ÌâÄ¿ÄÚÈÝ

º£ÑóÊÇÒ»¸ö·á¸»µÄ×ÊÔ´±¦¿â£¬Í¨¹ýº£Ë®µÄ×ÛºÏÀûÓÿɻñµÃÐí¶àÎïÖʹ©ÈËÀàʹÓá£
£¨1£© º£Ë®ÖÐÑεĿª·¢ÀûÓãº
¢ñ.º£Ë®ÖÆÑÎÄ¿Ç°ÒÔÑÎÌ﷨ΪÖ÷£¬½¨ÑÎÌï±ØÐëÑ¡ÔÚÔ¶Àë½­ºÓÈ뺣¿Ú£¬¶à·çÉÙÓ꣬³±Ï«Âä²î´óÇÒÓÖƽ̹¿Õ¿õµÄº£Ì²¡£Ëù½¨ÑÎÌï·ÖΪÖüË®³Ø¡¢Õô·¢³ØºÍ_______³Ø¡£
II.Ä¿Ç°¹¤ÒµÉϲÉÓñȽÏÏȽøµÄÀë×Ó½»»»Ä¤µç½â²Û·¨½øÐÐÂȼҵÉú²ú£¬ÔÚµç½â²ÛÖÐÑôÀë×Ó½»»»Ä¤Ö»ÔÊÐíÑôÀë×Óͨ¹ý£¬×èÖ¹ÒõÀë×ÓºÍÆøÌåͨ¹ý£¬Çë˵Ã÷ÂȼîÉú²úÖÐÑôÀë×Ó½»»»Ä¤µÄ×÷ÓÃ____________________________________________¡££¨Ð´Ò»µã¼´¿É£©
£¨2£©µçÉøÎö·¨ÊǽüÄêÀ´·¢Õ¹ÆðÀ´µÄÒ»ÖֽϺõĺ£Ë®µ­»¯¼¼Êõ£¬ÆäÔ­ÀíÈçÏÂͼËùʾ¡£Çë»Ø´ðºóÃæµÄÎÊÌ⣺

¢ñ.º£Ë®²»ÄÜÖ±½ÓͨÈëµ½¸Ã×°ÖÃÖУ¬ÀíÓÉÊÇ_____________________________________________¡£
¢ò. B¿ÚÅųöµÄÊÇ________(Ìî¡°µ­Ë®¡±»ò¡°Å¨Ë®¡±)¡£
£¨3£©Óÿà±£¨º¬Na+¡¢K+¡¢Mg2+¡¢Cl-¡¢Br-µÈÀë×Ó£©¿ÉÌáÈ¡ä壬ÆäÉú²úÁ÷³ÌÈçÏ£º

¢ñ.ÈôÎüÊÕËþÖеÄÈÜÒºº¬BrO3£­£¬ÔòÎüÊÕËþÖз´Ó¦µÄÀë×Ó·½³ÌʽΪ£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß£ß
_________________________________________¡£
¢ò.ͨ¹ý¢ÙÂÈ»¯ÒÑ»ñµÃº¬Br2µÄÈÜÒº£¬ÎªºÎ»¹Ðè¾­¹ý´µ³ö¡¢ÎüÊÕ¡¢ËữÀ´ÖØлñµÃº¬Br2µÄÈÜÒº£¿_____________________________________________________________________¡£
¢ó.ÏòÕôÁóËþÖÐͨÈëË®ÕôÆø¼ÓÈÈ£¬¿ØÖÆζÈÔÚ900C×óÓÒ½øÐÐÕôÁóµÄÔ­ÒòÊÇ___________________________________________________________________________¡£

£¨1£© ¢ñ ½á¾§   ¢ò ×èÖ¹H2ÓëCl2·¢Éú·´Ó¦ÉõÖÁ·¢Éú±¬Õ¨»ò×èÖ¹Cl2ÓëÉú³ÉµÄNaOHÈÜÒº·´Ó¦¶øʹÉÕ¼î²úÆ·²»´¿µÈ        £¨ÆäËûºÏÀí´ð°¸Ò²¼Æ·Ö£©
£¨2£©¢ñ º£Ë®Öк¬½Ï¶àMg2+ ºÍCa2+µÈÑôÀë×Ó£¬µç½âʱ»á²úÉúMg(OH)2¡¢Ca(OH)2µÈ³Áµí´Ó¶ø¶ÂÈûÑôÀë×Ó½»»»Ä¤    £¨ºÏÀí´ð°¸¾ù¼Æ·Ö£© £¨3·Ö£©    ¢ò  µ­Ë®
£¨3£©¢ñ3CO32£­+ 3Br2 £½ 5Br£­+BrO3£­+3CO2¡ü 
¢ò¸»¼¯ä壬Ìá¸ßBr2µÄŨ¶È £¨ºÏÀí´ð°¸¾ù¼Æ·Ö£©
¢óζȹýµÍÄÑÒÔ½«Br2Õô·¢³öÀ´£¬µ«Î¶ȹý¸ßÓֻὫ´óÁ¿µÄË®ÕôÁó³öÀ´£¨ºÏÀí´ð°¸¾ù¼Æ·Ö£©

½âÎöÊÔÌâ·ÖÎö£º£¨1£©¢ñ ÑδÓÖüË®³ØÖо­¹ýÕô·¢³Ø£¬µÃµ½Å¨¶È½Ï´óµÄ±ˮ£¬ÔÙ½«Â±Ë®×ªÒƵ½½á¾§³ØÖнᾧ£¬×îÖյôÖÑΣ»¢òÑôÀë×Ó½»»»Ä¤Ö»ÔÊÐíÑôÀë×Óͨ¹ý£¬×èÖ¹ÒõÀë×ÓºÍÆøÌåͨ¹ý£¬´Ó¶ø£¬ÔÚÑô¼«²úÉúµÄÂÈÆøÓëÔÚÒõ¼«²úÉúµÄÇâÆø¾Í²»»áÒò½Ó´¥·´Ó¦¶ø·¢Éú±¬Õ¨£¬Í¬Ê±Òõ¼«ÉϲúÉúµÄÇâÑõ»¯ÄÆÒ²²»»áÓëÂÈÆø·´Ó¦£¬´¿¶È½Ï¸ß£»
£¨2£©¢ñ º£Ë®Öи»º¬Ca2+¡¢Mg2+Àë×Ó£¬Èô½«º£Ë®Ö±½ÓͨÈëµ½¸Ã×°ÖÃÖУ¬Ôòº£Ë®ÖеÄCa2+¡¢Mg2+Àë×ÓÒ×ÓëÇâÑõ¸ùÀë×Ó½áºÏ²úÉúMg(OH)2¡¢Ca(OH)2µÈ³Áµí´Ó¶ø¶ÂÈûÑôÀë×Ó½»»»Ä¤ ¢ò Í¼ÖТڴ¦ÔÚÖ±Á÷µç³¡µÄ×÷ÓÃÏ£¬ÈÜÒºÖеÄÀë×Ó¾Í×÷¶¨ÏòǨÒÆ¡£ÑôĤֻÔÊÐíÑôÀë×Óͨ¹ý¶ø°ÑÒõÀë×Ó½ØÁôÏÂÀ´£»ÒõĤֻÔÊÐíÒõÀë×Óͨ¹ý¶ø°ÑÑôÀë×Ó½ØÁôÏÂÀ´¡£½á¹ûʹÕâЩСÊÒµÄÒ»²¿·Ö±ä³Éº¬Àë×ÓºÜÉٵĵ­Ë®ÊÒ£¬³öË®³ÆΪµ­Ë®¡£¶øÓëµ­Ë®ÊÒÏàÁÚµÄСÊÒÔò±ä³É¾Û¼¯´óÁ¿Àë×ÓµÄŨˮÊÒ£¬³öË®³ÆΪŨˮ¡£ËùÒÔB´¦³öˮΪµ­Ë®£»
£¨3£©¢ñ´¿¼îÊÇ̼ËáÄÆ£¬Óëäå·´Ó¦ÓÐBrO3£­Éú³É£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ3CO32£­+ 3Br2 £½ 5Br£­+BrO3£­+3CO2¡ü
¢ò´Ó¢Ù³öÀ´µÄÈÜÒºÖÐäåµÄº¬Á¿²»¸ß£¬Èç¹ûÖ±½ÓÕôÁ󣬲úÆ·³É±¾¸ß£¬ËùÒÔÐèÒª½øÒ»²½Å¨Ëõä壬Ìá¸ßäåµÄŨ¶È£»
¢ó Î¶ȹý¸ßË®ÕôÆøÕô³ö£¬äåÖк¬ÓÐË®·Ö£»Î¶ȹýµÍäå²»ÄÜÍêÈ«Õô³ö£¬²úÂʵ͡£
¿¼µã£º¿¼²éº£Ë®µÄ×ÛºÏÀûÓᢺ£Ë®É¹ÑΡ¢µçÉøÎö·¨µ­»¯º£Ë®¡¢´Óº£Ë®ÖÐÌáäåµÄÔ­Àí

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

£¨14·Ö£©¸õÌú¿óµÄÖ÷Òª³É·Ö¿É±íʾΪFeO¡¤Cr2O3£¬»¹º¬ÓÐMgO¡¢Al2O3¡¢Fe2O3µÈÔÓÖÊ£¬ÒÔÏÂÊÇÒÔ¸õÌú¿óΪԭÁÏÖƱ¸ÖظõËá¼Ø£¨K2Cr2O7£©µÄÁ÷³Ìͼ£º

ÒÑÖª£º¢Ù4FeO?Cr2O3+8Na2CO3+7O28Na2CrO4+2Fe2O3+8CO2¡ü£»
¢ÚNa2CO3+Al2O32NaAlO2+CO2¡ü£»
¢ÛCr2CO72-+H2O2CrO42-+2H+
¸ù¾ÝÌâÒâ»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¹ÌÌåXÖÐÖ÷Òªº¬ÓР     £¨Ìîд»¯Ñ§Ê½£©£»Òª¼ì²âËữ²Ù×÷ÖÐÈÜÒºµÄpHÊÇ·ñµÈÓÚ4£®5£¬Ó¦¸ÃʹÓà  ______________   £¨ÌîдÒÇÆ÷»òÊÔ¼ÁÃû³Æ£©¡£
£¨2£©Ëữ²½ÖèÓô×Ëáµ÷½ÚÈÜÒºpH£¼5£¬ÆäÄ¿µÄÊÇ  _______________________ ¡£
£¨3£©²Ù×÷¢óÓжಽ×é³É£¬»ñµÃK2Cr2O7¾§ÌåµÄ²Ù×÷ÒÀ´ÎÊÇ£º¼ÓÈëKCl¹ÌÌå¡¢Õô·¢Å¨Ëõ¡¢         ¡¢¹ýÂË¡¢       ¡¢¸ÉÔï¡£
£¨4£©Ï±íÊÇÏà¹ØÎïÖʵÄÈܽâ¶ÈÊý¾Ý£¬²Ù×÷III·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ£ºNa2Cr2O7 + 2KCl£½K2Cr2O7 ¡ý+ 2NaCl£¬¸Ã·´Ó¦ÔÚÈÜÒºÖÐÄÜ·¢ÉúµÄÀíÓÉÊÇ_____________________________¡£

ÎïÖÊ
KCl
NaCl
K2Cr2O7
Na2Cr2O7
Èܽâ¶È£¨g/100gË®£©
0¡æ
28
35£®7
4£®7
163
40¡æ
40£®1
36£®4
26£®3
215
80¡æ
51£®3
38
73
376
 
£¨5£©¸±²úÆ·YÖ÷Òªº¬ÇâÑõ»¯ÂÁ£¬»¹º¬ÉÙÁ¿Ã¾¡¢ÌúµÄÄÑÈÜ»¯ºÏÎï¼°¿ÉÈÜÐÔÔÓÖÊ£¬¾«È··ÖÎöYÖÐÇâÑõ»¯ÂÁº¬Á¿µÄ·½·¨ÊdzÆÈ¡n gÑùÆ·£¬¼ÓÈë¹ýÁ¿        ____ £¨ÌîдÊÔ¼Á£©¡¢Èܽ⡢¹ýÂË¡¢ÔÙ       £¨ÌîдÊÔ¼Á£©¡¢¡­¡­×ÆÉÕ¡¢ÀäÈ´¡¢³ÆÁ¿£¬µÃ¸ÉÔï¹ÌÌåm g¡£¼ÆËãÏéÆ·ÖÐÇâÑõ»¯ÂÁµÄÖÊÁ¿·ÖÊýΪ    £¨Óú¬m¡¢nµÄ´úÊýʽ±íʾ£©¡£

Ï¡ÍÁÔªËØÊÇÖÜÆÚ±íÖТó B×åîÖ¡¢îƺÍïçϵԪËصÄ×ܳƣ¬ËüÃǶ¼ÊǺܻîÆõĽðÊô£¬ÐÔÖʼ«ÎªÏàËÆ£¬³£¼û»¯ºÏ¼ÛΪ+3¡£ÆäÖÐîÆ£¨Y£©ÔªËØÊǼ¤¹âºÍ³¬µ¼µÄÖØÒª²ÄÁÏ¡£ÎÒ¹úÔ̲Ø×ŷḻµÄîÆ¿óʯ£¨ Y2FeBe2Si2O10£©£¬ÒÔ´Ë¿óʯΪԭÁÏÉú²úÑõ»¯îÆ£¨Y2O3£©µÄÖ÷ÒªÁ÷³ÌÈçÏ£º

ÒÑÖª£ºI£®ÓйؽðÊôÀë×ÓÐγÉÇâÑõ»¯Îï³ÁµíʱµÄpHÈçÏÂ±í£º

 
¿ªÊ¼³ÁµíʱµÄpH
ÍêÈ«³ÁµíʱµÄpH
Fe3+
2.7
3.7
Y3+
6.0
8.2
 
¢ò£®ÔÚÖÜÆÚ±íÖУ¬îë¡¢ÂÁÔªËØ´¦ÓÚµÚ¶þÖÜÆں͵ÚÈýÖÜÆڵĶԽÇÏßλÖ㬻¯Ñ§ÐÔÖÊÏàËÆ¡£
£¨1£©îÆ¿óʯ£¨Y2 FeBe2Si2O10£©µÄ×é³ÉÓÃÑõ»¯ÎïµÄÐÎʽ¿É±íʾΪ              ¡£
£¨2£©Óû´ÓNa2 SiO3ºÍNa2BeO2µÄ»ìºÏÈÜÒºÖÐÖƵÃBe(OH)2³Áµí¡£Ôò
¢Ù×îºÃÑ¡ÓÃÑÎËáºÍÏÂÁÐÑ¡ÏîÖеĠ      ÊÔ¼Á£¨Ìî×Öĸ´úºÅ£©£¬
a£®NaOHÈÜÒº    b£®°±Ë®    c£®CO2Æø    d£®HNO3
ÔÙͨ¹ý     ²Ù×÷·½¿ÉʵÏÖ£¨ÌîʵÑé²Ù×÷Ãû³Æ£©¡£
¢Úд³öNa2BeO2Óë×ãÁ¿ÑÎËá·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ£º                         £»
£¨3£©ÎªÊ¹Fe3+³ÁµíÍêÈ«£¬ÐëÓð±Ë®µ÷½ÚpH =a£¬ÔòaÓ¦¿ØÖÆÔÚ                  
µÄ·¶Î§ÄÚ£»¼ÌÐø¼Ó°±Ë®µ÷½ÚpH =b·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ                       £¬ÈÜÒºÖÐFe3+ÍêÈ«³ÁµíµÄÅж¨±ê×¼ÊÇ                          ¡£

Òø°±ÈÜÒº·Å¾Ãºó»á²úÉúµþµª»¯Òø£¨AgN3£©¶øÒýÆð±¬Õ¨£¬Ö±½ÓÅÅ·Å»áÎÛȾ»·¾³£¬ÇÒÔì³ÉÒø×ÊÔ´µÄÀË·Ñ¡£Ä³Ñо¿Ð¡×éÉè¼ÆÁË´ÓÒø¾µ·´Ó¦ºóµÄ·ÏÒºÖУ¨º¬¹ýÁ¿µÄÒø°±ÈÜÒº£¬¼ÙÉè²»º¬µ¥ÖÊÒø£©»ØÊÕÒøµÄÈçÏÂÁ½ÖÖʵÑéÁ÷³Ì£º
£¨ÒÑÖª£º[Ag£¨NH3£©2]£«ÔÚÈÜÒºÖдæÔÚƽºâ£º[Ag£¨NH3£©2]£«??Ag£«£«2NH3£©

£¨1£©Ð´³ö¼×·½°¸µÚ¢Ù²½·ÏÒºÓëÏ¡HNO3·´Ó¦µÄÀë×Ó·½³Ìʽ                                                                       ¡£
£¨2£©¼×·½°¸µÚ¢Ú²½¼ÓÈëµÄÌú·ÛÒª¹ýÁ¿µÄÄ¿µÄÊÇ                                                                                                                                               ¡£
¼×·½°¸Á÷³Ì¿ÉÄܲúÉúµÄ´óÆøÎÛȾÎïÊÇ                                                                       ¡£
£¨3£©ÒÒ·½°¸Èô×îÖյõ½Òø·ÛµÄÖÊÁ¿Æ«´ó£¬ÅųýδϴµÓ¸É¾»µÄÒòËØ£¬¿ÉÄܵÄÔ­ÒòÊÇ                                                                       ¡£
£¨4£©ÊµÑéÊÒÅäÖÆÒø°±ÈÜÒºµÄ²Ù×÷¹ý³ÌÊÇ                                                                                                                                                                                                                       ¡£
£¨5£©ÒÑÖªÒÒ·½°¸µÚ¢Û²½·´Ó¦ÓÐH2SÆøÌå²úÉú£¬Èô×îÖյõ½Òø·Û21.6 g£¬²»¿¼ÂÇÆäËûËðʧ£¬ÀíÂÛÉϸò½ÐèÒª¼ÓÈëÌú·Û       g¡£

º£Ë®ÊDZ¦¹óµÄ×ÔÈ»×ÊÔ´£®ÀûÓú£Ë®Ë®¿ÉÒԵõ½Ò»ÏµÁвúÆ·£®Ò²¿ÉÒÔ½øÐзÏÆø´¦Àí¡£
£¨1£©ÀûÓÃÂȼҵ²úÆ·´¦Àíº¬SO2µÄÑÌÆøµÄÁ÷³ÌÈçÏ£º

¢Ù¡°ÎüÊÕ×°Öá±Öз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ                   .
¢ÚÉÏÊöÁ÷³ÌÖÐÑ­»·ÀûÓõÄÎïÖÊÊÇ               ¡£
£¨2£©ÀûÓú£Ë®ÍÑÁò¿ÉÒÔÓÐЧµØ½â¾öúȼÉÕÅŷŵÄSO2Ôì³ÉµÄһϵÁл·¾³ÎÊÌâ¡£Æ乤ÒÕÁ÷³ÌÈçͼËùʾ£º

ÌìÈ»º£Ë®ÎüÊÕº¬ÁòµÄÑÌÆøºó£®ÐèÒªÓÃÑõÆø½øÐÐÑõ»¯´¦Àí£®Æä·´Ó¦Ô­ÀíµÄ»¯Ñ§·½³ÌʽÊÇ         £»Ñõ»¯ºóµÄº£Ë®ÐèÒª¼ÓÈëÇâÑõ»¯ÄÆ£®ÓëÖ®»ìºÏºó²ÅÄÜÅÅ·Å£®¸Ã²Ù×÷µÄÖ÷ҪĿµÄÊÇ        ¡£
£¨3£©´Óº£Ë®ÖÐ×½´¿´ÖÑκóµÄĸҺÖк¬ÓÐK+¡¢Na+¡¢Mg2+µÈÑôÀë×Ó£®¶ÔĸҺ½øÐÐһϵÁеļӹ¤¿ÉÖƵýðÊôþ¡£
¢Ù´ÓÀë×Ó·´Ï¯µÄ½Ç¶È˼¿¼£®ÔÚĸҺÖмÓÈëʯ»ÒÈéËùÆðµÄ×÷ÓÃÊÇ             ¡£
¢ÚÒªÀûÓÃMgCl2¡¤6H2OÖƵÃÞÌË®ÂÈ»¯Ã¾£®Ó¦²ÉÈ¡µÄ´ëÊ©ÊÇ             ¡£
¢Ûµç½âÈÛÈÚµÄÎÞË®ÂÈ»¯Ã¾ËùµÃµÄþÕôÆøÔÚÌض¨µÄ»·¾³ÖÐÀäÈ´ºó¼´Îª¹ÌÌåþ£®ÏÂÁÐÎïÖÊÖпÉÒÔÓÃ×÷þÕôÆøµÄÀäÈ´¼ÁµÄÊÇ      £¨Ìî×Öĸ£©¡£
A£®Ar   B£®CO2    C ¿ÕÆø    D£®O2    E£®Ë®ÕôÆø

º£ÑóÊÇÉúÃüµÄÒ¡Àº¡¢×ÊÔ´µÄ±¦¿â¡£ÖйúҪʵʩº£ÑóÇ¿¹úÕ½ÂÔ£¬ÊµÏÖÓɺ£Ñó´ó¹úÏòº£ÑóÇ¿¹úÂõ½øµÄÃÎÏë¡£º£Ñó¾­¼ÃÒѾ­³ÉΪÀ­¶¯ÎÒ¹ú¹úÃñ¾­¼Ã·¢Õ¹µÄÖØÒªÒýÇ棬º£Ë®µÄ×ۺϿª·¢¡¢ÀûÓÃÊǺ£Ñ󾭼õÄÒ»²¿·Ö£¬º£Ë®ÖпÉÌáÈ¡¶àÖÖ»¯¹¤Ô­ÁÏ£¬ÏÂÃæÊǹ¤ÒµÉ϶Ժ£Ë®µÄ¼¸Ïî×ÛºÏÀûÓõÄʾÒâͼ¡£ÆäÁ÷³ÌÈçÏÂͼËùʾ£º

£¨1£©Ð´³ö¢Ù¡¢¢Ú·´Ó¦µÄÀë×Ó·½³Ìʽ£º
¢Ù______________________£¬¢Ú______________________¡£
£¨2£©¹¤ÒµÉÏÀûÓõç½â±¥ºÍʳÑÎË®²úÉúµÄÇâÆøºÍÂÈÆøÖÆÈ¡ÑÎËᣬΪÁËÌåÏÖÂÌÉ«»¯Ñ§ÀíÄʹÂÈÆø³ä·Ö·´Ó¦£¬²ÉÈ¡½«ÂÈÆøÔÚÇâÆøÖÐȼÉյİ취£¬¿É±ÜÃâÂÈÆøȼÉÕ²»ÍêÈ«ÎÛȾ¿ÕÆø£¬Çëд³öÂÈÆøÔÚÇâÆøÖÐȼÉÕµÄʵÑéÏÖÏó£º______________________¡£
£¨3£©´ÖÑÎÖк¬ÓÐCa£²£«¡¢Mg£²£«¡¢SO4£²£­µÈÔÓÖÊ£¬¾«Öƺó¿ÉµÃµ½±¥ºÍNaC1ÈÜÒº¡£ÏÖÓÐÏÂÁгýÔÓÊÔ¼Á£ºA£®ÑÎËá   B£®ÇâÑõ»¯±µÈÜÒº   C£®Ì¼ËáÄÆÈÜÒº¡£¾«ÖÆʱ¼ÓÈë¹ýÁ¿³ýÔÓÊÔ¼ÁµÄÕýȷ˳ÐòÊÇ______________¡££¨ÌîÐòºÅ£©
£¨4£©½ðÊôþÔÚ¿ÕÆøÖÐȼÉÕʱ£¬³ýÉú³ÉMgOÍ⣬»¹ÓÐÉÙÁ¿Mg3N2Éú³É¡£°ÑµÈÎïÖʵÄÁ¿µÄ½ðÊôþ·Ö±ð·ÅÔÚ£ºA£®´¿ÑõÆø£¨O2£©ÖУ»B£®¶þÑõ»¯Ì¼ÆøÌåÖУ»C£®¿ÕÆøÖС£ÍêȫȼÉպ󣬵õ½µÄ¹ÌÌåÎïÖʵÄÖÊÁ¿ÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ______________¡££¨ÌîÐòºÅ£©
£¨5£©½«µç½â±¥ºÍNaClÈÜÒºÉú³ÉµÄÂÈÆøͨÈëÇâÑõ»¯ÄÆÈÜÒºÖпÉÒԵõ½NaClO¡£Ä³»¯Ñ§ÐËȤС×é̽¾¿NaClOÓëÄòËØCO(NH2)2µÄ·´Ó¦²úÎͨ¹ýʵÑé·¢ÏÖ²úÎï³ýijÖÖÑÎÍ⣬ÆäÓà²úÎﶼÊÇÄܲÎÓë´óÆøÑ­»·µÄÎïÖÊ£¬Ôò¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ____________¡£

º£Ë®ÖÐÖ÷ÒªÀë×ӵĺ¬Á¿ÈçÏ£º

³É·Ö
º¬Á¿/(mg/L)
³É·Ö
º¬Á¿/(mg/L)
Cl-
18980
Ca2+
400
Na+
10560
HCO3-
142
SO42-
2560
Mg2+
1272
 
£¨1£©³£ÎÂÏ£¬º£Ë®µÄpHÔÚ7.5~8.6Ö®¼ä£¬ÆäÔ­ÒòÊÇ(ÓÃÀë×Ó·½³Ìʽ±íʾ)______________________________¡£
£¨2£©µçÉøÎö·¨µ­»¯º£Ë®Ê¾ÒâͼÈçͼËùʾ£¬ÆäÖÐÒõ(Ñô)Àë×Ó½»»»Ä¤½öÔÊÐíÒõ(Ñô)Àë×Óͨ¹ý¡£Òõ¼«ÉϲúÉúÇâÆø£¬Òõ¼«¸½½ü»¹²úÉúÉÙÁ¿°×É«³Áµí£¬Æä³É·ÖÓÐ________ºÍCaCO3£¬Éú³ÉCaCO3µÄÀë×Ó·½³ÌʽÊÇ_______________¡£

£¨3£©Óú£Ë®¿ÉͬʱÉú²úÂÈ»¯ÄƺͽðÊôþ»òþµÄ»¯ºÏÎÆäÁ÷³ÌÈçÏÂͼËùʾ£º

¢ÙÔÚʵÑéÊÒÖÐÓÉ´ÖÑΡ°Öؽᾧ¡±Öƾ«ÑεIJÙ×÷°üÀ¨Èܽ⡢¹ýÂË¡¢Õô·¢¡­Ï´µÓµÈ²½Ö裻ÓйØÆäÖС°Õô·¢¡±²½ÖèµÄÐðÊöÕýÈ·µÄÊÇ_____¡£
a£®Õô·¢µÄÄ¿µÄÊǵõ½Èȱ¥ºÍÈÜÒº
b£®Õô·¢µÄÄ¿µÄÊÇÎö³ö¾§Ìå
c£®Ó¦ÓÃÓàÈÈÕô¸ÉÈÜÒº
d£®Ó¦Õô·¢ÖÁÓн϶ྦྷÌåÎö³öʱΪֹ
¢ÚÓÉMgCl2ÈÜÒºµÃµ½MgCl2?6H2O¾§Ìåʱ£¬Ò²ÐèÒªÕô·¢£¬Õô·¢µÄÄ¿µÄÊǵõ½Èȱ¥ºÍÈÜÒº£¬ÅжÏÈÜÒºÒѱ¥ºÍµÄÏÖÏóÊÇ_________________________¡£
£¨4£©25¡æʱ£¬±¥ºÍMg(OH)2ÈÜÒºµÄŨ¶ÈΪ5¡Á10-4 mol£¯L¡£
¢Ù±¥ºÍMg(OH)2ÈÜÒºÖеμӷÓ̪£¬ÏÖÏóÊÇ___________________________¡£
¢ÚijѧϰС×é²âº£Ë®ÖÐMg2+º¬Á¿(mg/L)µÄ·½·¨ÊÇ£ºÈ¡Ò»¶¨Ìå»ýµÄº£Ë®£¬¼ÓÈë×ãÁ¿_________£¬ÔÙ¼ÓÈë×ãÁ¿NaOH£¬½«Mg2+תΪMg(OH)2¡£25¡æ£¬¸Ã·½·¨²âµÃµÄMg2+º¬Á¿Óë±íÖÐ1272mg/LµÄ¡°ÕæÖµ¡±±È£¬Ïà¶ÔÎó²îԼΪ______£¥[±£Áô2λСÊý£¬º£Ë®Öб¥ºÍMg(OH)2ÈÜÒºµÄÃܶȶ¼ÒÔl g/cm3¼Æ]¡£

ÒÔ»ÆÌú¿óΪԭÁÏÖÆÁòËá²úÉúµÄÁòËáÔüÖк¬Fe2O3¡¢SiO2¡¢Al2O3¡¢MgOµÈ¡£ÊµÑéÊÒÄ£Ä⹤ҵÒÔÁòËáÔüÖƱ¸Ìúºì(Fe2O3)£¬¹ý³ÌÈçÏ£º

£¨1£©ÁòËáÔüµÄ³É·ÖÖÐÊôÓÚÁ½ÐÔÑõ»¯ÎïµÄÊÇ        £¬ д³öËáÈܹý³ÌFe2O3ÓëÏ¡ÁòËá·´Ó¦µÄÀë×Ó·´Ó¦·½³Ìʽ£º                                                        £»
£¨2£©Éú²ú¹ý³ÌÖУ¬ÎªÁËÈ·±£ÌúºìµÄ´¿¶È£¬Ñõ»¯¹ý³ÌÐèÒªµ÷½ÚÈÜÒºµÄpHµÄ·¶Î§ÊÇ_________£»£¨²¿·ÖÑôÀë×ÓÒÔÇâÑõ»¯ÎïÐÎʽ³ÁµíʱÈÜÒºµÄpH¼ûÏÂ±í£©

³ÁµíÎï
Fe(OH)3
Al(OH)3
Fe(OH)2
Mg(OH)2
¿ªÊ¼³Áµí
2.7
3.8
7.5
9.4
ÍêÈ«³Áµí
3.2
5.2
9.7
12.4
 
£¨3£©ÂËÔüAµÄÖ÷Òª³É·ÖΪ      £¬ÂËÒºB¿ÉÒÔ»ØÊÕµÄÎïÖÊÓÐ______________________£»
£¨4£©¼òÊöÏ´µÓ¹ý³ÌµÄʵÑé²Ù×÷                                                                £»
£¨5£©ÒÑÖªÁòËáÔüÖÊÁ¿Îªw kg£¬¼ÙÉèÌúºìÖƱ¸¹ý³ÌÖУ¬ÌúÔªËØËðºÄ25%£¬×îÖյõ½ÌúºìµÄÖÊÁ¿Îªm kg£¬ÔòÔ­À´ÁòËáÔüÖÐÌúÔªËØÖÊÁ¿·ÖÊýΪ         £¨ÓÃ×î¼ò·ÖÊý±í´ïʽ±íʾ£©¡£
£¨ÒÑÖªÏà¶ÔÔ­×ÓÖÊÁ¿£ºO 16  S 32  Fe 56 £©

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

¾«Ó¢¼Ò½ÌÍø