ÌâÄ¿ÄÚÈÝ
£¨14·Ö£¬Ã¿¿Õ2·Ö£©
I£®ÔÚ1LÈÝÆ÷ÖÐͨÈëCO2¡¢H2¸÷2mol£¬ÔÚÒ»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£ºCO2 + H2CO + H2O£¬
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÔÚ830¡æÌõ¼þÏ£¬·´Ó¦´ïµ½Æ½ºâʱCO2µÄת»¯ÂÊΪ50%¡£Çó¸ÃÌõ¼þÏÂƽºâ³£ÊýK1=________¡£
£¨2£©ÔÚ£¨1£©µÄ»ù´¡ÉÏ£¬°ÑÌåϵζȽµÖÁ800¡æ¡£ÒÑÖª¸ÃÌõ¼þϵÄƽºâ³£ÊýK2=0.81£¬¿ÉÒÔÍÆÖª ¸Ã·´Ó¦µÄÕý·´Ó¦Îª___________·´Ó¦£¨Ìî¡°ÎüÈÈ¡±¡¢¡°·ÅÈÈ¡±£©¡£
£¨3£©ÔÚ£¨1£©µÄ»ù´¡ÉÏ£¬Ñ¹ËõÈÝÆ÷Ìå»ýÖ®0.5L¡£¸ÃÌõ¼þϵÄƽºâ³£ÊýΪK3¡£ÔòK3________K1
£¨4£©T¡æʱ£¬Ä³Ê±¿Ì²âµÃÌåϵÖи÷ÎïÖʵÄÁ¿ÈçÏ£ºn£¨CO2£©=1.2mol£¬n£¨H2£©=1.5mol£¬
n£¨CO£©=0.9mol£¬n£¨H2O£©=0.9mol£¬Ôò´Ëʱ¸Ã·´Ó¦ ½øÐÐ.
£¨Ìî¡°ÏòÕý·´Ó¦·½Ïò¡±¡°ÏòÄæ·´Ó¦·½Ïò¡±»ò¡°´¦ÓÚƽºâ״̬¡±£©¡£
II£®ÏòÒ»ÈÝ»ýΪ1L µÄÃܱÕÈÝÆ÷ÖмÓÈëÒ»¶¨Á¿µÄX¡¢Y£¬·¢Éú»¯Ñ§·´Ó¦X(g)£«2Y(s) 2Z(g)£»¡÷H£¼0¡£ÓÒͼÊÇÈÝÆ÷ÖÐX¡¢ZµÄÎïÖʵÄÁ¿Å¨¶ÈËæʱ¼ä±ä»¯µÄÇúÏß¡£
(1)0¡«10min ÈÝÆ÷ÄÚÆøÌåµÄѹǿÖð½¥ ___________¡£
£¨Ìî¡°±ä´ó¡±¡¢¡°±äС¡±»ò¡°ÎÞ·¨È·¶¨¡±£©
(2)ÍƲâÔÚµÚ7minʱÇúÏ߱仯µÄÔÒò¿ÉÄÜÊÇ ___
µÚ13minʱÇúÏ߱仯µÄÔÒò¿ÉÄÜÊÇ __£¨ÌîÐòºÅ£©
¢ÙÔö¼ÓZµÄÁ¿ ¢ÚÔö¼ÓXµÄÁ¿ ¢ÛÉýÎÂ
¢Ü½µÎ ¢ÝʹÓô߻¯¼Á
I£®£¨1£© ¢Ù 1 ¢Ú ·ÅÈÈ ¢Û ="= " ¢ÜÕý·´Ó¦·½Ïò
II£®£¨1£©±ä´ó £¨2£© ¢Û¢Ý £¨3£©¢Û
½âÎö
£¨14·Ö£¬Ã¿¿Õ2·Ö£©Ä³ÊÔ¼Á³§ÓÃÒø£¨º¬ÔÓÖÊÍ£©ºÍÏõËᣨº¬Fe3+£©·´Ó¦ÖÆÈ¡ÏõËáÒø£¬²½ÖèÈçÏ£º
£¨1£©¹¤ÒµÉÏÒ»°ãÑ¡ÓÃÖеÈŨ¶ÈµÄÏõËáºÍÒø·´Ó¦À´ÖÆÈ¡ÏõËáÒø¡£ÇëÔÚϱí¿Õ¸ñ´¦Ìî¿Õ¡£
|
Óŵã |
ȱµã |
ʹÓÃŨÏõËá |
·´Ó¦ËÙÂÊ¿ì |
ËáºÄ½Ï´ó£¬²úÉúNOxµÄÁ¿½Ï¶à |
ʹÓÃÏ¡ÏõËá |
|
|
£¨2£©²½ÖèB¼ÓÈȱ£ÎµÄ×÷ÓÃÊÇ £º
a£® ÓÐÀûÓÚ¼Ó¿ì·´Ó¦ËÙÂÊ
b£®ÓÐÀûÓÚδ·´Ó¦µÄÏõËá»Ó·¢
c£®ÓÐÀûÓÚÏõËá³ä·Ö·´Ó¦£¬½µµÍÈÜÒºÖÐH+µÄŨ¶È
£¨3£©²½ÖèCÊÇΪÁ˳ýÈ¥Fe3+¡¢Cu2+µÈÔÓÖÊ£¬³åϡʱ²úÉú³ÁµíµÄÔÒòÊÇ £»
£¨4£©²½ÖèCÖмÓË®µÄÁ¿Ó¦¸ÃÊÊÁ¿£¬Èô¼ÓÈë¹ý¶àµÄË®£¬¶ÔºóÐø²½ÖèÔì³ÉµÄ²»Á¼Ó°ÏìÊÇ£º
£»
£¨5£©²½ÖèE½øÐеIJÙ×÷ÊÇ ¡£
£¨6£©ÖƵõÄÏõËáÒøÖк¬ÓÐÉÙÁ¿ÏõËáÍ£¬Í¨³£³ýÈ¥ÏõËá͵ķ½·¨ÊÇÔÚ²½ÖèE֮ǰ¼ÓÊÊÁ¿ÐÂÖƵÄAg2O£¬Ê¹Cu2+ת»¯ÎªCu(OH)2³Áµí£¬·´Ó¦ºó¹ýÂ˳ýÈ¥¡£¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º ¡£