ÌâÄ¿ÄÚÈÝ

ÂÈ»¯ï§¼ò³Æ¡°ÂÈ李±£¬ÓֳƱɰ£¬ÎªÎÞÉ«¾§Ìå»ò°×É«½á¾§ÐÔ·ÛÄ©£¬Ò×ÈÜÓÚË®ÖУ¬ÔÚ¹¤Å©ÒµÉú²úÖÐÓÃ;¹ã·º¡£ÒÔÂÈ»¯ÄƺÍÁòËáï§ÎªÔ­ÁÏÖƱ¸ÂÈ»¯ï§¼°¸±²úÆ·ÁòËáÄÆ£¬¹¤ÒÕÁ÷³ÌÈçÏ£º

ÂÈ»¯ï§ºÍÁòËáÄƵÄÈܽâ¶ÈËæζȱ仯ÈçͼËùʾ¡£»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÊµÑéÊÒ½øÐÐÕô·¢Å¨ËõÓõ½µÄÖ÷ÒªÒÇÆ÷ÓР         ¡¢ÉÕ±­¡¢²£Á§°ô¡¢¾Æ¾«µÆµÈ¡£
£¨2£©ÊµÑé¹ý³ÌÖгÃÈȹýÂ˵ÄÄ¿µÄÊÇ                                              ¡£
£¨3£©Ð´³ö¡°Õô·¢Å¨Ëõ¡±Ê±·¢ÉúµÄ»¯Ñ§·½³Ìʽ£º                                    ¡£
£¨4£©Ä³Ñо¿ÐÔѧϰС×éΪ²â¶¨¸ÃNH4Cl²úÆ·ÖеªµÄº¬Á¿£¬Éè¼ÆÁËÈçͼװÖ㬲¢½øÐÐÁËÌÖÂÛ¡£

¼×ͬѧ£º¸ù¾Ý´ËʵÑé²âµÃµÄÊý¾Ý£¬¼ÆËãµÄNH4Cl²úÆ·µÄº¬µªÁ¿¿ÉÄÜÆ«¸ß£¬ÒòΪʵÑé×°ÖÃÖдæÔÚÒ»¸öÃ÷ÏÔȱÏÝÊÇ£º                     ____              ¡£
ÒÒͬѧ£ºÊµÑé¹ý³ÌÖУ¬ÍùÉÕÆ¿ÖмÓÈëµÄŨÇâÑõ»¯ÄÆÈÜÒºµÄÀë×Ó·´Ó¦·½³ÌʽΪ            £¬·´Ó¦¹ý³ÌÖÐNaOHÒ»¶¨Òª×ãÁ¿²¢³ä·Ö¼ÓÈÈ£¬Ô­ÒòÊÇ                                  ¡£
ÓøĽøºóµÄʵÑé×°ÖÃÖØнøÐÐʵÑ飬³ÆÈ¡13.0gNH4Cl²úÆ·£¬²âµÃʵÑéºóB×°ÖÃÔöÖØ3.4g¡£Ôò¸Ã»¯·Êº¬µªÁ¿Îª        ¡£

£¨1£©Õô·¢Ãó
£¨2£©·ÀÖ¹ÂÈ»¯ï§¾§ÌåÎö³ö¶øËðºÄ
£¨3£©£¨NH4£©2SO4£«2NaCl= Na2SO4¡ý£«2NH4Cl
£¨4£©¢ÙA¡¢B×°Öüäȱһ¸ö¸ÉÔï×°Öà        ¢Ú NH4++OH-NH3¡ü+H2O
ʹÂÈ»¯ï§³ä·Ö·´Ó¦Íêȫת»¯ÎªNH3        21.5£¥

½âÎöÊÔÌâ·ÖÎö£º£¨1£©ÊµÑéÊÒ½øÐÐÕô·¢Å¨ËõÓõ½µÄÖ÷ÒªÒÇÆ÷ÓÐÕô·¢Ãó¡¢ÉÕ±­¡¢²£Á§°ô¡¢¾Æ¾«µÆµÈ£¬
£¨2£©ÓÉͼÏñ¿ÉÒÔ¿´³ö£¬ÔÚijһζȷ¶Î§ÄÚ£¬ÂÈ»¯ï§µÄÈܽâ¶ÈµÍÓÚÁòËá淋ÄÈܽâ¶È£¬ËùÒÔ³ÃÈȹýÂË·ÀÖ¹ÂÈ»¯ï§¾§ÌåÎö³ö¶øËðºÄ£¬¿ÉÒÔʹÁòËáï§ÒÔ³ÁµíÐÎʽÂ˳ö¶øÂÈ»¯ï§ÁôÔÚÂËÒºÀï¡£
£¨3£©Õô·¢Å¨ËõʱÁòËáï§ÒÔ¹ÌÌåÐÎʽ´æÔÚ£¬ËùÒÔ»¯Ñ§·½³ÌʽΪ£¨NH4£©2SO4£«2NaCl= Na2SO4¡ý£«2NH4Cl

¡÷

¸ßÎÂ

 
£¨4£©¢Ù°±ÆøÔÚ½øÈëB֮ǰδ¸ÉÔËùÒÔº¬µªÁ¿Æ«¸ß£¬Ó¦ÔÚA¡¢B×°Öüä¼ÓÒ»¸ö¸ÉÔï×°Öã¬

¢ÚÉÕÆ¿Öз¢ÉúµÄÀë×Ó·½³ÌʽÊÇÇ¿¼îÓëï§ÑÎÈÜÒºµÄ·´Ó¦£¬Àë×Ó·½³ÌʽΪNH4++OH-= NH3¡ü+H2O£»¼îÒª×ãÁ¿µÄÄ¿µÄ¾ÍÊÇʹ笠ùÀë×ÓÈ«²¿×ª»¯Îª°±Æø£»3.4g¼´Îª°±ÆøµÄÖÊÁ¿£¬Æ䵪ԪËصÄÖÊÁ¿Îª3.4g¡Á14¡Â17=2.8g,ËùÒÔº¬µªÁ¿Îª2.8g/13.0g¡Á100%=21.5%¡£
¿¼µã£º¿¼²é»¯Ñ§Ó빤ҵÉú²úµÄÁªÏµ¡¢¶ÔʵÑéµÄ·ÖÎö¡¢ÊµÑé×°ÖõÄÅжϣ¬Àë×Ó·½³ÌʽµÄÊéд¡¢»¯Ñ§¼ÆËãµÈ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¹¤ÒµÉϳ£ÓøõÌú¿ó£¨ÓÐЧ³É·ÝΪFeO¡¤Cr2O3£¬Ö÷ÒªÔÓÖÊΪSiO2¡¢Al2O3£©ÎªÔ­ÁÏÉú²úÖظõËá¼Ø£¨K2Cr2O7£©£¬ÊµÑéÊÒÄ£Ä⹤ҵ·¨ÓøõÌú¿óÖÆÖظõËá¼ØµÄÖ÷Òª¹¤ÒÕÁ÷³ÌÈçÏÂͼ£¬Éæ¼°µÄÖ÷Òª·´Ó¦ÊÇ£º6FeO¡¤Cr2O3£«24NaOH£«7KClO3=12Na2CrO4£«3Fe2O3£«7KCl£«12H2O£¬ÊԻشðÏÂÁÐÎÊÌ⣺

£¨1£©¢ÝÖÐÈÜÒº½ðÊôÑôÀë×ӵļìÑé·½·¨ÊÇ                                ¡£
£¨2£©²½Öè¢Û±»³ÁµíµÄÀë×ÓΪ£¨ÌîÀë×Ó·ûºÅ£©                             ¡£
£¨3£©ÔÚ·´Ó¦Æ÷¢ÙÖУ¬¶þÑõ»¯¹èÓë´¿¼î·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º                  ¡£
£¨4£©Ñ̵ÀÆøÖеÄCO2¿ÉÓëH2ºÏ³É¼×´¼¡£CH3OH¡¢H2µÄȼÉÕÈÈ·Ö±ðΪ£º¦¤H=£­725.5 kJ/mol¡¢¦¤H=£­285.8 kJ/mol£¬Ð´³ö¹¤ÒµÉÏÒÔCO2¡¢H2ºÏ³ÉCH3OHµÄÈÈ»¯Ñ§·½³Ìʽ£º            ¡£
£¨5£©2011ÄêÔÆÄÏÇú¾¸µÄ¸õÎÛȾʼþ£¬ËµÃ÷º¬¸õ·ÏÔü£¨·ÏË®£©µÄËæÒâÅŷŶÔÈËÀàÉú´æ»·¾³Óм«´óµÄΣº¦¡£µç½â·¨ÊÇ´¦Àí¸õÎÛȾµÄÒ»ÖÖ·½·¨£¬½ðÊôÌú×÷Ñô¼«¡¢Ê¯Ä«×÷Òõ¼«µç½âº¬Cr2O72-µÄËáÐÔ·ÏË®£¬Ò»¶Îʱ¼äºó²úÉúFe(OH)3ºÍCr(OH)3³Áµí¡£
¢Ùд³öµç½â·¨´¦Àí·ÏË®µÄ×Ü·´Ó¦µÄÀë×Ó·½³Ìʽ                               ¡£
¢ÚÒÑÖªCr(OH)3µÄKsp=6.3¡Á10¨C31£¬ÈôµØ±íË®¸õº¬Á¿×î¸ßÏÞÖµÊÇ0.1 mg/L£¬ÒªÊ¹ÈÜÒºÖÐc(Cr3+)½µµ½·ûºÏµØ±íË®ÏÞÖµ£¬Ðëµ÷½ÚÈÜÒºµÄc(OH-)¡Ý     mol/L£¨Ö»Ð´¼ÆËã±í´ïʽ£©¡£

¶þÑõ»¯ÃÌ¿ÉÓÃ×÷¸Éµç³ØÈ¥¼«¼Á£¬ºÏ³É¹¤ÒµµÄ´ß»¯¼ÁºÍÑõ»¯¼Á£¬²£Á§¹¤ÒµºÍÌ´ɹ¤ÒµµÄ×ÅÉ«¼Á¡¢ÏûÉ«¼Á¡¢ÍÑÌú¼ÁµÈ¡£
£¨1£©¶þÑõ»¯ÃÌÔÚËáÐÔ½éÖÊÖÐÊÇÒ»ÖÖÇ¿Ñõ»¯¼Á£¬ÇëÓû¯Ñ§·½³Ìʽ֤Ã÷£º______________________¡£
£¨2£©Ð¿¡ªÃ̼îÐÔµç³Ø¾ßÓÐÈÝÁ¿´ó¡¢·ÅµçµçÁ÷´óµÄÌص㣬Òò¶øµÃµ½¹ã·ºÓ¦Óᣵç³ØµÄ×Ü·´Ó¦Ê½ÎªZn£¨s£©£«2MnO2£¨s£©£«H2O£¨l£©=Zn£¨OH£©2£¨s£©£«Mn2O3£¨s£©¡£
¢Ùµç³Ø¹¤×÷ʱ£¬MnO2·¢Éú________·´Ó¦¡£
¢Úµç³ØµÄÕý¼«·´Ó¦Ê½Îª________¡£
£¨3£©¹¤ÒµÉÏÒÔÈíÃÌ¿óΪԭÁÏ£¬ÀûÓÃÁòËáÑÇÌúÖƱ¸¸ß´¿¶þÑõ»¯Ã̵ÄÁ÷³ÌÈçÏ£º

ÒÑÖª£ºÈíÃÌ¿óµÄÖ÷Òª³É·ÖΪMnO2£¬»¹º¬Si£¨16.27%£©¡¢Fe£¨5.86%£©¡¢Al£¨3.42%£©¡¢Zn£¨2.68%£©ºÍCu£¨0.86%£©µÈÔªËصĻ¯ºÏÎ²¿·ÖÑôÀë×ÓÒÔÇâÑõ»¯Îï»òÁò»¯ÎïµÄÐÎʽÍêÈ«³ÁµíʱÈÜÒºµÄpH¼ûÏÂ±í¡£

³ÁµíÎï
Al£¨OH£©3
Fe£¨OH£©3
Fe£¨OH£©2
Mn£¨OH£©2
Cu£¨OH£©2
pH
5.2
3.2
9.7
10.4
6.7
³ÁµíÎï
Zn£¨OH£©2
CuS
ZnS
MnS
FeS
pH
8.0
£­0.42
2.5
7
7
 
»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÙÁòËáÑÇÌúÔÚËáÐÔÌõ¼þϽ«MnO2»¹Ô­ÎªMnSO4£¬Ëá½þʱ·¢ÉúµÄÖ÷Òª·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ________________________¡£
¢ÚÊÔ¼ÁXΪ________¡£
¢ÛÂËÔüAµÄÖ÷Òª³É·ÖΪ________¡£
¢Ü¼ÓÈëMnSµÄÄ¿µÄÖ÷ÒªÊdzýÈ¥ÈÜÒºÖеÄ________¡£

¿ÕÆø´µ³ö·¨¹¤ÒÕ,ÊÇÄ¿Ç°¡°º£Ë®Ìáä塱µÄ×îÖ÷Òª·½·¨Ö®Ò»¡£Æ乤ÒÕÁ÷³ÌÈçÏÂ:

(1)äåÔÚÖÜÆÚ±íÖÐλÓÚ¡¡¡¡¡¡¡¡ÖÜÆÚ¡¡¡¡¡¡¡¡×å¡£ 
(2)²½Öè¢ÙÖÐÓÃÁòËáËữ¿ÉÌá¸ßCl2µÄÀûÓÃÂÊ,ÀíÓÉÊÇ                                                                            
(3)²½Öè¢ÜÀûÓÃÁËSO2µÄ»¹Ô­ÐÔ,·´Ó¦µÄÀë×Ó·½³ÌʽΪ                                                         
(4)²½Öè¢ÞµÄÕôÁó¹ý³ÌÖÐ,ζÈÓ¦¿ØÖÆÔÚ80~90 ¡æ,ζȹý¸ß»ò¹ýµÍ¶¼²»ÀûÓÚÉú²ú,Çë½âÊÍÔ­Òò¡¡                                                                                                        ¡£ 
(5)²½Öè¢àÖÐäåÕôÆøÀäÄýºóµÃµ½ÒºäåÓëäåË®µÄ»ìºÏÎï,¿ÉÀûÓÃËüÃǵÄÃܶÈÏà²îºÜ´óµÄÌصã½øÐзÖÀë¡£·ÖÀëÒÇÆ÷µÄÃû³ÆÊÇ¡¡                                                      ¡£ 
(6)²½Öè¢Ù¡¢¢ÚÖ®ºó²¢Î´Ö±½ÓÓá°º¬Br2µÄº£Ë®¡±½øÐÐÕôÁóµÃµ½Òºäå,¶øÊǾ­¹ý¡°¿ÕÆø´µ³ö¡±¡°SO2ÎüÊÕ¡±¡°Ñõ»¯¡±ºóÔÙÕôÁó,ÕâÑù²Ù×÷µÄÒâÒåÊÇ¡¡                                                                            ¡£ 

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

¾«Ó¢¼Ò½ÌÍø