ÌâÄ¿ÄÚÈÝ
£¨14·Ö£©Ä³ÊµÑéС×éÀûÓÃÏÂͼËùʾʵÑé×°Öã¬Ê¹ÇâÆøÆ½»ºµØÍ¨¹ý×°ÓнðÊô¸ÆµÄÓ²Öʲ£Á§¹ÜÖÆÈ¡Ç⻯¸Æ£¬²¢·ÖÎö²úÆ·µÄ³É·Ö¼°´¿¶È¡£
![]()
£¨1£©ÊÔ¹ÜAµÄ×÷ÓÃÓÐ £» ¡£
£¨2£©Ç뽫ÏÂÁÐÖÆ±¸Ç⻯¸ÆµÄ²Ù×÷²½Öè²¹³äÍêÕû£º
¢Ù´ò¿ª»îÈûKͨÈëH2£»
¢Ú £»
¢Ûµãȼ¾Æ¾«µÆ£¬½øÐз´Ó¦£»
¢Ü·´Ó¦½áÊøºó£¬ £»
¢Ý²ð³ý×°Öã¬È¡³ö²úÎï¡£
£¨3£©¾·ÖÎö£¬²úÆ·ÖÐÖ»º¬¸Æ¡¢ÇâÁ½ÖÖÔªËØ¡£È¡ÉÙÁ¿²úÆ·£¬Ð¡ÐļÓÈëË®ÖУ¬¹Û²ìµ½ÓÐÆøÅÝð³ö£¬µÎÈëÒ»µÎ·Ó̪ÊÔÒº£¬ÈÜÒº±äºì¡£·´Ó¦µÄ»¯Ñ§·½³Ìʽ¿ÉÄÜÓÐCaH2+2H2O=Ca(OH)2+2H2¡ü¡¢ ¡£
£¨4£©È¡2.30 g²úÆ·ÈÜÓÚÕôÁóË®£¬Åä³É500 mLÈÜÒº£»È¡25.00 mL¸ÃÈÜÒºÓÚ×¶ÐÎÆ¿ÖУ¬µÎÈë2µÎ·Ó̪ÊÔÒº£¬ÓÃ0.2500 mol/L ÑÎËáµÎ¶¨£»Èý´ÎƽÐÐʵÑ飬ƽ¾ùÏûºÄÑÎËá22.00 mL¡£
¢ÙÅäÖÆÈÜÒºËùÓõIJ£Á§ÒÇÆ÷ÓнºÍ·µÎ¹Ü¡¢ÉÕ±¡¢²£Á§°ô¡¢Á¿Í²¡¢ £»
¢ÚÅжϵζ¨ÖÕµãµÄ·½·¨ÊÇ £»
¢Û²úÆ·ÖÐÁ½ÖֳɷÖÎïÖʵÄÁ¿Ö®±ÈΪ ¡£