(16·Ö) ÏÂͼÖмס¢ÒÒ¡¢±ûµÄµç¼«²ÄÁ϶¼ÊÇʯīºÍÌú£¬ÆäÖбûÊÇÂȼҵÉú²úʾÒâͼ¡£

 

£¨1£©Èô¼×¡¢ÒÒÁ½ÉÕ±­¾ùÊ¢·ÅCuSO4ÈÜÒº£¬

¢Ù¼×ÖÐÌú°ôÉϵĵ缫·´Ó¦Ê½Îª_______________________________________¡£

¢ÚÒÒ×°Öù¤×÷Ò»¶Îʱ¼äºó£¬ÏòÉÕ±­ÖмÓÈëÊÊÁ¿µÄ¼îʽ̼ËáÍ­¡¾Cu2(OH)2CO3¡¿£¬ÄÜʹÈÜÒº»Ö¸´µ½Æðʼ״̬£¬Çëд³öÕâ¶Îʱ¼äÄÚÒÒ×°Ö÷¢ÉúµÄËùÓз´Ó¦µÄ»¯Ñ§·½³Ìʽ

________________________________________________________________________¡£

£¨2£©Èô¼×¡¢ÒÒÁ½ÉÕ±­¾ùÊ¢·Å±¥ºÍNaClÈÜÒº£¬

¢Ù¼×ÖÐʯī°ôÉϵĵ缫·´Ó¦Ê½Îª___________________¡£

¢Ú½«ÊªÈóµÄµí·Ûµâ»¯¼ØÊÔÖ½·ÅÔÚÒÒÉÕ±­______£¨Ìî¡°Fe¡±»ò¡°C¡±£©µç¼«µÄÉÏ·½£¬·¢ÏÖÊÔÖ½ÏȱäÀ¶ºóÍÊÉ«£¬ÕâÊÇÒòΪ¹ýÁ¿µÄCl2Ñõ»¯ÁËÉú³ÉµÄI2¡£Èô·´Ó¦ÎïCl2ºÍI2µÄÎïÖʵÄÁ¿Ö®±ÈΪ5¡Ã1£¬ÇÒÉú³ÉÁ½ÖÖËᣬÔòÆä¶ÔÓ¦µÄ»¯Ñ§·½³ÌʽΪ_________________________¡£

¢Û¼ÙÉè²úÉúµÄÆøÌåÈ«²¿ÒݳöÈÜÒº£¬µ±ÒÒ·´Ó¦ÓÐ0.01 molµç×ÓתÒƺóֹͣʵÑ飬´ËʱÉÕ±­ÖÐÈÜÒºµÄÌå»ýΪ100 mL£¬ÔòÈÜÒº»ìºÏ¾ùÔȺóµÄpH = ____________¡£

¢Üµ±ÒÒÖеķ´Ó¦ÓÃÓÚ¹¤ÒµÉú²úʱ£¬ÎªÁË×èÖ¹Á½¼«²úÎïÖ®¼äµÄ·´Ó¦£¬Í¨³£Ê¹ÓÃÈç±ûͼËùʾµÄ×°Öã¬ÆäÖÐÀë×Ó½»»»Ä¤Ö»ÔÊÐíNa£«Í¨¹ý£¬Na£«µÄÒƶ¯·½ÏòÈçͼÖбê×¢£¬ÔòH2µÄ³ö¿ÚÊÇ________(Ìî´úºÅ)¡£

 

£¨1£©¢ÙFe ¨C 2e£­= Fe2£«£¨2·Ö£©

¢Ú2CuSO4 + 2H2O2Cu + O2¡ü+ 2H2SO4£¬2H2O2H2¡ü+ O2¡ü£¨2·Ö£©

£¨2£©¢Ù O2 + 4e£­+ 2H2O = 4OH£­£¨2·Ö£©

¢Ú C£¨1·Ö£©£¬5Cl2 + I2+ 6H2O = 10HCl + 2HIO3£¨3·Ö£©

¢Û 13£¨3·Ö£©

¢Ü F£¨1·Ö£©

½âÎö:

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010ÄêÖØÇìÊÐÎ÷ÄÏʦ´ó¸½ÖиßÈýÉÏѧÆÚµÚÒ»´ÎÔ¿¼£¨Àí×Û£©»¯Ñ§²¿·Ö ÌâÐÍ£ºÌî¿ÕÌâ

(16·Ö) ÏÂͼÖмס¢ÒÒ¡¢±ûµÄµç¼«²ÄÁ϶¼ÊÇʯīºÍÌú£¬ÆäÖбûÊÇÂȼҵÉú²úʾÒâͼ¡£
 
£¨1£©Èô¼×¡¢ÒÒÁ½ÉÕ±­¾ùÊ¢·ÅCuSO4ÈÜÒº£¬
¢Ù¼×ÖÐÌú°ôÉϵĵ缫·´Ó¦Ê½Îª_______________________________________¡£
¢ÚÒÒ×°Öù¤×÷Ò»¶Îʱ¼äºó£¬ÏòÉÕ±­ÖмÓÈëÊÊÁ¿µÄ¼îʽ̼ËáÍ­¡¾Cu2(OH)2CO3¡¿£¬ÄÜʹÈÜÒº»Ö¸´µ½Æðʼ״̬£¬Çëд³öÕâ¶Îʱ¼äÄÚÒÒ×°Ö÷¢ÉúµÄËùÓз´Ó¦µÄ»¯Ñ§·½³Ìʽ
________________________________________________________________________¡£
£¨2£©Èô¼×¡¢ÒÒÁ½ÉÕ±­¾ùÊ¢·Å±¥ºÍNaClÈÜÒº£¬
¢Ù¼×ÖÐʯī°ôÉϵĵ缫·´Ó¦Ê½Îª___________________¡£
¢Ú½«ÊªÈóµÄµí·Ûµâ»¯¼ØÊÔÖ½·ÅÔÚÒÒÉÕ±­______£¨Ìî¡°Fe¡±»ò¡°C¡±£©µç¼«µÄÉÏ·½£¬·¢ÏÖÊÔÖ½ÏȱäÀ¶ºóÍÊÉ«£¬ÕâÊÇÒòΪ¹ýÁ¿µÄCl2Ñõ»¯ÁËÉú³ÉµÄI2¡£Èô·´Ó¦ÎïCl2ºÍI2µÄÎïÖʵÄÁ¿Ö®±ÈΪ5¡Ã1£¬ÇÒÉú³ÉÁ½ÖÖËᣬÔòÆä¶ÔÓ¦µÄ»¯Ñ§·½³ÌʽΪ_________________________¡£
¢Û¼ÙÉè²úÉúµÄÆøÌåÈ«²¿ÒݳöÈÜÒº£¬µ±ÒÒ·´Ó¦ÓÐ0.01 molµç×ÓתÒƺóֹͣʵÑ飬´ËʱÉÕ±­ÖÐÈÜÒºµÄÌå»ýΪ100 mL£¬ÔòÈÜÒº»ìºÏ¾ùÔȺóµÄpH = ____________¡£
¢Üµ±ÒÒÖеķ´Ó¦ÓÃÓÚ¹¤ÒµÉú²úʱ£¬ÎªÁË×èÖ¹Á½¼«²úÎïÖ®¼äµÄ·´Ó¦£¬Í¨³£Ê¹ÓÃÈç±ûͼËùʾµÄ×°Öã¬ÆäÖÐÀë×Ó½»»»Ä¤Ö»ÔÊÐíNa£«Í¨¹ý£¬Na£«µÄÒƶ¯·½ÏòÈçͼÖбê×¢£¬ÔòH2µÄ³ö¿ÚÊÇ________(Ìî´úºÅ)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄêÖØÇìÊиßÈý9ÔÂÔ¿¼Àí¿Æ×ÛºÏÊÔÌ⣨»¯Ñ§²¿·Ö£© ÌâÐÍ£ºÌî¿ÕÌâ

(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)=________

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010ÄêÖØÇìÊиßÈýÉÏѧÆÚµÚÒ»´ÎÔ¿¼£¨Àí×Û£©»¯Ñ§²¿·Ö ÌâÐÍ£ºÌî¿ÕÌâ

(16·Ö) ÏÂͼÖмס¢ÒÒ¡¢±ûµÄµç¼«²ÄÁ϶¼ÊÇʯīºÍÌú£¬ÆäÖбûÊÇÂȼҵÉú²úʾÒâͼ¡£

 

£¨1£©Èô¼×¡¢ÒÒÁ½ÉÕ±­¾ùÊ¢·ÅCuSO4ÈÜÒº£¬

¢Ù¼×ÖÐÌú°ôÉϵĵ缫·´Ó¦Ê½Îª_______________________________________¡£

¢ÚÒÒ×°Öù¤×÷Ò»¶Îʱ¼äºó£¬ÏòÉÕ±­ÖмÓÈëÊÊÁ¿µÄ¼îʽ̼ËáÍ­¡¾Cu2(OH)2CO3¡¿£¬ÄÜʹÈÜÒº»Ö¸´µ½Æðʼ״̬£¬Çëд³öÕâ¶Îʱ¼äÄÚÒÒ×°Ö÷¢ÉúµÄËùÓз´Ó¦µÄ»¯Ñ§·½³Ìʽ

________________________________________________________________________¡£

£¨2£©Èô¼×¡¢ÒÒÁ½ÉÕ±­¾ùÊ¢·Å±¥ºÍNaClÈÜÒº£¬

¢Ù¼×ÖÐʯī°ôÉϵĵ缫·´Ó¦Ê½Îª___________________¡£

¢Ú½«ÊªÈóµÄµí·Ûµâ»¯¼ØÊÔÖ½·ÅÔÚÒÒÉÕ±­______£¨Ìî¡°Fe¡±»ò¡°C¡±£©µç¼«µÄÉÏ·½£¬·¢ÏÖÊÔÖ½ÏȱäÀ¶ºóÍÊÉ«£¬ÕâÊÇÒòΪ¹ýÁ¿µÄCl2Ñõ»¯ÁËÉú³ÉµÄI2¡£Èô·´Ó¦ÎïCl2ºÍI2µÄÎïÖʵÄÁ¿Ö®±ÈΪ5¡Ã1£¬ÇÒÉú³ÉÁ½ÖÖËᣬÔòÆä¶ÔÓ¦µÄ»¯Ñ§·½³ÌʽΪ_________________________¡£

¢Û¼ÙÉè²úÉúµÄÆøÌåÈ«²¿ÒݳöÈÜÒº£¬µ±ÒÒ·´Ó¦ÓÐ0.01 molµç×ÓתÒƺóֹͣʵÑ飬´ËʱÉÕ±­ÖÐÈÜÒºµÄÌå»ýΪ100 mL£¬ÔòÈÜÒº»ìºÏ¾ùÔȺóµÄpH = ____________¡£

¢Üµ±ÒÒÖеķ´Ó¦ÓÃÓÚ¹¤ÒµÉú²úʱ£¬ÎªÁË×èÖ¹Á½¼«²úÎïÖ®¼äµÄ·´Ó¦£¬Í¨³£Ê¹ÓÃÈç±ûͼËùʾµÄ×°Öã¬ÆäÖÐÀë×Ó½»»»Ä¤Ö»ÔÊÐíNa£«Í¨¹ý£¬Na£«µÄÒƶ¯·½ÏòÈçͼÖбê×¢£¬ÔòH2µÄ³ö¿ÚÊÇ________(Ìî´úºÅ)¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

(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´óµ½0.05mol/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)=________

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸