ÌâÄ¿ÄÚÈÝ
¡°Ë®ÊÇÉúÃüÖ®Ô´¡±¡£Çë»Ø´ðÏÂÁÐÓëË®ÓйصÄÎÊÌ⣺
£¨1£©ÓãÀà¿ÉÒÔÔÚË®ÖкôÎü£¬ÊÇÒòΪˮÖÐÈÜÓУ¿ ¡£
£¨2£©Ë®ÌåÎÛȾµÄÀ´Ô´Ö÷ÒªÓй¤ÒµÎÛȾ¡¢Å©ÒµÎÛȾ¡¢________________¡£
£¨3£©ÕáÌÇÔÚÈÈË®ÖÐÈܽâ±ÈÔÚÀäË®Öп죬Ó÷Ö×ÓµÄÏà¹ØÖªÊ¶½âÊÍ ¡£
£¨4£©µç½âÒ»¶¨Á¿µÄË®£¬µ±ÆäÖÐÒ»¸öµç¼«²úÉú5 mLÆøÌåʱ£¬ÁíÒ»µç¼«²úÉúµÄÆøÌåÌå»ý¿ÉÄÜÊÇ ( ) mL »ò ( ) mL¡£
£¨5£©ÈÈˮƿµ¨±ÚÉϵÄË®¹¸µÄÖ÷Òª³É·ÖÊÇ̼Ëá¸ÆºÍÇâÑõ»¯Ã¾£¬¿ÉÓô×ËáÈܽâ³ýÈ¥¡£ÒÑÖª´×ËáÓëÇâÑõ»¯Ã¾·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º2CH3COOH + Mg£¨OH£©2 = £¨CH3COO£©2Mg +2H2O £¬Ôò´×ËáÓë̼Ëá¸Æ·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ¡¡£¿![]()
£¨1£©ÑõÆø£¨¡°Ñõ·Ö×Ó¡±¡¢¡°O2¡±µÈºÏÀí´ð°¸¾ù¸ø·Ö£©
£¨2£©Éú»îÎÛȾ£¨¡°·ÅÉäÐÔË®ÎÛȾ¡±¡¢¡°¸»ÓªÑø»¯ÎÛȾ¡±¡¢¡°²¡Ô´Î¢ÉúÎïÎÛȾ¡±µÈºÏÀí´ð°¸¸ø·Ö£©
£¨3£©Î¶ÈÉý¸ß£¬·Ö×ÓÔÚË®ÖеÄÀ©É¢ËÙÂʼӿ죬´Ó¶ø¼ÓËÙÕáÌÇÈܽ⣨ºÏÀí´ð°¸¾ù¸ø·Ö£©
£¨4£©2.5 10£¨´ð°¸²»ÒªÇó˳Ðò£©
£¨5£©2CH3COOH + CaCO3 = £¨ CH3COO£©2 Ca + CO2¡ü+ H2O