ÌâÄ¿ÄÚÈÝ

¢ñ.Óõ¼Ïß½«A¡¢BÁ½×°ÖõÄËĸöµç¼«½øÐÐÁ¬½Ó£¬Ê¹a¼«Îö³öÍ­¡£»Ø´ðÓйØÎÊÌâ¡£

£¨1£©µ¼Ïß½«ÈÝÆ÷AºÍBÁ¬½Óʱ£¬Zn½Ó £¬Cu½Ó  £¨Ìî¡°a¡±»ò¡°b¡±£©
£¨2£©ÈÝÆ÷AÖÐCu¼«·¢ÉúµÄµç¼«·´Ó¦Îª                                ¡£
£¨3£©B×°ÖýР                 £¬ÈÜÒºÖеÄNO3-Ïò_____¼«Òƶ¯£¨Ìî¡°a¡±»ò¡°b¡±£©¡£
£¨4£©Èôb¼«¹Û²ìµ½ÓÐÎÞÉ«ÎÞζÆøÅݲúÉú, ¾­¹ýÒ»¶Îʱ¼äºó£¬Í£Ö¹·´Ó¦²¢½Á°è¾ùÔÈ£¬ÈÜÒºµÄpHÖµ½«       £¨Ìî¡°Éý¸ß¡±¡¢¡°½µµÍ¡±»ò¡°²»±ä¡±£©£¬¼ÓÈëÒ»¶¨Á¿µÄ     ºó£¨Ìѧʽ£©£¬ÈÜÒºÄָܻ´ÖÁÓ뷴ӦǰÍêÈ«Ò»Ö¡£Èô·´Ó¦Ò»¶Îʱ¼äºó£¬²âµÃÈÜÒºÖÐCu2+Ũ¶ÈûÓÐÃ÷ÏÔϽµ£¬¿ÉÄܵÄÔ­ÒòÊÇ£º                               ¡£
¢ò.³´²ËµÄÌú¹øûÓÐÏ´¸É¾»ÈÝÒ×ÉúÐâ¡£ÓñØÒªµÄÎÄ×ÖºÍÓйػ¯Ñ§·½³Ìʽ˵Ã÷ÌúÐâÊÇÈçºÎÐγɵĠ                                                             
                                                                       
                                                                        
                                                                                                                                              ¡£
¢ñ.£¨1£©a£¬b £¨2£©2H+ + 2e- = H2¡ü£¨3£©µç½â³Ø£¬b£¨»òÑô£©
£¨4£©½µµÍ£¬CuO b¼«²ÄÁÏΪCu
¢ò.Ìú¹øÊÇÉúÌúÖÆÔìµÄ£¬ÉúÌúÖеÄFe¡¢CºÍ³´²Ëδϴ¾»µÄʳÑÎË®¹¹³ÉÔ­µç³Ø£¨1·Ö£©£¬Ê¹Ìú·¢Éúµç»¯Ñ§¸¯Ê´Éú³ÉFe(OH)2£¬ Fe(OH)2ÔÙ±»Ñõ»¯³ÉFe(OH)3£¬Fe(OH)3²¿·ÖʧˮµÃµ½ÌúÐâ¡£¸º¼«£º2Fe - 4e- = 2Fe2+£»Õý¼«£ºO2 + 2H2O +4e- = 4OH-£»
4Fe(OH) +O2 +2H2O = 4Fe(OH)3

ÊÔÌâ·ÖÎö£ºI.A¿ÉÐγÉÔ­µç³Ø£¬Zn×÷¸º¼«£¬Cu×÷Õý¼«£¬Ê¹a¼«Îö³öÍ­£¬a¼«ÎªÒõ¼«£¬ËùÒÔaÓëZnÁ¬½Ó£¬bÓëCuÁ¬½Ó¡£AÖÐCu¼«·¢ÉúµÄ·´Ó¦Îª2H++2e-=H2¡ü¡¢Zn¼«·¢ÉúµÄ·´Ó¦ÎªZn-2e-=Zn2+¡£B×°Öýеç½â³Ø£¬ÈÜÒºÖÐNO3-ÏòÑô¼«Òƶ¯¡£µç½âÏõËáÍ­ÈÜÒºÉú³ÉÍ­¡¢ÏõËáºÍÑõÆø£¬ÈÜÒºËáÐÔÔöÇ¿£¬pH¼õС£¬Ñô¼«Îö³öÑõÆø¡¢Òõ¼«Îö³öÍ­£¬¼ÓÈëÑõ»¯Í­ÄÜʹÈÜҺŨ¶È»Ö¸´µ½·´Ó¦Ç°Å¨¶È£¬Èô·´Ó¦Ò»¶Îʱ¼äºóCu2+Ũ¶ÈûÓÐÃ÷ÏÔϽµ£¬ËµÃ÷Ñô¼«ÈܽâCuÉú³ÉCu2+¡£II.ÌúÉúÐâµÄ»úÀíÊÇÌú¡¢Ì¿¡¢Ê³ÑÎË®ÐγÉÔ­µç³Ø£¬FeΪ¸º¼«£¬Fe-2e-=Fe2+¡£Ì¿ÎªÕý¼«£¬µç¼«·´Ó¦ÎªO2+2H2O+4e-=4OH-£¬×Ü·´Ó¦Îª2Fe+O2+2H2O=2Fe(OH)2£¬Fe(OH)2±»Ñõ»¯ÎªFe(OH)3¡¢Fe(OH)3ʧˮÉú³ÉFe2O3¡¤xH2O¡£
µãÆÀ£º¸ÖÌúµç»¯¸¯Ê´·ÖÁ½ÖÖ£ºÎüÑõ¸¯Ê´ºÍÎöÇⸯʴ£¬µ±µç½âÖÊÈÜÒºËáÐÔ½Ïǿʱ£¬·¢ÉúÎöÇⸯʴ£»µ±µç½âÖÊÈÜÒº³ÊÖÐÐÔ»ò¼îÐÔʱ·¢ÉúÎüÑõ¸¯Ê´£¬
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨12·Ö£©¡°ÉúÎïÖÊ"ÊÇÖ¸ÓÉÖ²Îï»ò¶¯ÎïÉúÃüÌåÑÜÉúµÃµ½µÄÎïÖʵÄ×ܳơ£¡°ÉúÎïÖÊÄÜ¡±ÊÇÈËÀà½â¾öÄÜԴΣ»úµÄÒ»ÖÖÖØÒªµÄÄÜÔ´¡£¡°ÉúÎïÖÊ¡±ÔÚÒ»¶¨Ìõ¼þÏ¿ÉÆø»¯ÎªCO¡¢H2µÈÔ­ÁÏÆø£¬½ø¶øºÏ³É¼×´¼ºÍ¶þ¼×ÃÑ£¨CH3OCH3),Ïà¹Ø·´Ó¦Îª£º
 
Çë¸ù¾Ýͼʾ»Ø´ðÎÊÌ⣺

(1) ÓÉH2ºÍCOºÏ³É¶þ¼×ÃѵÄÈÈ»¯Ñ§·½³ÌʽÊÇ                                      ¡£                      
(2) ÔÚT1¡¢T2ζÈÏ£¬ÏòÁ½¸öÈÝ»ýÏàͬµÄÃܱÕÈÝÆ÷ÖзֱðͨÈë1molCOºÍ2molH2ºÏ³É¼×´¼£¬ÔòT1¡¢T2ζÈ϶ÔÓ¦·´Ó¦µÄƽºâ³£ÊýK1_______K2(Ñ¡Ìî¡°<¡±¡¢¡°>¡±»ò¡°=¡±)¡£
(3) ÔÚÒ»¶¨Ìõ¼þÏ£¬ÏòÒ»¸öÈÝ»ý¿É±äµÄÃܱÕÈÝÆ÷ÖгäÈë4molH2¡¢2molCO¡¢1molCH3OCH3£¨g£©ºÍ1molH20(g),¾­Ò»¶¨Ê±¼ä·´Ó¦¢Ú´ïµ½Æ½ºâ״̬£¬´Ëʱ²âµÃ»ìºÏÆøÌåÃܶÈÊÇÏàͬÌõ¼þÏÂÆðʼʱµÄ1.6±¶¡£·´Ó¦¿ªÊ¼Ê±Õý¡¢Äæ·´Ó¦ËÙÂʵĴóС¹ØϵΪV(Õý)_______V(Ä棩£¨Ñ¡Ìî¡° >¡±¡¢¡°< ¡±»ò¡°=¡±)£¬Æ½ºâʱ=_________mol¡£
(4) ÓÒͼΪÂÌÉ«µçÔ´¡°¶þ¼×ÃÑȼÁϵç³Ø¡±µÄ¹¤×÷Ô­ÀíʾÒâͼ¡££¨a¡¢b¾ùΪ¶à¿×ÐÔPtµç¼«£©

bµç¼«ÊÇ_______¼«£º£¨Ìî¡°Õý¡±»ò¡°¸º¡±£©
aµç¼«Éϵĵ缫·´Ó¦Îª___         ____¡£

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

¾«Ó¢¼Ò½ÌÍø