ÌâÄ¿ÄÚÈÝ
(08ÉϺ£ËÉ½ÇøÄ£Äâ)ÔÚÑо¿ÏõËáµÄÐÔÖÊʱ£¬Ä³°à¼¶Ñ§Éú×öÁËÈçÏÂÈý¸öʵÑ飺
ʵÑéÒ»£ºÊÔ¹ÜÖмÓÈëÍÆ¬ºÍŨÏõËᣬƬ¿Ìºó£¬Á¢¼´ÈûÉÏ´øÓв£Á§µ¼¹ÜµÄÏðƤÈû£¬µ¼¹ÜÉìÈë×°ÓÐÇâÑõ»¯ÄÆÈÜÒºµÄÉÕ±ÖС£
ʵÑé¶þ£ºÊÔ¹ÜÖмÓÈëÍÆ¬ºÍÏ¡ÏõËᣨ¿ÉÊʵ±¼ÓÈÈ£©£¬Æ¬¿Ìºó£¬Á¢¼´ÈûÉÏ´øÓв£Á§µ¼¹ÜµÄÏðƤÈû£¬µ¼¹ÜÉìÈë×°ÓÐÇâÑõ»¯ÄÆÈÜÒºµÄÉÕ±ÖС£
ʵÑéÈý£ºÊÔ¹ÜÖмÓÈëºìÈȵÄľ̿ºÍŨÏõËᣬƬ¿Ìºó£¬Á¢¼´ÈûÉÏ´øÓв£Á§µ¼¹ÜµÄÏðƤÈû£¬µ¼¹ÜÉìÈë×°ÓÐÇâÑõ»¯ÄÆÈÜÒºµÄÉÕ±ÖС£
ÎÊ£º
£¨1£©ÊµÑéÈýÏÖÏó
£¨2£©ÊµÑéÒ»Öз¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ
(3) ʵÑé¶þÖÐÏõËáµÄ×÷ÓÃÊÇ
(4)ÈôÓÃÍÀ´ÖÆÈ¡ÏõËáÍÈÜÒº£¬²ÉÓÃʲô·½·¨·½·¨½ÏºÃ£¿£¨Ð´³ö»¯Ñ§·½³Ìʽ£©
£¨5£©Èô°ÑÏõËáÍÈÜÒºÖÆ³ÉÏõËá;§Ì壬Ӧ¿¼ÂÇÄÄЩÎÊÌ⣿
¡£
´ð°¸£º
£¨1£©²úÉúºì×ØÉ«ÆøÌ壬¹ÌÌåÈܽ⣨ 2·Ö£©
£¨2£©Cu + 2NO3- + 4 H+ ¡ú Cu2+ + 2NO2¡ü + 2H2O£¨ 2·Ö£©
(3) Ñõ»¯¼Á Ëá £¨ 2·Ö£©£¨¼ÓÈÈ£©
£¨4£© £¨4·Ö£© 2 Cu + O2 ¡ú 2CuO £»CuO + 2HNO3 ¡ú Cu£¨NO3£©2 + 2H2O
£¨5£©·Àֹˮ½â£¬Ëá¹ýÁ¿£»Î¶Ȳ»Äܹý¸ß£¬·ÀÖ¹·Ö½â¡££¨ 2·Ö£©
2(08Õã½Ê¡¿ª»¯ÖÐѧģÄâ)ʵÑéÊÒÖиù¾Ý2SO2£«O2
2SO3£»¦¤H=-393.2 kJ?mol-1Éè¼ÆÈçÏÂͼËùʾʵÑé×°ÖÃÀ´ÖƱ¸SO3¹ÌÌå¡£Çë»Ø´ðÏÂÁÐÎÊÌâ¡£
£¨1£©ÊµÑéǰ£¬±ØÐë½øÐеIJÙ×÷ÊÇ£¨Ìî²Ù×÷Ãû³Æ£¬²»±ØÐ´¾ßÌå¹ý³Ì£©¡¡¡¡¡¡¡¡¡¡¡¡
£¨2£©ÔÚA×°ÖÃÖмÓÈëNa2SO3¹ÌÌåµÄͬʱ£¬»¹Ðè¼Ó¼¸µÎË®£¬È»ºóÔٵμÓŨÁòËá¡£¼Ó¼¸µÎË®µÄ×÷ÓÃÊÇ ¡¡
£¨3£©Ð¡ÊÔ¹ÜCµÄ×÷ÓÃÊÇ
£¨4£©¹ã¿ÚÆ¿DÄÚÊ¢µÄÊÔ¼ÁÊÇ ¡£×°ÖÃDµÄÈý¸ö×÷ÓÃÊÇ ¢Ù¡¡¡¡ ¡¡¡¡
¢Ú ¢Û
£¨5£©ÊµÑéÖе±Cr2O3±íÃæºìÈÈʱ£¬Ó¦½«¾Æ¾«µÆÒÆ¿ªÒ»»á¶ùÔÙ¼ÓÈÈ£¬ÒÔ·Àζȹý¸ß£¬ÕâÑù×öµÄÔÒòÊÇ ¡¡ ¡¡
£¨6£©×°ÖÃFÖÐUÐ͹ÜÄÚÊÕ¼¯µ½µÄÎïÖʵÄÑÕÉ«¡¢×´Ì¬ÊÇ
£¨7£©×°ÖÃGµÄ×÷ÓÃÊÇ
£¨8£©´ÓG×°Öõ¼³öµÄÎ²Æø´¦Àí·½·¨ÊÇ
(08ÍîÄÏ8УµÚÈý´ÎÁª¿¼)1 Lij»ìºÏÈÜÒº£¬¿ÉÄܺ¬ÓеÄÀë×ÓÈçÏÂ±í£º
¿ÉÄÜ´óÁ¿»¹ÓеÄÑôÀë×Ó |
|
¿ÉÄÜ´óÁ¿»¹ÓеÄÒõÀë×Ó |
|
£¨1£©Íù¸ÃÈÜÒºÖÐÖðµÎ¼ÓÈë
ÈÜÒº²¢Êʵ±
¼ÓÈÈ£¬²úÉú³ÁµíºÍÆøÌåµÄÎïÖʵÄÁ¿£¨
£©
Óë¼ÓÈë
ÈÜÒºµÄÌå»ý£¨V£©µÄ¹ØÏµ
ÈçÓÒͼËùʾ¡£Ôò¸ÃÈÜÒºÖÐÈ·¶¨º¬ÓеÄÀë×Ó
ÓÐ_______________£»²»ÄÜÈ·¶¨ÊÇ·ñº¬ÓÐ
µÄÑôÀë×ÓÓÐ______________£¬ÒªÈ·¶¨Æä´æ
Ôڿɲ¹³ä×öµÄʵÑéÊÇ___________£»¿Ï¶¨²»´æÔÚµÄÒõÀë×ÓÓÐ_________________¡£
£¨2£©¾¼ì²â£¬¸ÃÈÜÒºÖк¬ÓдóÁ¿µÄ
£¬ÈôÏò1 L¸Ã»ìºÏÈÜÒºÖÐͨÈë¨D¶¨ÔεÄ
£¬ÈÜÒºÖÐ
µÄÎïÖʵÄÁ¿ÓëͨÈë
µÄÌå»ý£¨±ê×¼×´¿ö£©µÄ¹ØÏµÈçϱíËùʾ£¬·ÖÎöºó»Ø´ðÏÂÁÐÎÊÌ⣺
![]()