ÌâÄ¿ÄÚÈÝ
ÄÉÃײÄÁÏÔÚ¾ø´ó¶àÊýͬѧÑÛÀï¶¼·Ç³£ÉñÃØ¡£Ä³»¯Ñ§Ñ§Ï°Ð¡×é¾ö¶¨Í¨¹ý²éÔÄÓйØ×ÊÁÏ£¬×Ô¼º¶¯ÊÖʵÑ飬֯×öÄÉÃ×Ìú·Û¡£¡¾ÊµÑéÔÀí¡¿
(1)ÓÃÁòËáÑÇÌú¡¢²ÝËáÁ½ÖÖÈÜÒºÖÆ±¸²ÝËáÑÇÌú¾§Ìå(Èܽâ¶È½ÏС)¡£¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
__________________________________________________________________
(2)±ºÉÕ²ÝËáÑÇÌú¾§Ì壺FeC2O4¡¤2H2O
Fe+2CO2¡ü+2H2O¡ü
¡¾²Ù×÷²½Öè¡¿
(1)²ÝËáÑÇÌú¾§ÌåµÄÖÆ±¸
![]()
¢ÙÏÖÓÐÉÕ±¡¢²£Á§°ô¡¢Ò©³×¡¢ÂËÖ½¡¢ÍÐÅÌÌìÆ½¡¢FeSO4¡¢²ÝËáµÈʵÑéÓÃÆ·£¬ÎªÅäÖÆÉÏÊöÁ½ÖÖÈÜÒº£¬È±ÉÙµÄÒÇÆ÷ÓÐ______________________________________________________________¡£
¢ÚÈôÅäÖÆµÄFeSO2ÈÜÒºÂԳʻÆÉ«£¬¿ÉÄܵÄÔÒòÊÇ_____________________________________¡£
¢Û¼ìÑé²ÝËáÑÇÌú¾§ÌåÊÇ·ñÏ´¾»µÄ·½·¨ÊÇ_____________________________________________¡£
(2)ÄÉÃ×Ìú·ÛµÄÖÆ±¸
½«Ò»¸ù³¤¶ÈԼΪ12 cm¡«14 cmµÄ²£Á§¹ÜÔھƾ«ÅçµÆÉÏÉÕÖÆ³ÉÈçͼËùʾÐÎ×´£¬ÏòÆäÖмÓÈë»ÆÉ«²ÝËáÑÇÌú¾§Ìå¡£ÔÙ½«¹Ü¿Ú²¿·Ö¼ÓÈÈÀϸ£¬È»ºóÔھƾ«ÅçµÆÉϾùÔȼÓÈÈ¡£µ±²ÝËáÑÇÌú·ÛÄ©ÊÜÈÈÍêÈ«±äºÚ£¬Á¢¼´½«¹Ü¿ÚÀϸµÄ²¿·Ö·ÅÔÚ»ðÑæÉÕÈÛ·â±Õ£¬ÕâÑù±ãµÃµ½Á˸ߴ¿¶ÈµÄÄÉÃ×Ìú·Û¡£
¢ÙʵÑéÖзֽâ²úÉúCO2ËùÆðµÄ×÷ÓÃÊÇ________________¡£
¢ÚÖ¸µ¼ÀÏʦÌáÐÑͬѧÃÇӦ׼ȷ°ÑÎÕ²£Á§¹ÜµÄÉÕÈÛ·â±Õʱ¼ä¡£ÄãÈÏΪÉÕÈÛ¹ýÔç¡¢¹ýÍíµÄºó¹ûÊÇ________________¡£
![]()
ʵÑéÔÀí
FeSO4+H
²Ù×÷²½Öè
(1)²ÝËáÑÇÌú¾§ÌåµÄÖÆ±¸
¢Ù
¢ÚÑùÆ·ÖпÉÄܺ¬ÓÐFe3+¡¢ÕôÁóˮδÄÜÊÂÏÈ´¦Àí¡¢ÈÜÒºÅäÖÆÊ±¼ä½Ï³¤
¢ÛÈ¡×îºóµÄÏ´µÓÒº£¬²â¶¨Ï´µÓÒºµÄpH£¬ÈôÏ´µÓÒºµÄpHµÈÓÚ7£¬ËµÃ÷ÒѾϴµÓ¸É¾»
(2)ÄÉÃ×Ìú·ÛµÄÖÆ±¸
¢ÙÇý¸Ï×°ÖÃÖÐ¿ÕÆø£¬·ÀÖ¹ºìÈȵÄÌú±»Ñõ»¯
¢Ú¹ýÔç¹ýÍí¶¼½«µ¼ÖÂËùµÃÄÉÃ×Ìú·Û´¿¶È²»¹»(¹ýÔçδÄÜÍêÈ«·Ö½â¡¢¹ýÍíÔì³ÉÌú±»Ñõ»¯)