ÌâÄ¿ÄÚÈÝ
»¯Ñ§¿Îºó£¬»¯Ñ§ÐËȤС×éµÄͬѧÔÚÕûÀíʵÑé×Àʱ£¬·¢ÏÖÓÐһƿÇâÑõ»¯ÄÆÈÜҺûÓÐÈûÏðƤÈû£¬Õ÷µÃÀÏʦͬÒâºó£¬¿ªÕ¹ÁËÒÔÏÂ̽¾¿£º
[Ìá³öÎÊÌâ1]¸ÃÇâÑõ»¯ÄÆÈÜÒºÊÇ·ñ±äÖÊÁËÄØ£¿
[ʵÑé̽¾¿1]
[Ìá³öÎÊÌâ2]¸ÃÇâÑõ»¯ÄÆÈÜÒºÊÇÈ«²¿±äÖÊ»¹ÊDz¿·Ö±äÖÊÄØ£¿
[²ÂÏëÓë¼ÙÉè]²ÂÏë1£ºÇâÑõ»¯ÄÆÈÜÒº²¿·Ö±äÖÊ£® ²ÂÏë2£ºÇâÑõ»¯ÄÆÈÜҺȫ²¿±äÖÊ£®
[²éÔÄ×ÊÁÏ]
£¨1£©ÂÈ»¯¸ÆÈÜÒº³ÊÖÐÐÔ£®
£¨2£©ÂÈ»¯¸ÆÈÜÒºÄÜÓë̼ËáÄÆÈÜÒº·´Ó¦£ºCaCl2+Na2CO3=CaCO3¡ý+2NaCl
[ʵÑé̽¾¿2]
[ʵÑé½áÂÛ]¸ÃÇâÑõ»¯ÄÆÈÜÒº______£¨Ìî¡°²¿·Ö¡±»ò¡°È«²¿¡±£©±äÖÊ£®
[·´Ë¼ÓëÆÀ¼Û]£¨1£©ÇâÑõ»¯ÄÆÈÜҺ¶ÖÃÓÚ¿ÕÆøÖÐÈÝÒ×±äÖÊ£¬Çëд³öÏà¹Ø·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º______£®
£¨2£©ÔÚÉÏÊö[ʵÑé̽¾¿2]ÖУ¬Ð¡Ã÷Ìá³ö¿ÉÓÃÇâÑõ»¯¸ÆÈÜÒº´úÌæÂÈ»¯¸ÆÈÜÒº£¬ÄãÈÏΪ¸Ã·½°¸______£¨Ìî¡°¿ÉÐС±»ò¡°²»¿ÉÐС±£©£®
[Ìá³öÎÊÌâ1]¸ÃÇâÑõ»¯ÄÆÈÜÒºÊÇ·ñ±äÖÊÁËÄØ£¿
[ʵÑé̽¾¿1]
| ʵÑé²Ù×÷ | ʵÑéÏÖÏó | ʵÑé½áÂÛ |
| È¡ÉÙÁ¿¸ÃÈÜÒºÓÚÊÔ¹ÜÖУ¬ÏòÈÜÒºÖеμÓÏ¡ÑÎËᣬ²¢²»¶ÏÕñµ´£® | ÓÐÆøÅÝð³ö£® | ÇâÑõ»¯ÄÆÈÜÒºÒ»¶¨±äÖÊÁË£® |
[²ÂÏëÓë¼ÙÉè]²ÂÏë1£ºÇâÑõ»¯ÄÆÈÜÒº²¿·Ö±äÖÊ£® ²ÂÏë2£ºÇâÑõ»¯ÄÆÈÜҺȫ²¿±äÖÊ£®
[²éÔÄ×ÊÁÏ]
£¨1£©ÂÈ»¯¸ÆÈÜÒº³ÊÖÐÐÔ£®
£¨2£©ÂÈ»¯¸ÆÈÜÒºÄÜÓë̼ËáÄÆÈÜÒº·´Ó¦£ºCaCl2+Na2CO3=CaCO3¡ý+2NaCl
[ʵÑé̽¾¿2]
| ʵÑé²½Öè | ʵÑéÏÖÏó | ʵÑé½áÂÛ |
| £¨1£©È¡ÉÙÁ¿¸ÃÈÜÒºÓÚÊÔ¹ÜÖУ¬ÏòÈÜÒºÖеμӹýÁ¿µÄÂÈ»¯¸ÆÈÜÒº£¬²¢²»¶ÏÕñµ´£® | ÓÐ______Éú³É£® | ˵Ã÷ÔÈÜÒºÖÐÒ»¶¨ÓÐ̼ËáÄÆ£® |
| £¨2£©È¡²½Ö裨1£©ÊÔ¹ÜÖеÄÉÙÁ¿ÉϲãÇåÒº£¬µÎ¼Ó·Ó̪ÈÜÒº£® | ÈÜÒº±äºìÉ«£® | ˵Ã÷ÔÈÜÒºÖÐÒ»¶¨ÓÐ______£® |
[·´Ë¼ÓëÆÀ¼Û]£¨1£©ÇâÑõ»¯ÄÆÈÜҺ¶ÖÃÓÚ¿ÕÆøÖÐÈÝÒ×±äÖÊ£¬Çëд³öÏà¹Ø·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º______£®
£¨2£©ÔÚÉÏÊö[ʵÑé̽¾¿2]ÖУ¬Ð¡Ã÷Ìá³ö¿ÉÓÃÇâÑõ»¯¸ÆÈÜÒº´úÌæÂÈ»¯¸ÆÈÜÒº£¬ÄãÈÏΪ¸Ã·½°¸______£¨Ìî¡°¿ÉÐС±»ò¡°²»¿ÉÐС±£©£®
[ʵÑé̽¾¿2]£¨1£©È¡ÉÙÁ¿¸ÃÈÜÒºÓÚÊÔ¹ÜÖУ¬ÏòÈÜÒºÖеμӹýÁ¿µÄÂÈ»¯¸ÆÈÜÒº£¬Òò̼ËáÄÆÓëÂÈ»¯¸ÆÄÜÉú³É̼Ëá¸Æ³Áµí£¬ËùÒÔÓа×É«³ÁµíÉú³É£¬ËµÃ÷ÔÈÜÒºÖÐÒ»¶¨ÓÐ̼ËáÄÆ£»
£¨2£©È¡²½Ö裨1£©ÊÔ¹ÜÖеÄÉÙÁ¿ÉϲãÇåÒº£¬µÎ¼Ó·Ó̪ÈÜÒº£¬ÈÜÒº±äºìÉ«£®ÈÜÒºÏÔ¼îÐÔ£¬ÒòÈÜÒºÖÐ̼ËáÄÆÒѱ»ÂÈ»¯¸ÆÍêÈ«ÏûºÄ£¬ËùÒÔ˵Ã÷ÔÈÜÒºÖÐÒ»¶¨ÓÐÇâÑõ»¯ÄÆ£®´Ó¶øµÃ³ö[ʵÑé½áÂÛ]¸ÃÇâÑõ»¯ÄÆÈÜÒº²¿·Ö±äÖÊ£»
[·´Ë¼ÓëÆÀ¼Û]£¨1£©ÇâÑõ»¯ÄÆÈÜҺ¶ÖÃÓÚ¿ÕÆøÖÐÈÝÒ×Óë¶þÑõ»¯Ì¼·´Ó¦Éú³É̼ËáÄÆºÍË®£¬Ïà¹Ø·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºCO2+2NaOH=Na2CO3+H2O£»
£¨2£©ÈçÓÃÇâÑõ»¯¸ÆÈÜÒº´úÌæÂÈ»¯¸ÆÈÜÒºÊDz»Ðеģ¬ÒòΪÇâÑõ»¯¸ÆÓë̼ËáÄÆ·´Ó¦»áÉú³ÉÇâÑõ»¯ÄÆ£¬ÕâÑùÎÞ·¨È·¶¨£¨2£©²½ÖèµÄʵÑé½áÂÛ£¬µÃ²»³öÇâÑõ»¯ÄƲ¿·Ö±äÖʵĽáÂÛ£»
¹Ê´ð°¸Îª£º[ʵÑé̽¾¿2]°×É«³Áµí£»ÇâÑõ»¯ÄÆ£¨»òNaOH£©£»
[ʵÑé½áÂÛ]²¿·Ö
[·´Ë¼ÓëÆÀ¼Û]£¨1£©CO2+2NaOH=Na2CO3+H2O£»£¨2£©²»¿ÉÐУ®
£¨2£©È¡²½Ö裨1£©ÊÔ¹ÜÖеÄÉÙÁ¿ÉϲãÇåÒº£¬µÎ¼Ó·Ó̪ÈÜÒº£¬ÈÜÒº±äºìÉ«£®ÈÜÒºÏÔ¼îÐÔ£¬ÒòÈÜÒºÖÐ̼ËáÄÆÒѱ»ÂÈ»¯¸ÆÍêÈ«ÏûºÄ£¬ËùÒÔ˵Ã÷ÔÈÜÒºÖÐÒ»¶¨ÓÐÇâÑõ»¯ÄÆ£®´Ó¶øµÃ³ö[ʵÑé½áÂÛ]¸ÃÇâÑõ»¯ÄÆÈÜÒº²¿·Ö±äÖÊ£»
[·´Ë¼ÓëÆÀ¼Û]£¨1£©ÇâÑõ»¯ÄÆÈÜҺ¶ÖÃÓÚ¿ÕÆøÖÐÈÝÒ×Óë¶þÑõ»¯Ì¼·´Ó¦Éú³É̼ËáÄÆºÍË®£¬Ïà¹Ø·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºCO2+2NaOH=Na2CO3+H2O£»
£¨2£©ÈçÓÃÇâÑõ»¯¸ÆÈÜÒº´úÌæÂÈ»¯¸ÆÈÜÒºÊDz»Ðеģ¬ÒòΪÇâÑõ»¯¸ÆÓë̼ËáÄÆ·´Ó¦»áÉú³ÉÇâÑõ»¯ÄÆ£¬ÕâÑùÎÞ·¨È·¶¨£¨2£©²½ÖèµÄʵÑé½áÂÛ£¬µÃ²»³öÇâÑõ»¯ÄƲ¿·Ö±äÖʵĽáÂÛ£»
¹Ê´ð°¸Îª£º[ʵÑé̽¾¿2]°×É«³Áµí£»ÇâÑõ»¯ÄÆ£¨»òNaOH£©£»
[ʵÑé½áÂÛ]²¿·Ö
[·´Ë¼ÓëÆÀ¼Û]£¨1£©CO2+2NaOH=Na2CO3+H2O£»£¨2£©²»¿ÉÐУ®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿