ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÒÑÖªH£«¿ÉÒÔд³ÉH3O+£¬´ÓijЩÐÔÖÊ¿´£¬NH3ºÍH2O£¬ºÍH3O£«£¬OH£­ºÍ£¬N3£­ºÍO2£­Á½Á½ÏàËÆ£¬¾Ý´ËÅжÏÏÂÁз´Ó¦Ê½(·´Ó¦Ìõ¼þºöÂÔ)ÕýÈ·µÄÊÇ(¡¡¡¡)

¢Ù2Na£«2NH3=2NaNH2£«H2¡ü

¢ÚCaO£«2NH4Cl=CaCl2£«2NH3¡ü£«H2O

¢Û3Mg(NH2)2=Mg3N2£«4NH3¡ü

¢ÜNH4Cl£«NaNH2=NaCl£«2NH3¡ü

A.¢ÙB.¢Ú¢ÜC.È«²¿D.¢Ù¢Ú¢Ü

¡¾´ð°¸¡¿C

¡¾½âÎö¡¿

´ÓijЩÐÔÖÊ¿´£¬NH3ºÍH2O£¬NH4+ºÍH3O+£¬OH-ºÍNH2-£¬N3-ºÍO2-Á½Á½ÏàËÆ£¬¸ù¾ÝÐÅÏ¢ÀûÓÃÀàÍÆ·½·¨·ÖÎöÅжϡ£

¢ÙNH3ºÍH2OÏàËÆ£¬¿ÉÒÔÒÀ¾Ý2Na+2H2O¨T2NaOH+H2¡ü£¬ÀàÍƵõ½·´Ó¦Îª£º2Na+2NH3¨T2NaNH2+H2¡ü£¬¹Ê¢ÙÕýÈ·£»

¢ÚNH4+ºÍH3O+ÏàËÆ£¬H£«¿ÉÒÔд³ÉH3O+£¬ÒÀ¾ÝCaO+2HCl=CaCl2+H2O£¬ÀàÍƵõ½·´Ó¦£ºCaO+2NH4Cl=CaCl2+2NH3¡ü+H2O£¬¹Ê¢ÚÕýÈ·£»

¢ÛOH-ºÍNH2-ÏàËÆ£¬N3-ºÍO2-ÏàËÆ£¬NH3ºÍH2OÏàËÆ£¬ÒÀ¾ÝMg(OH)2=MgO+H2O£¬ÀàÍƵõ½£º3Mg(NH2)2=Mg3N2+4NH3¡ü£¬¹Ê¢ÛÕýÈ·£»

¢ÜOH-ºÍNH2-ÏàËÆ£¬NH3ºÍH2OÏàËÆ£¬ÒÀ¾ÝNH4Cl+NaOH=NaCl+NH3+H2O£¬ÀàÍƵõ½£ºNH4Cl+NaNH2¨TNaCl+2NH3£¬¹Ê¢ÜÕýÈ·£»

¹ÊÑ¡C¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿Ä³³ÇÊжԴóÆø½øÐмà²â£¬·¢ÏÖ¸ÃÊÐÊ×ÒªÎÛȾÎïΪ¿ÉÎüÈë¿ÅÁ£ÎïPM2.5(Ö±¾¶Ð¡ÓÚµÈÓÚ2.5¦ÌmµÄÐü¸¡¿ÅÁ£Îï)£¬ÆäÖ÷ÒªÀ´Ô´ÎªÈ¼Ãº¡¢»ú¶¯³µÎ²ÆøµÈ¡£Òò´Ë£¬¶ÔPM2.5¡¢SO2¡¢NOxµÈ½øÐÐÑо¿¾ßÓÐÖØÒªÒâÒå¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©PM2.5·ÖÉ¢ÔÚ¿ÕÆøÖÐÐγɵķÖɢϵ__(Ìî¡°ÊôÓÚ¡±»ò¡°²»ÊôÓÚ¡±)½ºÌå¡£

£¨2£©½«PM2.5Ñù±¾ÓÃÕôÁóË®´¦ÀíÖƳɴý²âÊÔÑù¡£Èô²âµÃ¸ÃÊÔÑùËùº¬Ë®ÈÜÐÔÎÞ»úÀë×ӵĻ¯Ñ§×é·Ö¼°Æäƽ¾ùŨ¶ÈÈçÏÂ±í£º

¸ù¾Ý±íÖÐÊý¾ÝÅжϴý²âÊÔÑùΪ__(Ìî¡°Ëᡱ»ò¡°¼î¡±)ÐÔ£¬±íʾ¸ÃÊÔÑùËá¼îÐÔµÄc(H£«)»òc(OH£­)=__mol¡¤L-1¡£

£¨3£©ÃºÈ¼ÉÕÅŷŵÄÑÌÆøº¬ÓÐSO2ºÍNOx£¬ÐγÉËáÓ꣬ÎÛȾ´óÆø£¬²ÉÓÃNaClO2ÈÜÒºÔÚ¼îÐÔÌõ¼þÏ¿ɶÔÑÌÆø½øÐÐÍÑÁò£¬ÍÑÏõ£¬Ð§¹û·Ç³£ºÃ¡£Íê³ÉÏÂÁжÔÑÌÆøÍÑÏõ¹ý³ÌµÄÀë×Ó·½³Ìʽ¡£

__ClO2-£«__NO£«__=__Cl-£«__NO3-£«__

£¨4£©Îª¼õÉÙSO2µÄÅÅ·Å£¬³£²ÉÈ¡µÄ´ëÊ©ÓУº

¢Ù½«Ãº×ª»¯ÎªÇå½àÆøÌåȼÁÏ¡£Ð´³ö½¹Ì¿ÓëË®ÕôÆø·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º__¡£

¢ÚÏ´µÓº¬SO2µÄÑÌÆø¡£ÒÔÏÂÎïÖÊ¿É×öÏ´µÓ¼ÁµÄÊÇ__(Ìî×Öĸ)¡£

a.Ca(OH)2 b.Na2CO3 c.CaCl2 d.NaHSO3

£¨5£©Æû³µÎ²ÆøÖÐNOxºÍCOµÄÉú³É¼°×ª»¯¡£

¢ÙÆû³µÆô¶¯ºó£¬Æû¸×ζÈÔ½¸ß£¬µ¥Î»Ê±¼äÄÚNOÅÅ·ÅÁ¿Ô½´ó£¬Ð´³öÆû¸×ÖÐÉú³ÉNOµÄ»¯Ñ§·½³Ìʽ£º__¡£

¢ÚÆû³µÈ¼ÓͲ»ÍêȫȼÉÕʱ²úÉúCO£¬Ä¿Ç°£¬ÔÚÆû³µÎ²ÆøϵͳÖÐ×°Öô߻¯×ª»¯Æ÷¿É¼õÉÙCOºÍNOµÄÎÛȾ£¬Æ仯ѧ·´Ó¦·½³ÌʽΪ__¡£

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

¾«Ó¢¼Ò½ÌÍø