ÌâÄ¿ÄÚÈÝ

9£®ÏÖÒªÅäÖÃÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄ100mL1mol•L-1NaOHÈÜÒº£¬ÏÂÁвÙ×÷ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®ÔÚÍÐÅÌÌìÆ½µÄÁ½¸öÍÐÅÌÉϸ÷·ÅÒ»ÕÅ´óСһÑùµÄÖ½£¬È»ºó½«NaOH¹ÌÌå·ÅÔÚÖ½ÉϽøÐгÆÁ¿
B£®°Ñ³ÆÁ¿µÄNaOH¹ÌÌå·ÅÈëÊ¢ÓÐÊÊÁ¿ÕôÁóË®µÄÉÕ±­ÖУ¬ÈܽâºóÁ¢¼´°ÑÈÜÒº×ªÒÆµ½ÈÝÁ¿Æ¿ÖÐ
C£®ÓÃÕôÁóˮϴµÓÉÕ±­¡¢²£Á§°ô2¡«3´Î£¬Ã¿´ÎÏ´µÓºóµÄÈÜÒº¶¼×¢ÈëÈÝÁ¿Æ¿ÖÐ
D£®ÑØ×Ų£Á§°ôÍùÈÝÁ¿Æ¿ÖмÓÈëÕôÁóË®£¬Ö±µ½ÈÜÒº°¼ÒºÃæÇ¡ºÃÓë¿Ì¶ÈÏàÇÐ

·ÖÎö A£®NaOHÒ׳±½â£¬¾ßÓи¯Ê´ÐÔ£»
B£®Ó¦ÀäÈ´ºó×ªÒÆ£»
C£®ÉÕ±­¡¢²£Á§°ôÏ´µÓ£¬²»»á¼õÉÙÈÜÖÊ£»
D£®¶¨ÈÝʱ¼ÓË®ÖÁ¿Ì¶ÈÏß1¡«2cm´¦£¬¸ÄÓýºÍ·µÎ¹ÜµÎ¼Ó£®

½â´ð ½â£ºA£®NaOHÒ׳±½â£¬¾ßÓи¯Ê´ÐÔ£¬ÔòÓ¦½«NaOHÔÚСÉÕ±­ÖгÆÁ¿£¬¹ÊA´íÎó£»
B£®Ó¦ÀäÈ´ºó×ªÒÆ£¬²»ÄÜÈܽâºóÁ¢¼´°ÑÈÜÒº×ªÒÆµ½ÈÝÁ¿Æ¿ÖУ¬¹ÊB´íÎó£»
C£®ÉÕ±­¡¢²£Á§°ôÏ´µÓ£¬²»»á¼õÉÙÈÜÖÊ£¬ÔòÓÃÕôÁóˮϴµÓÉÕ±­¡¢²£Á§°ô2¡«3´Î£¬Ã¿´ÎÏ´µÓºóµÄÈÜÒº¶¼×¢ÈëÈÝÁ¿Æ¿ÖУ¬¹ÊCÕýÈ·£»
D£®¶¨ÈÝʱ¼ÓË®ÖÁ¿Ì¶ÈÏß1¡«2cm´¦£¬¸ÄÓýºÍ·µÎ¹ÜµÎ¼Ó£¬µ½ÈÜÒº°¼ÒºÃæÇ¡ºÃÓë¿Ì¶ÈÏàÇУ¬¹ÊD´íÎó£»
¹ÊÑ¡C£®

µãÆÀ ±¾Ì⿼²éÈÜÒºµÄÅäÖÆ£¬Îª¸ßƵ¿¼µã£¬°ÑÎÕÅäÖÆÈÜÒºµÄ²Ù×÷¡¢ÒÇÆ÷¡¢ÊµÑé¼¼ÄÜΪ½â´ðµÄ¹Ø¼ü£¬²àÖØ·ÖÎöÓëʵÑéÄÜÁ¦µÄ¿¼²é£¬×¢ÒâʵÑé»ù±¾²Ù×÷£¬ÌâÄ¿ÄѶȲ»´ó£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
20£®ÔÚÒ»¶¨Ìå»ýµÄÃܱÕÈÝÆ÷ÖУ¬½øÐÐÈçÏ»¯Ñ§·´Ó¦£º
CO£¨g£©+H2O£¨g£©$\stackrel{´ß»¯¼Á}{?}$CO2£¨g£©+H2£¨g£©£¬Æä»¯Ñ§Æ½ºâ³£ÊýKºÍζÈtµÄ¹ØÏµÈç±í£º
    t¡æ    700    800    830    1000   1200
    K    1.7    1.1    1.0    0.6    0.4
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¸Ã·´Ó¦µÄ»¯Ñ§Æ½ºâ³£Êý±í´ïʽΪK=$\frac{c£¨C{O}_{2}£©£®c£¨{H}_{2}£©}{c£¨CO£©£®c£¨{H}_{2}O£©}$£¬¸Ã·´Ó¦Îª·ÅÈÈ·´Ó¦£¨Ñ¡Ìî¡°ÎüÈÈ¡±¡¢¡°·ÅÈÈ¡±£©£®
£¨2£©ÄÜÅжϸ÷´Ó¦ÊÇ·ñ´ïµ½»¯Ñ§Æ½ºâ״̬µÄÒÀ¾ÝÊÇBC£»
A£®ÈÝÆ÷ÖÐѹǿ²»±ä  B£®»ìºÏÆøÌåÖÐc£¨CO£©²»±ä  C£®vÄæ£¨H2£©=vÕý£¨H2O£© D£®c£¨CO2£©=c£¨CO£©
£¨3£©830¡æÊ±£¬ÈÝÆ÷Öеķ´Ó¦ÒѴﵽƽºâ£®ÔÚÆäËûÌõ¼þ²»±äµÄÇé¿öÏ£¬ÔÚ´ËζÈÏ£¬Èô¸ÃÈÝÆ÷Öк¬ÓÐ1molCO2¡¢1.2molH2¡¢0.75molCO¡¢1.5molH2O£¬Õâ״̬·ñ£¨ÊÇ»ò·ñ£©´¦ÓÚÆ½ºâ״̬£¿Èô²»ÊÇ£¬·´Ó¦ÏòÄĸö·½Ïò½øÐУ¿ÏòÄæ·´Ó¦·½Ïò£®£¨Ñ¡Ìî¡°ÏòÕý·´Ó¦·½Ïò¡±¡¢¡°ÏòÄæ·´Ó¦·½Ïò¡±£©£®
£¨4£©Èô830¡æÊ±£¬ÏòÈÝÆ÷ÖгäÈëlmolCO¡¢5molH2O£¬·´Ó¦´ïµ½Æ½ºâºó£¬COµÄת»¯ÂÊΪ83.3%£¨»ò83%£¬»ò5/6£©£®

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

¾«Ó¢¼Ò½ÌÍø