ÌâÄ¿ÄÚÈÝ
Âí¶ûÊÏÑÎAΪdzÀ¶ÂÌÉ«½á¾§£¬ÊÇÒ»ÖÖÖØÒªµÄ»¯Ñ§ÊÔ¼Á£¬¿ÉÓÃ×÷¾ÛºÏ´ß»¯¼Á¡£AµÄÈÜÒº¾ßÓÐÒÔÏÂת»¯¹Øϵ
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©°×É«³ÁµíCµÄ»¯Ñ§Ê½Îª_____________£¬ÎÞÉ«ÆøÌåEµÄµç×ÓʽΪ_________________¡£
£¨2£©ÓÉDת»¯³ÉΪFµÄ»¯Ñ§·´Ó¦·½³ÌʽΪ_________________¡£
£¨3£©³ÆÈ¡Ò»¶¨Á¿AµÄ¾§Ì壬¾²â¶¨Æä½á¾§Ë®º¬Á¿Ô¼Îª27.6%¡£½«ÆäÍêÈ«ÈÜÓÚË®Åä³ÉÈÜÒº£¬¾¹ýÉÏÊöת»¯×î¶à¿ÉµÃµ½µÄCºÍFµÄÎïÖʵÄÁ¿Ö®±ÈΪ2©U1¡£ÓÉ´Ë¿ÉÈ·¶¨AµÄ»¯Ñ§Ê½Îª_______________¡£
£¨4£©ÉÏÊöת»¯¹ØϵÖУ¬ÈôÓÃÏ¡ÏõËá´úÌæÏ¡ÑÎËᣬ_________£¨Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±£©Í¨¹ý¸Ãת»¯¹Øϵȷ¶¨AµÄ×é³É£¬ÀíÓÉÊÇ_______________________¡£Òª×¼È·³Æ³öCµÄÖÊÁ¿£¬ÐèÒªÖظ´½øÐиÉÔï¡¢ÀäÈ´¡¢³ÆÁ¿µÄ²Ù×÷£¬Ä¿µÄÊÇ_____________________¡£
£¨5£©AµÄ±ê×¼ÈÜÒº£¬³£ÓÃÓڱ궨¸ßÃÌËá¼ØÈÜÒºµÈ¡£Çëд³öÔÚ¸ßÃÌËá¼ØÈÜÒºÖеμÓAµÄ±ê×¼ÈÜҺʱ£¬Ëù·¢ÉúµÄÀë×Ó·´Ó¦·½³Ìʽ£¬²¢±êÃ÷µç×ÓתÒÆÇé¿ö¡£________________________¡£
£¨1£©°×É«³ÁµíCµÄ»¯Ñ§Ê½Îª_____________£¬ÎÞÉ«ÆøÌåEµÄµç×ÓʽΪ_________________¡£
£¨2£©ÓÉDת»¯³ÉΪFµÄ»¯Ñ§·´Ó¦·½³ÌʽΪ_________________¡£
£¨3£©³ÆÈ¡Ò»¶¨Á¿AµÄ¾§Ì壬¾²â¶¨Æä½á¾§Ë®º¬Á¿Ô¼Îª27.6%¡£½«ÆäÍêÈ«ÈÜÓÚË®Åä³ÉÈÜÒº£¬¾¹ýÉÏÊöת»¯×î¶à¿ÉµÃµ½µÄCºÍFµÄÎïÖʵÄÁ¿Ö®±ÈΪ2©U1¡£ÓÉ´Ë¿ÉÈ·¶¨AµÄ»¯Ñ§Ê½Îª_______________¡£
£¨4£©ÉÏÊöת»¯¹ØϵÖУ¬ÈôÓÃÏ¡ÏõËá´úÌæÏ¡ÑÎËᣬ_________£¨Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±£©Í¨¹ý¸Ãת»¯¹Øϵȷ¶¨AµÄ×é³É£¬ÀíÓÉÊÇ_______________________¡£Òª×¼È·³Æ³öCµÄÖÊÁ¿£¬ÐèÒªÖظ´½øÐиÉÔï¡¢ÀäÈ´¡¢³ÆÁ¿µÄ²Ù×÷£¬Ä¿µÄÊÇ_____________________¡£
£¨5£©AµÄ±ê×¼ÈÜÒº£¬³£ÓÃÓڱ궨¸ßÃÌËá¼ØÈÜÒºµÈ¡£Çëд³öÔÚ¸ßÃÌËá¼ØÈÜÒºÖеμÓAµÄ±ê×¼ÈÜҺʱ£¬Ëù·¢ÉúµÄÀë×Ó·´Ó¦·½³Ìʽ£¬²¢±êÃ÷µç×ÓתÒÆÇé¿ö¡£________________________¡£
£¨1£©BaSO4£»
£¨2£©4Fe(OH)2+O2+2H2O = 4Fe(OH)3
£¨3£©(NH4)2Fe(SO4)2¡¤6H2O
£¨4£©²»ÄÜ£»ÒòΪ²»ÄÜÈ·¶¨ËüÊÇÁòËáÑλ¹ÊÇÑÇÁòËáÑΣ»³¹µ×³ýÈ¥C±íÃæµÄË®·Ö
£¨5£©
£¨2£©4Fe(OH)2+O2+2H2O = 4Fe(OH)3
£¨3£©(NH4)2Fe(SO4)2¡¤6H2O
£¨4£©²»ÄÜ£»ÒòΪ²»ÄÜÈ·¶¨ËüÊÇÁòËáÑλ¹ÊÇÑÇÁòËáÑΣ»³¹µ×³ýÈ¥C±íÃæµÄË®·Ö
£¨5£©
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿