ÌâÄ¿ÄÚÈÝ
£¨2012?»Ý³ÇÇøÄ£Ä⣩A¡¢B¡¢C¡¢D¡¢E¡¢FÁùÖÖÎïÖʶ¼ÊdzõÖл¯Ñ§Ñ§Ï°¹ý³ÌÖÐÉæ¼°µ½µÄÓйػ¯ºÏÎËüÃÇÖ®¼äµÄת»¯¹ØϵÈçÏÂͼËùʾ£¬ÆäÖÐAÊdz£ÓõĽ¨Öþ²ÄÁÏ£¬DÊÇ×î³£¼ûµÄÈܼÁ£®
£¨1£©¸ù¾ÝÉÏͼÍƶϣ¬AÊÇ»¯Ñ§Ê½
£¨2£©Ð´³ö·´Ó¦¢ÚµÄ»¯Ñ§·½³Ìʽ
£¨3£©·´Ó¦¢ÛÊôÓÚ
£¨1£©¸ù¾ÝÉÏͼÍƶϣ¬AÊÇ»¯Ñ§Ê½
CaCO3
CaCO3
£¬EµÄË׳ÆÊÇÊìʯ»Ò
Êìʯ»Ò
£»£¨2£©Ð´³ö·´Ó¦¢ÚµÄ»¯Ñ§·½³Ìʽ
CO2+2NaOH=Na2CO3+H2O
CO2+2NaOH=Na2CO3+H2O
£»·´Ó¦¢ÜµÄ»¯Ñ§·½³ÌʽCa£¨OH£©2+Na2CO3=CaCO3¡ý+2NaCl
Ca£¨OH£©2+Na2CO3=CaCO3¡ý+2NaCl
£»£¨3£©·´Ó¦¢ÛÊôÓÚ
»¯ºÏ
»¯ºÏ
·´Ó¦£¨Óá°»¯ºÏ¡±¡¢¡°·Ö½â¡±¡¢¡°Öû»¡±»ò¡°¸´·Ö½â¡±Ìî¿Õ£©£®ÊµÑéÊÒÖÆÈ¡ÎïÖÊC²»ÓÃÎïÖÊFµÄÔÒòÊÇ·´Ó¦ËÙ¶ÈÌ«¿ì£¬²»ÀûÓÚÆøÌåµÄÊÕ¼¯
·´Ó¦ËÙ¶ÈÌ«¿ì£¬²»ÀûÓÚÆøÌåµÄÊÕ¼¯
£®·ÖÎö£º±¾ÌâÊôÓÚÍƶÏÌ⣬¸ù¾ÝÌâÄ¿¸ø³öµÄÁ÷³ÌͼºÍÐÅÏ¢£ºAÊdz£ÓõĽ¨Öþ²ÄÁÏ£¬DÊÇ×î³£¼ûµÄÈܼÁ£¬Òò´ËAÊÇ̼Ëá¸Æ£¬DÊÇË®£»Ì¼Ëá¸ÆÔÚ¸ßεÄÌõ¼þÏÂÉú³ÉÑõ»¯¸ÆºÍË®£»¶þÑõ»¯Ì¼ÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦Éú³É̼ËáÄƺÍË®£¬Òò´ËBÊÇÑõ»¯¸Æ£¬CÊǶþÑõ»¯Ì¼£»Ñõ»¯¸ÆºÍË®·´Ó¦Éú³ÉÇâÑõ»¯¸Æ£¬ÇâÑõ»¯¸ÆµÄË×ÃûÊÇÊìʯ»Ò»òÏûʯ»Ò£»ÇâÑõ»¯¸ÆºÍ̼ËáÄÆÈÜÒº·´Ó¦Éú³É̼Ëá¸Æ°×É«³ÁµíºÍÇâÑõ»¯ÄÆ£¬Òò´ËEÊÇÇâÑõ»¯¸Æ£¬FÊÇ̼ËáÄÆ£®»¯ºÏ·´Ó¦ÊÇ¡°¶à±äÒ»¡±£®
½â´ð£º½â£º£¨1£©AÊdz£ÓõĽ¨Öþ²ÄÁÏ£¬DÊÇ×î³£¼ûµÄÈܼÁ£¬Òò´ËAÊÇ̼Ëá¸Æ£¬ÇâÑõ»¯¸ÆµÄË×ÃûÊÇÊìʯ»Ò»òÏûʯ»Ò£¬¹Ê´ð°¸Îª£ºCaCO3£¬Êìʯ»Ò£¨»òÏûʯ»Ò£©
£¨2£©¶þÑõ»¯Ì¼ÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦Éú³É̼ËáÄƺÍË®£¬ÇâÑõ»¯¸ÆºÍ̼ËáÄÆÈÜÒº·´Ó¦Éú³É̼Ëá¸Æ°×É«³ÁµíºÍÇâÑõ»¯ÄÆ£¬Åäƽ¼´¿É£¬¹Ê´ð°¸Îª£ºCO2+2NaOH=Na2CO3+H2O£»Ca£¨OH£©2+Na2CO3=CaCO3¡ý+2NaCl£»
£¨3£©Ñõ»¯¸ÆºÍË®·´Ó¦Éú³ÉÇâÑõ»¯¸Æ£¬ÊôÓÚ»¯ºÏ·´Ó¦£»Ì¼ËáÄÆÊÇ·ÛÄ©£¬·´Ó¦ËÙ¶ÈÌ«¿ì£¬²»ÀûÓÚÆøÌåµÄÊÕ¼¯£»¹Ê´ð°¸Îª£º»¯ºÏ£¬·´Ó¦ËÙ¶ÈÌ«¿ì£¬²»ÀûÓÚÆøÌåµÄÊÕ¼¯
£¨2£©¶þÑõ»¯Ì¼ÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦Éú³É̼ËáÄƺÍË®£¬ÇâÑõ»¯¸ÆºÍ̼ËáÄÆÈÜÒº·´Ó¦Éú³É̼Ëá¸Æ°×É«³ÁµíºÍÇâÑõ»¯ÄÆ£¬Åäƽ¼´¿É£¬¹Ê´ð°¸Îª£ºCO2+2NaOH=Na2CO3+H2O£»Ca£¨OH£©2+Na2CO3=CaCO3¡ý+2NaCl£»
£¨3£©Ñõ»¯¸ÆºÍË®·´Ó¦Éú³ÉÇâÑõ»¯¸Æ£¬ÊôÓÚ»¯ºÏ·´Ó¦£»Ì¼ËáÄÆÊÇ·ÛÄ©£¬·´Ó¦ËÙ¶ÈÌ«¿ì£¬²»ÀûÓÚÆøÌåµÄÊÕ¼¯£»¹Ê´ð°¸Îª£º»¯ºÏ£¬·´Ó¦ËÙ¶ÈÌ«¿ì£¬²»ÀûÓÚÆøÌåµÄÊÕ¼¯
µãÆÀ£º±¾¿¼µãÊôÓÚÎïÖʵÄÍƶÏÌ⣬ÊÇͨ¹ý¶ÔʵÑé·½·¨ºÍ¹ý³ÌµÄ̽¾¿£¬ÔڱȽϼø±ðµÄ»ù´¡ÉÏ£¬µÃ³öÁËÕýÈ·µÄʵÑé½áÂÛ£®±¾¿¼µãÊÇÖп¼µÄÖØÒªÄÚÈÝÖ®Ò»£¬Ò»°ãÓÐÁ½ÖÖÀàÐÍ£ºÒ»ÊÇͼ¿òʽÍƶÏÌ⣻¶þÊÇÎÄ×ÖÃèÊöÐÍÍƶÏÌ⣻±¾ÌâÊôÓÚµÚÒ»ÖÖÀàÐÍ£®²»ÂÛÄÄÒ»ÖÖÀàÐÍ£¬¶¼ÊÇͨ¹ýʵÑéÏÖÏ󣬴ӶøµÃ³öÎïÖʵÄ×é³É£®´Ë¿¼µãÖ÷Òª³öÏÖÔÚÌî¿ÕÌâºÍʵÑéÌâÖУ®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿