ÌâÄ¿ÄÚÈÝ
(08Õã½Ê¡ÑÏÖÝÖÐѧ¶þÄ£)ÏÂÁÐʵÑéÖÐËùÑ¡ÓõÄÒÇÆ÷»ò²Ù×÷ÒÔ¼°½áÂÛ²»ºÏÀíµÄÊÇ ¡£
A¡¢ÓÃÍÐÅÌÌìÆ½³ÆÁ¿11.7gÂÈ»¯Äƾ§Ìå
B¡¢ÓüîʽµÎ¶¨¹ÜÁ¿È¡25mL̼ËáÄÆÈÜÒº
C¡¢²»ÒËÓôÉÛáÛöׯÉÕÇâÑõ»¯ÄÆ¡¢Ì¼ËáÄÆ
D¡¢²â¶¨ÈÜÒºµÄpHʱ£¬Óýྻ¡¢¸ÉÔïµÄ²£Á§°ôպȡÈÜÒº£¬µÎÔÚÓÃÕôÁóË®Èóʪ¹ýµÄpHÊÔÖ½ÉÏ£¬ÔÙÓë±ê×¼±ÈÉ«¿¨±È½Ï
E¡¢ÓÃÕôÁóË®ºÍpHÊÔÖ½£¬¾Í¿ÉÒÔ¼ø±ðpHÏàµÈµÄH2SO4ºÍCH2COOHÈÜÒº
F¡¢Á¿Í²ÄÚÒºÌåÌå»ýÕýÈ·¶ÁÊýΪ10.0mLʱ£¬È«²¿µ¹ÈëÉÕ±ÄÚµÄʵ¼ÊÌå»ý10.0mL
G¡¢100mLÈÝÁ¿Æ¿ÄÚÒºÃæÕýºÃ´ïµ½¿Ì¶ÈÏߣ¬È«²¿µ¹ÈëÉÕ±ÄÚµÄʵ¼ÊÌå»ýСÓÚ100mL
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⣺
![]()