ÌâÄ¿ÄÚÈÝ

¹¤ÒµÉϲâÁ¿¶þÑõ»¯Áò¡¢µªÆø¡¢ÑõÆø»ìºÏÆøÌåÖжþÑõ»¯Áòº¬Á¿µÄ×°ÖÃÈçÏÂͼËùʾ£¬·´Ó¦¹ÜÖÐ×°ÓеâµÄµí·ÛÈÜÒº¡£¶þÑõ»¯ÁòºÍµâ·¢ÉúµÄ·´Ó¦Îª£¨µªÆø¡¢ÑõÆø²»Óëµâ·´Ó¦£©£ºSO2+I2+2H2O=H2SO4+2HI
(1)»ìºÏÆøÌå½øÈë·´Ó¦¹Üºó£¬Á¿Æø¹ÜÄÚÔö¼ÓµÄË®µÄÌå»ýµÈÓÚ ______________£¨Ìîд»¯Ñ§Ê½£©µÄÌå»ý¡£
(2)·´Ó¦¹ÜÄÚÈÜÒºÀ¶É«Ïûʧºó£¬Ã»Óм°Ê±Í£Ö¹Í¨Æø£¬Ôò²âµÃµÄ¶þÑõ»¯Áòº¬Á¿___________£¨Ìî¡°Æ«¸ß¡±¡°Æ«µÍ¡±»ò¡°²»ÊÜÓ°Ï족£©¡£
(3)·´Ó¦¹ÜÄڵĵâµÄµí·ÛÈÜÒºÒ²¿ÉÒÔÓÃ__________ £¨ÌîдÎïÖÊÃû³Æ£©´úÌæ¡£
(4)ÈôµâÈÜÒºÌå»ýΪVamL£¬Å¨¶ÈΪc mol¡¤L-1£¬N2ÓëO2µÄÌå»ýΪVbmL(ÒÑÕÛËãΪ±ê×¼×´¿öϵÄÌå»ý)¡£ÓÃc¡¢Va¡¢Vb±í ʾ¶þÑõ»¯ÁòµÄÌå»ý·ÖÊýΪ___________¡£
(5)½«ÉÏÊö×°ÖøÄΪ¼òÒ×ʵÑé×°Ö㬳ýµ¼¹ÜÍ⣬»¹ÐèÑ¡ÓõÄÒÇÆ÷Ϊ__________£¨ÌîÏÂÁÐÒÇÆ÷µÄ±àºÅ£©¡£
a.ÉÕ±­ b£®ÊÔ¹Ü c£®¹ã¿ÚÆ¿ d£®ÈÝÁ¿Æ¿ e£®Á¿Í² f.µ¥¿×Èû g£®Ë«¿×Èû
(1) N2 ºÍO2
(2)Æ«µÍ
(3)ËáÐÔ¸ßÃÌËá¼ØÈÜÒº
(4)
(5)b¡¢c¡¢e¡¢g»òb.e.g»òc¡¢e¡¢g£¨´ð°¸²»Î¨Ò»£©
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨2010?¸£½¨£©¹¤ÒµÉϳ£ÓÃÌúÖÊÈÝÆ÷Ê¢×°ÀäŨËᣮΪÑо¿ÌúÖʲÄÁÏÓëÈÈŨÁòËáµÄ·´Ó¦£¬Ä³Ñ§Ï°Ð¡×é½øÐÐÁËÒÔÏÂ̽¾¿»î¶¯£º
[̽¾¿Ò»]
£¨1£©½«ÒÑÈ¥³ý±íÃæÑõ»¯ÎïµÄÌú¶¤£¨Ì¼Ëظ֣©·ÅÈëÀäŨÁòËáÖУ¬10·ÖÖÓºóÒÆÈËÁòËáÍ­ÈÜÒºÖУ¬Æ¬¿ÌºóÈ¡³ö¹Û²ì£¬Ìú¶¤±íÃæÎÞÃ÷ÏԱ仯£¬ÆäÔ­ÒòÊÇ
Ìú¶¤±íÃæ±»Ñõ»¯
Ìú¶¤±íÃæ±»Ñõ»¯
£®
£¨2£©Áí³ÆÈ¡Ìú¶¤6.0g·ÅÈë15.0mlŨÁòËáÖУ¬¼ÓÈÈ£¬³ä·ÖÓ¦ºóµÃµ½ÈÜÒºX²¢ÊÕ¼¯µ½ÆøÌåY£®
¢Ù¼×ͬѧÈÏΪXÖгýFe3+Í⻹¿ÉÄܺ¬ÓÐFe2+£®ÈôҪȷÈÏÆäÖеÄFe2+£¬Ó¦ÏÈÓÃ
d
d
Ñ¡ÌîÐòºÅ£©£®
a£®KSCNÈÜÒººÍÂÈË®     b£®Ìú·ÛºÍKSCNÈÜÒº    c£®Å¨°±Ë®      d£®ËáÐÔKMnO4ÈÜÒº
¢ÚÒÒͬѧȡ336ml£¨±ê×¼×´¿ö£©ÆøÌåYͨÈë×ãÁ¿äåË®ÖУ¬·¢Éú·´Ó¦£ºSO2+Br2+2H2O¨T2HBr+H2SO4È»ºó¼ÓÈë×ãÁ¿BaCl2ÈÜÒº£¬¾­Êʵ±²Ù×÷ºóµÃ¸ÉÔï¹ÌÌå2.33g£®ÓÉÓÚ´ËÍÆÖªÆøÌåYÖÐSO2µÄÌå»ý·ÖÊýΪ
66.7%
66.7%
£®
[̽¾¿¶þ]
·ÖÎöÉÏÊöʵÑéÖÐSO2Ìå»ý·ÖÊýµÄ½á¹û£¬±ûͬѧÈÏÎªÆøÌåYÖл¹¿ÉÄܺ¬Á¿ÓÐH2ºÍQÆøÌ壮Ϊ´ËÉè¼ÆÁËÏÂÁÐ̽¾¿ÊµÑé×´Öã¨Í¼ÖмгÖÒÇÆ÷Ê¡ÂÔ£©£®

£¨3£©×°ÖÃBÖÐÊÔ¼ÁµÄ×÷ÓÃÊÇ
¼ìÑé¶þÑõ»¯ÁòÊÇ·ñ³ý¾¡
¼ìÑé¶þÑõ»¯ÁòÊÇ·ñ³ý¾¡
£®
£¨4£©ÈÏÎªÆøÌåYÖл¹º¬ÓÐQµÄÀíÓÉÊÇ
C+2H2SO4£¨Å¨ÁòËᣩ
  ¡÷  
.
 
CO2 ¡ü+2SO2¡ü+2H2O
C+2H2SO4£¨Å¨ÁòËᣩ
  ¡÷  
.
 
CO2 ¡ü+2SO2¡ü+2H2O
£¨Óû¯Ñ§·½³Ìʽ±íʾ£©£®
£¨5£©ÎªÈ·ÈÏQµÄ´æÔÚ£¬ÐèÔÚ×°ÖÃÖÐÌí¼ÓMÓÚ
c
c
£¨Ñ¡ÌîÐòºÅ£©£®
a£®A֮ǰ      b£®A-B¼ä       c£®B-C¼ä       d£®C-D¼ä
£¨6£©Èç¹ûÆøÌåYÖк¬ÓÐH2£¬Ô¤¼ÆÊµÑéÏÖÏóÓ¦ÊÇ
DÖйÌÌåÓɺÚÉ«±äºìºÍEÖйÌÌåÓɰױäÀ¶£®
DÖйÌÌåÓɺÚÉ«±äºìºÍEÖйÌÌåÓɰױäÀ¶£®
£®
£¨7£©ÈôÒª²â¶¨ÏÞ¶¨Ìå»ýÆøÌåYÖÐH2µÄº¬Á¿£¨±ê×¼×´¿öÏÂÔ¼ÓÐ28ml H2£©£¬³ý¿ÉÓòâÁ¿H2Ìå»ýµÄ·½·¨Í⣬¿É·ñÑ¡ÓÃÖÊÁ¿³ÆÁ¿µÄ·½·¨£¿×ö³öÅжϲ¢ËµÃ÷ÀíÓÉ
·ñ£¬ÓÃÍÐÅÌÌìÆ½ÎÞ·¨³ÆÁ¿D»òEµÄ²îÁ¿£®
·ñ£¬ÓÃÍÐÅÌÌìÆ½ÎÞ·¨³ÆÁ¿D»òEµÄ²îÁ¿£®
£®
¿ÕÆøÖеÄSO2º¬Á¿ºÍ¿ÉÎüÈë¿ÅÁ£µÄº¬Á¿£¨¿ÉÓÃg?cm-3±íʾ£©¶¼ÊÇÖØÒªµÄ¿ÕÆøÖÊÁ¿Ö¸±ê£®ÔÚ¹¤ÒµÉú²úÉϹ涨£º¿ÕÆøÖжþÑõ»¯ÁòµÄ×î´óÔÊÐíÅÅ·ÅŨ¶È²»µÃ³¬¹ý0.02mg?L-1£®
£¨1£©Îª²â¶¨Ä³µØ·½µÄ¿ÕÆøÖÐSO2ºÍ¿ÉÎüÈë¿ÅÁ£µÄº¬Á¿£¬¼×ͬѧÉè¼ÆÁËÈçͼ1ËùʾµÄʵÑé×°Öãº
¾«Ó¢¼Ò½ÌÍø
×¢£ºÆøÌåÁ÷ËÙ¹ÜÊÇÓÃÀ´²âÁ¿µ¥Î»Ê±¼äÄÚͨ¹ýÆøÌåµÄÌå»ýµÄ×°ÖÃ
¢ÙÓ¦ÓÃÉÏÊö×°Öòⶨ¿ÕÆøÖеÄSO2º¬Á¿ºÍ¿ÉÎüÈë¿ÅÁ£µÄº¬Á¿£¬³ý²â¶¨ÆøÌåÁ÷ËÙ£¨µ¥Î»£ºcm3?min-1£©Í⣬»¹ÐèÒª²â¶¨
 
£®
¢ÚÒÑÖª£ºµâµ¥ÖÊ΢ÈÜÓÚË®£¬KI¿ÉÒÔÔö´óµâÔÚË®ÖеÄÈܽâ¶È£®ÇëÄãЭÖú¼×ͬѧÍê³É100mL 5¡Á10-4mol?L-1µâÈÜÒºµÄÅäÖÆ£º
µÚÒ»²½£º×¼È·³ÆÈ¡1.27gµâµ¥ÖʼÓÈëÉÕ±­ÖУ¬
 
£®
µÚ¶þ²½£º
 
£®
µÚÈý²½£º´ÓµÚ¶þ²½ËùµÃÈÜÒºÖУ¬È¡³ö10.00mLÈÜÒºÓÚ100mLÈÝÁ¿Æ¿ÖУ¬¼ÓˮϡÊÍÖÁ¿Ì¶ÈÏߣ®
£¨2£©ÒÒͬѧÄâÓÃÈçͼ1¼òÒ××°Öòⶨ¿ÕÆøÖеÄSO2º¬Á¿£º×¼È·ÒÆÈ¡50mL 5¡Á10-4 mol?L-1µÄµâÈÜÒº£¬×¢ÈëÈçͼËùʾ¹ã¿ÚÆ¿ÖУ¬¼Ó2¡«3µÎµí·Ûָʾ¼Á£¬´ËʱÈÜÒº³ÊÀ¶É«£®ÔÚÖ¸¶¨µÄ²â¶¨µØµã³éÆø£¬Ã¿´Î³éÆø100mL£¬Ö±µ½ÈÜÒºµÄÀ¶É«È«²¿Íʾ¡ÎªÖ¹£¬¼Ç¼³éÆø´ÎÊý£¨n£©£®
¢Ù¼ÙÉèÒÒͬѧµÄ²âÁ¿ÊÇ׼ȷµÄ£¬ÒÒͬѧ³éÆøµÄ´ÎÊýÖÁÉÙΪ
 
´Î£¬·½¿É˵Ã÷¸ÃµØ¿ÕÆøÖеÄSO2º¬Á¿·ûºÏÅŷűê×¼£®
¢ÚÈç¹ûÒÒͬѧÓø÷½·¨²âÁ¿¿ÕÆøÖÐSO2µÄº¬Á¿Ê±£¬Ëù²âµÃµÄÊýÖµ±Èʵ¼Êº¬Á¿µÍ£¬ÇëÄã¶ÔÆä¿ÉÄܵÄÔ­Òò£¨¼ÙÉèÈÜÒºÅäÖÆ¡¢³ÆÁ¿»òÁ¿È¡¼°¸÷ÖÖ¶ÁÊý¾ùÎÞ´íÎó£©Ìá³öÁ½ÖÖºÏÀí¼ÙÉ裺
 
£»
 
£®
¢Û±ûͬѧÈÏΪ£ºÒÒͬѧµÄʵÑé·½°¸ÐèÒª³éÆøµÄ´ÎÊýÌ«¶à£¬²Ù×÷Âé·³£®ÓëÒÒÌÖÂۺ󣬾ö¶¨½«³éÆø´ÎÊý½µµ½100´ÎÒÔÏ£¬ÇëÄãÉè¼ÆºÏÀíµÄ¸Ä½ø·½°¸£º
 
£®

¿ÕÆøÖеÄSO2º¬Á¿ºÍ¿ÉÎüÈë¿ÅÁ£µÄº¬Á¿(¿ÉÓÃg/cm3±íʾ)¶¼ÊÇÖØÒªµÄ¿ÕÆøÖÊÁ¿Ö¸±ê¡£ÔÚ¹¤ÒµÉú²úÉϹ涨£º¿ÕÆøÖжþÑõ»¯ÁòµÄ×î´óÔÊÐíÅÅ·ÅŨ¶È²»µÃ³¬¹ý0.02 mg/L¡£

(1)Ϊ²â¶¨Ä³µØ·½µÄ¿ÕÆøÖÐSO2ºÍ¿ÉÎüÈë¿ÅÁ£µÄº¬Á¿£¬¼×ͬѧÉè¼ÆÁËÈçÏÂͼËùʾµÄʵÑé×°Öãº

×¢£ºÆøÌåÁ÷ËÙ¹ÜÊÇÓÃÀ´²âÁ¿µ¥Î»Ê±¼äÄÚͨ¹ýÆøÌåµÄÌå»ýµÄ×°ÖÃ

¢ÙÓ¦ÓÃÉÏÊö×°Öòⶨ¿ÕÆøÖеÄSO2º¬Á¿ºÍ¿ÉÎüÈë¿ÅÁ£µÄº¬Á¿£¬³ý²â¶¨ÆøÌåÁ÷ËÙ(µ¥Î»£ºcm3/min)Í⣬»¹ÐèÒª²â¶¨___________________________________________________

¢ÚÒÑÖª£ºµâµ¥ÖÊ΢ÈÜÓÚË®£¬KI¿ÉÒÔÔö´óµâÔÚË®ÖеÄÈܽâ¶È¡£

ÇëÄãЭÖú¼×ͬѧÍê³É100 mL 5¡Á10£­4 mol/LµâÈÜÒºµÄÅäÖÆ£º

µÚÒ»²½£º×¼È·³ÆÈ¡1.27gµâµ¥ÖʼÓÈëÉÕ±­ÖУ¬___________________________

µÚ¶þ²½£º___________________________________________________________£»

µÚÈý²½£º´ÓµÚ¶þ²½ËùµÃÈÜÒºÖУ¬È¡³ö10.00 mLÈÜÒºÓÚ100 mLÈÝÁ¿Æ¿ÖУ¬¼ÓˮϡÊÍÖÁ¿Ì¶ÈÏß¡£

(2)ÒÒͬѧÄâÓÃÈçͼËùʾ¼òÒ××°Öòⶨ¿ÕÆøÖеÄSO2º¬Á¿£º×¼È·ÒÆÈ¡50 mL 5¡Á10£­4 mol/LµÄµâÈÜÒº£¬×¢ÈëͼÖÐËùʾ¹ã¿ÚÆ¿ÖУ¬¼Ó2¡«3µÎµí·Ûָʾ¼Á£¬´ËʱÈÜÒº³ÊÀ¶É«¡£ÔÚÖ¸¶¨µÄ²â¶¨µØµã³éÆø£¬Ã¿´Î³éÆø100 mL£¬Ö±µ½ÈÜÒºµÄÀ¶É«È«²¿Íʾ¡ÎªÖ¹£¬¼Ç¼³éÆø´ÎÊý(n)¡£

¢Ù¼ÙÉèÒÒͬѧµÄ²âÁ¿ÊÇ׼ȷµÄ£¬ÒÒͬѧ³éÆøµÄ´ÎÊýÖÁÉÙΪ________´Î£¬·½¿É˵Ã÷¸ÃµØ¿ÕÆøÖеÄSO2º¬Á¿·ûºÏÅŷűê×¼¡£

¢ÚÈç¹ûÒÒͬѧÓø÷½·¨²âÁ¿¿ÕÆøÖÐSO2µÄº¬Á¿Ê±£¬Ëù²âµÃµÄÊýÖµ±Èʵ¼Êº¬Á¿µÍ£¬ÇëÄã¶ÔÆä¿ÉÄܵÄÔ­Òò(¼ÙÉèÈÜÒºÅäÖÆ¡¢³ÆÁ¿»òÁ¿È¡¼°¸÷ÖÖ¶ÁÊý¾ùÎÞ´íÎó)Ìá³öÁ½ÖÖºÏÀí¼ÙÉ裺__________________¡¢______________________¡£

¢Û±ûͬѧÈÏΪ£ºÒÒͬѧµÄʵÑé·½°¸ÐèÒª³éÆøµÄ´ÎÊýÌ«¶à£¬²Ù×÷Âé·³¡£ÓëÒÒÌÖÂۺ󣬾ö¶¨½«³éÆø´ÎÊý½µµ½100´ÎÒÔÏ£¬ÇëÄãÉè¼ÆºÏÀíµÄ¸Ä½ø·½°¸£º_________________________¡£

 

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

¾«Ó¢¼Ò½ÌÍø