ÌâÄ¿ÄÚÈÝ

ÒÑÖªA¡¢B¡¢D¡¢E¡¢F¡¢GΪÖÐѧ»¯Ñ§Öг£¼ûµÄ»¯ºÏÎËüÃÇÖ®¼äÓÐÈçͼËùʾµÄת»¯¹Øϵ£¨·´Ó¦Ìõ¼þ¼°²¿·Ö²úÎïÒÑÂÔÈ¥£©¡£AΪÂÌÉ«·ÛÄ©£¬º¬H¡¢C¡¢O¡¢CuËÄÖÖÔªËØ¡£³£ÎÂÏ£¬DΪÎÞÉ«ÎÞζÆøÌ壬BΪºÚÉ«·ÛÄ©£¬E·Ö×ӽṹÖк¬ÓÐÈ©»ù¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©D¸úG·´Ó¦µÄ»¯Ñ§·½³Ìʽ                                                ¡£

£¨2£©FÖÐÒ»¶¨º¬ÓеĹÙÄÜÍÅÃû³Æ               ¡£

£¨3£©Ä³¿ÎÍâС×éÉè¼ÆÁËÏÂÁÐʵÑé×°Öã¬Í¨¹ý²â¶¨×°Öü׺ÍÒÒÖÐÊÔ¼ÁµÄÖÊÁ¿±ä»¯£¬Ì½¾¿AµÄ»¯Ñ§Ê½¡£

¢ÙΪʹÊý¾Ý׼ȷ£¬»¹Ðè²¹³ä×°Öã¬ÇëÄãÔÚ·½¿òÄÚ»æ³ö×°ÖÃͼ²¢Ð´³öÊÔ¼ÁÃû³Æ¡£

¢ÚÏò×°ÖÃÖйÄÈë¿ÕÆøµÄÄ¿µÄÊÇ                         £»±û×°ÖÃÖÐÒ©Æ·µÄÃû³ÆÊÇ           £¬ÊµÑéʱ£¬¸ÃҩƷδ¼ûÃ÷ÏԱ仯£¬Ö¤Ã÷                            ¡£

¢ÛÈçºÎÅжÏAÒÑÍêÈ«·Ö½â£¿                                                 ¡£

¢ÜʵÑé²âµÃ³öÈçÏÂÊý¾Ý£ºAÊÜÈȺóÍêÈ«·Ö½â£¬¹ÌÌåÓÉ8.0 g±äΪ6.0 g£¬×°ÖÃÒÒÔöÖØ0.9 g¡£Ð´³öAµÄ»¯Ñ§Ê½£¨±íʾΪ¼îʽÑΣ©£º                                            ¡£

£¨1£©2Na2O2£«2CO2£½2Na2CO3£«O2£¨2·Ö£©

£¨2£©ôÇ»ù£¨2·Ö£©

£¨3£©¢Ù»ò £¨2·Ö£©

¢Ú ½«A·Ö½â²úÉúµÄË®ÕôÆøËÍÈëÊ¢ÓÐŨÁòËáµÄÏ´ÆøÆ¿ÖУ¨2·Ö£©£»ÎÞË®ÁòËáÍ­£¨2·Ö£©£»A·Ö½â²úÉúµÄË®ÕôÆøÈ«²¿±»Å¨ÁòËáÎüÊÕ¡££¨2·Ö£©

¢Û Á¬ÐøÁ½´Î¼ÓÈÈ¡¢¹ÄÆø¡¢ÀäÈ´£¬³ÆÁ¿¼××°ÖõÄÖÊÁ¿£¬ÖÊÁ¿²î²»³¬¹ý0.1 g £¨2·Ö£©

¢ÜCu3(OH)4CO3»òCuCO3¡¤2Cu(OH)2£¨2·Ö£©

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÒÑÖªA¡¢B¡¢D¡¢EËÄÖÖÎïÖÊÖоùº¬ÓÐͬһÖÖ¶ÌÖÜÆÚÔªËØ£¬¸ÃÔªËØÔ­×ÓµÄ×îÍâ²ãµç×ÓÊýÊÇÄÚ²ãµç×ÓÊýµÄ3±¶£¬DΪÆøÌåµ¥ÖÊ£¬EΪºÚÉ«·ÛÄ©£®Èçͼת»¯¹ØϵÖÐE¾ùÆð´ß»¯×÷Óã¨Ä³Ð©²úÎïÒÑÂÔÈ¥£©£®Ôò£º
£¨1£©AµÄ»¯Ñ§Ê½Îª
KClO3
KClO3
£»B·Ö×ӵĵç×ÓʽΪ
£»
£¨2£©ÒÑÖªEÓ뺬AÖÐijԪËصÄŨËáÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦£¬²úÉúÒ»ÖÖº¬ÓиÃÔªËصÄÆøÌåX£®¼×ͬѧΪ̽¾¿¸ÃÆøÌåµÄÐÔÖÊ£¬Éè¼ÆÁËÈçͼװÖã¨IIIÖмгÖ×°ÖÃÒÑÂÔÈ¥£©£®

¢ÙIÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ
MnO2+4H++2Cl-
  ¡÷  
.
 
Mn2++Cl2¡ü+2H2O
MnO2+4H++2Cl-
  ¡÷  
.
 
Mn2++Cl2¡ü+2H2O
£»
¢ÚʵÑ鿪ʼ²»¾Ã£¬¹Û²ìµ½×°ÖÃIVÖеÄÏÖÏóΪ
ÈÜÒº±äÀ¶
ÈÜÒº±äÀ¶
£»
¢ÛʵÑé½áÊøºó£¬¸ÃͬѧÔÚ×°ÖÃIIIÖй۲쵽bµÄºìÉ«ÍÊÈ¥£¬¶øûÓгöÏÖ¡°aÎÞÃ÷ÏÔÏÖÏó¡±ÕâÒ»Ô¤ÆÚÏÖÏó£®Îª´ïµ½ÊµÑéÄ¿µÄ£¬ËûÔÚ×°ÖÃ
×°ÖÃIIºÍIII
×°ÖÃIIºÍIII
£¨Ìî×°ÖÃÐòºÅ£©Ö®¼äÓÖÌí¼ÓÁËÏ´ÆøÆ¿£¬¸Ã×°ÖõÄ×÷ÓÃÊÇ
³ýÈ¥ÂÈÆøÖеÄË®ÕôÆø£¨»ò£º¸ÉÔïÂÈÆø£©
³ýÈ¥ÂÈÆøÖеÄË®ÕôÆø£¨»ò£º¸ÉÔïÂÈÆø£©
£»
£¨3£©½«ÆøÌåXͨÈëµ½×ÏÉ«µÄʯÈïÊÔÒºÖУ¬¹Û²ìµ½µÄÏÖÏóÊÇ
×ÏÉ«µÄʯÈïÊÔÒºÏȱäºìºóÍÊÉ«
×ÏÉ«µÄʯÈïÊÔÒºÏȱäºìºóÍÊÉ«
£»
£¨4£©ÆøÌåXÄÜÓëÈÜÒºB·´Ó¦£¬Éú³ÉÆøÌåD£¬ÊÔд³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
Cl2+H2O2=2HCl+O2
Cl2+H2O2=2HCl+O2
£®
ÒÑÖªA¡¢B¡¢D¡¢EËÄÖÖÎïÖÊÖоùº¬ÓÐͬһÖÖ¶ÌÖÜÆÚÔªËØ£¬DΪÆøÌåµ¥ÖÊ£¬AÑæÉ«³Êdz»ÆÉ«£¨Í¸¹ýÀ¼É«îܲ£Á§£©£¬EΪºÚÉ«·ÛÄ©£¬ÔÚÏÂÊöת»¯ÖÐE¾ùÆð´ß»¯×÷Óã¨Ä³Ð©²úÎïÒÑÂÔÈ¥£©£¨Èçͼ1£©£®
¾«Ó¢¼Ò½ÌÍø
Çë»Ø´ð£º
£¨1£©AµÄÃû³ÆΪ
 
£»BµÄ½á¹¹Ê½Îª
 
£»
£¨2£©ÒÑÖªAÓëijËᣨº¬AÖÐÔªËØM£©ÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦£¬²úÉúÆøÌåµ¥ÖÊX£¨º¬ÔªËØM£©£®XÄÜÓëÒºÌåB·´Ó¦Éú³ÉÆøÌåD£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
 
£»
£¨3£©Ä³Ð£»¯Ñ§ÐËȤС×éΪÑо¿Xµ¥ÖʵÄÐÔÖÊ£¬Éè¼ÆÈçͼ2ËùʾװÖýøÐÐʵÑ飮װÖâóÖмгÖ×°ÖÃÒÑÂÔÈ¥£¬ÆäÖÐaΪ¸ÉÔïµÄÆ·ºìÊÔÖ½£¬bΪʪÈóµÄÆ·ºìÊÔÖ½£®
¢Ù¼ÓÈëÒ©Æ·Ç°£¬¼ì²é¢ñÖÐÆøÌå·¢Éú×°ÖÃÆøÃÜÐԵIJÙ×÷ÊÇ£ºÓÃֹˮ¼Ð¼ÐסC´¦£¬È»ºóÏò·ÖҺ©¶·ÖмÓË®£¬´ò¿ª
 
£¬Ò»¶Îʱ¼äºó
 
£¬ÔòÆøÃÜÐÔÁ¼ºÃ£»
¢Úд³ö×°ÖÃIÖз¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ
 
£»
¢ÛʵÑé¹ý³ÌÖУ¬×°ÖâôÖеÄʵÑéÏÖÏóΪ
 
£»
¢ÜʵÑé½áÊøºó£¬¸Ã×éͬѧÔÚ×°ÖâóÖй۲쵽bµÄºìÉ«ÍÊÈ¥£¬µ«ÊDz¢Î´¹Û²ìµ½¡°aÎÞÃ÷ÏԱ仯¡±ÕâÒ»Ô¤ÆÚÏÖÏó£®ÎªÁË´ïµ½ÕâһʵÑéÄ¿µÄ£¬ÄãÈÏΪӦÓÐ
 
Ö®¼ä»¹ÐèÌí¼ÓÏ´ÆøÆ¿£¨Ñ¡Ìî×°ÖÃÐòºÅ£©£¬¸Ã×°ÖõÄ×÷ÓÃÊÇ
 
£®

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

¾«Ó¢¼Ò½ÌÍø