ÌâÄ¿ÄÚÈÝ
ij»¯Ñ§»î¶¯Ð¡×éÓûÓô¿¼îÖÆÈ¡Éռ×öÁËÈçÏÂʵÑ飺
ÏÖÈ¡20g´¿¼îÑùÆ·(¼ÙÉèÔÓÖʲ»ÈÜÓÚË®£¬Ò²²»ÓëÆäËûÎïÖÊ·´Ó¦)£¬ÏòÆäÖмÓÈë34.1gˮʹÆäÍêÈ«Èܽ⣬ȻºóÔÙ¼ÓÈë100gµÄÈÜÖÊÖÊÁ¿·ÖÊýΪl7.1%µÄÇâÑõ»¯±µÈÜÒº£¬ÖÁÇ¡ºÃÍêÈ«·´Ó¦£¬¹ýÂ˵õ½ÂËÔüºÍÒ»ÖÖ²»±¥ºÍÈÜÒº¡£Çë»Ø´ðÎÊÌ⣺
(1)ÉÏÊöʵÑé²½ÖèÖУ¬ÇëÖ¸³ö²Ù×÷ÖеĴíÎó £»
(2)Çëд³ö·¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ £»
(3)Áгö²Î¼Ó·´Ó¦¹ÌÌåÎïÖÊÖÊÁ¿(x)µÄ±ÈÀýʽΪ £»
(4)Çó´¿¼îÑùÆ·µÄ´¿¶ÈΪ £»
(5)·´Ó¦ºóËùµÃÈÜÒºµÄÈÜÖÊÖÊÁ¿·ÖÊýΪ ¡£
£¨1£©¹ýÂËʱûÓò£Á§°ôÒýÁ÷ £¨2£©Na2CO3+ Ba(OH)2==BaCO3¡ý+2NaOH
£¨3£© £¨4£©53% £¨5£©6.4%
½âÎöÊÔÌâ·ÖÎö£º(1)¹ýÂËʱ²£Á§°ôµÄ×÷ÓÃÊÇÒýÁ÷£»£¨3£©¸ù¾Ý»¯Ñ§·½³ÌʽµÄ¼ÆË㣬±ÈÀýʽΪ£»£¨4£©¸ù¾Ý£¨2£©ÎÊÇó³ö²Î¼Ó·´Ó¦µÄ̼ËáÄƵÄÖÊÁ¿£¬Îª20g´¿¼îÑùÆ·Öд¿¾»µÄ̼ËáÄƵÄÖÊÁ¿£»£¨5£©¸ù¾ÝÈÜÒºÖÊÁ¿·ÖÊýµÄ¶¨Ò壬µÃÓÃÈÜÖÊÖÊÁ¿³ýÒÔÈÜÒº×ÜÖÊÁ¿È»ºó³ËÒÔ°Ù·ÖÖ®°Ù£¬ÈÜÖÊÖÊÁ¿¼´Îª·´Ó¦Éú³ÉµÄÇâÑõ»¯ÄƵÄÖÊÁ¿£¬¸ù¾Ý£¨2£©Öл¯Ñ§·½³Ìʽ¿ÉÒÔËã³ö£¬ÈÜÒºÖÊÁ¿¼´Îª20g´¿¼îÑùÆ·Öд¿¾»µÄ̼ËáÄƵÄÖÊÁ¿¼ÓÉÏ34.1gË®£¬ÔÙ¼ÓÉÏ100gµÄÈÜÖÊÖÊÁ¿·ÖÊýΪl7.1%µÄÇâÑõ»¯±µÈÜÒºµÄÖÊÁ¿£¬×îºó¼õÈ¥Éú³ÉµÄ̼Ëá±µµÄÖÊÁ¿£¨¸ù¾Ý£¨2£©Öл¯Ñ§·½³Ìʽ¿ÉÒÔËã³ö£©¡£
¿¼µã£º»¯Ñ§·½³ÌʽµÄ¼ÆËã
µãÆÀ£ºÕâÀàÌâÄ¿ÒªÓвã´ÎµÄ×ö£¬²»Òª¸¡Ô꣬ÊôÓÚÖп¼±Ø¿¼ÌâÐÍ£¬×¢Òâ¼ÆËãϸÐÄ¡£