ÌâÄ¿ÄÚÈÝ

ij»¯Ñ§¿ÎÍâС×éÄ£Ä⹤ҵÉú²úÖÆÈ¡HNO3£¬Éè¼ÆÏÂͼËùʾװÖã¬ÆäÖÐaΪһ¸ö¿É³ÖÐø¹ÄÈë¿ÕÆøµÄÏðƤÇò¡£Çë»Ø´ðÏÂÁÐÎÊÌâ:

(1)д³ö×°ÖÃAÖÐÖ÷Òª·´Ó¦µÄ»¯Ñ§·½³Ìʽ                           

(2)ÈçºÎ¼ì²é×°ÖÃAµÄÆøÃÜÐÔ                                     

(3)ÒÑÖª1molNO2ÓëҺ̬ˮ·´Ó¦Éú³ÉHNO3ÈÜÒººÍNOÆøÌå·Å³öÈÈÁ¿45.5kJ£¬Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ                                             

¸Ã·´Ó¦ÊÇÒ»¸ö¿ÉÄæ·´Ó¦£¬ÓûÒªÌá¸ßNO2µÄת»¯ÂÊ£¬¿É²ÉÈ¡µÄ´ëÊ©ÊÇ________

 A£®Éý¸ßζȠ   B£®½µµÍζȠ    C£®Ôö´óѹǿ    D£®Ôö´óѹǿ

(4)ʵÑé½áÊøºó£¬¹Ø±Õֹˮ¼Ðb¡¢c£¬½«×°ÖÃD£¬½þÈë±ùË®ÖУ¬ÏÖÏóÊÇ______

(5)×°ÖÃCÖÐŨH2SO4µÄ×÷ÓÃÊÇ                           

(6)ÇëÄã°ïÖú¸Ã»¯Ñ§Ð¡×éÉè¼ÆʵÑéÊÒÖÆÈ¡NH3µÄÁíÒ»·½°¸              (Óû¯Ñ§·´Ó¦·½³Ìʽ±íʾ)

(1)2NH4Cl+Ca(OH)22NH3¡ü+CaCl2+2H2O(2·Ö)

£¨2£©½«AÖе¼¹Ü϶˽þÈëË®ÖУ¬ÓÃË«ÊÖ½ôÎÕ´óÊԹܵײ¿£¬µ¼¹Ü¿ÚÓÐÆøÅݲúÉú£¬Ë«ÊÖÀ뿪£¬µ¼¹Ü²úÉúÒ»¶ÎË®Öô£¬ËµÃ÷×°ÖÃA²»Â©Æø¡£(2·Ö)

(3)3NO2(g)+H2O(l)=2HNO3(aq)+NO(g) ¡÷H=£­136.5kJ¡¤mol-1(2·Ö)BC  (2·Ö)

(4)ÑÕÉ«±ädz£»(2·Ö)(5)ÎüÊÕ¶àÓàµÄNH3£»(2·Ö)

(6) NH3¡¤H2O = NH3¡ü+H2O£» (ÆäËûºÏÀí´ð°¸Ò²¿ÉÒÔ) (2·Ö)


½âÎö:

ÂÔ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ij»¯Ñ§¿ÎÍâС×éÄ£Ä⹤ҵÉú²úÖÆÈ¡HNO3£¬Éè¼ÆÏÂͼËùʾװÖã¬ÆäÖÐaΪһ¸ö¿É³ÖÐø¹ÄÈë¿ÕÆøµÄÏðƤÇò£®
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³ö×°ÖÃAÖÐÖ÷Òª·´Ó¦µÄ»¯Ñ§·½³Ìʽ
2NH4Cl+Ca£¨OH£©2
  ¡÷  
.
 
2NH3¡ü+CaCl2+2H2O
2NH4Cl+Ca£¨OH£©2
  ¡÷  
.
 
2NH3¡ü+CaCl2+2H2O
£®
£¨2£©¼ì²é×°ÖÃAµÄÆøÃÜÐÔ£ºÏȽ«AÖе¼¹Ü϶˽þÈëË®ÖУ¬ÔÙ
ÓÃË«ÊÖ½ôÎÕ´óÊԹܵײ¿£¬µ¼¹Ü¿ÚÓÐÆøÅݲúÉú£¬Ë«ÊÖÀ뿪£¬µ¼¹Ü²úÉúÒ»¶ÎË®Öô£¬ËµÃ÷×°ÖÃA²»Â©Æø
ÓÃË«ÊÖ½ôÎÕ´óÊԹܵײ¿£¬µ¼¹Ü¿ÚÓÐÆøÅݲúÉú£¬Ë«ÊÖÀ뿪£¬µ¼¹Ü²úÉúÒ»¶ÎË®Öô£¬ËµÃ÷×°ÖÃA²»Â©Æø
£®
£¨3£©ÒÑÖª1molNO2ÓëҺ̬ˮ·´Ó¦Éú³ÉHNO3ÈÜÒººÍNOÆøÌå·Å³öÈÈÁ¿45.5kJ£¬Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ
3NO2£¨g£©+H2O£¨l£©=2HNO3£¨aq£©+NO£¨g£©¡÷H=-136.5kJ?mol-1
3NO2£¨g£©+H2O£¨l£©=2HNO3£¨aq£©+NO£¨g£©¡÷H=-136.5kJ?mol-1
£¬¸Ã·´Ó¦ÊÇÒ»¸ö¿ÉÄæ·´Ó¦£¬ÓûÒªÌá¸ßNO2µÄת»¯ÂÊ£¬¿É²ÉÈ¡µÄ´ëÊ©ÊÇ
BC
BC

A£®Éý¸ßζȠ     B£®½µµÍζȠ       C£®Ôö´óѹǿ       D£®¼õСѹǿ
£¨4£©ÊµÑé½áÊøºó£¬¹Ø±Õֹˮ¼Ðb¡¢c£¬½«×°ÖÃD½þÈë±ùË®ÖУ¬ÏÖÏóÊÇ
ÑÕÉ«±ädz
ÑÕÉ«±ädz
£®
£¨5£©×°ÖÃCÖÐŨH2SO4µÄ×÷ÓÃÊÇ
¸ÉÔïÆøÌå¡¢ÎüÊÕ¶àÓàµÄ°±Æø
¸ÉÔïÆøÌå¡¢ÎüÊÕ¶àÓàµÄ°±Æø
£®
£¨6£©ÇëÄã°ïÖú¸Ã»¯Ñ§Ð¡×éÉè¼ÆʵÑéÊÒÖÆÈ¡NH3µÄÁíÒ»·½°¸
NH3?H2O
  ¡÷  
.
 
NH3¡ü+H2O
NH3?H2O
  ¡÷  
.
 
NH3¡ü+H2O
£¨Óû¯Ñ§·´Ó¦·½³Ìʽ±íʾ£©£®

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

¾«Ó¢¼Ò½ÌÍø