ÌâÄ¿ÄÚÈÝ

(15·Ö) ÒÑ֪ijÓлúÎïAµÄÖÊÆ×ͼÏÔʾÆäÏà¶Ô·Ö×ÓÖÊÁ¿Îª84£¬Ôò

(1) ÌþAµÄ·Ö×ÓʽΪ_____________¡£

(2) ÈôÌþA²»ÄÜʹäåË®ÍÊÉ«£¬µ«ÔÚÒ»¶¨Ìõ¼þÏÂÄÜÓëÂÈÆø·¢ÉúÈ¡´ú·´Ó¦£¬ÆäÒ»ÂÈÈ¡´úÎïÖ»ÓÐÒ»ÖÖ£¬ÔòÌþAµÄ½á¹¹¼òʽΪ__________________¡£

(3) ÈôÌþAÄÜʹäåË®ÍÊÉ«£¬ÔÚ´ß»¯¼Á×÷ÓÃÏ£¬ÓëH2¼Ó³É£¬Æä¼Ó³É²úÎï¾­²â¶¨·Ö×ÓÖк¬ÓÐ4¸ö¼×»ù£¬ÌþA¿ÉÄÜÓеĽṹ¼òʽΪ                                      ¡£

(4) ±ÈÌþAÉÙÒ»¸ö̼ԭ×ÓÇÒÄÜʹäåË®ÍÊÉ«µÄAµÄͬϵÎïÓÐ________ÖÖͬ·ÖÒì¹¹Ìå¡£

(5)ÈôÌþAºìÍâ¹âÆ×±íÃ÷·Ö×ÓÖк¬ÓÐ̼̼˫¼ü£¬ºË´Å¹²ÕñÆ×±íÃ÷·Ö×ÓÖÐÖ»ÓÐÒ»ÖÖÀàÐ͵ÄÇâ¡£ÔÚÏÂͼÖУ¬D1¡¢D2»¥ÎªÍ¬·ÖÒì¹¹Ì壬E1¡¢E2»¥ÎªÍ¬·ÖÒì¹¹Ìå¡£

·´Ó¦¢ÚµÄ»¯Ñ§·½³ÌʽΪ   ______________________     £»CµÄ»¯Ñ§Ãû³ÆΪ   ___________   £»

E2µÄ½á¹¹¼òʽÊÇ   ________________     £»¢ÞµÄ·´Ó¦ÀàÐÍÊÇ   _____________    ¡£

 

¡¾´ð°¸¡¿

£¨15·Ö£©¢Å C6H12£»  £¨2·Ö£©     (2) £»   £¨2·Ö£©

(3) (CH3)3CCH=CH2»òCH2=C(CH3)CH(CH3)2»ò  (CH3)2C=C(CH3)2£»   £¨3·Ö£©

(4) 5 (2·Ö)

£¨5£©£¨2·Ö£©

2£¬3-¶þ¼×»ù-1£¬3-¶¡¶þÏ©£¨1·Ö£©£¨2·Ö£©   È¡´ú·´Ó¦£¨¸÷1·Ö£©

¡¾½âÎö¡¿ÂÔ

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

(16·Ö)(1)ÒÑÖª£º»¹Ô­ÐÔHSO3£­>I£­£¬Ñõ»¯ÐÔIO3£­>I2¡£ÔÚNaIO3ÈÜÒºÖеμÓÉÙÁ¿NaHSO3ÈÜÒº£¬·¢ÉúÏÂÁз´Ó¦£ºNaIO3+NaHSO3¡úI2+Na2SO4+H2SO4+H2O

 ¢ÙÅäƽÉÏÊö·´Ó¦µÄ»¯Ñ§·½³Ìʽ(½«»¯Ñ§¼ÆÁ¿ÊýÌîÔÚ·½¿òÄÚ)£»²¢Ð´³öÆäÑõ»¯²úÎï____________¡£

¢ÚÔÚNaIO3ÈÜÒºÖеμӹýÁ¿NaHSO3ÈÜÒº£¬·´Ó¦ÍêÈ«ºó£¬ÍƲⷴӦºóÈÜÒºÖеĻ¹Ô­²úÎïΪ____________ (Ìѧʽ)£»

(2)ÏòijÃܱÕÈÝÆ÷ÖмÓÈË0.15 mol/L  A¡¢0.05 mol/L CºÍÒ»¶¨Á¿µÄBÈýÖÖÆøÌå¡£Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£¬¸÷ÎïÖÊŨ¶ÈËæʱ¼ä±ä»¯ÈçÏÂͼÖм×ͼËùʾ[t0ʱc(B)δ»­³ö£¬t1ʱÔö´óµ½0.05 mol/L]¡£ÒÒͼΪt2ʱ¿Ìºó¸Ä±ä·´Ó¦Ìõ¼þ£¬Æ½ºâÌåϵÖÐÕý¡¢Äæ·´Ó¦ËÙÂÊËæʱ¼ä±ä»¯µÄÇé¿ö¡£

¢ÙÈôt4ʱ¸Ä±äµÄÌõ¼þΪ¼õСѹǿ£¬ÔòBµÄÆðʼÎïÖʵÄÁ¿Å¨¶ÈΪ________mol/L£»

¢ÚÈôt1=15 s£¬Ôòt0¡«t1½×¶ÎÒÔCŨ¶È±ä»¯±íʾµÄƽ¾ù·´Ó¦ËÙÂÊΪv(C)=_______mol/(L¡¤s)¡£

¢Ût3ʱ¸Ä±äµÄijһ·´Ó¦Ìõ¼þ¿ÉÄÜÊÇ_______(Ñ¡ÌîÐòºÅ)¡£

aʹÓô߻¯¼Á  bÔö´óѹǿ  cÔö´ó·´Ó¦ÎïŨ¶È

¢ÜÓмס¢ÒÒÁ½¸öÈÝ»ý¾ùΪ2LµÄÃܱÕÈÝÆ÷£¬ÔÚ¿ØÖÆÁ½ÈÝÆ÷ζÈÏàͬÇҺ㶨Çé¿öÏ£¬Ïò¼×ÖÐͨÈë3mol A£¬´ïµ½Æ½ºâʱ£¬BµÄÌå»ý·ÖÊýΪ20£¥£¬ÔòÏòÒÒÈÝÆ÷ÖгäÈë1 mol CºÍ0.5mol B£¬´ïµ½Æ½ºâʱ£¬CµÄŨ¶Èc(C)=________

 

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

¾«Ó¢¼Ò½ÌÍø