ÌâÄ¿ÄÚÈÝ

ÒÑÖªKNO3¡¢Cu(NO3)2¡¢AgNO3ÈýÖÖÏõËáÑÎÈÈ·Ö½âµÄ·´Ó¦·½³ÌʽÈçÏÂ(Ìõ¼þ¶¼ÊǼÓÈÈ)£º
¢Ù2KNO3=2KNO2+O2¡ü
¢Ú2Cu(NO3)2=2CuO+4NO2¡ü+O2¡ü
¢Û2AgNO3=2Ag+2NO2¡ü+O2¡ü
½ñÓÐij¹ÌÌå¿ÉÄÜÓÉÉÏÊöÈýÖÖÏõËáÑÎÖеÄÒ»ÖÖ»ò¼¸ÖÖ×é³É¡£È¡ÊÊÁ¿¸Ã¹ÌÌå³ä·Ö¼ÓÈÈ£¬µÃµ½Ò»¶¨Á¿µÄÆøÌ壬½«ÕâЩÆøÌåͨÈëË®ÖУ¬½á¹ûÈ«²¿±»ÎüÊÕ£¬Ã»ÓÐÊ£Ó࣬»Ø´ð£º
£¨1£©¸Ã¹ÌÌåÊÇ·ñ¿ÉÄÜÖ»ÓÉÒ»ÖÖÑÎ×é³É£¿Èô¿ÉÄÜ£¬Ö¸³öÊÇÄÄÖÖÑΣ¬Èô²»¿ÉÄÜ£¬ËµÃ÷Ô­Òò¡£___________________________________________________________________________________.
£¨2£©Èô¹ÌÌåÊÇ»ìºÏÎָ³öËüµÄ¿ÉÄÜ×é³É(¼´ÎïÖʵÄÁ¿Ö®±È)£¬ÈôûÓÐÕâÖÖ×é³É£¬Ò²Çë˵Ã÷__________________________________________________________________________________¡£
(1)¿ÉÄÜ£¬Cu(NO3)2·Ö½âËùµÃNO2¡¢O2ÎïÖʵÄÁ¿Ö®±ÈΪ4:1£¬Í¨ÈëË®Öз¢Éú·´Ó¦£º4NO2+O2+2H2O=4HNO3,È«²¿ÎüÊÕ¡£
(2)ûÓÐÕâÖÖ×é³É¡£ÓÉ(1)¿ÉÖª£¬Ö»Óе±NO2¡¢O2ÎïÖʵÄÁ¿Ö®±ÈΪ4:1ʱ²Å¿ÉÄܱ»ÍêÈ«ÎüÊÕ£¬ÈôΪ»ìºÏÎÔò²»¿ÉÄܳöÏÖ4:1µÄÇé¿ö,Òò´Ë²»»áÓÐÕâÖÖ×é³É.
¸½:ÁíÓÐÒ»ÖÖÌâÐÍ£­£­£­¡­¡­Èô¸ÃÆøÌå¾­Ë®³ä·ÖÎüÊÕºó£¬Ê£ÓàÆøÌåµÄÌå»ýÔÚͬÎÂͬѹÏÂΪÎüÊÕÇ°µÄ1/6£¬£¨ºóÃæÎÊÌâÏàͬ£©
£­£­£­´ð£º
£¨1£©¿ÉÄÜ£»AgNO3.
£¨2£©KNO3ºÍCu(NO3)2ÎïÖʵÄÁ¿Ö®±ÈΪ1:1£»
»òÕßÓÉÈýÖÖÑÎ×é³É£¬ÆäÖÐKNO3ºÍCu(NO3)2ÎïÖʵÄÁ¿Ö®±ÈΪ1:1£¬¶øAgNO3ÎïÖʵÄÁ¿ÊÇÈÎÒâµÄ¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
º¬µª·ÏË®½øÈëË®Ìåºó¶ø¶Ô»·¾³Ôì³ÉµÄÎÛȾԽÀ´Ô½ÑÏÖØ£¬¶Ôº¬µª·ÏË®½øÐÐÓÐЧµÄ¼ì²âºÍºÏÀíµÄ´¦ÀíÊÇÈËÃÇÑо¿ºÍ¹ØÐĵÄÖØÒªÎÊÌâ¡£
¢Å»·¾³×¨¼ÒÈÏΪ¿ÉÒÔÓýðÊôÂÁ½«Ë®ÌåÖеÄNO3£­»¹Ô­ÎªN2£¬´Ó¶øÏû³ýÎÛȾ¡£Æä·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ£º6NO3£­£«10Al£«18H2O£½3N2¡ü£«10Al(OH)3£«6OH¡ª¡£ÓÖÓÐÈËÈÏΪ½ðÊôþ±ÈÂÁÄܸü¿ìÏû³ýµªµÄÎÛȾ£¬Æä·´Ó¦Ô­ÀíºÍ½ðÊôÂÁÏàͬ¡£
д³öþºÍº¬NO3£­µÄ·ÏË®·´Ó¦µÄÀë×Ó·½³Ìʽ
____________________                     ¡£
¢ÆË®ÖеÄNO2£­ÊǺ¬µªÓлúÎï·Ö½âµÄ²úÎÆäŨ¶ÈµÄ´óСÊÇˮԴÎÛȾ³Ì¶ÈµÄ±êÖ¾Ö®Ò»¡£¼ì²âË®ÖеÄNO2£­¿ÉÓÃÄ¿ÊÓ±ÈÉ«·¨£¨±È½ÏÈÜÒºÑÕÉ«Éîdz¶ÈÒԲⶨËùº¬ÓÐÉ«ÎïÖÊŨ¶ÈµÄ·½·¨£©£¬¼ì²â²½ÖèÈçÏ£º
²½ÖèÒ»£ºÅäÖƱê×¼ÈÜÒº£º³ÆÈ¡0.69gNaNO2£¬ÈÜÓÚË®ºóÔÚÈÝÁ¿Æ¿ÖÐÏ¡ÊÍÖÁ1LµÃÈÜÒºA£¬ÒÆÈ¡5mLÈÜÒºA£¬Ï¡ÊÍÖÁ1L£¬µÃÈÜÒºB¡£
²½Öè¶þ£ºÅäÖƱê׼ɫ½×£ºÈ¡6Ö»¹æ¸ñΪ10mLµÄ±ÈÉ«¹Ü£¨¼´Öʵء¢´óС¡¢ºñ±¡ÏàͬÇÒ¾ßÓÐÈûµÄƽµ×ÊԹܣ©£¬·Ö±ð¼ÓÈëÌå»ý²»µÈµÄÈÜÒºB£¬²¢Ï¡ÊÍÖÁ10mL£¬ÔÙ¼ÓÈëÉÙÐí£¨Ô¼0.30g£©°±»ù±½»ÇËá·ÛÄ©£¬ÊµÑé½á¹ûÈçϱíËùʾ¡£
É«½×ÐòºÅ
1
2
3
4
5
6
¼ÓÈëÈÜÒºBµÄÌå»ý/mL
0.0
2.0
4.0
6.0
8.0
10.0
·´Ó¦ºóÈÜÒºÑÕÉ«
ÓÉÎÞÉ«±äΪÓÉdzµ½ÉîµÄÓ£ÌÒºìÉ«
 
²½ÖèÈý£ºÈ¡10mLË®Ñùµ¹Èë±ÈÉ«¹ÜÖУ¬¼ÓÉÙÐí°±»ù±½»ÇËᣬÏÔÉ«ºóÓë±ê׼ɫ½×¶Ô±È¡£
¢ÙÀûÓò½ÖèÈýÔÚÒ°Íâ¼ì²âË®ÑùÎÛȾ³Ì¶ÈµÄÓŵã
ÊÇ                                   ¡£
¢Ú²½Öè¶þÖÐÉè¼ÆÉ«½×ÐòºÅ1µÄ×÷ÓÃ
ÊÇ                                             ¡£
¢ÛÈç¹ûË®ÑùÏÔÉ«ºó±È6ºÅ»¹ÉӦ²ÉÈ¡µÄ´ëÊ©
ÊÇ                                   ¡£
¢ÇÏÖÓк¬NH3Ϊ3.4mg¡¤L¡ª1µÄ·ÏË®150m3£¨ÃܶÈΪ1g¡¤cm-3£©£¬¿É²ÉÓÃÈçÏ·½·¨½øÐд¦Àí£º½«¼×´¼¼ÓÈ뺬°±µÄ·ÏË®ÖУ¬ÔÚÒ»ÖÖ΢ÉúÎï×÷ÓÃÏ·¢Éú·´Ó¦£º
2O2 + NH3=== NO3£­ + H+ + H2O    6NO3£­+ 6H+ + 5CH3OH¡ú 3N2¡ü + 5CO2¡ü + 13H2O
ÈôÓô˷½·¨´¦Àí£¬¼ÙÉèÿ²½µÄת»¯ÂÊΪ100%£¬ÐèÒª¼×´¼             g¡£

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

¾«Ó¢¼Ò½ÌÍø