ËÉ»¨µ°ÓÖÃûƤµ°£¬Î¢É½ºþ³ö²úµÄËÉ»¨µ°Òòɫζ¾ß¼ÑÏíÓþÆë³£®ËÉ»¨µ°µÄÖÆ×÷¹¤ÒÕÒ»°ãÊÇÓÃË®½«»ÒÁÏ[Ö÷ÒªÊÇÉúʯ»Ò(CaO)¡¢²Ýľ»Ò(³É·ÖÖк¬K2CO3)¡¢´¿¼î(Na2CO3)¡¢Ê³ÑÎ(NaCl)µÈ]µ÷³Éºý×´£¬Í¿ÓÚÐÂÏÊѼµ°£¬ÃÜ·â±£´æÒ»¶Îʱ¼äºó£¬¼´¿ÉµÃµ½ËÉ»¨µ°£®Çë¸ù¾ÝÒÔÉÏÐÅÏ¢»Ø´ð£º

¢ÙËÉ»¨µ°ëçÖÆÖз¢ÉúµÄ»¯Ñ§·´Ó¦ÓÐ________¸ö£»

¢Ú°þ³ýËÉ»¨µ°Íâ±ß»ÒÁϺ󣬳£³£·¢ÏÖµ°¿ÇÉÏÓÐһЩ¼áÓ²µÄ°×É«°ßµã£¬Õâ°ßµãµÄÖ÷Òª³É·ÖÊÇ________£»

¢ÛʳÓÃÁÓÖÊËÉ»¨µ°Ê±ÍùÍùÓÐÂéɬ¸Ð£¬¿ÉÄܵÄÔ­ÒòÊÇ________£®

´ð°¸£º
½âÎö£º

¡¡¡¡¢Ù3

¡¡¡¡¢Ú̼Ëá¸Æ(»òCaCO3)

¡¡¡¡¢Û»ÒÁÏÖмîÐÔÎïÖʵÄÅä±È¹ý¶à(»òëçÖÆʱ¼ä¹ý¶ÌµÈ)


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

24¡¢£¨³ýÓÐ˵Ã÷µÄÍ⣬ÿ¿Õ1·Ö£¬¹²14·Ö£©ËÉ»¨µ°ÓÖÃûƤµ°£¬Î¢É½ºþ³ö²úµÄËÉ»¨µ°Òòɫζ¾ß¼ÑÏíÓþÆë³£®ËÉ»¨µ°µÄÖÆ×÷¹¤ÒÕÒ»°ãÊÇÓÃË®½«»ÒÁÏ[Ö÷ÒªÊÇÉúʯ»Ò£¨CaO£©¡¢´¿¼î£¨Na2CO3£©¡¢Ê³ÑΣ¨NaCl£©µÈ]µ÷³Éºý×´£¬Í¿ÓÚÐÂÏÊѼµ°£¬ÃÜ·â±£´æÒ»¶Îʱ¼äºó£¬¼´¿ÉµÃµ½ËÉ»¨µ°£®Çë¸ù¾ÝÒÔÉÏÐÅÏ¢»Ø´ð£º
¢ÙËÉ»¨µ°ëçÖÆÖй²·¢Éú
2
¸ö»¯Ñ§·´Ó¦£¬Ð´³öÆäÖÐÊôÓÚ»¯ºÏ·´Ó¦µÄ»¯Ñ§·´Ó¦»¯Ñ§·½³Ìʽ£º
CaO+H2O¨TCa£¨OH£©2
£®
¢Ú°þ³ýËÉ»¨µ°Íâ±ß»ÒÁϺ󣬳£³£·¢ÏÖµ°¿ÇÉÏÓÐһЩ¼áÓ²µÄ²»ÈÜÓÚË®µÄ°×É«°ßµã£¬Õâ°ßµãµÄÖ÷Òª³É·ÖÊÇ
CaCO3
£¬Ð´³öÕâÖֳɷÖÓëÑÎËá·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
Ca£¨OH£©2+Na2CO3¨TCaCO3¡ý+2NaOH
£®
¢ÛʳÓÃÁÓÖÊËÉ»¨µ°Ê±ÍùÍùÓÐɬζ£¬Ê³ÓÃʱÈçºÎ³ýȥɬζ
¼Óµãʳ´×
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

22¡¢ËÉ»¨µ°ÓÖÃûƤµ°£¬Î¢É½ºþ³ö²úµÄËÉ»¨µ°Òòɫζ¾ß¼ÑÏíÓþÆë³£®ËÉ»¨µ°µÄÖÆ×÷¹¤ÒÕÒ»°ãÊÇÓÃË®½«»ÒÁÏ[Ö÷ÒªÊÇÉúʯ»Ò£¨CaO£©¡¢´¿¼î£¨Na2CO3£©¡¢Ê³ÑΣ¨NaCl£©µÈ]µ÷³Éºý×´£¬Í¿ÓÚÐÂÏÊѼµ°£¬ÃÜ·â±£´æÒ»¶Îʱ¼äºó£¬¼´¿ÉµÃµ½ËÉ»¨µ°£®Çë¸ù¾ÝÒÔÉÏÐÅÏ¢»Ø´ð£º
¢ÙËÉ»¨µ°ëçÖÆÖй²·¢Éú
2
¸ö»¯Ñ§·´Ó¦£¬Ð´³öÆäÖÐÊôÓÚ»¯ºÏ·´Ó¦µÄ»¯Ñ§·´Ó¦»¯Ñ§·½³Ìʽ
CaO+H2O¨TCa£¨OH£©2
£®
¢Ú°þ³ýËÉ»¨µ°Íâ±ß»ÒÁϺ󣬳£³£·¢ÏÖµ°¿ÇÉÏÓÐһЩ¼áÓ²µÄ²»ÈÜÓÚË®µÄ°×É«°ßµã£¬Õâ°ßµãµÄÖ÷Òª³É·ÖÊÇ
CaCO3£¨Ì¼Ëá¸Æ£©
£¬Ð´³öÕâÖֳɷÖÓëÑÎËá·´Ó¦µÄ»¯Ñ§·½³Ìʽ
CaCO3 +2HCl¨TCaCl2+H2O+CO2¡ü
£®
¢ÛʳÓÃÁÓÖÊËÉ»¨µ°Ê±ÍùÍùÓÐɬζ£¬Ê³ÓÃʱÈçºÎ³ýȥɬζ
¼Óµãʳ´×
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ËÉ»¨µ°ÓÖÃûƤµ°£¬Î¢É½ºþ³ö²úµÄËÉ»¨µ°Òòɫζ¾ß¼ÑÏíÓþÆë³£®ËÉ»¨µ°µÄÖÆ×÷¹¤ÒÕÒ»°ãÊÇÓÃË®½«»ÒÁÏ[Ö÷ÒªÊÇÉúʯ»Ò£¨CaO£©¡¢´¿¼î£¨Na2CO3£©¡¢Ê³ÑΣ¨NaCl£©µÈ]µ÷³Éºý×´£¬Í¿ÓÚÐÂÏÊѼµ°£¬ÃÜ·â±£´æÒ»¶Îʱ¼äºó£¬¼´¿ÉµÃµ½ËÉ»¨µ°£®Çë¸ù¾ÝÒÔÉÏÐÅÏ¢»Ø´ð£º
£¨1£©ËÉ»¨µ°ëçÖÆÖй²·¢Éú2¸ö»¯Ñ§·´Ó¦£¬Ð´³öÆäÖÐÊôÓÚ»¯ºÏ·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
CaO+H2O¨TCa£¨OH£©2
CaO+H2O¨TCa£¨OH£©2
£®
£¨2£©°þ³ýËÉ»¨µ°Íâ±ß»ÒÁϺ󣬳£³£·¢ÏÖµ°¿ÇÉÏÓÐһЩ¼áÓ²µÄ²»ÈÜÓÚË®µÄ°×É«°ßµã£¬Õâ°ßµãµÄÖ÷Òª³É·ÖÊÇ̼Ëá¸Æ£¬Ð´³öËüÓëÑÎËá·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
CaCO3+2HCl¨TCaCl2+H2O+CO2¡ü
CaCO3+2HCl¨TCaCl2+H2O+CO2¡ü
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ËÉ»¨µ°ÓÖÃûƤµ°£¬Î¢É½ºþ³ö²úµÄËÉ»¨µ°Òòɫζ¾ß¼ÑÏíÓþÆë³£®ËÉ»¨µ°µÄÖÆ×÷¹¤ÒÕÒ»°ãÊÇÓÃË®½«»ÒÁÏ[Ö÷ÒªÊÇÉúʯ»Ò£¨CaO£©¡¢´¿¼î£¨Na2CO3£©¡¢Ê³ÑΣ¨NaCl£©µÈ]µ÷³Éºý×´£¬Í¿ÓÚÐÂÏÊѼµ°£¬ÃÜ·â±£´æÒ»¶Îʱ¼äºó£¬¼´¿ÉµÃµ½ËÉ»¨µ°£®Çë¸ù¾ÝÒÔÉÏÐÅÏ¢»Ø´ð£º
¢ÙËÉ»¨µ°ëçÖÆÖй²·¢Éú
2
2
¸ö»¯Ñ§·´Ó¦£¬Ð´³öÆäÖÐÊôÓÚ»¯ºÏ·´Ó¦µÄ»¯Ñ§·´Ó¦»¯Ñ§·½³Ìʽ£º
CaO+H2O¨TCa£¨OH£©2
CaO+H2O¨TCa£¨OH£©2
£®
¢Ú°þ³ýËÉ»¨µ°Íâ±ß»ÒÁϺ󣬳£³£·¢ÏÖµ°¿ÇÉÏÓÐһЩ¼áÓ²µÄ²»ÈÜÓÚË®µÄ°×É«°ßµã£¬Õâ°ßµãµÄÖ÷Òª³É·ÖÊÇ
CaCO3£¨Ì¼Ëá¸Æ£©
CaCO3£¨Ì¼Ëá¸Æ£©
£¬Ð´³öÕâÖֳɷÖÓëÑÎËá·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
CaCO3+2HCl¨TCaCl2+H2O+CO2¡ü
CaCO3+2HCl¨TCaCl2+H2O+CO2¡ü
£®
¢ÛʳÓÃÁÓÖÊËÉ»¨µ°Ê±ÍùÍùÓÐɬζ£¬Ê³ÓÃʱÈçºÎ³ýȥɬζ£¿¿ÉÒÔ¼Ó
¼Óµãʳ´×
¼Óµãʳ´×
£®
¢ÜÈ¡ÖÆËÉ»¨µ°µÄºý×´Îï·ÅÈëË®Öгä·ÖÈܽâºó¹ýÂ˵ÃÂËÒº¼×£¬ÔòÂËÒº¼×ÖÐÒ»¶¨º¬
ÂÈ»¯ÄƺÍÇâÑõ»¯ÄÆ
ÂÈ»¯ÄƺÍÇâÑõ»¯ÄÆ
£¬¿ÉÄܺ¬
ÇâÑõ»¯¸Æ»ò̼ËáÄÆ
ÇâÑõ»¯¸Æ»ò̼ËáÄÆ
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ËÉ»¨µ°ÓÖÃûƤµ°£¬Î¢É½ºþ³ö²úµÄËÉ»¨µ°Òòɫζ¾ß¼ÑÏíÓþÆ볡£ËÉ»¨µ°µÄÖÆ×÷¹¤ÒÕÒ»°ãÊÇÓÃË®½«»ÒÁÏ[Ö÷ÒªÊÇÉúʯ»Ò(CaO)¡¢´¿¼î(Na2CO3)¡¢Ê³ÑÎ(NaCl)µÈ]µ÷³Éºý×´£¬Í¿ÓÚÐÂÏÊѼµ°£¬ÃÜ·â±£´æÒ»¶Îʱ¼äºó£¬¼´¿ÉµÃµ½ËÉ»¨µ°¡£Çë¸ù¾ÝÒÔÉÏÐÅÏ¢»Ø´ð£º

¢ÙËÉ»¨µ°ëçÖÆÖй²·¢Éú             ¸ö»¯Ñ§·´Ó¦£¬Ð´³öÆäÖÐÊôÓÚ»¯ºÏ·´Ó¦µÄ»¯Ñ§·´Ó¦»¯Ñ§·½³Ìʽ£º                                                    ¡£

¢Ú°þ³ýËÉ»¨µ°Íâ±ß»ÒÁϺ󣬳£³£·¢ÏÖµ°¿ÇÉÏÓÐһЩ¼áÓ²µÄ²»ÈÜÓÚË®µÄ°×É«°ßµã£¬Õâ°ßµãµÄÖ÷Òª³É·ÖÊÇ      £¬Ð´³öÕâÖֳɷÖÓëÑÎËá·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º___________________          _________¡£

¢ÛʳÓÃÁÓÖÊËÉ»¨µ°Ê±ÍùÍùÓÐɬζ£¬Ê³ÓÃʱÈçºÎ³ýȥɬζ_____________¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸