ÌâÄ¿ÄÚÈÝ


£¨1£©ÏÂÁи÷×éÖеÄÁ½ÖÖÓлúÎ¿ÉÄÜÊÇÏàͬµÄÎïÖÊ¡¢Í¬ÏµÎï»òͬ·ÖÒì¹¹ÌåµÈ£¬ÇëÅжÏËüÃÇÖ®¼äµÄ¹ØÏµ
¢Ù2£­¼×»ù¶¡ÍéºÍ¶¡Íé                            ¡£
¢Ú1£­ÒÑÏ©ºÍ»·ÒÑÍé                              ¡£
£¨2£©Ö§Á´Ö»ÓÐÒ»¸öÒÒ»ùÇÒʽÁ¿×îСµÄÍéÌþµÄ½á¹¹¼òʽ                              ¡£
£¨3£©Ð´³öÒÒÈ©ÈÜÒºÓë×ãÁ¿µÄÒø°±ÈÜÒº¹²ÈȵĻ¯Ñ§·½³Ìʽ£º                 £»
£¨4£©Ð´³ö1,3-¶¡¶þÏ©Óëäåµ¥ÖÊ·¢Éú1,4-¼Ó³ÉµÄ·´Ó¦·½³Ìʽ                
£¨5£©Ð´³öÓÉÒÒ´¼Ò»²½ÖÆäåÒÒÍéµÄ»¯Ñ§·½³Ìʽ                   

£¨10·Ö£©
£¨1£©¢ÙͬϵÎ1·Ö£©   ¢Úͬ·ÖÒì¹¹Ì壨1·Ö£©
£¨2£©3£¨2·Ö£©
£¨3£©CH3CHO£«2[Ag(NH3)2]£«£«2OH£­ CH3COO£­£«NH£«2Ag¡ý£«3NH3£«H2O£¨2·Ö£©
£¨4£©CH2=CH£­CH=CH2£«Br2¡úCH2BrCH=CHCH2Br£¨2·Ö£©
£¨5£©CH3CH2OH+HBr  CH3CH2Br+H2O£¨2·Ö£©

½âÎöÊÔÌâ·ÖÎö£º¡£¢Ù2£­¼×»ù¶¡Íé·Ö×ÓʽÊÇC5H12¡¢¶¡Íé·Ö×ÓʽÊÇC4H10£¬¶þÕß»¥ÎªÍ¬ÏµÎï¡£
¢Ú1£­ÒÑÏ©·Ö×ÓʽÊÇC6H12¡¢»·ÒÑÍé·Ö×ÓʽÊÇC6H12¶þÕß»¥ÎªÍ¬·ÖÒì¹¹Ìå¡£
£¨2£©Ö§Á´Ö»ÓÐÒ»¸öÒÒ»ùµÄÍéÌþ£¬ÒÒ»ùÁ¬µÄ̼ԭ×Ó±àºÅ×îСÊÇ3(´ÓÈκÎÒ»¶ËÆð¶¼ÊÇ)£¬¹ÊʽÁ¿×îСµÄÍéÌþÊÇ£º¡£
£¨3£©ÒÒÈ©ÈÜÒºÓë×ãÁ¿µÄÒø°±ÈÜÒº¹²ÈȵĻ¯Ñ§·½³Ìʽ£ºCH3CHO£«2[Ag(NH3)2]£«£«2OH£­ CH3COO£­£«NH£«2Ag¡ý£«3NH3£«H2O£»
£¨4£©1,3-¶¡¶þÏ©Óëäåµ¥ÖÊ·¢Éú1,4-¼Ó³ÉµÄ·´Ó¦·½³ÌʽCH2=CH£­CH=CH2£«Br2¡úCH2BrCH=CHCH2Br
£¨5£©ÓÉÒÒ´¼Ò»²½ÖÆäåÒÒÍéµÄ»¯Ñ§·½³ÌʽCH2=CH£­CH=CH2£«Br2¡úCH2BrCH=CHCH2Br¡£
¿¼µã£º×¼È·µÄ±í´ïÓлúÎïµÄ»¯Ñ§ÓÃÓï¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

£¨12·Ö£©ÒÒȲÓëÒÒÏ©ÀàËÆ£¬ÔÚ´ß»¯¼Á£¨10%ÁòËáºÍ5%ÁòËṯµÄË®ÈÜÒº£©µÄ×÷ÓÃÏ£¬Ò²¿ÉÒÔÓëµÈÎïÖʵÄÁ¿µÄË®·¢Éú¼Ó³É·´Ó¦¡£
£¨1£©ÒÑÖªµ±ôÇ»ùºÍË«¼ü̼ԭ×ÓÖ±½ÓÏàÁ¬£¬³ÊÏ©´¼½á¹¹²»Îȶ¨£¬»áת»¯ÎªôÊ»ù»¯ºÏÎ

ÔòÒÒȲºÍË®¼Ó³ÉµÄ»¯Ñ§·½³ÌʽÊÇ                                            ¡£
ÓÃÈçÏÂ×°ÖÃÍê³ÉÒÒȲºÍË®µÄ¼Ó³É·´Ó¦£¬¼Ó³É²úÎïÊÕ¼¯ÔÚD×°ÖÃÖС££¨Ê¡ÂÔ²¿·Ö¼ÓÈȺͼгÖÒÇÆ÷£©

¢ÙʵÑéÊÒÖÆ±¸ÒÒȲµÄ»¯Ñ§·½³ÌʽÊÇ                                             ¡£
¢ÚA×°ÖÃÖÆ³öµÄÒÒȲÓÐÄÑÎŵįøÎ¶£¬Îª³ýÈ¥ÕâЩÔÓÖÊ£¬×°ÖÃBÖÐÊ¢·ÅµÄÊÔ¼ÁÊÇ                  ¡£
¢ÛC×°ÖòÉÓ÷ÐˮԡµÄ×÷ÓÃÊÇ           £¨Ìî×Öĸ£©¡£
a£®±ãÓÚ¿ØÎ£¬Ê¹ÊÜÈȾùÔÈ    b£®¼ÓÈÈ·´Ó¦ÎÌá¸ß»¯Ñ§·´Ó¦ËÙÂÊ    c£®Õô³öÉú³ÉÎï
£¨3£©Îª¼ìÑé¼Ó³É²úÎïÖеĹÙÄÜÍÅ£¬Ñ¡ÔñµÄÒ©Æ·ÊÇ            £¨Ìî×Öĸ£©¡£
a£®½ðÊôÄÆ           b£®10%ÇâÑõ»¯ÄÆÈÜÒº      c£®2%ÁòËáÍ­ÈÜÒº
d£®2%ÏõËáÒøÈÜÒº    e£®±¥ºÍ̼ËáÄÆÈÜÒº
¼ìÑé¹ÙÄÜÍŵÄʵÑéÖУ¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ                                   ¡£

¶õ¶û¶à˹ÊÐÒÑ̽Ã÷ú̿´¢Á¿1496ÒÚ¶à¶Ö£¬Ô¼Õ¼È«¹ú×Ü´¢Á¿µÄ1/6¡£ÃºÊÇÖØÒªµÄÄÜÔ´£¬Ò²ÊÇÉú²ú»¯¹¤²úÆ·µÄÖØÒªÔ­ÁÏ¡£ÊÔÓÃËùѧ֪ʶ£¬½â´ðÏÂÁÐÎÊÌâ¡£
£¨1£©ÃºµÄת»¯¼¼Êõ°üÀ¨ÃºµÄÆø»¯¼¼ÊõºÍÒº»¯¼¼Êõ¡£ÃºµÄÒº»¯¼¼ÊõÓÖ·ÖΪ       ºÍ               ¡£
£¨2£©ÔÚúȼÉÕǰÐè¶Ôú½øÐÐÍÑÁò´¦Àí¡£ÃºµÄijÖÖÍÑÁò¼¼ÊõµÄÔ­ÀíΪ
FeS2Fe2++SO42-Fe3+
ÕâÖÖÍÑÁò¼¼Êõ³ÆÎªÎ¢ÉúÎïÍÑÁò¼¼Êõ¡£¸Ã¼¼ÊõµÄµÚÒ»²½·´Ó¦µÄÀë×Ó·½³ÌʽΪ               £¬µÚ¶þ²½·´Ó¦µÄÀë×Ó·½³ÌʽΪ                                                ¡£
£¨3£©¹¤ÒµÃº¸ÉÁóµÃµ½µÄ²úÆ·Óн¹Ì¿¡¢                                    ¡£
£¨4£©¹¤ÒµÉÏÖ÷Òª²ÉÓð±Ñõ»¯·¨Éú²úÏõËᣬÈçͼÊǰ±Ñõ»¯ÂÊÓë°±£­¿ÕÆø»ìºÏÆøÖÐÑõ°±±ÈµÄ¹ØÏµ¡£ÆäÖÐÖ±Ïß±íʾ·´Ó¦µÄÀíÂÛÖµ£»ÇúÏß±íʾÉú²úʵ¼ÊÇé¿ö¡£µ±°±Ñõ»¯ÂÊ´ïµ½100%£¬ÀíÂÛÉÏr[n(O2)/n(NH3)]£½          £¬Êµ¼ÊÉú²úÒª½«rֵά³ÖÔÚ1.7¡«2.2Ö®¼ä£¬Ô­ÒòÊÇ                                         ¡£

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

¾«Ó¢¼Ò½ÌÍø