ÌâÄ¿ÄÚÈÝ

ϱíÖУ¬¶Ô³ÂÊö¢ñ¡¢¢òµÄÕýÈ·ÐÔ¼°Á½Õß¼äÒò¹û¹ØÏµµÄÅжϣ¬ÍêÈ«ÕýÈ·µÄÊÇ

Ñ¡Ïî

³ÂÊö¢ñ

³ÂÊö¢ò

ÅжÏ

A

ÂÁÖÆ´¶¾ß×îºÃ²»ÒªÊ¢·ÅËáÐÔ»ò¼îÐÔ½ÏÇ¿µÄÒºÌåʳÎï

ÒòΪAlºÍAl2O3¼È¿ÉÒÔÓëËá·´Ó¦¡¢ÓÖ¿ÉÒÔÓë¼î·´Ó¦

¢ñ¶Ô£¬¢ò¶Ô£¬ÓÐ

B

ÂÁ²­Ôھƾ«µÆ»ðÑæÉϼÓÈÈÈÛ»¯µ«²»µÎÂä

ÂÁ²­¶ÔÈÛ»¯µÄÂÁÓнÏÇ¿µÄÎü¸½×÷ÓÃ

¢ñ¶Ô£¬¢ò¶Ô£¬ÓÐ

C

º£Ð¥Ê¹Ë®Ô´·¢ÉúÎÛȾ¿ÉÓÃÃ÷·¯½øÐÐÏû¶¾ºÍ¾»»¯

ÒòΪÃ÷·¯Ë®½âÉú³ÉAl(OH)3½ºÌ壬¾ßÓÐÎü¸½ÐÔ

¢ñ¶Ô£¬¢ò´í£¬ÎÞ

D

Ìú»òÂÁÖÆ³ÉµÄ²Û³µ¿ÉÒÔÃÜ·âÖüÔËŨÁòËá»òŨÏõËá

ÒòΪÌúºÍÂÁ²»ÄÜÓëŨÁòËá»òŨÏõËá·´Ó¦

¢ñ´í£¬¢ò¶Ô£¬ÎÞ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

̼¡¢µª¼°Æä»¯ºÏÎïÔÚ¹¤Å©ÒµÉú²úÉú»îÖÐÓÐ×ÅÖØÒª×÷Óá£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÓÃCH4 ´ß»¯»¹Ô­NOx ¿ÉÒÔÏû³ýµªÑõ»¯ÎïµÄÎÛȾ¡£ÀýÈ磺

CH4(g) + 4NO2(g) = 4NO(g)£«CO2(g) + 2H2O(g) ¦¤H1£½£­574 kJ¡¤mol£­1

CH4(g) + 4NO(g) = 2 N2(g)£«CO2(g) + 2H2O(g) ¦¤H2

Èô2 mol CH4 »¹Ô­NO2 ÖÁN2£¬Õû¸ö¹ý³ÌÖзųöµÄÈÈÁ¿Îª1734 kJ£¬Ôò¦¤H2£½ £»

£¨2£©¾Ý±¨µÀ£¬¿ÆÑ§¼ÒÔÚÒ»¶¨Ìõ¼þÏÂÀûÓÃFe2O3Óë¼×Íé·´Ó¦¿ÉÖÆÈ¡¡°ÄÉÃ×¼¶¡±µÄ½ðÊôÌú¡£Æä·´Ó¦ÈçÏ£º Fe2O3(s) + 3CH4(g) 2Fe(s) + 3CO(g) +6H2(g) ¦¤H>0

¢Ù Èô·´Ó¦ÔÚ5LµÄÃܱÕÈÝÆ÷ÖнøÐУ¬1minºó´ïµ½Æ½ºâ£¬²âµÃFe2O3ÔÚ·´Ó¦ÖÐÖÊÁ¿¼õÉÙ3.2g¡£Ôò¸Ã¶Îʱ¼äÄÚCOµÄƽ¾ù·´Ó¦ËÙÂÊΪ ________________ ¡£

¢Ú Èô¸Ã·´Ó¦ÔÚºãκãѹÈÝÆ÷ÖнøÐУ¬ÄܱíÃ÷¸Ã·´Ó¦´ïµ½Æ½ºâ״̬µÄÊÇ_____(Ñ¡ÌîÐòºÅ)

a£®CH4µÄת»¯ÂʵÈÓÚCOµÄ²úÂÊ

b£®»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿²»±ä

c£®v£¨CO£©Óëv£¨H2£©µÄ±ÈÖµ²»±ä

d£®¹ÌÌåµÄ×ÜÖÊÁ¿²»±ä

¢Û ¸Ã·´Ó¦´ïµ½Æ½ºâʱijÎïÀíÁ¿ËæÎ¶ȱ仯ÈçͼËùʾ£¬µ±Î¶ÈÓÉT1Éý¸ßµ½T2ʱ£¬Æ½ºâ³£ÊýKA ______KB£¨Ìî¡°>¡±¡¢¡° <¡±»ò¡°=¡±£©¡£×Ý×ø±ê¿ÉÒÔ±íʾµÄÎïÀíÁ¿ÓÐÄÄЩ ¡£

a£®H2µÄÄæ·´Ó¦ËÙÂÊ

b£®CH4µÄµÄÌå»ý·ÖÊý

c£®»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿

£¨3£©ÈôÍù20mL 0.0lmol¡¤L-lµÄÈõËáHNO2ÈÜÒºÖÐÖðµÎ¼ÓÈëÒ»¶¨Å¨¶ÈµÄÉÕ¼îÈÜÒº£¬²âµÃ»ìºÏÈÜÒºµÄζȱ仯ÈçͼËùʾ£¬ÏÂÁÐÓйØËµ·¨ÕýÈ·µÄÊÇ________

¢Ù¸ÃÉÕ¼îÈÜÒºµÄŨ¶ÈΪ0.02mol¡¤L-1

¢Ú¸ÃÉÕ¼îÈÜÒºµÄŨ¶ÈΪ0.01mol¡¤L-1

¢Û HNO2µÄµçÀëÆ½ºâ³£Êý£ºbµã>aµã

¢Ü´Óbµãµ½cµã£¬»ìºÏÈÜÒºÖÐÒ»Ö±´æÔÚ£ºc(Na+)>c(NO2-)>c(OH)> c(H+)

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

¾«Ó¢¼Ò½ÌÍø