¢ñ.Éú»îÖг£ÓÃÒ©Æ·ºÜ¶à£¬È磺¢Ùµâ¾Æ¡¢¢ÚÇàùËØ¡¢¢Û°¢Ë¾Æ¥ÁÖ¡¢¢ÜÆÏÌÑÌÇ×¢ÉäÒº¡¢¢Ý¿¹ËáÒ©£¨Ö÷Òª³É·ÖΪ̼ËáÇâÄÆ£©¡£

¢ÅÉÏÊöÎïÖÊÖÐÊôÓÚ¿¹ÉúËصÄÊÇ                £¨ÌîдÐòºÅ£©

¢Æ̼ËáÇâÄÆ¿É·ÀÖÎθËá·ÖÃÚ¹ý¶à£¬Æä·¢»Ó¹¦Ð§Ê±µÄÀë×Ó·½³ÌʽΪ¡¡¡¡¡¡   ¡¡¡¡¡¡                     ¡¡¡¡¡£

¢ÇÁÚôÇ»ù±½¼×ËáµÄ½á¹¹Ê½Îª £¬Ë×ÃûΪˮÑîËᣬÔÚҽѧÉϾßÓнâÈÈÕòÍ´ºÍ¿¹Ñס¢¿¹·çʪ×÷Óã¬ÓÉÓÚËü¾ßÓÐËáÐÔ¶ø´Ì¼¤Î¸³¦µÀ²úÉúÑÏÖصķ´Ó¦£¬Òò´Ëͨ³£½«ËüÏÈת»¯ÎªÒÒõ£Ë®ÑîËᣨ°¢Ë¾Æ¥ÁÖ£©£¬·þÓúóÔÚÈËÌåÄڻỺÂý·´Ó¦Éú³ÉË®ÑîË᣺

¾Ý´Ë£¬Çë»Ø´ð£º

Õâ¸ö·´Ó¦µÄ·´Ó¦ÀàÐÍΪ                       ¡£

ÓÐʱ£¬Éú²ú³§¼Ò½«ÒÒõ£Ë®ÑîËáÖƳÉÆäÄÆÑλò¸ÆÑΣ¬Ä¿µÄÊÇ                            £¬ÒÔ·ÀÖ¹´Ì¼¤Î¸ð¤Ä¤¡£

(4)ÏÂÁйØÓÚÒ©ÎïʹÓõÄ˵·¨ÖУ¬ÕýÈ·µÄÊÇ__________¡££¨ÌîдÐòºÅ£©

A£®µâ¾ÆÄÜʹµ°°×ÖʱäÐÔ£¬³£ÓÃÓÚÍâ·óÏû¶¾

B£®³¤ÆÚ´óÁ¿·þÓð¢Ë¾Æ¥ÁÖ¿ÉÔ¤·À¼²²¡£¬Ã»Óж¾¸±×÷ÓÃ

C£®Ê¹ÓÃÇàùËØ¿ÉÖ±½Ó¾²Âö×¢É䣬²»Ðè½øÐÐƤ·ôÃô¸ÐÊÔÑé

D£®Ëæ×Åƽ¼ÛÒ©·¿µÄ¿ªÉ裬Éú²¡Á˶¼¿ÉÒÔµ½Ò©µê×Ô¼ºÂòÒ©·þÓ㬲»Óõ½Ò½Ôº¾ÍÕï

¢ò.ÌÇÀàÊÇÈËÌåÐèÒªµÄÖØÒªÓªÑøËØ£¬µí·ÛÊÇÖØÒªµÄÒ»ÀàÌÇ¡£

ÔÚÊÔ¹ÜÖмÓÈë0.5gµí·ÛºÍ4mL20%µÄH2SO4ÈÜÒº£¬¼ÓÈÈ3¡«4min£¬È»ºóÓüîÒºÖкÍÊÔ¹ÜÖеÄH2SO4ÈÜÒº¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÏÂÁÐÎïÖʸ»º¬µí·ÛµÄÊÇ             £¨Ìî¡°´óÃס±»ò¡°Çà²Ë¡±»ò¡°ÖíÈ⡱£©

£¨2£©µí·ÛÍêÈ«Ë®½âÉú³ÉÓлúÎïµÄ»¯Ñ§Ê½Îª             ¡£µí·ÛË®½â²úÎïÔÚÌåÄÚ±»Ñõ»¯µÄ»¯Ñ§·½³ÌʽΪ___________________________________¡£

£¨3£©ÈôÒª¼ìÑéµí·ÛûÓÐÍêÈ«Ë®½â£¬¿ÉÈ¡ÉÙÁ¿ÉÏÊöÈÜÒº¼ÓÈë        £¨ÌîÊÔ¼ÁµÄÃû³Æ£©£¬Ó¦¹Û²ìµ½                            ¡£

¢ó.»ñµÃ½à¾»°²È«µÄÒûÓÃË®ÊÇÿ¸öÈ˵ÄÕý³£ÐèÒª¡£Ä³Å©´åµØÇøΪ»ñµÃÒûÓÃË®£¬ÔÚ½«µØ±íˮȡ»Ø¼Òºó£¬³£Ê¹ÓÃƯ°×·Û»òƯ°×¾«Æ¬½øÐÐɱ¾úÏû¶¾£¬ÆäÔ­Àí¿ÉÓû¯Ñ§·½³Ìʽ±íʾΪ                                                ¡£

¢ô.¸ÖÌúµÄÉú²úÓëʹÓÃÊÇÈËÀàÎÄÃ÷ºÍÉú»î½ø²½µÄÒ»¸öÖØÒª±êÖ¾¡£

£¨1£©¹¤ÒµÁ¶ÌúµÄÖ÷Òª»¯Ñ§·´Ó¦·½³ÌʽΪ                                 ¡£

£¨2£©³´¹ý²ËµÄÌú¹øδ¼°Ê±Ï´¾»£¨²ÐÒºÖк¬NaCl£©,µÚ¶þÌì±ã»áÒò¸¯Ê´³öÏÖºìºÖÉ«Ðâ°ß¡£ÊԻشð£º

¢ÙÌú¹øµÄ¸¯Ê´Ö÷ÒªÊÇÓÉ           ¸¯Ê´Ôì³ÉµÄ¡£

¢ÚÌú¹øÐâÊ´µÄ¸º¼«·´Ó¦Ê½Îª                                          ¡£

£¨3£©Îª·ÀÖ¹ÂÖ´¬µÄ´¬ÌåÔÚº£Ë®Öи¯Ê´£¬Ò»°ãÔÚ´¬ÉíÁ¬½Ó            £¨Ìп¿é¡±»ò¡°Í­¿é¡±£©¡£

£¨15·Ö£©

¢ñ.£¨1£©¢Ú  

£¨2£©HCO3-+H+=CO2¡ü+H2O

£¨3£©Ë®½â·´Ó¦£¨»òÈ¡´ú·´Ó¦£©    ¼õÈõËáÐÔ

£¨4£©A

¢ò.£¨1£©´óÃ×£¨2£© C6H12O6  £¬C6H12O6+6O2=6CO2+6H2O £¨3£©µâË®£¬±äÀ¶    

¢ó.Ca(ClO)2+H2O+CO2=CaCO3¡ý+2HClO

¸ßÎÂ

 

¢ô(1)Fe2O3+3CO  ===2Fe+3CO2    (2)ÎüÑõ¸¯Ê´»òµç»¯Ñ§¸¯Ê´£¬Fe-2e-= Fe2+£¨3£©Ð¿¿é

(ÒÔÉÏÿ¿Õ1·Ö)

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012ÄêËս̰æ¸ßÖл¯Ñ§Ñ¡ÐÞ2 4.1 ²ÄÁϵļӹ¤´¦ÀíÁ·Ï°¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

½ðÊô¼°Æ仯ºÏÎïµÄÓ¦Óúܹ㷺£¬ÏÖ¾Ù¼¸ÖÖ³£¼û½ðÊô¼°Æ仯ºÏÎïµÄÓÃ;£¬»Ø´ðÓйØÎÊÌ⣺

(1)FeCl3ÈÜÒº³£ÓÃÓÚ¸¯Ê´Ó¡Ë¢µç·ͭ°å£¬ÆäÀë×Ó·½³ÌʽΪ£º_______________________________¡£

(2)»ÆÍ­¿óÈÛÁ¶ºóµÃµ½µÄ´ÖÍ­Öк¬ÉÙÁ¿Fe¡¢Ag¡¢AuµÈ½ðÊôÔÓÖÊ£¬Ðè½øÒ»²½²ÉÓõç½â·¨¾«ÖÆ¡£Çë¼òÊö´ÖÍ­µç½âµÃµ½¾«Í­µÄÔ­Àí£º_________________________________________¡£

(3)ij¿ÆÑÐÈËÔ±·¢ÏÖÁÓÖʲ»Ðâ¸ÖÔÚËáÖи¯Ê´»ºÂý£¬µ«ÔÚijЩÑÎÈÜÒºÖи¯Ê´ÏÖÏóÃ÷ÏÔ¡£Çë´ÓϱíÌṩµÄÒ©Æ·ÖÐÑ¡ÔñÁ½ÖÖ(Ë®¿ÉÈÎÑ¡)£¬Éè¼Æ×î¼ÑʵÑ飬ÑéÖ¤ÁÓÖʲ»Ðâ¸ÖÒ×±»¸¯Ê´¡£

ŨÁòËᡢϡÁòËá¡¢CuO¡¢NaOHÈÜÒº

Óйط´Ó¦µÄ»¯Ñ§·½³Ìʽ____________________________________________________¡£

ÁÓÖʲ»Ðâ¸Ö¸¯Ê´µÄʵÑéÏÖÏó________________________________________________¡£

(4)ºÏ½ðÔÚÉú»îºÍÉú²úÉÏÓÃ;ԽÀ´Ô½¹ã·º¡£ÇëÓÃÏà¹ØµÄ֪ʶ½âÊÍÔÚº½¿ÕÒµÉÏÂÁºÏ½ð±ÈÌúºÏ½ðÓÃ;¸ü¹ã·ºµÄÔ­Òò¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

½ðÊô¼°Æ仯ºÏÎïµÄÓ¦Óúܹ㷺£¬ÏÖ¾Ù¼¸ÖÖ³£¼û½ðÊô¼°Æ仯ºÏÎïµÄÓÃ;£¬»Ø´ðÓйØÎÊÌ⣺
(1)FeCl3ÈÜÒº³£ÓÃÓÚ¸¯Ê´Ó¡Ë¢µç·ͭ°å£¬ÆäÀë×Ó·½³ÌʽΪ£º________¡£
(2)»ÆÍ­¿óÈÛÁ¶ºóµÃµ½µÄ´ÖÍ­Öк¬ÉÙÁ¿Fe¡¢Ag¡¢AuµÈ½ðÊôÔÓÖÊ£¬Ðè½øÒ»²½²ÉÓõç½â·¨¾«ÖÆ¡£Çë¼òÊö´ÖÍ­µç½âµÃµ½¾«Í­µÄÔ­Àí£º________¡£
(3)ij¿ÆÑÐÈËÔ±·¢ÏÖÁÓÖʲ»Ðâ¸ÖÔÚËáÖи¯Ê´»ºÂý£¬µ«ÔÚijЩÑÎÈÜÒºÖи¯Ê´ÏÖÏóÃ÷ÏÔ¡£Çë´ÓϱíÌṩµÄÒ©Æ·ÖÐÑ¡ÔñÁ½ÖÖ(Ë®¿ÉÈÎÑ¡)£¬Éè¼Æ×î¼ÑʵÑ飬ÑéÖ¤ÁÓÖʲ»Ðâ¸ÖÒ×±»¸¯Ê´¡£
ŨÁòËᡢϡÁòËá¡¢CuO¡¢NaOHÈÜÒº
Óйط´Ó¦µÄ»¯Ñ§·½³Ìʽ________¡£
ÁÓÖʲ»Ðâ¸Ö¸¯Ê´µÄʵÑéÏÖÏó________¡£
(4)ºÏ½ðÔÚÉú»îºÍÉú²úÉÏÓÃ;ԽÀ´Ô½¹ã·º¡£ÇëÓÃÏà¹ØµÄ֪ʶ½âÊÍÔÚº½¿ÕÒµÉÏÂÁºÏ½ð±ÈÌúºÏ½ðÓÃ;¸ü¹ã·ºµÄÔ­Òò¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÔÚËÄ´¨µÄ¿¹Õð¾ÈÔÖÖУ¬´óÁ¿µÄ»¯Ñ§ÊÔ¼ÁºÍÒ©Æ··¢»ÓÁ˺ܺõı£ÕÏÉú»î¡¢·ÀÒßÖβ¡µÄ×÷Óá£

£¨1£©È˵ÄÉú»îºÍÉúÃü¶¼Àë²»¿ªÊ³ÑΡ£½«ÏÂÁйØÓÚÂÈ»¯ÄƵÄÃèÊö½øÐзÖÀࣺ

¢ÙÂÈ»¯ÄÆÊÇÓÉÄÆÔªËغÍÂÈÔªËØ×é³ÉµÄ

¢ÚÂÈ»¯ÄÆÔÚά³ÖÈËÌåÄڵĵç½âÖÊƽºâÖÐÆðÖØÒª×÷ÓÃ

¢ÛÂÈ»¯ÄÆÊdz£Óõĵ÷ζƷ

¢ÜÂÈ»¯ÄÆÖÐÄÆÀë×ÓºÍÂÈÀë×ӵĸöÊýÏàµÈ

¿É½«ÉÏÊöËÄÏî·Ö³ÉÁ½À࣬Ôò·ÖÀàµÄÒÀ¾Ý·Ö±ðÊÇ______________¡¢_______________¡£

£¨2£©¾ÈÔÖÊ×ÏÈÒª±£Ö¤ÈËÐóµÄÒûÓÃË®Çå½àÎÞ¾ú¡£¸ßÌúËá¼Ø£¨K2FeO4£©ÊÇÒ»ÖÖ±ÈCl2¡¢O3»¯Ñ§ÐÔÖʸüÇ¿¡¢ÓÃ;¸ü¹ã¡¢ÎÞ¶þ´ÎÎÛȾµÄ¶à¹¦ÄÜÂÌÉ«Ë®´¦Àí¼Á£¬ÆäÖƱ¸·´Ó¦£º

2Fe(OH)3+3ClO£­+4OH£­===2FeO42£­+3Cl£­+5H2O¡£

ÆäÏû¶¾¾»Ë®µÄ·´Ó¦Ô­ÀíΪ£ºFeO42£­ Fe3£« Fe(OH)3¡£Çë»Ø´ð£º

¢ÙÿÉú³É1 mol K2FeO4תÒƵç×Ó                ¡£

¢ÚK2FeO4ÄÜÏû¶¾ÊÇÓÉÓÚÆä¾ßÓР                 ¡£

¢ÛK2FeO4Äܾ»Ë®ÊÇÓÉÓÚ²úÎïÄÜÐγɠ                 £¬Îü¸½Ðü¸¡ÎÆ侻ˮ¹ý³Ì_________

a£®Ö»ÓÐÎïÀí¹ý³Ì    b£®Ö»ÊÇ»¯Ñ§¹ý³Ì     c£®ÊÇÎïÀíºÍ»¯Ñ§¹ý³Ì

£¨3£©AlCl3Ò²Êdz£Óõľ»Ë®¼Á£¬Æä²úÎïAl(OH)3¼ÈÄÜÈÜÓÚÇ¿ËáÈÜÒºÒ²ÄÜÈÜÓÚÇ¿¼îÈÜÒº£¬ÇëÓñØÒªµÄ·½³ÌʽºÍÎÄ×Ö´Ó·´Ó¦Ô­ÀíµÄ½Ç¶ÈÓèÒÔ˵Ã÷¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨08×Ͳ©ÊÐÄ£Ä⣩£¨10·Ö£©ÔÚËÄ´¨µÄ¿¹Õð¾ÈÔÖÖУ¬´óÁ¿µÄ»¯Ñ§ÊÔ¼ÁºÍÒ©Æ··¢»ÓÁ˺ܺõı£ÕÏÉú»î¡¢·ÀÒßÖβ¡µÄ×÷Óá£

£¨1£©È˵ÄÉú»îºÍÉúÃü¶¼Àë²»¿ªÊ³ÑΡ£½«ÏÂÁйØÓÚÂÈ»¯ÄƵÄÃèÊö½øÐзÖÀࣺ

¢ÙÂÈ»¯ÄÆÊÇÓÉÄÆÔªËغÍÂÈÔªËØ×é³ÉµÄ

¢ÚÂÈ»¯ÄÆÔÚά³ÖÈËÌåÄڵĵç½âÖÊƽºâÖÐÆðÖØÒª×÷ÓÃ

¢ÛÂÈ»¯ÄÆÊdz£Óõĵ÷ζƷ

¢ÜÂÈ»¯ÄÆÖÐÄÆÀë×ÓºÍÂÈÀë×ӵĸöÊýÏàµÈ

¿É½«ÉÏÊöËÄÏî·Ö³ÉÁ½À࣬Ôò·ÖÀàµÄÒÀ¾Ý·Ö±ðÊÇ______________¡¢_______________¡£

£¨2£©¾ÈÔÖÊ×ÏÈÒª±£Ö¤ÈËÐóµÄÒûÓÃË®Çå½àÎÞ¾ú¡£¸ßÌúËá¼Ø£¨K2FeO4£©ÊÇÒ»ÖÖ±ÈCl2¡¢O3»¯Ñ§ÐÔÖʸüÇ¿¡¢ÓÃ;¸ü¹ã¡¢ÎÞ¶þ´ÎÎÛȾµÄ¶à¹¦ÄÜÂÌÉ«Ë®´¦Àí¼Á£¬ÆäÖƱ¸·´Ó¦£º

2Fe(OH)3+3ClO£­+4OH£­===2FeO42£­+3Cl£­+5H2O¡£

ÆäÏû¶¾¾»Ë®µÄ·´Ó¦Ô­ÀíΪ£ºFeO42£­ Fe3£« Fe(OH)3¡£Çë»Ø´ð£º

¢ÙÿÉú³É1 mol K2FeO4תÒƵç×Ó                ¡£

¢ÚK2FeO4ÄÜÏû¶¾ÊÇÓÉÓÚÆä¾ßÓР                 ¡£

¢ÛK2FeO4Äܾ»Ë®ÊÇÓÉÓÚ²úÎïÄÜÐγɠ                 £¬Îü¸½Ðü¸¡ÎÆ侻ˮ¹ý³Ì_________

a£®Ö»ÓÐÎïÀí¹ý³Ì    b£®Ö»ÊÇ»¯Ñ§¹ý³Ì     c£®ÊÇÎïÀíºÍ»¯Ñ§¹ý³Ì

£¨3£©AlCl3Ò²Êdz£Óõľ»Ë®¼Á£¬Æä²úÎïAl(OH)3¼ÈÄÜÈÜÓÚÇ¿ËáÈÜÒºÒ²ÄÜÈÜÓÚÇ¿¼îÈÜÒº£¬ÇëÓñØÒªµÄ·½³ÌʽºÍÎÄ×Ö´Ó·´Ó¦Ô­ÀíµÄ½Ç¶ÈÓèÒÔ˵Ã÷¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸