê1Ô£¬È«¹ú¶à¸öÊ¡ÊгöÏÖÑÏÖØµÄÎíö²ÌìÆø£®µ¼ÖÂÎíö²ÐγɵÄÖ÷ÒªÎÛȾÎïÊÇc
a£®SO2                  b£®NO2                 c£®PM2.5
£¨2£©À¬»øÎÞº¦»¯´¦ÀíÓÐÀûÓÚ±£»¤Éú̬»·¾³£®ÏÂÁÐÓйØÉú»îÀ¬»øµÄ´¦ÀíºÏÀíµÄÊÇa
a£®ÓóøÓàÀ¬»øÉú²úÕÓÆø    b£®·Ï¾ÉËÜÁ϶Ìì·ÙÉÕ    c£®·Ï¾Éµç³Ø¾ÍµØÌîÂñ
£¨3£©Ñ¡Ôñ»·±£µÄ×°ÐÞ²ÄÁÏ¿ÉÒÔÓÐЧµØ¼õÉÙ¾ÓÊÒÎÛȾ£®ÁÓÖʽººÏ°åÊͷųöµÄÖ÷ÒªÎÛȾÎïÊÇc
a£®NO                   b£®CO                  c£®¼×È©
£¨4£©ÎÒ¹úÉú»îÀ¬»øÒ»°ã¿É·ÖΪÒÔÏÂËÄ´óÀࣺ¿É»ØÊÕÀ¬»ø¡¢³øÓàÀ¬»ø¡¢Óк¦À¬»øºÍÆäËûÀ¬»ø£®ÒÔÏÂÎïÖÊÊôÓÚ¿É»ØÊÕÀ¬»øµÄÊÇa
a£®·Ï±¨Ö½                b£®ÔüÍÁ                  c£®¹ýÆÚÒ©Æ·£®

·ÖÎö £¨1£©PM2.5¡±ÊÇÖ¸´óÆø²ãÖÐÖ±¾¶¡Ý2.5¦ÌmµÄ¿ÅÁ£ÎÄܱ»·ÎÎüÊÕ²¢½øÈëѪҺ£¬¶ÔÈËÌåΣº¦ºÜ´ó£¬ÊÇÐγÉÎíö²µÄÖ÷ÒªÎÛȾÎ
£¨2£©ÓóøÓàÀ¬»øÉú²úÕÓÆø ±ä·ÏΪ±¦£»·Ï¾ÉËÜÁ϶Ìì·ÙÉÕ²úÉúÓж¾Óк¦ÆøÌ壬·Ï¾Éµç³ØÖк¬ÓÐÖØ½ðÊô£¬¾ÍµØÌîÂñ»á¶ÔË®ÍÁÔì³ÉÑÏÖØÎÛȾ£»
£¨3£©ÁÓÖʽººÏ°åÊͷŵÄÖ÷ÒªÎÛȾÎïÊǼ×È©£¬ÊͷźóÔì³ÉÊÒÄÚÎÛȾ£»
£¨4£©¿É»ØÊÕÀ¬»ø¾ÍÊÇ¿ÉÒÔÔÙÉúÑ­»·µÄÀ¬»ø£¬±¾Éí»ò²ÄÖÊ¿ÉÔÙÀûÓõÄÖ½Àà¡¢Ó²Ö½°å¡¢²£Á§¡¢ËÜÁÏ¡¢½ðÊô¡¢ÈËÔìºÏ³É²ÄÁϰü×°£®

½â´ð ½â£º£¨1£©µ¼ÖÂÎíö²ÐγɵÄÖ÷ÒªÎÛȾÎïÊÇpM2.5£¬¹ÊÑ¡c£»
£¨2£©ÓóøÓàÀ¬»øÉú²úÕÓÆø ±ä·ÏΪ±¦£»·Ï¾ÉËÜÁ϶Ìì·ÙÉÕ²úÉúÓж¾Óк¦ÆøÌ壬·Ï¾Éµç³ØÖк¬ÓÐÖØ½ðÊô£¬¾ÍµØÌîÂñ»á¶ÔË®ÍÁÔì³ÉÑÏÖØÎÛȾ£¬¹ÊÑ¡a£»
£¨3£©¼×È©ÊÇÁÓÖʽººÏ°åÊͷŵÄÖ÷ÒªÎÛȾÎÔì³ÉÊÒÄÚÎÛȾ£¬¹ÊÑ¡c£»
£¨4£©·Ï±¨Ö½Äܹ»»ØÊÕ´¦Àíºó»¹¿ÉÒÔ×÷ÎªÖÆÖ½µÄÔ­ÁÏ£¬¿ÉÒÔ»ØÊÕ£»ÔüÍÁÖ»Êǽ¨ÖþÀ¬»øµÄÒ»ÖÖ£»¹ýÆÚÒ©Æ·ÊôÓÚÓк¦À¬»ø£¬¹ÊÑ¡a£®

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éÁ˳£¼ûµÄÎÛȾÎÓëÉú²úÉú»îÃÜÇÐÏà¹Ø£¬ÄѶȲ»´ó£¬×¢Òâ֪ʶµÄ»ýÀÛ£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
2£®Êª·¨Á¶Ð¿µÄÒ±Á¶¹ý³Ì¿ÉÓÃÈçͼ1¼òÂÔ±íʾ£º

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©NH3µÄ¿Õ¼ä¹¹ÐÍÊÇÈý½Ç×¶ÐΣ®°±ÆøÒ×Òº»¯£¬Òº°±³£×öÖÆÀä¼Á£¬°±ÆøÒ×Òº»¯µÄÔ­ÒòÊǰ±·Ö×ÓÖ®¼äÄÜÐγÉÇâ¼ü£®
£¨2£©ÒÑÖªZnOÊôÓÚÁ½ÐÔÑõ»¯Îд³öZnOÓëNaOHÈÜÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º2NaOH+ZnO=Na2ZnO2+H2O£®
£¨3£©ÉÏÊöµç½â¹ý³ÌÖÐÎö³öпµÄµç¼«·´Ó¦Ê½Îª[Zn£¨NH3£©4]2++2e-=Zn+4NH3¡ü£®
£¨4£©²úÉúµÄSO2¿ÉÓÃBa£¨NO3£©2ÈÜÒºÎüÊÕ£¬²¿·Ö²úÎï¿É×÷Ϊ¹¤ÒµÔ­ÁÏ£¬Æä·´Ó¦µÄÀë×Ó·½³ÌʽΪ3SO2+2H2O+2NO3-+3Ba2+=3BaSO4¡ý+2NO¡ü+4H+£®
£¨5£©°±ÊÇ×îÖØÒªµÄ»¯¹¤²úÆ·Ö®Ò»£®ºÏ³É°±ÓõÄÇâÆø¿ÉÒÔ¼×ÍéΪԭÁÏÖÆµÃ£º
CH4£¨g£©+H2O£¨g£©?CO£¨g£©+3H2£¨g£©£®Óйػ¯Ñ§·´Ó¦µÄÄÜÁ¿±ä»¯ÈçͼËùʾ£¬ÔòCH4£¨g£©ÓëH2O£¨g£©·´Ó¦Éú³ÉCO£¨g£©ºÍH2£¨g£©µÄÈÈ»¯Ñ§·½³ÌʽΪCH4£¨g£©+H2O£¨g£©=CO£¨g£©+3H2£¨g£©¡÷H=+161.1kJ•mol-1£®
£¨6£©CO¶ÔºÏ³É°±µÄ´ß»¯¼ÁÓж¾º¦×÷Ó㬳£ÓÃÒÒËá¶þ°±ºÏÍ­£¨¢ñ£©ÈÜÒºÀ´ÎüÊÕÔ­ÁÏÆøÖÐCO£¬Æä·´Ó¦Ô­ÀíΪ£º£¨l£©+CO£¨g£©+NH3£¨g£©?CH3COO•CO£¨l£©¡÷H£¼0£®ÎüÊÕCOºóµÄÒÒËáÍ­°±Òº¾­¹ýÊʵ±´¦ÀíºóÓÖ¿ÉÔÙÉú£¬»Ö¸´ÆäÎüÊÕCOµÄÄÜÁ¦ÒÔ¹©Ñ­»·Ê¹Óã¬ÔÙÉúµÄÊÊÒËÌõ¼þÊÇB£¨ÌîдѡÏî±àºÅ£©£®
A£®¸ßΡ¢¸ßѹ  B£®¸ßΡ¢µÍѹ C£®µÍΡ¢µÍѹ  D£®µÍΡ¢¸ßѹ
£¨7£©Óð±ÆøÖÆÈ¡ÄòËØµÄ·´Ó¦Îª£º2NH3£¨g£©+CO2£¨g£©?CO£¨NH2£©2£¨l£©+H2O£¨g£©¡÷H£¼0£¬ºãκãÈÝÃܱÕÈÝÆ÷ÖУ¬ÏÂÁÐÒÀ¾ÝÄÜ˵Ã÷¸Ã·´Ó¦´ïµ½Æ½ºâ״̬µÄÊÇABDE£®
£¨1£©ÈÝÆ÷ÖÐÆøÌåÃܶȲ»±ä
B¡¢ÈÝÆ÷ÖÐÆøÌåѹǿ²»±ä
C¡¢n£¨NH3£©£ºn£¨CO2£©=1£º2
D¡¢µ¥Î»Ê±¼äÄÚÏûºÄ1molCO2£¬Í¬Ê±ÏûºÄ1molH2O
E¡¢ÈÝÆ÷ÄÚζȱ£³Ö²»±ä
£¨8£©Ä³Î¶ÈÏ£¬ÏòÈÝ»ýΪ100LµÄÃܱÕÈÝÆ÷ÖÐͨÈë4mol NH3ºÍ2molCO2£¬·¢Éú2NH3£¨g£©+CO2£¨g£©?CO£¨NH2£©2£¨l£©+H2O£¨g£©·´Ó¦£¬¸Ã·´Ó¦½øÐе½40sʱ´ïµ½Æ½ºâ£¬´ËʱCO2µÄת»¯ÂÊΪ50%£®¸ÃζÈÏ´˷´Ó¦Æ½ºâ³£ÊýKµÄֵΪ2500£®Í¼2ÖеÄÇúÏß±íʾ¸Ã·´Ó¦ÔÚǰ25sÄڵķ´Ó¦½ø³ÌÖеÄNH3Ũ¶È±ä»¯£®Èô·´Ó¦ÑÓÐøÖÁ70s£¬±£³ÖÆäËüÌõ¼þ²»±äÇé¿öÏ£¬ÇëÔÚͼ3ÖÐÓÃʵÏß»­³öʹÓô߻¯¼Áʱ¸Ã·´Ó¦´Ó¿ªÊ¼ÖÁƽºâʱµÄÇúÏߣ®

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

¾«Ó¢¼Ò½ÌÍø