ÌâÄ¿ÄÚÈÝ

ϱíÁгöÁËÈýÖÖÑÀ¸àĦ²Á¼ÁËùÊôµÄÎïÖÊÀà±ð£®
£¨1£©
Á½ÃæÕëÑÀ¸à ÕäÖéÍõÑÀ¸à ÖлªÑÀ¸à
Ħ²Á¼Á ÇâÑõ»¯ÂÁ ̼Ëá¸Æ ¶þÑõ»¯¹è
Ħ²Á¼ÁµÄÎïÖÊÀà±ð
£¨Ö¸Ëá¡¢¼î¡¢ÑΡ¢Ñõ»¯Î
£¨2£©¸ù¾ÝÄãµÄÍÆ²â£¬ÑÀ¸àĦ²Á¼ÁµÄÈܽâÐÔÊÇ
 
£¨Ìî¡°Ò×ÈÜ¡±»ò¡°ÄÑÈÜ¡±£©
£¨3£©ÑÀ¸àÖеÄĦ²Á¼Á֮һ̼Ëá¸Æ¿ÉÒÔÓÃʯ»ÒʯÀ´ÖƱ¸£®Ä³Ñ§ÉúÉè¼ÆÁËÒ»ÖÖÖÆ±¸Ì¼Ëá¸ÆµÄʵÑé·½°¸£¬ÆäÁ÷³ÌͼΪ
ʯ»Òʯ   
¸ßÎÂ
 Éúʯ»Ò 
¼ÓË®
 Ê¯»ÒË® 
¼Ó̼ËáÄÆÈÜÒº
 Ì¼Ëá¸Æ
Çëд³öÉÏÊö·½°¸ÖÐÓйط´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
¢Ù
 
£»
¢Ú
 
£»
¢Û
 
£®
£¨4£©ÇëÄãÈÔÓÃʯ»ÒʯΪԭÁÏ£¨ÆäËûÊÔ¼Á×ÔÑ¡£©£¬Éè¼ÆÁíÒ»ÖÖÖÆ±¸Ì¼Ëá¸ÆµÄʵÑé·½°¸£®²¢·ÂÕÕ£¨3£©ËùʾÁ÷³Ìͼ£¬ÇëÔÚÏÂͼÖеÄÀ¨ºÅÄÚÌîÈëÊʵ±µÄÎïÖÊ
ʯ»Òʯ
¸ßÎÂ
Éúʯ»Ò
()
   Ê¯»ÒË®
()
 Ì¼Ëá¸Æ
ÄãÉè¼ÆµÄ·½°¸µÄÓŵãÊÇ
 
£®
£¨5£©¼ìÑéÑÀ¸àÖÐÊÇ·ñº¬ÓÐ̼ËáÑεÄʵÑé·½·¨ÊÇ
 
£®
·ÖÎö£ºÌ¼Ëá¸ÆÄÜ·Ö½âÉú³ÉÑõ»¯¸Æ£¬Ñõ»¯¸ÆÄÜÓëË®·´Ó¦Éú³ÉÇâÑõ»¯¸Æ£¬ÇâÑõ»¯¸ÆÄÜÓë¶þÑõ»¯Ì¼»ò̼ËáÄÆ·´Ó¦Éú³É̼Ëá¸Æ£¬¼ìÑé̼ËáÑÎͨ³£Ê¹ÓüÓËá»¯ÆøµÄ·½·¨£®
½â´ð£º½â£º£¨1£©ÇâÑõ»¯ÂÁÊôÓڼ̼Ëá¸ÆÊôÓÚÑΣ¬¶þÑõ»¯¹èÊôÓÚÑõ»¯ÎËùÒÔ±¾Ìâ´ð°¸Îª£º
  Á½ÃæÕëÑÀ¸à ÕäÖéÍõ·À³ôÑÀ¸à ÖлªÍ¸Ã÷ÑÀ¸à
       
  ¼î ÑÎ Ñõ»¯Îï
£¨2£©ÇâÑõ»¯ÂÁ¡¢Ì¼Ëá¸Æ¡¢¶þÑõ»¯¹è¶¼ÊÇÄÑÈÜÓÚË®µÄÎïÖÊ£¬ËùÒÔ±¾Ìâ´ð°¸Îª£ºÄÑÈÜ£»
£¨3£©Ì¼Ëá¸ÆÄܸßηֽâÉú³ÉÑõ»¯¸ÆºÍ¶þÑõ»¯Ì¼£¬Ñõ»¯¸ÆÄÜÓëË®·´Ó¦Éú³ÉÇâÑõ»¯¸Æ£¬ÇâÑõ»¯¸ÆÄÜÓë̼ËáÄÆ·¢Éú¸´·Ö½â·´Ó¦Éú³É̼Ëá¸ÆºÍÇâÑõ»¯ÄÆ£¬ËùÒÔ±¾Ìâ´ð°¸Îª£º
¢ÙCaCO3
 ¸ßΠ
.
 
CaO+CO2¡ü£»
¢ÚCaO+H2O¨TCa£¨OH£©2£»
¢ÛCa£¨OH£©2+Na2CO3¨TCaCO3¡ý+2NaOH£»
£¨4£©Ì¼Ëá¸Æ»¹¿ÉÒÔÓÉÇâÑõ»¯¸ÆÓë¶þÑõ»¯Ì¼·´Ó¦Éú³É£¬Õâʱ¿ÉÒÔÀûÓÃ̼Ëá¸Æ·Ö½âÉú³ÉµÄ¶þÑõ»¯Ì¼£¬´ïµ½½ÚÔ¼µÄÄ¿µÄ£¬ËùÒÔ±¾Ìâ´ð°¸Îª£º¾«Ó¢¼Ò½ÌÍø
¶þÑõ»¯Ì¼µÃµ½³ä·ÖÀûÓ㬽ÚÔ¼Ô­ÁÏ£»
£¨5£©¼ìÑé̼ËáÑγ£Ê¹ÓüÓËá»¯ÆøµÄ·½·¨£¬¹Û²ì²úÉúµÄÆøÌåÄÜ·ñʹʯ»ÒË®±ä»ë×Ç£¬ËùÒÔ±¾Ìâ´ð°¸Îª£ºÏòÑÀ¸àÖмÓÈëÏ¡ÑÎËᣬ¿´ÊÇ·ñÓÐÆøÌå²úÉú£®ÈçÓÐÆøÌå·Å³öÇÒÉú³ÉµÄÆøÌåÄÜʹ³ÎÇåµÄʯ»ÒË®±ä»ë×Ç£¬ÔòÑÀ¸àÖк¬ÓÐ̼ËáÑΣ»·ñÔò²»º¬Ì¼ËáÑΣ®
µãÆÀ£º±¾Ì⿼²éÁË̼Ëá¸Æ¡¢Ñõ»¯¸ÆºÍÇâÑõ»¯¸ÆµÄÏ໥ת»¯ÒÔ¼°Ì¼ËáÑεļìÑ飬Íê³É´ËÌ⣬¿ÉÒÔÒÀ¾ÝÒÑÓеÄ֪ʶ½øÐУ®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
СԶͨ¹ý²éÔÄ×ÊÁÏÖªµÀÁËÑÀ¸àµÄÖ÷Òª³É·Ö£®Ëü½«Ò»¶¨±ÈÀýµÄ̼Ëá¸Æ·ÛÄ©ºÍʳÑε¹ÈëСÉÕ±­ÖУ¬È»ºó¼ÓÈëÊÊÁ¿¸ÊÓÍ£¨±£Êª¼Á£©¡¢Ìðζ¼ÁºÍÏ㾫µÈ£¬½Á°è¾ùÔȺóÖÆµÃÑÀ¸à£®
£¨1£©ÑÀ¸àÊÇ
»ìºÏÎï
»ìºÏÎï
£¨Ìî´¿¾»Îï¡¢»ìºÏÎ£®
£¨2£©Ð¡Ô¶²â¶¨×ÔÖÆÑÀ¸à¼°ÆäËûÇå½àÓÃÆ·µÄpH£¬¼Ç¼ÈçÏ£º
ÎïÖÊ ×ÔÖÆÑÀ¸à ½à²ÞÁé ÓÍÎÛ¾»
pH 8 2 12
×ÔÖÆÑÀ¸àÏÔ
¼îÐÔ
¼îÐÔ
£¨Ìî¡°ËáÐÔ¡±¡¢¡°¼îÐÔ¡±¡¢¡°ÖÐÐÔ¡±£©£¬½à²ÞÁé¿Éʹ×ÏɫʯÈïÊÔÒº±ä³É
ºì
ºì
É«£®
£¨3£©Ï±íÁгöÁËÈýÖÖÑÀ¸àĦ²Á¼ÁËùÊôµÄÎïÖÊÀà±ð£®
Á½ÃæÕëÑÀ¸à ÕäÖéÍõÑÀ¸à ÖлªÑÀ¸à
Ħ²Á¼Á ÇâÑõ»¯ÂÁ ̼Ëá¸Æ ¶þÑõ»¯¹è
Ħ²Á¼ÁµÄÎïÖÊÀà±ð
£¨Ö¸Ëá¡¢¼î¡¢ÑΡ¢Ñõ»¯Î
¼î
¼î
ÑÎ
ÑÎ
Ñõ»¯Î
Ñõ»¯Î
£¨4£©ÑÀ¸àÖеÄĦ²Á¼Á֮һ̼Ëá¸Æ¿ÉÒÔÓÃʯ»ÒʯÀ´ÖƱ¸£®Ä³Ñ§ÉúÉè¼ÆÁËÒ»ÖÖÖÆ±¸Ì¼Ëá¸ÆµÄʵÑé·½°¸£¬ÆäÁ÷³ÌͼΪ£º

Çëд³öÉÏÊö·½°¸ÖÐÓйط´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
¢Ù
CaCO3
 ¸ßΠ
.
 
CaO+CO2¡ü
CaCO3
 ¸ßΠ
.
 
CaO+CO2¡ü
£»¢Ú
CaO+H2O=Ca£¨OH£©2
CaO+H2O=Ca£¨OH£©2
£»¢Û
Ca£¨OH£©2+Na2CO3=CaCO3¡ý+2NaOH
Ca£¨OH£©2+Na2CO3=CaCO3¡ý+2NaOH
£®

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

¾«Ó¢¼Ò½ÌÍø