ÌâÄ¿ÄÚÈÝ

Ë®·Ö×Ó¼ä´æÔÚÒ»ÖֽС°Çâ¼ü¡±µÄ×÷ÓÃ(½éÓÚ·¶µÂ»ªÁ¦Ó뻯ѧ¼üÖ®¼ä)£¬ÇÒÆä±Ë´Ë½áºÏ¶øÐγÉ(H2O)n¡£ÔÚ±ùÖÐÿ¸öË®·Ö×Ó±»4¸öË®·Ö×Ó°üΧÐγɱäÐεÄÕýËÄÃæÌ壬ͨ¹ý¡°Çâ¼ü¡±Ï໥Á¬½Ó³ÉÅÓ´óµÄ·Ö×Ó¾§Ì塪¡ª±ù£¬Æä½á¹¹Ê¾ÒâͼÈçÏÂËùʾ£º

(1)1 molË®ÖÐÓÐ______________ mol¡°Çâ¼ü¡±¡£

(2)Ë®·Ö×ÓµçÀëÉú³ÉÁ½ÖÖº¬ÓÐÏàͬµç×ÓÊýµÄ΢Á££¬ÆäµçÀë·½³ÌʽΪ______________________¡£

(3)ÔÚ±ùµÄ½á¹¹ÖУ¬Ã¿¸öË®·Ö×ÓÓëÏàÁÚµÄ4¸öË®·Ö×ÓÒÔÇâ¼üÏàÁ¬½Ó£¬ÔÚ±ùÖгýÇâ¼üÍ⣬»¹´æÔÚ·Ö×Ó¼ä×÷ÓÃÁ¦(11 kJ¡¤mol-1)¡£ÒÑÖª±ùµÄÉý»ªÈÈÊÇ51 kJ¡¤mol-1£¬Ôò±ùÖÐÇâ¼üµÄÄÜÁ¿ÊÇ______________ kJ¡¤mol-1¡£

(4)ÓÃx¡¢y¡¢z·Ö±ð±íʾH2O¡¢H2S¡¢H2SeµÄ·Ðµã(¡æ)£¬Ôòx¡¢y¡¢zµÄ´óС¹ØϵÊÇ______________£¬ÆäÅжÏÒÀ¾ÝÊÇ____________________________________________________________¡£

(1)2

(2)2H2OH3O++OH-

(3)20

(4)x£¾z£¾y  Ë®·Ö×Ó¼ä´æÔÚÇâ¼ü£¬H2SeµÄ·Ö×Ó¼ä×÷ÓÃÁ¦´óÓÚH2S

½âÎö£º(1)ÿ¸öË®·Ö×ÓÏàÁÚµÄ4¸öË®·Ö×ÓÒÔÇâ¼üÏàÁ¬½Ó£¬ÐγÉ4¸öÇâ¼ü£¬µ«Ã¿¸öÇâ¼ü¹é2¸öË®·Ö×ÓËùÓУ¬ËùÒÔ1 mol±ùÖÐÓÐ2 molÇâ¼ü¡£

(2)Ë®ÊǼ«ÈõµÄµç½âÖÊ£¬¿ÉµçÀëÉú³ÉOH-ºÍH3O+¡£

(3)±ùÖÐÇâ¼üµÄÄÜÁ¿Îª£º=20 kJ¡¤mol-1

(4)¶ÔÓÚ×é³ÉºÍ½á¹¹ÏàËƵÄÎïÖÊ£¬Ïà¶Ô·Ö×ÓÖÊÁ¿Ô½´ó£¬·Ö×Ó¼ä×÷ÓÃÁ¦Ô½´ó£¬·ÐµãÔ½¸ß£¬¶øË®·Ö×Ӽ仹´æÔÚÇâ¼ü£¬·Ö×Ó¼ä×÷ÓÃÁ¦×î´ó£¬·Ðµã×î¸ß¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø