ÌâÄ¿ÄÚÈÝ

£¨13·Ö£©Ê³ÑÎÊÇÈÕ³£Éú»îµÄ±ØÐèÆ·£¬Ò²ÊÇÖØÒªµÄ»¯¹¤Ô­ÁÏ¡£

 

¢Å´ÖʳÑγ£º¬ÓÐÉÙÁ¿K+¡¢Ca2+¡¢Mg2+¡¢Fe3+¡¢SO42-µÈÔÓÖÊÀë×Ó£¬ÊµÑéÊÒÌá´¿NaClµÄÁ÷³ÌÈçÏ£º

ÌṩµÄÊÔ¼Á£º±¥ºÍNa2CO3ÈÜÒº   ±¥ºÍK2CO3ÈÜÒº  NaOHÈÜÒº  BaCl2ÈÜÒº  Ba(NO3)2ÈÜÒº      ËÄÂÈ»¯Ì¼

Óû³ýÈ¥ÈÜÒº¢ñÖеÄCa2+¡¢Mg2+¡¢Fe3+¡¢SO42-Àë×Ó£¬Ñ¡³öaËù´ú±íµÄÊÔ¼Á£¬°´µÎ¼Ó˳ÐòÒÀ´ÎΪ                       ¡££¨Ö»Ìѧʽ£©

¢ÆÓÃÌá´¿µÄNaClÅäÖÆ500mL4.00 mol?L-1NaClÈÜÒº£¬ËùÓÃÒÇÆ÷³ýÒ©³×¡¢²£Á§°ôÍ⻹ÓР                                         

                £¨ÌîÒÇÆ÷Ãû³Æ£©¡£

¢Çµç½â±¥ºÍʳÑÎË®µÄ×°ÖÃÈçͼËùʾ£¬Èç¹ûÔÚ±¥ºÍʳÑÎË®ÖеμӷÓ̪£¬Í¨µçºó        (ÌîX»òY)¼«¸½½üÈÜÒº±äºì£¬Ð´³öµç½â±¥ºÍʳÑÎË®µÄ»¯Ñ§·½³Ìʽ                               ¡£

¢ÈʵÑéÊÒÖƱ¸H2ºÍCl2ͨ³£²ÉÓÃÏÂÁз´Ó¦£º

Zn£«H2SO4====ZnSO4£«H2¡ü   MnO2£«4HCl(Ũ)MnCl2£«Cl2¡ü£«2H2O

¾Ý´Ë£¬´ÓÏÂÁÐËù¸øÒÇÆ÷×°ÖÃÖÐÑ¡ÔñÖƱ¸²¢ÊÕ¼¯H2µÄ×°Öà            £¨Ìî´úºÅ£©ºÍÖƱ¸²¢ÊÕ¼¯¸ÉÔï¡¢´¿¾»Cl2µÄ×°Öà            £¨Ìî´úºÅ£©¡£

¿ÉÑ¡ÓÃÖƱ¸ÆøÌåµÄ×°Öãº

£¨1£©NaOH¡¢BaCl2¡¢Na2CO3»ò£¨BaCl2¡¢NaOH¡¢Na2CO3£©  £¨2·Ö£©

   £¨´íÑ¡¡¢¶àÑ¡²»¸ø·Ö¡¢NaOHÈÜÒºµÄ¼ÓÈë˳Ðò¼°ÊÇ·ñ´ðNaOHÈÜÒº²»Ó°ÏìµÃ·Ö£©

£¨2£©Ììƽ¡¢ÉÕ±­¡¢500mLÈÝÁ¿Æ¿£¬½ºÍ·µÎ¹Ü(4·Ö)£¨ÉÙÒ»¸ö¿Û1·Ö£»Ð´´í1¸ö¿Û1·Ö£¬¿ÛÍêΪֹ£©

£¨3£©   X  £¨1·Ö£©

    £¨2·Ö£©

£¨4£© e  £¨2·Ö£©    d    £¨2·Ö£©

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ʳÑÎÊÇÈÕ³£Éú»îµÄ±ØÐèÆ·£¬Ò²ÊÇÖØÒªµÄ»¯¹¤Ô­ÁÏ¡£
£¨1£©´ÖʳÑγ£º¬ÓÐÉÙÁ¿K+¡¢Ca2+¡¢Mg2+¡¢Fe3+¡¢SO42-µÈÔÓÖÊÀë×Ó£¬ÊµÑéÊÒÌá´¿NaClµÄÁ÷³ÌÈçÏÂ
ÌṩµÄÊÔ¼Á£º±¥ºÍNa2CO3ÈÜÒº ±¥ºÍK2CO3ÈÜÒº NaOHÈÜÒº BaCl2ÈÜÒº Ba(NO3)2ÈÜÒº 75%ÒÒ´¼ ËÄÂÈ»¯Ì¼
¢Ù Óû³ýÈ¥ÈÜÒºIÖеÄCa2+¡¢Mg2+¡¢Fe3+¡¢SO42-Àë×Ó£¬Ñ¡³öaËù´ú±íµÄÊÔ¼Á£¬°´µÎ¼Ó˳ÐòÒÀ´ÎΪ__________
£¨Ö»Ìѧʽ£©¡£
¢ÚÏ´µÓ³ýÈ¥NaCl¾§Ìå±íÃ渽´øµÄÉÙÁ¿KCl£¬Ñ¡ÓõÄÊÔ¼ÁΪ__________________¡£
£¨2£©ÓÃÌá´¿µÄNaClÅäÖÆ500mL4.00 mol¡¤L-1NaClÈÜÒº
¢Ù ±¾´ÎʵÑéËùÓÃÒÇÆ÷³ýÒ©³×¡¢²£Á§°ôÍ⻹ÓÐ_____________£¨ÌîÒÇÆ÷³Æ£©¡£
¢ÚÓÃÍÐÅÌÌìƽ³ÆÈ¡µÄNaCl¾§ÌåµÄÖÊÁ¿Îª£º_____________£»
¢ÛÔÚÅäÖÆÈÜÒºµÄ¹ý³ÌÖжà´ÎÓõ½²£Á§°ô£¬ÔÚÈܽâʱ²£Á§°ôµÄ×÷ÓÃÊÇ£º_____________£¬ ÔÚÒÆҺʱ²£Á§°ôµÄ×÷ÓÃÊÇ£º__________________¡£
¢Ü¹Û²ìÒºÃæʱ£¬Èô¸©Êӿ̶ÈÏߣ¬»áʹËùÅäÖƵÄÈÜÒºµÄŨ¶È_________£¨Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡± »ò¡°ÎÞÓ°Ï족£¬ÏÂͬ)£»¼ÓÕôÁóˮʱ²»É÷³¬¹ýÁ˿̶ÈÏߺ󵹳ö²¿·ÖÈÜÒº£¬Ê¹ÒºÃæÓë¿Ì¶ÈÏßÏàÇУ¬»á__________£»
£¨3£©ÓæÑ=1.84g¡¤mL-1£¬ÖÊÁ¿·ÖÊýΪ98%µÄŨÁòËáÅäÖÆ200mL1mol¡¤L-1µÄÏ¡ÁòËáÓëÉÏÊöÅäÖÆÈÜÒºµÄ²½ÖèÉϵIJî±ðÖ÷ÒªÓÐÈýµã£º
¢Ù¼ÆË㣺ÀíÂÛÉÏӦȡŨÁòËáµÄÌå»ýV=___________mL(¾«È·µ½Ð¡ÊýµãºóÁ½Î»)£»
¢ÚÁ¿È¡£ºÓÉÓÚÁ¿Í²ÊÇÒ»ÖÖ´ÖÂÔµÄÁ¿¾ß£¬ÈçÏ뾫ȷÁ¿È¡£¬±ØÐèÑ¡ÓÃ_____________£¨ÌîÒÇÆ÷Ãû³Æ£©¡£
¢ÛÈܽ⣺ϡÊÍŨÁòËáµÄ·½·¨_________________________________________¡£

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

¾«Ó¢¼Ò½ÌÍø