ÌâÄ¿ÄÚÈÝ

£¨2011?ÈÕÕÕ£©ÏÂÁÐ×°ÖÃÓÃÓÚʵÑéÊÒÀïCO2µÄÖƱ¸¡¢¾»»¯¡¢¼ìÑ飬×îºóÊÕ¼¯Ò»Æ¿¸ÉÔïµÄCO2£®Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©¶ÔʵÑéÊÒÖÆÈ¡¶þÑõ»¯Ì¼Ò©Æ·Ñ¡ÔñµÄ̽¾¿ÊµÑ飬¼Ç¼ÈçÏ£º
×é±ðҩƷʵÑéÏÖÏó
¢Ù̼ËáÄÆ·ÛÄ©ºÍÏ¡ÑÎËá²úÉúÆøÅÝËÙÂʺܿì
¢Ú¿é״ʯ»ÒʯºÍÏ¡ÁòËá²úÉúÆøÅÝËÙÂÊ»ºÂý²¢Öð½¥Í£Ö¹
¢Û¿é״ʯ»ÒʯºÍÏ¡ÑÎËá²úÉúÆøÅÝËÙÂÊÊÊÖÐ
    ´ÓÖÆÈ¡ºÍÊÕ¼¯µÄ½Ç¶È·ÖÎö£¬Ò»°ãÑ¡ÔñµÚ
¢Û
¢Û
£¨ÌîÐòºÅ£©×éÒ©Æ·£¬Ëù·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
CaCO3+2HCl¨TCaCl2+H2O+CO2¡ü
CaCO3+2HCl¨TCaCl2+H2O+CO2¡ü
£®
£¨2£©°ÑC¡¢DÖеÄÒºÌåÃû³ÆÌîÈëÏÂ±í£º
B£¨ÑΣ©C£¨¼î£©D£¨Ëᣩ
ÒºÌåÃû³Æ̼ËáÇâÄÆÈÜÒº
³ÎÇåʯ»ÒË®
³ÎÇåʯ»ÒË®
ŨÁòËá
ŨÁòËá
£¨3£©BÖз´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
NaHCO3+HCl¨TNaCl+H2O+CO2¡ü
NaHCO3+HCl¨TNaCl+H2O+CO2¡ü
£®
£¨4£©CÖз´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
CO2+Ca£¨OH£©2¨TCaCO3¡ý+H2O
CO2+Ca£¨OH£©2¨TCaCO3¡ý+H2O
£®
£¨5£©·´Ó¦¹ý³ÌÖн«µ¯»É¼Ð¹Ø±Õ£¬ÔÚAÖп´µ½µÄÏÖÏóÊÇ
AÖÐ׶ÐÎÆ¿ÄÚÒºÃæϽµ£»³¤¾±Â©¶·ÄÚÒºÃæÉÏÉý
AÖÐ׶ÐÎÆ¿ÄÚÒºÃæϽµ£»³¤¾±Â©¶·ÄÚÒºÃæÉÏÉý
£®
£¨6£©EÊÕ¼¯·½·¨ËµÃ÷¶þÑõ»¯Ì¼¾ßÓеÄÎïÀíÐÔÖÊÊÇ
¶þÑõ»¯Ì¼µÄÃܶȱȿÕÆø´ó
¶þÑõ»¯Ì¼µÄÃܶȱȿÕÆø´ó
£®
·ÖÎö£º£¨1£©¶ÔʵÑéÊÒÖÆÈ¡¶þÑõ»¯Ì¼Ò©Æ·Ñ¡Ôñ£ºÑ¡ÔñÒ©Æ·¼´¾­¼Ãʵ»ÝÓÖÄܳÖÐø¡¢Îȶ¨µÄ·¢Éú»¯Ñ§·´Ó¦£¬²úÉúÆøÌåËÙ¶ÈÊÊÖУ¬±ãÓÚÊÕ¼¯£¬ÄÜÓû¯Ñ§·½³Ìʽ±íʾÆä·´Ó¦Ô­Àí£®
£¨2£©¸ù¾ÝÌâÒâËù¸ø×°ÖÃÓÃÓÚʵÑéÊÒÀïCO2µÄÖƱ¸¡¢¾»»¯¡¢¼ìÑ飬×îºóÊÕ¼¯Ò»Æ¿¸ÉÔïµÄCO2µÄÒªÇó£¬Ñ¡Ôñ¼ìÑé¡¢¸ÉÔïCO2ÆøÌåµÄÒºÌåÒ©Æ·£®
£¨3£©Ñ¡ÓÃ̼ËáÇâÄÆÈÜÒº¾»»¯ÂÈ»¯ÇâÆøÌ壬д³ö·¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ£®
£¨4£©¼ìÑé¶þÑõ»¯Ì¼ÆøÌåµÄ·½·¨£ºÍù³ÎÇåµÄʯ»ÒË®ÖÐͨ¶þÑõ»¯Ì¼£¬±ä»ë×Ç£®Ð´³öÆä·´Ó¦µÄ»¯Ñ§·½³Ìʽ£®
£¨5£©·´Ó¦¹ý³ÌÖн«µ¯»É¼Ð¹Ø±Õ£¬×¶ÐÎÆ¿ÄÚ²úÉúµÄÆøÌåÎÞ·¨Åųö£¬ÆøѹÔö´ó£¬ÔÚAÖп´µ½µÄÏÖÏóÊÇAÖÐ׶ÐÎÆ¿ÄÚÒºÃæϽµ£»³¤¾±Â©¶·ÄÚÒºÃæÉÏÉý£®
£¨6£©EÊÕ¼¯·½·¨ÀûÓöþÑõ»¯Ì¼ÃܶȱȿÕÆø´óµÄÎïÀíÐÔÖÊ£®
½â´ð£º½â£º£¨1£©ÊµÑéÊÒÖÆÈ¡¶þÑõ»¯Ì¼Ñ¡ÔñҩƷʱ¼´×¢Öؾ­¼Ãʵ»ÝÓÖÄܳÖÐø¡¢Îȶ¨µÄ·¢Éú»¯Ñ§·´Ó¦£¬²úÉúÆøÌåËÙ¶ÈÊÊÖУ¬±ãÓÚÊÕ¼¯£»
¹Ê´ð°¸Îª£º¢Û£»  Ëù·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºCaCO3+2HCl¨TCaCl2+H2O+CO2¡ü
   £¨2£©ÒòC×°ÖÃÊǼìÑé¶þÑõ»¯Ì¼ÆøÌåµÄ¼îÒº£¬¹Ê±íÄÚÌî ³ÎÇåʯ»ÒË®£¬ÒòD×°ÖÃÊǸÉÔï¶þÑõ»¯Ì¼ÆøÌåµÄËáÒº£¬¹Ê±íÄÚÌîŨÁòËᣮ
   £¨3£©ÒòB×°ÖÃÊǾ»»¯ÂÈ»¯ÇâÆøÌåµÄÑÎÒºNaHCO3£¬¹Ê·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºNaHCO3+HCl¨TNaCl+H2O+CO2¡ü£®
   £¨4£©ÒòC×°ÖÃÊǼìÑé¶þÑõ»¯Ì¼ÆøÌåµÄ¼îÒº£¬¹ÊÆä·´Ó¦µÄ·½³ÌʽΪ£ºCO2+Ca£¨OH£©2¨TCaCO3¡ý+H2O
   £¨5£©ÔÚA×°Öõķ´Ó¦¹ý³ÌÖн«µ¯»É¼Ð¹Ø±Õ£¬×¶ÐÎÆ¿ÄÚ²úÉúµÄÆøÌåÎÞ·¨Åųö£¬ÆøѹÔö´ó£¬¹ÊÔÚAÖоͻá³öÏÖ׶ÐÎÆ¿ÄÚÒºÃæϽµ£»³¤¾±Â©¶·ÄÚÒºÃæÉÏÉý£®
  £¨6£©ÓÃE×°ÖÃÏòÉÏÅÅ¿ÕÆø·¨ÊÕ¼¯¶þÑõ»¯Ì¼ÆøÌå˵Ã÷ËüµÄÃܶȱȿÕÆø´ó£®
¹Ê´ð°¸Îª£º£¨1£©¢Û¡¢CaCO3+2HCl¨TCaCl2+H2O+CO2¡ü
£¨2£©
              B£¨ÑΣ©            C£¨¼î£©            D£¨Ëᣩ
ÒºÌåÃû³Æ      Ì¼ËáÇâÄÆÈÜÒº       ³ÎÇåʯ»ÒË®        Å¨ÁòËá
£¨3£©NaHCO3+HCl¨TNaCl+H2O+CO2¡ü
£¨4£©CO2+Ca£¨OH£©2¨TCaCO3¡ý+H2O
£¨5£©AÖÐ׶ÐÎÆ¿ÄÚÒºÃæϽµ£»³¤¾±Â©¶·ÄÚÒºÃæÉÏÉý
£¨6£©¶þÑõ»¯Ì¼µÄÃܶȱȿÕÆø´ó
µãÆÀ£º´ËÌ⿼²éÁËʵÑéÊÒÖÆÈ¡¶þÑõ»¯Ì¼ÆøÌåµÄÒ©Æ·Ñ¡ÓÃ̽¾¿¡¢·´Ó¦Ô­Àí¼°Æä¾»»¯¡¢¼ìÑé¡¢¸ÉÔïµÄÒ©Æ·Ñ¡ÓÃ̽¾¿ºÍÊÕ¼¯·½·¨µÈ֪ʶµãµÄ¿¼²é£¬É漰֪ʶÃæ¹ã£¬Ì½¾¿ÎÊÌâ¶à£¬×ÛºÏÐÔÇ¿£¬ÊµÊǶÍÁ¶Ñ§ÉúµÄÒ»µÀºÃÌ⣮
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø