ÌâÄ¿ÄÚÈÝ

ÔÚÒ»Ö§½à¾»µÄÊÔ¹ÜÖмÓÈëÉÙÁ¿µÄAgNO3ÈÜÒº£¬È»ºó£¬µÎ¼ÓÏ¡°±Ë®£¬Ö±µ½Éú³ÉµÄ³Áµí¸ÕºÃÈܽâΪֹ£¬µÃµ½µÄÎÞÉ«ÈÜÒº³ÆÎªÒø°±ÈÜÒº£®ÔÚ´ËÈÜÒºÖеμӼ¸µÎÒÒÈ©ÈÜÒº£¬¾­Ë®Ô¡¼ÓÈÈ£¬¿É¹Û²ìµ½µÄÏÖÏóÊÇ
 
´Ë·´Ó¦¿ÉÒÔÓÃÀ´¼ìÑéÈ©»ùµÄ´æÔÚ£®·´Ó¦µÄÖ÷Òª»¯Ñ§·½³ÌʽΪ
 
£®ÔÚNaOHÈÜÒºÖмÓÈ뼸µÎCuSO4ÈÜÒº£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ
 
£»ÔÚÆäÖмÓÈëÒÒÈ©ÈÜÒº²¢¼ÓÈÈÖó·Ð£¬¿É¹Û²ìµ½µÄÏÖÏóÊÇ
 
£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
 
£®´Ë·´Ó¦Ò²¿ÉÓÃÓÚ¼ìÑéÈ©»ù£®
¿¼µã£ºÒÒÈ©µÄÒø¾µ·´Ó¦
רÌ⣺ʵÑéÌâ
·ÖÎö£ºÒÒÈ©º¬ÓÐÈ©»ù£¬¾ßÓл¹Ô­ÐÔ£¬ÒÒÈ©ÓëÒø°±ÈÜÒºÔÚˮԡ¼ÓÈȵÄÌõ¼þÏ·¢ÉúÑõ»¯»¹Ô­·´Ó¦£¬Éú³ÉÒÒËá狀ÍÒø¡¢°±ÆøºÍË®£¬·´Ó¦ºóÔÙÔÚÊԹܱÚÉÏÎö³ö¹âÁÁµÄÒø£¬¸Ã·´Ó¦½Ð×öÒø¾µ·´Ó¦£¬·²ÊǺ¬ÓÐÈ©»ùµÄÎïÖʾùÄÜ·¢ÉúÒø¾µ·´Ó¦£¬¿ÉÒÔÓÃÀ´¼ìÑéÈ©»ùµÄ´æÔÚ£»
NaOHÈÜÒºÖмÓÈ뼸µÎCuSO4ÈÜÒº£¬·¢Éú¸´·Ö½â·´Ó¦Éú³ÉÇâÑõ»¯Í­³ÁµíºÍÁòËáÄÆ£¬ÇâÑõ»¯Í­¾ßÓÐÑõ»¯ÐÔ£¬¿ÉÒÔÑõ»¯È©»ù±¾Éí±»»¹Ô­Îª×©ºìÉ«µÄÑõ»¯ÑÇÍ­£¬ÀûÓø÷´Ó¦Ò²¿É¼ìÑéÈ©»ùµÄ´æÔÚ£®
½â´ð£º ½â£ºÒÒÈ©º¬ÓÐÈ©»ù£¬¾ßÓл¹Ô­ÐÔ£¬ÔÚˮԡ¼ÓÈȵÄÌõ¼þÏÂÓëÒø°±ÈÜÒº·¢ÉúÑõ»¯»¹Ô­·´Ó¦£¬ÔÚÊÔ¹ÜÉÏÎö³öÒø£»Äܹ»ÓëÐÂÖÆµÄÇâÑõ»¯Í­ÔÚ¼ÓÈȵÄÌõ¼þÏ·¢ÉúÑõ»¯»¹Ô­·´Ó¦£¬Éú³ÉשºìÉ«µÄ³Áµí£¬ÀûÓÃÕâÁ½¸ö·´Ó¦¿ÉÒÔ¼ìÑéÈ©»ùµÄ´æÔÚ£®
ËùÒÔ£ºÔÚÒ»Ö§½à¾»µÄÊÔ¹ÜÖмÓÈëÉÙÁ¿µÄAgNO3ÈÜÒº£¬È»ºó£¬µÎ¼ÓÏ¡°±Ë®£¬Ö±µ½Éú³ÉµÄ³Áµí¸ÕºÃÈܽâΪֹ£¬µÃµ½µÄÎÞÉ«ÈÜÒº³ÆÎªÒø°±ÈÜÒº£®ÔÚ´ËÈÜÒºÖеμӼ¸µÎÒÒÈ©ÈÜÒº£¬¾­Ë®Ô¡¼ÓÈÈ£¬¿É¹Û²ìµ½µÄÏÖÏóÊÇ£ºÊԹܱÚÉϸ½×ÅÒ»²ã¹âÁÁµÄÒø£¬´Ë·´Ó¦¿ÉÒÔÓÃÀ´¼ìÑéÈ©»ùµÄ´æÔÚ£®·´Ó¦µÄÖ÷Òª»¯Ñ§·½³ÌʽΪ£ºCH3CHO+2Ag£¨NH3£©2OH
ˮԡ¼ÓÈÈ
CH3COONH4+H2O+2Ag¡ý+3NH3¡ü£¬ÔÚNaOHÈÜÒºÖмÓÈ뼸µÎCuSO4ÈÜÒº£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ£ºCu2++2OH-=Cu£¨OH£©2¡ý£¬ÔÚÆäÖмÓÈëÒÒÈ©ÈÜÒº²¢¼ÓÈÈÖó·Ð£¬¿É¹Û²ìµ½µÄÏÖÏóÊÇ£ºÈÜÒºÖгöÏÖשºìÉ«µÄ³Áµí£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºCH3CHO+2Cu£¨OH£©2
¡÷
CH3COOH+2H2O+Cu2O¡ý£¬´Ë·´Ó¦Ò²¿ÉÓÃÓÚ¼ìÑéÈ©»ù£¬
¹Ê´ð°¸Îª£ºÊԹܱÚÉϸ½×ÅÒ»²ã¹âÁÁµÄÒø£»CH3CHO+2Ag£¨NH3£©2OH
ˮԡ¼ÓÈÈ
CH3COONH4+H2O+2Ag¡ý+3NH3¡ü£»Cu2++2OH-=Cu£¨OH£©2¡ý£»ÈÜÒºÖгöÏÖשºìÉ«µÄ³Áµí£»
CH3CHO+2Cu£¨OH£©2
¡÷
CH3COOH+2H2O+Cu2O¡ý£®
µãÆÀ£º±¾ÌâΪÓлúʵÑéÌ⣬¿¼²éÁËÒø¾µ·´Ó¦ºÍÈ©ÓëÐÂÖÆÇâÑõ»¯Í­·´Ó¦ÊµÑéµÄ²Ù×÷¹ý³Ì£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽµÄÊéд£¬ÊµÑéµÄÏÖÏó£¬ÊµÑéµÄÓÃ;£¬ÖªµÀÒÒÈ©µÄ½á¹¹ÓÈÆäÊÇÆä¹ÙÄÜÍÅΪȩ»ùÊǽâÌâµÄ¹Ø¼ü£¬ÕÆÎÕ¸ÃʵÑéʱ»¹Ó¦¸Ã×¢ÒâÒø°±ÈÜÒºµÄÅäÖã¬ÊµÑéµÄ×¢ÒâÊÂÏ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÆßË®ÁòËáþ£¨MgSO4?7H2O£©ÔÚӡȾ¡¢ÔìÖ½ºÍÒ½Ò©µÈ¹¤ÒµÉ϶¼Óй㷺µÄÓ¦Óã¬ÀûÓû¯¹¤³§Éú²úÅðɰµÄ·ÏÔü-Ò»ÅðþÄà¿ÉÖÆÈ¡ÆßË®ÁòËáþ£®ÅðþÄàµÄÖ÷Òª³É·ÖÊÇMgCO3£¬»¹º¬ÓÐÆäËûÔÓÖÊ£¨MgO¡¢SiO2¡¢Fe2O3¡¢FeO¡¢CaO¡¢Al2O3¡¢MnOµÈ£©£®
±í1¡¡²¿·ÖÑôÀë×ÓÒÔÇâÑõ»¯ÎïÐÎʽÍêÈ«³ÁµíʱÈÜÒºµÄpH
³ÁµíÎïAl£¨OH£©3Fe£¨OH£©3Fe£¨OH£©2Mn£¨OH£©2Mg£¨OH£©2
pHÖµ5.23.29.710.411.2
±í2¡¡Á½ÖÖÑεÄÈܽâ¶È£¨µ¥Î»Îªg/100gË®£©
ζÈ/¡æ1030405060
CaSO40.190.210.210.200.19
MgSO4?7H2O30.935.540.845.6/
ÅðþÄàÖÆÈ¡ÆßË®ÁòËáþµÄ¹¤ÒÕÁ÷³ÌÈçÏ£º

¸ù¾ÝÒÔÉÏÁ÷³Ìͼ²¢²Î¿¼±í¸ñpHÊý¾ÝºÍÈܽâ¶ÈÊý¾Ý£¬ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©³ÁµíCµÄ»¯Ñ§Ê½ÊÇ
 

£¨2£©¹ýÂË¢óÐè³ÃÈȹýÂ˵ÄÀíÓÉÊÇ
 

£¨3£©²Ù×÷¢ñºÍ²Ù×÷¢òµÄÃû³Æ·Ö±ðΪ
 
¡¢
 

£¨4£©ÂËÒºIÖмÓÈëÅðþÄ࣬µ÷½ÚÈÜÒºµÄpH=5¡«6£¬¿É³ýÈ¥ÂËÒº¢ñÖÐ
 
£¨ÌîÀë×Ó·ûºÅ£©£»¼ÓÈëNaClOÈÜÒº¼ÓÈÈÖó·Ð£¬¿É½«Fe2+Ñõ»¯³ÉFe3+µÄͬʱ£¬»¹»á½«ÈÜÒºÖеÄMn2+Ñõ»¯³ÉMnO2£¬Ð´³öMn2+±»Ñõ»¯³ÉMnO2µÄÀë×Ó·´Ó¦·½³ÌʽΪ
 

£¨5£©ÅðþÄàÊÇÅðþ¿óÉú²úÅðɰ£¨Na2B4O7?10H2O£©Ê±µÄ·ÏÔü£®½«Åðþ¿ó£¨Mg2B2O5?H2O£©ìÑÉÕ¡¢·ÛËéºó¼ÓË®ºÍ´¿¼î£¬ÔÚ¼ÓÈȼÓѹÏÂͨÈëCO2¿ÉµÃÅðɰ£®´Ë·´Ó¦ÎªìؼõС·´Ó¦£¬ÊÔд³ö·´Ó¦·½³Ìʽ
 
£®

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

¾«Ó¢¼Ò½ÌÍø