ÌâÄ¿ÄÚÈÝ

19£®ÈéËáÊÇÊÀ½çÉϹ«ÈϵÄÈý´óÓлúËáÖ®Ò»£¬ÆäÓ¦Ó÷dz£¹ã·º£¬¿ÉÓÃ×÷ʳƷËáζ¼Á¡¢Ò½Ò©·À¸¯¼Á¡¢¹¤ÒµpHµ÷½Ú¼ÁµÈ£®ÈéËáµÄ½á¹¹¼òʽÈçͼËùʾ£®
Çë»Ø´ð£º
£¨1£©ÈéËá·Ö×ÓÖк¬Ñõ¹ÙÄÜÍŵÄÃû³ÆÊÇôÇ»ù¡¢ôÈ»ù£®
£¨2£©ÈéËáµÄ·Ö×ÓʽÊÇC3H6O3£®
£¨3£©ÈéËá¾ÛºÏµÃµ½µÄ¾ÛÈéËá¿ÉÒÔ³é³ÉË¿Ïߣ¬ÕâÖÖÏßÊÇÁ¼ºÃµÄÒ½ÓÃÊÖÊõ·ìºÏÏߣ®ÈéËáÔÚÒ»¶¨Ìõ¼þϾۺÏÉú³É¾ÛÈéËáµÄ»¯Ñ§·½³ÌʽÊÇ£®
£¨4£©ÈéËáÓжàÖÖͬ·ÖÒì¹¹Ì壬Çëд³öÈÎÒâÒ»ÖÖÂú×ãÏÂÁÐÌõ¼þµÄͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ£ºHCOOCH£¨OH£©CH3£®
¢Ùº¬õ¥»ùÇÒÄÜ·¢ÉúÒø¾µ·´Ó¦
¢ÚÄÜÓë½ðÊôÄÆ·´Ó¦²úÉúÇâÆø£®

·ÖÎö £¨1£©¸ù¾Ý¹ÙÄÜÍÅµÄ½á¹¹ÌØµãºÍÃû³Æ½áºÏÓлúÎïµÄ½á¹¹¼òʽÀ´»Ø´ð£»
£¨2£©¸ù¾ÝÓлúÎïµÄ½á¹¹¼òʽ½áºÏ·Ö×ÓʽµÄÊéдÀ´»Ø´ð£»
£¨3£©ÈéËáͨ¹ýõ¥»¯·´Ó¦½øÐеÄËõ¾Û·´Ó¦Éú³É¾ÛÈéË᣻
£¨4£©¢Ùº¬õ¥»ùÇÒÄÜ·¢ÉúÒø¾µ·´Ó¦£¬ËµÃ÷ÊǼ×Ëáijõ¥µÄ½á¹¹£»
¢ÚÄÜÓë½ðÊôÄÆ·´Ó¦²úÉúÇâÆø£¬ËµÃ÷º¬ÓÐôÇ»ù£®

½â´ð ½â£º£¨1£©ÈéËá·Ö×ÓÖк¬Ñõ¹ÙÄÜÍŵÄÃû³ÆÊÇôÇ»ù¡¢ôÈ»ù£¬¹Ê´ð°¸Îª£ºôÇ»ù¡¢ôÈ»ù£»
£¨2£©¸ù¾ÝÓлúÎïÈéËáµÄ½á¹¹¼òʽ£¬µÃµ½ÈéËá·Ö×ÓʽΪ£ºC3H6O3£¬¹Ê´ð°¸Îª£ºC3H6O3£»
£¨3£©ÈéËáÔÚÒ»¶¨Ìõ¼þÏ·¢Éú·´Ó¦µÃµ½¾ÛÈéËáµÄ»¯Ñ§·½³ÌʽÊÇ£º£¬
¹Ê´ð°¸Îª£º£»
£¨4£©¢Ùº¬õ¥»ùÇÒÄÜ·¢ÉúÒø¾µ·´Ó¦£¬ËµÃ÷ÊǼ×Ëáijõ¥HCOO-µÄ½á¹¹£»
¢ÚÄÜÓë½ðÊôÄÆ·´Ó¦²úÉúÇâÆø£¬ËµÃ÷º¬ÓÐôÇ»ù-OH£¬·Ö×ÓʽΪC3H6O3£¬Æä½á¹¹¼òʽ¿ÉÒÔÊÇHCOOCH£¨OH£©CH3£¬¹Ê´ð°¸Îª£ºHCOOCH£¨OH£©CH3£®

µãÆÀ ±¾Ì⿼²éѧÉúÓлúÎï½á¹¹¼òʽºÍ·Ö×ÓʽµÄÊéд¡¢Í¬·ÖÒì¹¹ÌåµÄÊéд¡¢ÎïÖʵÄÐÔÖʺ͹ÙÄÜÍŵĹØÏµµÈ֪ʶ£¬ÄѶÈÖеȣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
11£®½«·Ï¾ÉпÃÌµç³Ø´¦ÀíµÃµ½º¬ÃÌ»ìºÏÎ¼ÈÄܼõÉÙËü¶Ô»·¾³µÄÎÛȾ£¬ÓÖÄÜʵÏÖ·Ïµç³ØµÄ×ÊÔ´»¯ÀûÓã®
¢ñ£®»ØÊÕ¶þÂÈ»¯ÃÌ£º½«·Ï¾ÉпÃ̵çºÜ´¦Àí£¬µÃµ½º¬ÃÌ»ìºÏÎÏò¸Ã»ìºÏÎï¼ÓÈëŨÑÎËá²¢¼ÓÈÈ£®
£¨1£©Ð´³öMnO£¨OH£©ÓëŨÑÎËá·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º2MnO£¨OH£©+6HCl $\frac{\underline{\;\;¡÷\;\;}}{\;}$2MnCl2+Cl2¡ü+4H2O£®
II£®ÃÌ»ØÊÕз½·¨£ºÏòº¬ÃÌ»ìºÏÎï¼ÓÈëÒ»¶¨Á¿µÄÏ¡ÁòËᡢϡ²ÝËᣬ²¢²»¶Ï½Á°èÖÁÎÞÆøÅÝΪֹ£¬ÆäÖ÷Òª·´Ó¦Îª£º2MnOOH+MnO2+2H2C2O4+3H2SO4=3MnSO4+4CO2¡ü+6H2O
£¨2£©ÓëʹÓÃŨÑÎËá»ØÊÕÃÌÏà±È£¬Ð·½·¨µÄÓŵãÊǹ¤ÒÕÁ÷³Ì¼òµ¥£»Éú³ÉCO2ºÍH2O²»Ó°ÏìMnSO4´¿¶È£»·´Ó¦¹ý³ÌÎÞÓж¾Óк¦ÎïÖÊÉú³É£¬²»Ôì³É¶þ´ÎÎÛȾ£»·ÏÎï×ÊÔ´»¯µÈ£¨´ð1µã¼´¿É£©£®
£¨3£©ÓûÒÔMnSO4ÈÜҺΪԭÁÏÖÆ±¸MnCO3£¬Ñ¡ÓõļÓÁÏ·½Ê½C£¨Ìî×Öĸ£©£¬Ô­ÒòÊDZÜÃâÏÈÉú³ÉMn£¨OH£©2£® ¼ºÖª£ºKsp£¨MnCO3£©=2.3¡Á10-11¡¢Ksp[Mn£¨OH£©2]=2£®l¡Á10-13
A£®½«MnSO4ÈÜÒºÓëNa2CO3ÈÜҺͬʱ¼ÓÈëµ½·´Ó¦ÈÝÆ÷ÖÐ
B£®½«MnSO4ÈÜÒº»ºÂý¼ÓÈ뵽ʢÓÐNa2CO3ÈÜÒ´µÄ·´Ó¦ÈÝÆ÷ÖÐ
C£®½«Na2CO3ÈÜÒº»ºÂý¼ÓÈ뵽ʢÓÐMnSO4ÈÜÒºµÄ·´Ó¦ÈÝÆ÷ÖÐ
D£®ÒÔÉÏÈýÖÖ·½Ê½¶¼ÎÞ·¨µÃµ½MnCO3
III£®ÏÖÒÔº¬ÃÌ»ìºÏÎïΪԭÁÏÖÆ±¸ÃÌпÌúÑõÌåµÄÖ÷Ë£Á÷³ÌÈçͼl Ëùʾ£º

»Ø´ðÏÂÁÐÎÊÌ⣺
£¨4£©MnxZn1-xFe2O4ÖÐÌúÔªËØ»¯ºÏ¼ÛΪ+3£¬ÔòÃÌÔªËØµÄ»¯ºÏ¼ÛΪ+2£®
£¨5£©»îÐÔÌú·Û³ý¹¯Ê±£¬Ìú·ÛµÄ×÷ÓÃÊÇ»¹Ô­¼Á £¨Ìî¡°Ñõ»¯¼Á¡±»ò¡°»¹Ô­¼Á¡±£©£®
£¨6£©Ëá½þʱ£¬MnO2ÓëH2SO4¡¢H2O2·´Ó¦µÄÀë×Ó·½³ÌʽΪMnO2+H2O2 +2H+=Mn2++O2¡ü+2H2O£®
£¨7£©³ý¹¯ÊÇÒÔµªÆøÎªÔØÆø´µÈëÂËÒºÖУ¬´ø³ö¹¯ÕôÆø¾­KMnO4ÈÜÒº½øÐÐÎüÊÕ¶øÊµÏֵģ®ÔÚºãÎÂϲ»Í¬pHʱ£¬KMnO4ÈÜÒº¶Ô·ÉµÄµ¥Î»Ê±¼äÈ¥³ýÂʼ°Ö÷Òª²úÎïÈçͼ2 Ëùʾ£º
¢Ùд³öpH=10ʱKMnO4ÈÜÒºÎüÊÕ¹¯ÕôÆøµÄÀë×Ó·½³Ìʽ5Hg+2MnO4-+16H+¨T2Mn2++5Hg2++8H2O£®
¢ÚÔÚÇ¿ËáÐÔ»·¾³Öй¯µÄµ¥Î»Ê±¼äÈ¥³ýÂʸߣ¬ÆäÔ­Òò³ýÇâÀë×ÓŨ¶ÈÔö´óʹKMnO4ÈÜÒºµÄÑõ»¯ÐÔÔöÇ¿Í⣬»¹¿ÉÄÜÊÇMn2+¶Ô·´Ó¦Æð´ß»¯×÷Ó㬵¥Î»Ê±¼äÄÚÈ¥³ýÂʸߣ®

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

¾«Ó¢¼Ò½ÌÍø