ÌâÄ¿ÄÚÈÝ


Ó¡Ë¢µç·°åÊÇÓÉËÜÁϺÍÍ­²­¸´ºÏ¶ø³É£¬£¬¿ÌÖÆÓ¡Ë¢µç·ʱҪÓÃÈÜÒº×÷Ϊ¡°¸¯Ê´Òº¡±

ÈܽâÍ­¡£

£¨1£©Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º________________________________________;

£¨2£©Ð´³öFeCl3Óë×ãÁ¿µÄZn·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º__________________________________;

£¨3£©¸¯Ê´ÒºÓþûáʧЧ£¬µ«¿ÉÒÔ»ØÊÕÀûÓãº

¢ÙÈôÒªµÃµ½µ¥ÖÊÍ­£¬ÏÂÁÐÊÔ¼ÁÄÜʵÏÖµÄÊÇ£¨ÌîÐòºÅ£©

A£®ÂÈÆø      B£®ÂÁ      C£®ÏõËá      D£®Ï¡ÁòËá      E£®Ìú

¢ÚÈôÒª½«×ª»¯Îª£¬ÏÂÁÐÊÔ¼ÁÄÜʵÏÖµÄÊÇ£¨ÌîÐòºÅ£©

A£®ÂÈÆø      B¡¢ÂÁ      C¡¢ÏõËá      D£®Ï¡ÁòËá      E£®Ìú


¡¾´ð°¸¡¿

£¨1£© Cu +2FeCl3= Cu Cl2+ 2FeCl2

£¨2£© 2FeCl3£«3Zn===2Fe+3Zn Cl2

£¨3£©¢ÙB.E;¢ÚA.C


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ʵÑéÊÒ²ÉÓÃMgCl2¡¢AlCl3µÄ»ìºÏÈÜÒºÓë¹ýÁ¿°±Ë®·´Ó¦ÖƱ¸MgAl2O4¶þÖ÷ÒªÁ÷³ÌÈçÏÂ:

¢ÅÖÆ±¸MgAl2O4¹ý³ÌÖУ¬¸ßαºÉÕʱ·¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ                           ¡£

¢ÆÈçͼËùʾ£¬¹ýÂ˲Ù×÷ÖеÄÒ»´¦´íÎóÊÇ                 ¡£

ÅжÏÁ÷³ÌÖгÁµíÊÇ·ñÏ´¾»ËùÓõÄÊÔ¼ÁÊÇ            ¡£¸ßαºÉÕʱ£¬ÓÃÓÚÊ¢·Å¹ÌÌåµÄÒÇÆ÷Ãû³ÆÊÇ            ¡£

¢ÇÔÚ25¡æÏ£¬ÏòŨ¶È¾ùΪ0£®01 mol¡¤L-1µÄMgCl2ºÍAlCl3»ìºÏÈÜÒºÖÐÖðµÎ¼ÓÈ백ˮ£¬ÏÈÉú³É_________________³Áµí(Ìѧʽ)£¬Éú³É¸Ã³ÁµíµÄÀë×Ó·½³Ìʽ_____________________£¨ÒÑÖª25¡æÊ±Ksp[Mg(OH)2]=1.8¡Á10-11£¬Ksp[Al(OH)3]=3¡Á10 -34¡££©

¢ÈÎÞË®AlCl3(183¡æÉý»ª)Óö³±Êª¿ÕÆø¼´²úÉú´óÁ¿°×Îí£¬ÊµÑéÊÒ¿ÉÓÃÏÂÁÐ×°ÖÃÖÆ±¸¡£

×°ÖÃBÖÐÊ¢·Å±¥ºÍNaClÈÜÒº,¸Ã×°ÖõÄÖ÷Òª×÷ÓÃÊÇ          £»FÖÐÊÔ¼ÁµÄ×÷ÓÃÊÇ         £»ÓÃÒ»¼þÒÇÆ÷×°ÌîÊʵ±ÊÔ¼ÁºóÒ²¿ÉÆðµ½FºÍGµÄ×÷Óã¬Ëù×°ÌîµÄÊÔ¼ÁΪ                    ¡£

¢É½«Mg¡¢Cu×é³ÉµÄ3.92g»ìºÏÎïͶÈë¹ýÁ¿Ï¡ÏõËáÖУ¬³ä·Ö·´Ó¦ºó£¬¹ÌÌåÍêÈ«ÈܽâʱÊÕ¼¯µ½»¹Ô­²úÎïNOÆøÌå1.792L£¨±ê×¼×´¿ö£©£¬Ïò·´Ó¦ºóµÄÈÜÒºÖмÓÈë4mol/LµÄNaOHÈÜÒº80mLʱ½ðÊôÀë×ÓÇ¡ºÃÍêÈ«³Áµí¡£ÔòÐγɳÁµíµÄÖÊÁ¿Îª                        g¡£

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

¾«Ó¢¼Ò½ÌÍø