ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿ÈçÏÂÁи÷ͼËùʾÊÇÖÐѧ»¯Ñ§Öг£ÓÃÓÚ»ìºÏÎï·ÖÀëºÍÌá´¿µÄ×°Öãº
A. B.
C. D.
Çë¸ù¾Ý×°ÖÃͼ»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÅAͼÖÐÓÃÓÚÊ¢·ÅÒºÌå»ìºÏÎïµÄÒÇÆ÷µÄÃû³ÆΪ£º__________________
¢ÆÑ¡ÔñºÏÊʵÄʵÑé×°Ö÷ÖÀëÒÔÏ»ìºÏÎÌî´ú±í×°ÖõÄ×Öĸ£©£¬
¢Ù´ÓµâË®ÖзÖÀë³öI2 £º ___________
¢Ú½øÐоƾ«ºÍË®µÄ·ÖÀ룺 ___________
¢Û¶þÑõ»¯Ã̺ÍÕáÌÇÈÜÒº£º ___________
¢ÇÎÒ¹ú¾ßÓÐÀúÊ·ÓƾõľÆÎÄ»¯£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÙÒªÌá¸ß¾Æ¾«µÄ¶ÈÊý£¬¿É²ÉÓõķ½·¨ÊÇ________________£¨Ìî²Ù×÷·½·¨£¬ÏÂͬ£©
¢ÚÖÐÒ©³£³£½þÅÝÔÚÃÀ¾ÆÖÐÖƳÉÒ©¾Æ£¬ÄãÈÏΪÀûÓõķ½·¨ÊÇ_____________¡£
¡¾´ð°¸¡¿ÕôÁóÉÕÆ¿BACÕôÁóÝÍÈ¡
¡¾½âÎö¡¿
£¨1£©AͼΪÕôÁó×°Öã¬ÓÃÓÚÊ¢·ÅÒºÌå»ìºÏÎïµÄÒÇÆ÷ΪÕôÁóÉÕÆ¿¡£
£¨2£©¸ù¾Ý»ìºÏÎïÖи÷ÎïÖÊÐÔÖʵIJîÒìÑ¡Ôñ·ÖÀë·½·¨ºÍ×°ÖÃͼ¡£
£¨3£©¸ù¾Ý¸÷ÎïÖÊÐÔÖʵIJîÒìÑ¡Ôñ·ÖÀë·½·¨¡£
£¨1£©AͼΪÕôÁó×°Öã¬ÓÃÓÚÊ¢·ÅÒºÌå»ìºÏÎïµÄÒÇÆ÷µÄÃû³ÆΪÕôÁóÉÕÆ¿¡£
£¨2£©AͼΪÕôÁó×°Öã¬BͼΪ·ÖҺװÖã¬CͼΪ¹ýÂË×°Öã¬DͼΪÕô·¢×°Öã»
¢ÙµâÒ×ÈÜÓÚCCl4¡¢±½µÈÓлúÈܼÁ£¬´ÓµâË®ÖзÖÀë³öI2£¬Ó¦ÏòµâË®ÖмÓÈëÝÍÈ¡¼Á£¨ÈçCCl4¡¢±½µÈ£©ÝÍÈ¡³öµâ£¬È»ºó·ÖÒº£¬´ð°¸Ñ¡B¡£
¢Ú¾Æ¾«ºÍË®ÊǷе㲻ͬ»¥Ïà»ìÈܵÄÒºÌå»ìºÏÎ·ÖÀë¾Æ¾«ºÍˮӦÓÃÕôÁ󷨣¬´ð°¸Ñ¡A¡£
¢ÛMnO2ΪÄÑÈÜÓÚË®µÄ¹ÌÌ壬·ÖÀëMnO2ºÍÕáÌÇÈÜÒºÓ¦ÓùýÂË·¨£¬´ð°¸Ñ¡C¡£
£¨3£©¢ÙÒªÌá¸ß¾Æ¾«µÄ¶ÈÊý£¬¼´¼õÉپƾ«ºÍË®µÄ»ìºÏÎïÖÐH2O£¬¾Æ¾«ºÍË®ÊǷе㲻ͬ»¥Ïà»ìÈܵÄÒºÌå»ìºÏÎ¿É²ÉÓõķ½·¨ÊÇÕôÁó¡£
¢ÚÖÐÒ©ÖжàÖֳɷÖÒ×ÈÜÓھƾ«£¨»òÒÒ´¼£©ÖУ¬ÓÃÃÀ¾Æ½þÅÝÖÐÒ©¿É½«ÖÐÒ©ÖжàÖֳɷֽþÈ¡µ½¾ÆÖУ¬ÖÐÒ©³£³£½þÅÝÔÚÃÀ¾ÆÖÐÖƳÉÒ©¾ÆÀûÓõķ½·¨ÊÇÝÍÈ¡¡£
![](http://thumb2018.1010pic.com/images/loading.gif)
¡¾ÌâÄ¿¡¿Îíö²ÌìÆøÑÏÖØÓ°ÏìÈËÃǵÄÉú»îºÍ½¡¿µ¡£ÆäÖÐÊ×ÒªÎÛȾÎïΪ¿ÉÎüÈë¿ÅÁ£ÎïPM2.5£¬ÆäÖ÷ÒªÀ´Ô´ÎªÈ¼Ãº¡¢»ú¶¯³µÎ²ÆøµÈ¡£Òò´Ë¸ÄÉÆÄÜÔ´½á¹¹¡¢»ú¶¯³µÏ޺ŵȴëÊ©ÄÜÓÐЧ¼õÉÙPM2.5¡£SO2¡¢NOxµÈÎÛȾ¡£
Çë»Ø´ðÏÂÁÐÎÊÌâ:
(1)½«Ò»¶¨Á¿µÄijPM2.5 ÑùÆ·ÓÃÕôÁóË®ÈܽâÖƳɴý²âÊÔÑù(ºöÂÔOH-)¡£³£ÎÂϲâµÃ¸ÃÊÔÑùµÄ×é³É¼°ÆäŨ¶ÈÈçϱí:¸ù¾Ý±íÖÐÊý¾ÝÅжϸÃÊÔÑùµÄpH=________¡£
Àë×Ó | K+ | Na+ | NH4+ | SO42- | NO3- | CI- |
Ũ¶Èmol/L | 4¡Á10-6 | 6¡Á10-6 | 2¡Á10-5 | 4¡Á10-5 | 3¡Á10-5 | 2¡Á10-5 |
(2)ÒÑÖªÆû³µÆû¸×ÖÐNOµÄÉú³É·´Ó¦Îª:N2(g)+ O2(g) 2NO(g) ¡÷H>0ºãΣ¬ºãÈÝÃܱÕÈÝÆ÷ÖУ¬ÏÂÁÐ˵·¨ÖУ¬ÄÜ˵Ã÷¸Ã·´Ó¦´ïµ½»¯Ñ§Æ½ºâ״̬µÄÊÇ______________¡£
A.»ìºÏÆøÌåµÄÃܶȲ»Ôٱ仯 B.»ìºÏÆøÌåµÄѹǿ²»Ôٱ仯
C.ÑõÆøµÄת»¯Âʲ»Ôٱ仯 D.N2¡¢O2¡¢NOµÄÎïÖʵÄÁ¿Ö®±ÈΪ1: 1: 2
(3)Ϊ¼õÉÙSO2 µÄÅÅ·Å£¬³£²ÉÈ¡µÄ´ëÊ©ÓÐ:
¢Ù½«Ãº×ª»¯ÎªÇå½àÆøÌåȼÁÏ¡£
ÒÑÖª:H2(g)+1/2O2(g)==H2O(g) ¡÷H= -241.8kJ/mol
C(s)+1/2O2(g)==CO(g) ¡÷H=-110.5kJ/mol
д³ö½¹Ì¿ÓëË®ÕôÆø·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ:__________________________¡£
¢ÚÏ´µÓº¬SO2µÄÑÌÆø¡£ÏÂÁпÉ×÷ΪϴµÓº¬SO2ÑÌÆøµÄÏ´µÓ¼ÁÊÇ________¡£
A.Ũ°±Ë® B.̼ËáÇâÄƱ¥ºÍÈÜÒº
C.FeCl2±¥ºÍÈÜÒº D.ËáÐÔCaCl2±¥ºÍÈÜÒº
(4)Æû³µÊ¹ÓÃÒÒ´¼ÆûÓͲ¢²»ÄܼõÉÙNOxµÄÅÅ·Å£¬ÕâʹNOxµÄÓÐЧÏû³ý³ÉΪ»·±£ÁìÓòµÄÖØÒª¿ÎÌ⡣ijÑо¿ÐÔС×éÔÚʵÑéÊÒÒÔAg-ZSM-5Ϊ´ß»¯¼Á£¬²âµÃNOת»¯ÎªN2µÄת»¯ÂÊËæζȱ仯Çé¿öÈçͼËùʾ¡£Èô²»Ê¹ÓÃCO£¬Î¶ȳ¬¹ý775K£¬·¢ÏÖNOµÄ·Ö½âÂʽµµÍ£¬Æä¿ÉÄܵÄÔÒòΪ____________£¬ÔÚn(NO)/n(CO)=1µÄÌõ¼þÏ£¬Îª¸üºÃµÄ³ýÈ¥NOxÎïÖÊ£¬Ó¦¿ØÖƵÄ×î¼ÑζÈÔÚ______K×óÓÒ¡£
(5)³µÁ¾ÅŷŵĵªÑõ»¯ÎúȼÉÕ²úÉúµÄ¶þÑõ»¯ÁòÊǵ¼ÖÂÎíö²ÌìÆøµÄ¡°×ï¿ý»öÊס±¡£»îÐÔÌ¿¿É´¦Àí´óÆøÎÛȾÎïNO¡£ÔÚ5LÃܱÕÈÝÆ÷ÖмÓÈëNO ºÍ»îÐÔÌ¿(¼ÙÉèÎÞÔÓÖÊ)£¬Ò»¶¨Ìõ¼þÏÂÉú³ÉÆøÌåEºÍF¡£µ±Î¶ȷֱðÔÚT1¡æºÍT2¡æʱ£¬²âµÃ¸÷ÎïÖÊƽºâʱÎïÖʵÄÁ¿(n/mol)Èçϱí:
¢Ùд³öNOÓë»îÐÔÌ¿·´Ó¦µÄ»¯Ñ§·½³Ìʽ____________________________¡£
¢ÚÈôT12£¬Ôò¸Ã·´Ó¦µÄ¡÷H_____________0(Ì>¡±¡¢¡°<¡±»ò¡°=¡±)¡£
¢ÛÉÏÊö·´Ó¦T1¡æʱ´ïµ½»¯Ñ§Æ½ºâºóÔÙͨÈë0.lmol NOÆøÌ壬Ôò´ïµ½Ð»¯Ñ§Æ½ºâʱNOµÄת»¯ÂÊΪ___________________¡£