£¨8·Ö£©¢ñ äå±»³ÆΪº£ÑóÔªËØ£¬ÏòÑαÖÐͨÈëÂÈÆø¿ÉÖƵÃä壺Cl2 + 2NaBr = 2NaCl + Br2£¬¸Ã·´Ó¦±»Ñõ»¯µÄÔªËØΪ          £¨Ð´ÔªËØ·ûºÅ£©£»ÈôÉÏÊö·´Ó¦ÖƵÃ16g Br2 £¬ÔòתÒƵĵç×ÓÊýÄ¿ÊÇ          ¸ö¡£

  ¢ò Çë°´ÒªÇóÊéд»¯Ñ§·½³Ìʽ»òÀë×Ó·½³Ìʽ

 £¨1£©Ð¡Ã÷ÔËÓû¯Ñ§ÊµÑéÖ¤Ã÷Á˾ÃÖÃÓÚ¿ÕÆøÖеÄƯ°×·ÛÒѱäÖÊ£¬ÇëÓû¯Ñ§·½³Ìʽ±íʾƯ°×·Û±äÖʵÄÔ­Òò                                                                              ¡£

£¨2£©FeSO4ÈÜÒºÓÃÏ¡H2SO4Ëữ£¬·ÅÖÃÒ»¶Îʱ¼äºóÂÔÏÔ»ÆÉ«£¬Ð´³ö±ä»¯¹ý³ÌµÄÀë×Ó·½³Ìʽ            

                                                              £»

È»ºóÏòÆäÖеμÓKI-µí·ÛÈÜÒº±äÀ¶É«£¬Ð´³ö±ä»¯¹ý³ÌµÄÀë×Ó·½³Ìʽ                            _¡£

 

£¨8·Ö£©¢ñBr£¨1·Ö£©   £»  0.2NA»ò1.204¡Á1023  £¨1·Ö£©

  ¢ò £¨1£©  Ca(ClO)2+CO2+H2O£½CaCO3+2HClO    2HClO 2HCl+O2¡ü  £¨Ã¿¿Õ1·Ö£©

 £¨2£©4Fe2£«£« 4H£« £«O2 £½4Fe3£«£« 2H2O£¨2·Ö£©  £»2Fe3£«£«2I£­ £½ 2Fe2£« £« I£¨2·Ö£©

½âÎö:ÂÔ

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2010?ÁijǶþÄ££©¡¾»¯Ñ§--»¯Ñ§Óë¼¼Êõ¡¿
äåÊǺ£Ë®ÖÐÖØÒªµÄ·Ç½ðÊôÔªËØ£¬µØÇòÉÏ99%µÄäåÔªËØÒÔBr-µÄÐÎʽ´æÔÚÓÚº£Ë®ÖУ¬Òò´Ëäå±»³ÆΪ¡°º£ÑóÔªËØ¡±£®Ä¿Ç°£¬´Óº£Ë®ÖÐÌáÈ¡äåÊÇÊÀ½ç¸÷¹úÉú²úäåµÄÖ÷Òª·½·¨£¬Æ乤ÒÕÁ÷³ÌÈçÏ£º
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³ö²½Öè¢ÙÖз´Ó¦µÄÀë×Ó·½³Ìʽ
2Br-+Cl2=2Cl-+Br2
2Br-+Cl2=2Cl-+Br2
£®
£¨2£©²½Öè¢ÛÔÚ
ÎüÊÕËþ
ÎüÊÕËþ
 £¨ÌîÉ豸Ãû³Æ£©ÖÐÍê³É£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
SO2+Br2+2H2O=2HBr+H2SO4
SO2+Br2+2H2O=2HBr+H2SO4
£®
£¨3£©ÔÚ²½Öè¢ÝÕôÁóµÄ¹ý³ÌÖУ¬Î¶ȿØÖÆÔÚ80¡æ¡«90¡æ£¬Î¶ȹý¸ß»ò¹ýµÍÓÚ²»ÀûÓÚÉú²ú£¬ÆäÔ­ÒòÊÇ
ζȹý¸ß£¬´óÁ¿Ë®ÕôÆøËæÖ®Åųö£¬äåÆøÖÐË®·ÖÔö¼Ó£»Î¶ȹýµÍ£¬äå²»ÄÜÍêÈ«Õô³ö£¬»ØÊÕÂÊÌ«µÍ
ζȹý¸ß£¬´óÁ¿Ë®ÕôÆøËæÖ®Åųö£¬äåÆøÖÐË®·ÖÔö¼Ó£»Î¶ȹýµÍ£¬äå²»ÄÜÍêÈ«Õô³ö£¬»ØÊÕÂÊÌ«µÍ
£®
£¨4£©äåµ¥ÖÊÔÚËÄÂÈ»¯Ì¼ÖеÄÈܽâ¶È±ÈÔÚË®ÖдóµÃ¶à£¬ËÄÂÈ»¯Ì¼ÓëË®²»»¥ÈÜ£¬¹Ê¿ÉÓÃÓÚÝÍÈ¡ä壬µ«ÔÚÉÏÊö¹¤ÒÕÖÐÈ´²»ÓÃËÄÂÈ»¯Ì¼£¬Ô­ÒòÊÇ
ËÄÂÈ»¯Ì¼ÝÍÈ¡·¨¹¤ÒÕ¸´ÔÓ¡¢É豸Ͷ×ʴ󣬾­¼ÃЧÒæµÍ¡¢»·¾³ÎÛȾÑÏÖØ
ËÄÂÈ»¯Ì¼ÝÍÈ¡·¨¹¤ÒÕ¸´ÔÓ¡¢É豸Ͷ×ʴ󣬾­¼ÃЧÒæµÍ¡¢»·¾³ÎÛȾÑÏÖØ
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¢ñ£®äå±»³ÆΪº£ÑóÔªËØ£¬ÏòÑαÖÐͨÈëÂÈÆø¿ÉÖƵÃä壺Cl2+2NaBr=2NaCl+Br2£¬¸Ã·´Ó¦±»Ñõ»¯µÄÔªËØΪ
Br
Br
£¨Ð´ÔªËØ·ûºÅ£©£»ÈôÉÏÊö·´Ó¦ÖƵÃ16g Br2£¬ÔòתÒƵĵç×ÓÊýÄ¿ÊÇ
0.2NA»ò1.204¡Á1023
0.2NA»ò1.204¡Á1023
¸ö£®
¢ò£®Çë°´ÒªÇóÊéд»¯Ñ§·½³Ìʽ»òÀë×Ó·½³Ìʽ
£¨1£©Ð¡Ã÷ÔËÓû¯Ñ§ÊµÑéÖ¤Ã÷Á˾ÃÖÃÓÚ¿ÕÆøÖеÄƯ°×·ÛÒѱäÖÊ£¬ÇëÓû¯Ñ§·½³Ìʽ±íʾƯ°×·Û±äÖʵÄÔ­Òò
Ca£¨ClO£©2+CO2+H2O=CaCO3+2HClO¡¢2HClO
 ¹âÕÕ 
.
 
2HCl+O2¡ü
Ca£¨ClO£©2+CO2+H2O=CaCO3+2HClO¡¢2HClO
 ¹âÕÕ 
.
 
2HCl+O2¡ü
£®
£¨2£©FeSO4ÈÜÒºÓÃÏ¡H2SO4Ëữ£¬·ÅÖÃÒ»¶Îʱ¼äºóÂÔÏÔ»ÆÉ«£¬Ð´³ö±ä»¯¹ý³ÌµÄÀë×Ó·½³Ìʽ
4Fe2++4H++O2=4Fe3++2H2O
4Fe2++4H++O2=4Fe3++2H2O
£»
È»ºóÏòÆäÖеμÓKI-µí·ÛÈÜÒº±äÀ¶É«£¬Ð´³ö±ä»¯¹ý³ÌµÄÀë×Ó·½³Ìʽ
2Fe3++2I-=2Fe2++I2
2Fe3++2I-=2Fe2++I2
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄêɽ¶«¼ÃÄþÓą̃¶þÖиßÒ»ÉÏѧÆÚÆÚÄ©Ä£Ä⿼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ

£¨8·Ö£©¢ñ äå±»³ÆΪº£ÑóÔªËØ£¬ÏòÑαÖÐͨÈëÂÈÆø¿ÉÖƵÃä壺Cl2 + 2NaBr =" 2NaCl" + Br2£¬¸Ã·´Ó¦±»Ñõ»¯µÄÔªËØΪ         £¨Ð´ÔªËØ·ûºÅ£©£»ÈôÉÏÊö·´Ó¦ÖƵÃ16g Br2 £¬ÔòתÒƵĵç×ÓÊýÄ¿ÊÇ         ¸ö¡£
¢ò Çë°´ÒªÇóÊéд»¯Ñ§·½³Ìʽ»òÀë×Ó·½³Ìʽ
£¨1£©Ð¡Ã÷ÔËÓû¯Ñ§ÊµÑéÖ¤Ã÷Á˾ÃÖÃÓÚ¿ÕÆøÖеÄƯ°×·ÛÒѱäÖÊ£¬ÇëÓû¯Ñ§·½³Ìʽ±íʾƯ°×·Û±äÖʵÄÔ­Òò                                                                             ¡£
£¨2£©FeSO4ÈÜÒºÓÃÏ¡H2SO4Ëữ£¬·ÅÖÃÒ»¶Îʱ¼äºóÂÔÏÔ»ÆÉ«£¬Ð´³ö±ä»¯¹ý³ÌµÄÀë×Ó·½³Ìʽ           
                                                             £»
È»ºóÏòÆäÖеμÓKI-µí·ÛÈÜÒº±äÀ¶É«£¬Ð´³ö±ä»¯¹ý³ÌµÄÀë×Ó·½³Ìʽ                           _¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014½ì½­ËÕÊ¡¸ßÈýÉÏѧÆÚµÚÒ»´ÎÔ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÊµÑéÌâ

äå±»³ÆΪ¡°º£ÑóÔªËØ¡±¡£ÒÑÖªBr2µÄ·ÐµãΪ59¡æ£¬Î¢ÈÜÓÚË®£¬Óж¾ÐÔºÍÇ¿¸¯Ê´ÐÔ¡£ÊµÑéÊÒÄ£Äâ´Óº£Ë®ÖÐÌáÈ¡äåµÄÖ÷Òª²½ÖèΪ£º

²½Öè1£º½«º£Ë®Õô·¢Å¨Ëõ³ýÈ¥´ÖÑÎ

²½Öè2£º½«³ýÈ¥´ÖÑκóµÄĸҺËữºó£¬Í¨ÈëÊÊÁ¿µÄÂÈÆø£¬Ê¹Br£­×ª»¯ÎªBr2¡£

²½Öè3£ºÏò²½Öè2ËùµÃË®ÈÜÒºÖÐͨÈëÈÈ¿ÕÆø»òË®ÕôÆø£¬½«äåµ¥ÖÊ´µÈëÊ¢ÓжþÑõ»¯ÁòË®ÈÜÒºµÄÈÝÆ÷ÖÐ

²½Öè4£ºÔÙÏò¸ÃÈÝÆ÷ÖÐͨÈëÊÊÁ¿µÄÂÈÆø£¬Ê¹Br£­×ª»¯ÎªBr2

²½Öè5£ºÓÃËÄÂÈ»¯Ì¼ÝÍÈ¡äåµ¥ÖÊ£¬¾­·ÖÒº¡¢ÕôÁóµÃ´Öäå¡£

£¨1£©²½Öè3Öеķ´Ó¦µÄÀë×Ó·½³Ìʽ                                              ¡£

£¨2£©²½Öè2ÖÐÒѾ­ÖƵÃÁËä壬»¹Òª½øÐв½Öè3ºÍ²½Öè4µÄÔ­ÒòÊÇ       äåÔªËØ¡£

£¨3£©²½Öè5ÖÐÝÍÈ¡ºÍ·ÖÒºËùÐèÒªµÄÖ÷Òª²£Á§ÒÇÆ÷Ϊ           ¡£

£¨4£©¿ÉÓÃÈçͼʵÑé×°Öþ«ÖÆ´Öäå¡£

¢ÙͼÖÐÀäȴˮӦ´ÓBµÄ   ¿Ú½øÈë(Ìî¡°a¡±»ò¡°b¡±) ¡£

¢ÚCÖмӱùµÄÄ¿µÄÊǽµÎ£¬¼õÉÙäåµÄ        ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸