¿´Í¼»Ø´ðÏÂÁÐÎÊÌ⣨²¿·ÖÒÇÆ÷Òѱ»ÐéÏßÕÚµ²£¬ÇÒÐéÏß²¿·ÖÒÇÆ÷¿ÉÒÔ¸ù¾ÝÐèÒªÌí¼Ó£©
£¨1£©Èç¹ûÔÚÐéÏß²¿·ÖÔö¼ÓÊÕ¼¯×°Öã¬ÉÏͼÖÐ×°ÖÿÉÓÃÓÚʵÑéÊÒÖÆÈ¡°±Æø£¬Çë»Ø´ðÒÔÏÂÎÊÌ⣺
¢Ù¿ÉÒÔÓÃ
Æ¿¿ÚÏòÏÂÅÅ¿ÕÆø·¨
Æ¿¿ÚÏòÏÂÅÅ¿ÕÆø·¨
·½·¨ÊÕ¼¯°±Æø£¬Èç¹ûÓÃÉÕÆ¿ÊÕ¼¯°±Æø£¬ÇëÓÃÎÄ×ÖÐðÊöÈçºÎ¼ìÑé°±Æø¼ºÊÕ¼¯Âú
ÓÃʪÈóµÄºìɫʯÈïÊÔÖ½¿¿½üÆ¿¿Ú£¬ÊÔÖ½±äÀ¶£¬ËµÃ÷ÒÑÊÕ¼¯Âú
ÓÃʪÈóµÄºìɫʯÈïÊÔÖ½¿¿½üÆ¿¿Ú£¬ÊÔÖ½±äÀ¶£¬ËµÃ÷ÒÑÊÕ¼¯Âú
£»
¢ÚÀûÓâÙÖÐÊÕ¼¯µÄ°±Æø¿ÉÒÔÍê³ÉÅçȪʵÑ飬ÊÇÒòΪ°±Æø¾ßÓÐ
¼«Ò×ÈÜÓÚË®
¼«Ò×ÈÜÓÚË®
µÄÎïÀíÐÔÖÊ£»
¢Ûд³öʵÑéÊÒÖÆ°±ÆøµÄ»¯Ñ§·´Ó¦·½³Ìʽ
2NH4Cl+Ca£¨OH£©2¨TCaCl2+2NH3¡ü+2H2O
2NH4Cl+Ca£¨OH£©2¨TCaCl2+2NH3¡ü+2H2O
£»
£¨2£©ÉÏͼװÖû¹¿ÉÒÔÓÃÓÚÍê³ÉijЩ¹ÌÌåÎïÖÊÊÜÈÈ·Ö½âµÄʵÑ飮Çë»Ø´ð£º
¢ÙNa
2CO
3ºÍNaHCO
3ÕâÁ½ÖÖÎïÖÊÖÐÓÐÒ»ÖÖ¿ÉÒÔÓÃÉÏͼװÖýøÐÐÊÜÈÈ·Ö½âµÄʵÑ飮¸ÃÎïÖÊÊÜÈÈ·Ö½âµÄ»¯Ñ§·´Ó¦·½³ÌʽΪ£º
£»
¢ÚΪÁËÑéÖ¤¢ÙÖиÃÎïÖÊÊÜÈÈ·Ö½â²úÉúµÄËùÓÐÆøÌå²úÎÔÚÐéÏß´¦Ó¦Á¬½Ó£¨±íÊöʱÇëÖ¸Ã÷ʹÓÃÒÇÆ÷£¬ÒÇÆ÷ÖÐÊ¢×°µÄÊÔ¼ÁºÍÒÇÆ÷µÄÁ¬½Ó˳Ðò£©
ÏÈÁ¬½Ó×°Óа×É«ÎÞË®ÁòËá͹ÌÌåµÄ¸ÉÔï¹Ü£¬ÔÙÁ¬½Ó×°ÓгÎÇåʯ»ÒË®µÄÉÕ±
ÏÈÁ¬½Ó×°Óа×É«ÎÞË®ÁòËá͹ÌÌåµÄ¸ÉÔï¹Ü£¬ÔÙÁ¬½Ó×°ÓгÎÇåʯ»ÒË®µÄÉÕ±
£»
¢ÛÇëÌá³öÒ»µãÓйØʵÑé¹ý³ÌÖа²È«ÎÊÌâµÄ½¨Ò飺
·´Ó¦ºóÏȳ·µ¼Æø¹Ü£¬ÔÙ³·¾Æ¾«µÆ
·´Ó¦ºóÏȳ·µ¼Æø¹Ü£¬ÔÙ³·¾Æ¾«µÆ
£»
¢ÜÈç¹û½«12.6gµÄ¸ÃÎïÖʼÓÈȷֽ⣬¼ÓÈÈÒ»¶Îʱ¼äºó£¬²âµÃÊ£Óà¹ÌÌåÖÊÁ¿Îª9.5g£¬ÔòÒѾ·Ö½âµÄ¸ÃÎïÖʵÄÖÊÁ¿Îª
8.4
8.4
g£®