ÌâÄ¿ÄÚÈÝ

¹¤ÒµÉÏÉú²ú¸ßÂÈËᣨ·Ðµã£º90oC£©Ê±»¹Í¬Ê±Éú²úÁËÑÇÂÈËáÄÆ£¬Æ乤ÒÕÁ÷³ÌÈçÏ£º

£¨1£©ÀäÈ´¹ýÂ˵ÄÄ¿µÄÊǽµµÍNaHSO4µÄ          £¬²¢·ÖÀë³öNaHSO4¾§Ìå¡£
£¨2£©·´Ó¦Æ÷2Öз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ                                           £¬SO2µÄ
×÷ÓÃÊÇ×÷      ¼Á¡£
£¨3£©ÉÏÊö¹¤ÒµÉú²ú¸ßÂÈËáµÄ»¯Ñ§·´Ó¦Îª:3NaClO3+3H2SO4£¨Å¨£©£½3NaHSO4+HClO4+2ClO2+H2O£¬Ñõ»¯²úÎïÓ뻹ԭ²úÎïµÄÎïÖʵÄÁ¿Ö®±ÈΪ       ¡£
£¨4£©¿ÉÒÔͨ¹ýÕôÁóÂËÒºµÄ·½·¨µÃµ½¸ßÂÈËáµÄÔ­Òò¿ÉÄÜÊǸßÂÈËáµÄ·Ðµã±È½Ï    £¨Ìî¡°¸ß¡±»ò¡°µÍ¡±£©£¬ÈÝÒ×´ÓÈÜÒºÖÐÒݳö£¬Ñ­»·Ê¹ÓõÄÎïÖÊÊÇ        ¡£
£¨1£©Èܽâ¶È      £¨2£©2ClO2£«SO2£«4OH£­£½2ClO2£­£«SO42£­£«2H2O£»   »¹Ô­
£¨3£©1£º2        £¨4£©µÍ       H2SO4

ÊÔÌâ·ÖÎö£º£¨1£©ÔÚ·´Ó¦Æ÷1ÖУ¬ÁòËáÄƺÍÁòËá·´Ó¦»ñµÃµÄÁòËáÇâÄƵÄÈܽâ¶ÈËæ×ÅζȵĽµµÍ¶ø¼õС£¬ÕâÑùÀäÈ´¹ýÂË£¬¿ÉÒÔ½µµÍNaHSO4µÄÈܽâ¶È²¢·ÖÀë³öNaHSO4¾§Ìå¡£
£¨2£©ÔÚ·´Ó¦Æ÷2ÖУ¬¿ÉÒÔʵÏÖ¶þÑõ»¯ÂÈÏòNaClO2µÄת»¯£¬¶þÑõ»¯Áò¿ÉÒÔ×÷Ϊ»¹Ô­¼Á°ÑClO2»¹Ô­ÎªNaClO2£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ2ClO2£«SO2£«4OH£­£½2ClO2£­£«SO42£­£«2H2O¡£
£¨3£©¸ù¾Ý·½³Ìʽ3NaClO3+3H2SO4£¨Å¨£©£½3NaHSO4+HClO4+2ClO2+H2O¿ÉÖª£¬NaClO3¼ÈÊÇÑõ»¯¼Á£¬Ò²ÊÇ»¹Ô­¼Á£¬ÆäÖÐÂÈÔªËصĻ¯ºÏ¼Û´Ó£«5¼Û²¿·ÖÉý¸ßµ½£«7¼Û£¬²¿·Ö½µµÍµ½£«4¼Û£¬ËùÒÔ¸ßÂÈËáÊÇÑõ»¯²úÎ¶þÑõ»¯ÂÈÊÇ»¹Ô­²úÎï¡£¸ù¾Ýµç×ÓµÃʧÊغã¿ÉÖª£¬Ñõ»¯²úÎïÓ뻹ԭ²úÎïµÄÎïÖʵÄÁ¿Ö®±ÈΪ1:2¡£
£¨4£©¿ÉÒÔͨ¹ýÕôÁóÂËÒºµÄ·½·¨µÃµ½¸ßÂÈËᣬÕâ˵Ã÷¸ßÂÈËáµÄ·Ðµã±È½ÏµÍ£¨·Ðµã£º90oC£©£»¸ù¾ÝÑ­»·Í¼¿ÉÒÔ·¢ÏÖÁòËá×÷Ϊ·´Ó¦Îï½øÈë·´Ó¦Æ÷1ÖУ¬ÓÖ×÷ΪÉú³ÉÎïÔÚ·´Ó¦Æ÷2ÖÐÉú³É£¬¿ÉÒÔÑ­»·Ê¹Óá£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
Ë®ÊÇÒ»ÖÖÖØÒªµÄ×ÔÈ»×ÊÔ´£¬ÊÇÈËÀàÀµÒÔÉú´æ²»¿ÉȱÉÙµÄÎïÖÊ¡£Ë®ÖÊÓÅÁÓÖ±½ÓÓ°ÏìÈËÌ彡¿µ
£¨1£©º¬Óн϶à______________µÄË®³ÆΪӲˮ£¬Ó²Ë®¼ÓÈȺó²úÉú³ÁµíµÄÀë×Ó·½³ÌʽΪ__________________________( д³öÉú³ÉÒ»ÖÖ³ÁµíÎïµÄ¼´¿É) ¡£
£¨2£©ÏÂͼΪijÒûÓÃË®³§´ÓÌìȻˮÖƱ¸´¿¾»Ë®(È¥Àë×ÓË®)µÄ¹¤ÒÕÁ÷³ÌʾÒâͼ£º

¢Ù»îÐÔÌ¿µÄ×÷ÓÃÊÇ__________________£»O3Ïû¶¾µÄÓŵãÊÇ________________¡£
¢ÚA¡¢BÖзÅÖõÄÎïÖÊÃû³Æ·Ö±ðÊÇ£ºA__________________£»B_______________¡£A¡¢BÖзÅÖõÄÎïÖÊÊÇ·ñ¿ÉÒÔ»¥»»?Çë˵Ã÷Ô­Òò________________________________________¡£
£¨3£©Í¨¹ýÊ©¼ÓÒ»¶¨Ñ¹Á¦Ê¹Ë®·Ö×Óͨ¹ý°ë͸Ĥ¶ø½«´ó·Ö×Ó»òÀë×Ó½ØÁô, ´Ó¶ø»ñµÃ´¿¾»Ë®µÄ·½·¨³ÆΪ                   ¡£µçÉøÎö·¨¾»»¯Ë®Ê±, ʹÀë×Óͨ¹ý°ë͸ĤµÄÍƶ¯Á¦ÊÇ                     ¡£
£¨4£©¼ìÑéÕôÁóË®µÄ´¿¶Èʱ, ×î¼òµ¥Ò×Ðеķ½·¨ÊDzⶨˮµÄ                ¡£
£¨5£©Ä³ ³Ç ÊÐ Óà ˮ ÖÐc(Ca2+)Ϊ1.0¡Á10-3mol/L, c(Mg2+)Ϊ5.0¡Á10-4mol/L,c(HCO3_)Ϊ8.0¡Á10-4mol/L¡£ÈçÓÃÒ©¼ÁÈí»¯¸Ã1000L ,Ó¦¼ÓÈëCa(OH) 2            g , Na2CO 3 __________ g ¡£
ÁòËáÑÇÎý£¨SnSO4£©¿ÉÓÃÓÚ¶ÆÎý¹¤Òµ£®Ä³Ð¡×éÉè¼ÆSnSO4ÖƱ¸Â·ÏßΪ£º

²éÔÄ×ÊÁÏ£º
¢ñ£®ËáÐÔÌõ¼þÏ£¬ÎýÔÚË®ÈÜÒºÖÐÓÐSn2+¡¢Sn4+Á½ÖÖÖ÷Òª´æÔÚÐÎʽ£¬Sn2+Ò×±»Ñõ»¯£®
¢ò£®SnCl2Ò×Ë®½âÉú³É¼îʽÂÈ»¯ÑÇÎý
£¨1£©ÎýÔ­×ӵĺ˵çºÉÊýΪ50£¬Óë̼ԪËØͬ´¦¢ôA×壬ÎýλÓÚÖÜÆÚ±íµÄµÚ     ÖÜÆÚ£¨1·Ö£©
£¨2£©²Ù×÷¢ñÊÇ                ¹ýÂË¡¢Ï´µÓµÈ£¨2·Ö£©
£¨3£©ÈܽâSnCl2·ÛÄ©Ðè¼ÓŨÑÎËᣬԭÒòÊÇ                                                  
£¨4£©¼ÓÈëSn·ÛµÄ×÷ÓÃÓÐÁ½¸ö£º¢Ùµ÷½ÚÈÜÒºpH ¢Ú                                   
£¨5£©·´Ó¦¢ñµÃµ½³ÁµíÊÇSnO£¬µÃµ½¸Ã³ÁµíµÄÀë×Ó·´Ó¦·½³ÌʽÊÇ                                      
£¨6£©ËáÐÔÌõ¼þÏ£¬SnSO4ÓëË«Ñõˮȥ·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ                                          
£¨7£©¸ÃС×éͨ¹ýÏÂÁз½·¨²â¶¨ËùÓÃÎý·ÛµÄ´¿¶È£¨ÔÓÖʲ»²ÎÓë·´Ó¦£©£º
¢Ù½«ÊÔÑùÈÜÓÚÑÎËáÖУ¬·¢ÉúµÄ·´Ó¦Îª£ºSn+2HCl¨TSnCl2+H2¡ü£»
¢Ú¼ÓÈë¹ýÁ¿µÄFeCl3£»
¢ÛÓÃÒÑ֪Ũ¶ÈµÄK2Cr2O7µÎ¶¨¢ÚÉú³ÉµÄFe2+£¬ÔÙ¼ÆËãÎý·ÛµÄ´¿¶È£¬ÇëÅäƽ·½³Ìʽ£º
FeCl2 +   K2Cr2O7 +   HCl =      FeCl3 +      KCl +     CrCl2+    
£¨11·Ö£©¢ñ.¡°Öж«ÓÐʯÓÍ£¬ÖйúÓÐÏ¡ÍÁ¡±¡£Ï¡ÍÁ½ðÊôÊÇÎÒ¹úÕ½ÂÔÐÔ×ÊÔ´£¬Ó¦¼ÓÒÔ±£»¤¡£Ï¡ÍÁ½ðÊôÊÇÖÜÆÚ±íÖТóB×åÖÐîÖ¡¢îƺÍïçϵʮÆßÖÖÔªËصÄ×ܳƣ¬¶¼ÊǺܻîÆõĽðÊô£¬ÐÔÖʼ«ÎªÏàËÆ£¬³£¼û»¯ºÏ¼ÛΪ+3¼Û¡£îÆ£¨Y£©ÔªËØÊǼ¤¹âºÍ³¬µ¼µÄÖØÒª²ÄÁÏ¡£ÎÒ¹úÔ̲Ø×ŷḻµÄîÆ¿óʯ£¨Y2FeBe2Si2O10£©£¬ÒÔ´Ë¿óʯΪԭÁÏÉú²úÑõ»¯îÆ£¨Y2O3£©µÄÖ÷ÒªÁ÷³ÌÈçÏ£º

ÒÑÖª£º¢ÙÓйؽðÊôÀë×ÓÐγÉÇâÑõ»¯Îï³ÁµíʱµÄpHÈçÏÂ±í£º
Àë×Ó
¿ªÊ¼³ÁµíʱµÄpH
ÍêÈ«³ÁµíʱµÄpH
Fe3+
2.7
3.7
Y3+
6.0
8.2
 
¢ÚÔÚÖÜÆÚ±íÖУ¬îë¡¢ÂÁÔªËØ´¦ÓÚµÚ¶þÖÜÆں͵ÚÈýÖÜÆڵĶԽÇÏßλÖ㬻¯Ñ§ÐÔÖÊÏàËÆ¡£
£¨1£©Óû´ÓNa2SiO3ºÍNa2BeO2µÄ»ìºÏÈÜÒºÖÐÖƵÃBe(OH)2³Áµí¡£Ôò×îºÃÑ¡ÓÃÑÎËáºÍ     £¨Ìî×Öĸ£©
Á½ÖÖÊÔ¼Á£¬Í¨¹ý±ØÒªµÄ²Ù×÷¼´¿ÉʵÏÖ¡£
A£®NaOHÈÜÒº      B¡¢°±Ë®     C¡¢CO2     D¡¢HNO3
£¨2£©Á÷³ÌͼÖÐÓð±Ë®µ÷½ÚpH£½aʱÉú³É³ÁµíµÄ»¯Ñ§Ê½Îª              £¬¼ÌÐø¼Ó°±Ë®µ÷ÖÁpH£½b£¬´Ëʱ·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ                                        ¡£
£¨3£©³ÁµíCΪ²ÝËáîÆ£¬Ð´³ö²ÝËáîƸô¾ø¿ÕÆø¼ÓÈÈÉú³ÉY2O3µÄ»¯Ñ§·½³Ìʽ£º                        ¡£
¢òïçϵԪËØî棨Ce£©ÊǵؿÇÖк¬Á¿×î¸ßµÄÏ¡ÍÁÔªËØ£¬ÔÚ¼ÓÈȵÄÌõ¼þÏÂCeCl3Ò×·¢ÉúË®½â¡£
£¨4£©ÎÞË®CeCl3¿ÉÓüÓÈÈCeCl3¡¤6H2OºÍNH4Cl¹ÌÌå»ìºÏÎïµÄ·½·¨À´ÖƱ¸£¬ÆäÖÐNH4ClµÄ×÷ÓÃÊÇ___
                                                                       ¡£
£¨5£©ÔÚijǿËáÐÔ»ìºÏÏ¡ÍÁÈÜÒºÖмÓÈëH2O2£¬µ÷½ÚpH¡Ö3£¬Ce3+ͨ¹ý·´Ó¦ÐγÉCe(OH)4³ÁµíµÃÒÔ·ÖÀ룬 Ð´³ö·´Ó¦µÄÀë×Ó·½³Ìʽ                                                            ¡£
·Ö×Óɸ¾ßÓоùÔȵÄ΢¿×½á¹¹£¬·Ö×Óɸɸ·Ö×÷ÓüûÏÂͼ¡£ÓÉÓÚ·Ö×Óɸ¾ßÓÐÎü¸½ÄÜÁ¦¸ß£¬ÈÈÎȶ¨ÐÔÇ¿µÈÆäËüÎü¸½¼ÁËùûÓеÄÓŵ㣬ʹµÃ·Ö×Óɸ»ñµÃ¹ã·ºµÄÓ¦Óá£Ä³ÖÖÐͺŵķÖ×ÓɸµÄ¹¤ÒµÉú²úÁ÷³Ì¿É¼òµ¥±íʾÈçÏ£º


ÔÚ¼ÓNH3¡¤H2Oµ÷½ÚpHµÄ¹ý³ÌÖУ¬ÈôpH¿ØÖƲ»µ±»áÓÐAl(OH)3Éú³É£¬¼ÙÉèÉú²úÁ÷³ÌÖÐÂÁÔªËغ͹èÔªËؾùûÓÐËðºÄ£¬ÄÆÔ­×ÓµÄÀûÓÃÂÊΪ10£¥¡£
£¨1£©·Ö×ÓɸµÄ¿×µÀÖ±¾¶Îª4A(1 A=10-10m)³ÆΪ4AÐÍ·Ö×Óɸ£¬µ±Na+±»Ca2+È¡´úʱ¾ÍÖƵÃ5AÐÍ·Ö×Óɸ£¬µ±Na+±»K+È¡´úʱ¾ÍÖƵÃ3AÐÍ·Ö×Óɸ¡£Òª¸ßЧ·ÖÀëÕý¶¡Íé(·Ö×ÓÖ±¾¶Îª4.65A)ºÍÒ춡Íé(·Ö×ÓÖ±¾¶Îª5.6A)Ó¦¸ÃÑ¡Óà         ÐÍ·Ö×Óɸ¡£
£¨2£©A12(SO4)3ÈÜÒºÓëNa2SiO3ÈÜÒº·´Ó¦Éú³É½ºÌåµÄÀë×Ó·½³ÌʽΪ                    
£¨3£©¸ÃÉú²úÁ÷³ÌÖÐËùµÃÂËÒºÀﺬÓеÄÀë×Ó³ýH+¡¢OH-Í⣬Ö÷ҪΪ                       £»¼ìÑéÆäÖнðÊôÑôÀë×ӵIJÙ×÷·½·¨ÊÇ                                       
£¨4£©¼ÓNH3¡¤H2Oµ÷½ÚpHºó£¬¼ÓÈȵ½90¡æ²¢³ÃÈȹýÂ˵ÄÔ­Òò¿ÉÄÜÊÇ                      
£¨5£©¸ÃÉú²úÁ÷³ÌÖÐËùµÃ·Ö×ÓɸµÄ»¯Ñ§Ê½Îª                             
îÑÌú¿óµÄÖ÷Òª³É·ÖΪFeTiO3£¨¿É±íʾΪFeO¡¤TiO2£©£¬º¬ÓÐÉÙÁ¿MgO¡¢CaO¡¢SiO2µÈÔÓÖÊ¡£ÀûÓÃîÑÌú¿óÖƱ¸ï®Àë×Óµç³Øµç¼«²ÄÁÏ£¨îÑËáï®Li4Ti5O12ºÍÁ×ËáÑÇÌúï®LiFePO4£©µÄ¹¤ÒµÁ÷³ÌÈçÏÂͼËùʾ£º   

ÒÑÖª£ºFeTiO3ÓëÑÎËá·´Ó¦µÄÀë×Ó·½³ÌʽΪ£ºFeTiO3£«4H+£«4Cl-£½Fe2+£«TiOCl42-£«2H2O
£¨1£©»¯ºÏÎïFeTiO3ÖÐÌúÔªËصĻ¯ºÏ¼ÛÊÇ       ¡£
£¨2£©ÂËÔüAµÄ³É·ÖÊÇ           ¡£
£¨3£©ÂËÒºBÖÐTiOCl42- ת»¯Éú³ÉTiO2µÄÀë×Ó·½³ÌʽÊÇ                         ¡£
£¨4£©·´Ó¦¢ÚÖйÌÌåTiO2ת»¯³É(NH4)2Ti5O15ÈÜҺʱ£¬TiÔªËصĽþ³öÂÊÓ뷴ӦζȵĹØϵÈçÏÂͼËùʾ¡£·´Ó¦Î¶ȹý¸ßʱ£¬TiÔªËؽþ³öÂÊϽµµÄÔ­ÒòÊÇ                    ¡£

£¨5£©·´Ó¦¢ÛµÄ»¯Ñ§·½³ÌʽÊÇ                                                 ¡£
£¨6£©ÓÉÂËÒºDÖƱ¸LiFePO4µÄ¹ý³ÌÖУ¬ËùÐè17%Ë«ÑõË®ÓëH2C2O4µÄÖÊÁ¿±ÈÊÇ      ¡£
£¨7£©Èô²ÉÓÃîÑËáﮣ¨Li4Ti5O12£©ºÍÁ×ËáÑÇÌúﮣ¨LiFePO4£©×÷µç¼«×é³Éµç³Ø£¬Æ乤×÷Ô­ÀíΪ£ºLi4Ti5O12£«3LiFePO4Li7Ti5O12£«3FePO4
¸Ãµç³Ø³äµçʱÑô¼«·´Ó¦Ê½ÊÇ                                           ¡£

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

¾«Ó¢¼Ò½ÌÍø