ÌâÄ¿ÄÚÈÝ
(9·Ö)ÈçͼÊÇijͬѧÉè¼ÆµÄ·ÅÈÈ·´Ó¦µÄ¹Û²ì×°Öá£![]()
ÆäʵÑé˳ÐòÊÇ£º¢Ù°´Í¼Ëùʾ½«ÊµÑé×°ÖÃÁ¬½ÓºÃ¡£
¢ÚÔÚUÐιÜÄÚ¼ÓÈëÉÙÁ¿ºìīˮ(»òÆ·ºì)ÈÜÒº¡£´ò¿ªTÐÍ¹Ü ÂÝÐý¼Ð£¬Ê¹UÐιÜÄÚÁ½±ßµÄÒºÃæ´¦ÓÚÍ¬Ò»Ë®Æ½Ãæ£¬ÔټнôÂÝÐý¼Ð¡£¢ÛÔÚÖмäµÄÊÔ¹ÜÀïÊ¢1 gÑõ»¯¸Æ£¬µ±µÎÈë2 mL×óÓÒµÄÕôÁóË®ºó£¬Í¬Ê±´ò¿ªÂÝÐý¼Ð¼´¿É¹Û²ì¡£
ÊԻشð£º(1)ʵÑéÖй۲쵽µÄÏÖÏóÊÇ
(2)¸ÃʵÑéÖбØÐë½øÐеÄÒ»²½ÊµÑé²Ù×÷ÊÇ
(3)¸ÃʵÑéµÄÔÀíÊÇ
________________________________________________________________________
(4)ʵÑéÖз´Ó¦µÄ»¯Ñ§·´Ó¦·½³ÌʽÊÇ
(5)˵Ã÷CaO¡¢H2OµÄÄÜÁ¿ÓëCa(OH)2µÄÄÜÁ¿Ö®¼äµÄ¹ØÏµ
(6)Èô¸ÃʵÑéÖÐCaO»»³ÉNaCl£¬ÊµÑ黹ÄÜ·ñ¹Û²ìµ½ÏàͬÏÖÏó£¿____¡£
£¨9·Ö£©(1)UÐβ£Á§¹ÜÀïµÄºìīˮ(»òÆ·ºì)»áÑØ¿ª¿Ú¶ËÉÏÉý£¨2·Ö£©
(2)¼ì²é×°ÖÃÆøÃÜÐÔ £¨1·Ö£©
(3)CaOºÍË®·´Ó¦·Å³öÈÈÁ¿Ê¹´óÊÔ¹ÜÖÐ¿ÕÆøÅòÕÍ£¬Ñ¹Ç¿Ôö´ó£¬ÒýÆðºìīˮ(»òÆ·ºì)ÔÚUÐιÜÖеÄÒºÃæ²»ÔÙÏàÆ½ £¨2·Ö£©
(4)CaO£«H2O===Ca(OH)2 £¨1·Ö£©
(5)CaOºÍH2OµÄÄÜÁ¿ºÍ´óÓÚCa(OH)2µÄÄÜÁ¿ £¨2·Ö£© (6)·ñ £¨1·Ö£©
½âÎö
| ÒÒËáÒÒõ¥ | ÒÒ´¼ | ÒÒËá | |
| ·Ðµã | 77.1¡æ | 78.5¡æ | 117.9¡æ |
£¨I£©×¼È·³ÆÁ¿20.0gÒÒËáÒÒõ¥´ÖÆ·ÓÚ×¶ÐÎÆ¿ÖУ¬ÓÃ0.50mol?L-1NaOHµÎ¶¨£¨·Ó̪×öָʾ¼Á£©£®ÖÕµãʱÏûºÄNaOHÈÜÒºµÄÌå»ýΪ40.0mL
£¨II£©ÁíÈ¡20.0gÒÒËáÒÒõ¥´Ö²úÆ·ÓÚ250mL×¶ÐÎÆ¿ÖУ¬¼ÓÈë100mL 2.1mol?L-1NaOHÈÜÒº»ìºÏ¾ùÔȺó£¬×°ÉÏÀäÄýÏ䣬ÔÚˮԡÉϼÓÈÈ»ØÁ÷Ô¼1Сʱ£¬×°ÖÃÈçͼ2Ëùʾ£®´ýÀäÈ´ºó£¬ÓÃ0.50mol?L-1HClµÎ¶¨¹ýÁ¿µÄNaOH£®ÖÕµãʱÏûºÄÑÎËáµÄÌå»ýΪ20.0mL£®
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÊµÑ飨II£©ÖÐÀäË®´ÓÀäË®Äý¹ÜµÄ
£¨2£©ÀûÓÃʵÑ飨I£©¡¢£¨II£©²âÁ¿µÄÊý¾Ý¼ÆËã´Ö²úÎïÖÐÒÒËáÒÒõ¥µÄÖÊÁ¿·ÖÊýΪ
£¨3£©ÊµÑé½áÊøºó£¬Í¬Ñ§ÃǶԴֲúÆ·ÖÐÒÒËáÒÒõ¥µÄº¬Á¿²»¸ß½øÐÐÌÖÂÛ£®
¢ÙÓÐÈËÈÏΪÊÇʵÑ飨II£©´øÀ´µÄÎó²î£®½¨Ò齫ͼ2ÖеÄ×¶ÐÎÆ¿¸ÄΪÈý¾±Æ¿£¬×°ÖÃÈçͼ3£¬ÔÚÈý¾±Æ¿µÄc¡¢d¿Ú×°ÅäÏà¹ØµÄÒÇÆ÷²¢½øÐÐÇ¡µ±µÄ²Ù×÷£¬¿ÉÒÔÌá¸ß²â¶¨µÄ¾«È·¶È£®ÄãÈÏΪÔÚÈý¾±Æ¿µÄc¡¢d¿Ú×°ÅäÏà¹ØµÄÒÇÆ÷»ò²Ù×÷ÊÇ£º
A£®×°ÉÏζȼƣ¬Ñϸñ¿ØÖÆ·´Ó¦Î¶È
B£®ÊµÑéÖо³£´ò¿ªÆ¿¿Ú£¬Óò£Á§½øÐнÁ°è
C£®ÔÚ·´Ó¦ºóÆÚ£¬ÓÃÊÂÏȰ²×°µÄ·ÖҺ©¶·Ìí¼ÓÒ»¶¨Á¿µÄNaOHÈÜÒº
¢Ú»¹ÓÐͬѧÈÏΪ¸Ä½øÒÒËáÒÒõ¥µÄÖÆÈ¡×°Öã¨Í¼1£©²ÅÄÜÌá¸ß²úÂÊ£®Äâ³öÄãµÄÒ»Ìõ¸Ä½ø½¨Òé