ÌâÄ¿ÄÚÈÝ

9£®ÅжϺ¬ÑõËáÇ¿ÈõµÄÒ»Ìõ¾­Ñé¹æÂÉÊÇ£ºº¬ÑõËá·Ö×ӽṹÖк¬·ÇôÇ»ùÑõÔ­×ÓÊýÔ½¶à£¬¸Ãº¬ÑõËáµÄËáÐÔԽǿ£®ÈçϱíËùʾ
Ãû³Æ´ÎÂÈËáÁ×ËáÁòËá¸ßÂÈËá
½á¹¹Ê½Cl-OH
·ÇôÇ»ùÑõÔ­×ÓÊý0123
ËáÐÔÈõËáÖÐÇ¿ËáÇ¿Ëá×îÇ¿Ëá
£¨1£©ÑÇÁ×ËᣨH3PO3£©ºÍÑÇÉéËᣨH3AsO3£©·Ö×ÓʽÏàËÆ£¬µ«ËüÃǵÄËáÐÔ²î±ðºÜ´ó£¬H3PO3ÊÇÖÐÇ¿ËᣬH3AsO3¼ÈÓÐÈõËáÐÔÓÖÓÐÈõ¼îÐÔ£®ÓÉ´Ë¿ÉÍƳöËüÃǵĽṹʽ·Ö±ðΪ ºÍ£¬¶þÕßËáÐÔÇ¿ÈõÔ­ÒòÑÇÁ×ËẬÓÐ1¸ö·ÇôÇ»ùÑõÔ­×Ó£¬¶øÑÇÉéËáûÓзÇôÇ»ùÑõÔ­×Ó£¬¹ÊÑÇÁ×ËáµÄËáÐÔÇ¿ÓÚÑÇÉéËáµÄËáÐÔ
£¨2£©ÑÇÁ×ËáºÍÑÇÉéËáÓë¹ýÁ¿µÄÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ·Ö±ðΪ£º¢ÙH3PO3+2NaOH=Na2HPO3+2H2O£» ¢ÚH3AsO3+3NaOH=Na3AsO3+3H2O£®

·ÖÎö £¨1£©¸ù¾Ýº¬ÑõËá·Ö×ӽṹÖк¬·ÇôÇ»ùÑõÔ­×ÓÊýÔ½¶à£¬¸Ãº¬ÑõËáµÄËáÐÔԽǿ·ÖÎö£»
£¨2£©ËáºÍ¼î·´Ó¦Éú³ÉÑκÍË®£¬½áºÏ¼¸ÔªËá·ÖÎö£®

½â´ð ½â£º£¨1£©º¬ÑõËá·Ö×ӽṹÖк¬·ÇôÇ»ùÑõÔ­×ÓÊýÔ½¶à£¬¸Ãº¬ÑõËáµÄËáÐÔԽǿ£¬ÑÇÁ×ËẬÓÐ1¸ö·ÇôÇ»ùÑõÔ­×Ó£¬ÊÇÖÐÇ¿ËᣬÑÇÉéËáûÓзÇôÇ»ùÑõÔ­×Ó£¬¼ÈÓÐÈõËáÐÔÓÖÓÐÈõ¼îÐÔ£»
¹Ê´ð°¸Îª£ºÑÇÁ×ËẬÓÐ1¸ö·ÇôÇ»ùÑõÔ­×Ó£¬¶øÑÇÉéËáûÓзÇôÇ»ùÑõÔ­×Ó£¬¹ÊÑÇÁ×ËáµÄËáÐÔÇ¿ÓÚÑÇÉéËáµÄËáÐÔ£»
£¨2£©ËáºÍ¼î·´Ó¦Éú³ÉÑκÍË®£¬ÑÇÁ×ËáÖк¬ÓÐ2¸öôÇ»ù£¬ÊôÓÚ¶þÔªËᣬÑÇÉéËáÊôÓÚÈýÔªËᣬÑÇÁ×ËáºÍÑÇÉéËá·Ö±ðºÍÇâÑõ»¯ÄƵķ´Ó¦·½³ÌʽΪ£ºH3PO3+2NaOH=Na2HPO3+2H2O¡¢H3AsO3+3NaOH=Na3AsO3+3H2O£¬
¹Ê´ð°¸Îª£ºH3PO3+2NaOH=Na2HPO3+2H2O£»H3AsO3+3NaOH=Na3AsO3+3H2O£®

µãÆÀ ±¾Ì⿼²éÎïÖʵÄÐÔÖʼ°·´Ó¦·½³ÌʽµÄÊéд£¬Îª¸ßƵ¿¼µã£¬×¢Òâ°ÑÎÕÏ°ÌâÖеÄÐÅÏ¢¼°ËáÐÔµÄÅжÏΪ½â´ðµÄ¹Ø¼ü£¬²àÖØÓÚѧÉúµÄ·ÖÎöÄÜÁ¦µÄ¿¼²é£¬ÌâÄ¿ÄѶȲ»´ó£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
1£®½ðÈÚΣ»ú֮ϣ¬ÖйúÂÁ¹¤ÒµÊܵ½Á˾޴óµÄ³å»÷£®2009Äê6ÔÂ30ÈÕÖÁ7ÔÂ2ÈÕÔÚÉϺ£Ð¹ú¼Ê²©ÀÀÖÐÐľٰìµÄÖйú¹ú¼ÊÂÁ¹¤ÒµÕ¹ÀÀ»áÈ´Ìá³öÁË¡°Õ¹ÏÖÂÁÒµ°»È»Éú»ú¡±µÄ¿ÚºÅ£¬³ÉΪȫÇòÂÁÒµµÄ½¹µãºÍÁÁµã£®Ä³»¯Ñ§ÐËȤС×éµÄͬѧÃǾö¶¨ÂÁ¼°Æ仯ºÏÎïµÄÐÔÖʱȽÏÕ¹¿ªÌ½¾¿»î¶¯£®ËÑË÷µ½µÄһЩʵÑéÊý¾ÝÈçÏ£º
ÂÁ¼°ÆäÑõ»¯ÎïµÄÓйØʵÑéÊý¾ÝÈçÏ£º
ÈÛµã/¡æ·Ðµã/¡æȼÉÕÈÈ/kJ•mol-1
ÂÁ6602467602
Ñõ»¯ÂÁ20502980/
£¨1£©½«ÂÁƬ²åÈëÏõËṯÈÜÒºÖУ¬ÂÁƬ±íÃæ»Ò°µ£¬ÓÃʪ²¼²ÁÊÔºó£¬ÂÁƬÉϸ½×ÅÁËҺ̬Òø°×É«µÄÎïÖÊ£®½«¸ÃÂÁƬ¾²ÖÃÒ»¶Îʱ¼ä£¬ÂÁƬ±íÃ泤³ö°×É«Ðë×´ÎÊÖ³ÖÂÁƬʱ£¬°×É«Ðë×´ÎïÍÑÂ䣬ÂÁƬ·¢ÌÌ£®¾­ÊµÑé°×É«Ðë×´ÎïÄÜÈÜÓÚÏ¡ÑÎËáÇÒÎÞÆøÌå·Å³ö£®ÊÔд³öÖ±½Óµ¼ÖÂÂÁƬ·¢Ì̵ÄÈÈ»¯Ñ§·´Ó¦·½³Ìʽ4Al£¨s£©+3O2£¨g£©=2Al2O3£¨s£©£¬¡÷H=-2408KJ/mol£®
£¨2£©¼ô³¤Ô¼6cm¡¢¿í2cmµÄͭƬ¡¢ÂÁƬ¸÷һƬ£¬·Ö±ðÓýÓÏßÖùƽÐеع̶¨ÔÚÒ»¿éËÜÁÏ°åÉÏ£¨¼ä¸ô2cm£©£®½«Í­Æ¬ÓëÂÁƬ·Ö±ðºÍµçÁ÷±íµÄ¡°+¡±¡¢¡°-¡±¶ËÏàÁ¬½Ó£¬µçÁ÷±íÖ¸Õëµ÷µ½ÖмäλÖã®È¡Á½¸ö50mLµÄСÉÕ±­£¬ÔÚÒ»¸öÉÕ±­ÖÐ×¢ÈëÔ¼40mLµÄŨÏõËᣬÔÚÁíÒ»Ö»ÉÕ±­ÖÐ×¢Èë40mL0.5mol/LµÄÁòËáÈÜÒº£®ÊԻشðÏÂÁÐÎÊÌ⣺
¢ÙÁ½µç¼«Í¬Ê±²åÈëÏ¡ÁòËáÖУ¬µçÁ÷±íÖ¸ÕëÆ«ÏòÂÁ£¨Ìî¡°ÂÁ¡±»ò¡°Í­¡±£©¼«£¬ÂÁƬÉϵ缫·´Ó¦Ê½ÎªAl-3e-=Al3+£»
¢ÚÁ½µç¼«Í¬Ê±²åÈëŨÏõËáʱ£¬µçÁ÷±íÖ¸ÕëÆ«ÏòÍ­£¨Ìî¡°ÂÁ¡±»ò¡°Í­¡±£©¼«£¬´ËʱÂÁÊÇÕý£¨Ìî¡°Õý¡±»ò¡°¸º¡±£©¼«£¬ÂÁƬÉϵ缫·´Ó¦Ê½Îª2NO3-+2e-+4H+=2NO2¡ü+2H2O£®
£¨3£©ÄÜÔ´ÎÊÌâÊÇÈËÃǹØÐĵÄÈȵ㣬ÓÐÈËÌá³öÓýðÊôÂÁ×÷ȼÁÏ£¬ÕâÕæÊÇÒ»Öִ󵨶øÐÂÓ±µÄÉèÏ룮¶Ô´Ë£¬ÄãµÄ¹ÛµãÊÇB£¨Ìî¡°A¡±»ò¡°B¡±£©£¬ÆäÀíÓÉÊǹ¤ÒµÉÏÓõç½âÑõ»¯ÂÁµÄ·½·¨ÖÆÈ¡ÂÁ£¬ÒªÏûºÄ´óÁ¿µÄµçÄÜ£®A£®¿ÉÐÐB£®²»¿ÉÐУ®

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

¾«Ó¢¼Ò½ÌÍø