ÌâÄ¿ÄÚÈÝ
Èçͼ±íʾÅäÖÆ 100mL 0.100mol?L-1 Na2CO3ÈÜÒºµÄ¼¸¸ö¹Ø¼üʵÑé²½ÖèºÍ²Ù×÷£¬¾Ýͼ»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÓÃÍÐÅÌÌìÆ½³ÆÈ¡Na2CO3?10H2O µÄÖÊÁ¿ÊÇ
£¨2£©²½ÖèEÖн«Ò»²£Á§ÒÇÆ÷ÉÏϵߵ¹Êý´Î£¬¸ÃÒÇÆ÷µÄÃû³ÆÊÇ
£¨3£©²½ÖèBͨ³£³ÆÎª×ªÒÆ£¬²½ÖèAͨ³£³ÆÎª
£¨4£©½«ÉÏÊöʵÑé²½ÖèA¡úF°´ÊµÑé¹ý³ÌÏȺó´ÎÐòÅÅÁÐ
£¨1£©ÓÃÍÐÅÌÌìÆ½³ÆÈ¡Na2CO3?10H2O µÄÖÊÁ¿ÊÇ
2.9g
2.9g
£®£¨2£©²½ÖèEÖн«Ò»²£Á§ÒÇÆ÷ÉÏϵߵ¹Êý´Î£¬¸ÃÒÇÆ÷µÄÃû³ÆÊÇ
ÈÝÁ¿Æ¿
ÈÝÁ¿Æ¿
£®£¨3£©²½ÖèBͨ³£³ÆÎª×ªÒÆ£¬²½ÖèAͨ³£³ÆÎª
¶¨ÈÝ
¶¨ÈÝ
£®£¨4£©½«ÉÏÊöʵÑé²½ÖèA¡úF°´ÊµÑé¹ý³ÌÏȺó´ÎÐòÅÅÁÐ
DCBFAE
DCBFAE
£®·ÖÎö£º£¨1£©¸ù¾Ým=nM=cvM¼ÆË㣻
£¨2£©ÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºµÄÒÇÆ÷ÊÇÈÝÁ¿Æ¿£»
£¨3£©ÓýºÍ·µÎ¹ÜµÎ¼ÓÈÜÒºµÄ²Ù×÷Ãû³ÆÊǶ¨ÈÝ£»
£¨4£©¸ù¾ÝÅäÖÆÈÜÒºµÄʵÑé²Ù×÷¹ý³Ì½øÐÐʵÑé²½ÖèÅÅÐò£®
£¨2£©ÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºµÄÒÇÆ÷ÊÇÈÝÁ¿Æ¿£»
£¨3£©ÓýºÍ·µÎ¹ÜµÎ¼ÓÈÜÒºµÄ²Ù×÷Ãû³ÆÊǶ¨ÈÝ£»
£¨4£©¸ù¾ÝÅäÖÆÈÜÒºµÄʵÑé²Ù×÷¹ý³Ì½øÐÐʵÑé²½ÖèÅÅÐò£®
½â´ð£º½â£º£¨1£©ÊµÑéÊÒÅäÖÆ100mL 0.100mol?L-1 Na2CO3ÈÜÒºÐèÒªNa2CO3?10H2OµÄÖÊÁ¿Îª£º0.1L¡Á0.1mol/L¡Á286g/mol=2.9g£¬¹Ê´ð°¸Îª£º2.9g£»
£¨2£©ÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºµÄÒÇÆ÷ÊÇÈÝÁ¿Æ¿£¬ËùÒÔ¸ÃÒÇÆ÷µÄÃû³ÆÊÇÈÝÁ¿Æ¿£¬¹Ê´ð°¸Îª£ºÈÝÁ¿Æ¿£»
£¨3£©¼ÓË®ÖÁÒºÃæ¾àÀë¿Ì¶ÈÏß1¡«2cmʱ£¬ÓýºÍ·µÎ¹ÜÏòÈÝÁ¿Æ¿ÖеμÓÈÜÒº£¬¸Ã²Ù×÷µÄÃû³ÆÊǶ¨ÈÝ£¬¹Ê´ð°¸Îª£º¶¨ÈÝ£»
£¨4£©²Ù×÷²½ÖèÓмÆËã¡¢³ÆÁ¿¡¢ÈÜ½â¡¢ÒÆÒº¡¢Ï´µÓÒÆÒº¡¢¶¨ÈÝ¡¢Ò¡ÔȵȲÙ×÷£¬ËùÒÔʵÑé¹ý³ÌÏȺó´ÎÐòÅÅÁÐΪ£ºDCBFAE£¬¹Ê´ð°¸Îª£ºDCBFAE£»
£¨2£©ÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºµÄÒÇÆ÷ÊÇÈÝÁ¿Æ¿£¬ËùÒÔ¸ÃÒÇÆ÷µÄÃû³ÆÊÇÈÝÁ¿Æ¿£¬¹Ê´ð°¸Îª£ºÈÝÁ¿Æ¿£»
£¨3£©¼ÓË®ÖÁÒºÃæ¾àÀë¿Ì¶ÈÏß1¡«2cmʱ£¬ÓýºÍ·µÎ¹ÜÏòÈÝÁ¿Æ¿ÖеμÓÈÜÒº£¬¸Ã²Ù×÷µÄÃû³ÆÊǶ¨ÈÝ£¬¹Ê´ð°¸Îª£º¶¨ÈÝ£»
£¨4£©²Ù×÷²½ÖèÓмÆËã¡¢³ÆÁ¿¡¢ÈÜ½â¡¢ÒÆÒº¡¢Ï´µÓÒÆÒº¡¢¶¨ÈÝ¡¢Ò¡ÔȵȲÙ×÷£¬ËùÒÔʵÑé¹ý³ÌÏȺó´ÎÐòÅÅÁÐΪ£ºDCBFAE£¬¹Ê´ð°¸Îª£ºDCBFAE£»
µãÆÀ£º±¾Ì⿼²éÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºµÄÅäÖÆ£¬ÄѶȲ»´ó£¬×¢ÒâʵÑé²½Öè¡¢Îó²î·ÖÎö¡¢ÈÝÁ¿Æ¿µÄѡȡ£¬ÎªÒ×´íµã£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿