ÒÑ֪ijÈÜÒºX¿ÉÄÜÓÉK+¡¢SO42-¡¢I-¡¢SiO32-¡¢MnO4-¡¢Ag+¡¢Ba2+¡¢Al3+¡¢Fe2+¡¢ALO2-¡¢CO32-ÖеÄÈô¸ÉÖÖ×é³É£®Ä³»¯Ñ§ÐËȤС×飬ͨ¹ýÏÂÁÐʵÑéÈ·¶¨ÁËÆä×é³É£®
I£®¹Û²ìÈÜÒº£ºÎÞɫ͸Ã÷£®
¢ò£®È¡ÊÊÁ¿¸ÃÈÜÒº£¬¼ÓÈë¹ýÁ¿µÄÏõËᣬÓÐÆøÌåÉú³É£¬²¢µÃµ½ÎÞÉ«ÈÜÒº£®
¢ó£®ÔÚ¢òËùµÃÈÜÒºÖÐÔÙ¼ÓÈë¹ýÁ¿NH4HCO3ÈÜÒº£¬ÓÐÆøÌåÉú³É£¬Í¬Ê±Îö³ö°×É«³ÁµíA£®
IV£ºÔÚ¢óËùµÃÈÜÒºÖмÓÈë¹ýÁ¿Ba£¨OH£©2ÈÜÒºÖÁ¹ýÁ¿£¬¼ÓÈÈÒ²ÓÐÆøÌåÉú³É£¬²¢Óа×É«³ÁµíBÎö³ö£®
¸ù¾ÝÉÏÊöʵÑé»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÓÉʵÑéI¿ÉÖªÔ­ÈÜÒºÒ»¶¨²»º¬ÓеÄÀë×ÓÊÇ
 

£¨2£©ÓÉʵÑé¢ò¿ÉÖªÔ­ÈÜÒºÒ»¶¨²»º¬ÓеÄÀë×ÓÊÇ
 
£¬Ò»¶¨º¬ÓеÄÀë×ÓÊÇ
 
£®
£¨3£©ÓÉʵÑé¢ó¿ÉÖªÔ­ÈÜÒºÖл¹Ò»¶¨º¬ÓеÄÀë×ÓÊÇ
 
£¬Éú³É³ÁµíAµÄÀë×Ó·½³ÌʽΪ
 

£¨4£©ÊµÑé¢ôÖпªÊ¼½×¶Î·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽһ¶¨ÓÐ
 

£¨5£©Í¨¹ýÉÏÊöʵÑ飬°×É«³ÁµíBµÄ×é³É¿ÉÄÜÊÇ
 
£®
¸Ã»¯Ñ§ÐËȤС×éµÄͬѧÓÖÉè¼ÆÁËÒ»¸ö¼òµ¥µÄºóÐøʵÑ飬ȷ¶¨Á˸óÁµíµÄ×é³É£¬¸Ã·½·¨ÊÇ£º
 
£®
¿¼µã£º³£¼ûÀë×ӵļìÑé·½·¨,Àë×Ó¹²´æÎÊÌâ
רÌ⣺Àë×Ó·´Ó¦×¨Ìâ
·ÖÎö£ºI£®¹Û²ìÈÜÒº£ºÎÞɫ͸Ã÷£¬ÔòÒ»¶¨²»´æÔÚÓÐÉ«Àë×Ó£ºMnO4-¡¢Fe2+£»
¢ò£®È¡ÊÊÁ¿¸ÃÈÜÒº£¬¼ÓÈë¹ýÁ¿µÄÏõËᣬÓÐÆøÌåÉú³É£¬¸ÃÆøÌåΪ¶þÑõ»¯Ì¼£¬ÔòÈÜÒºÖÐÒ»¶¨´æÔÚCO32-£¬Ò»¶¨²»´æÔÚÓëCO32-Àë×Ó·´Ó¦µÄÀë×Ó£»µÃµ½ÎÞÉ«ÈÜÒº£¬ÔòÒ»¶¨²»´æÔÚµâÀë×Ӻ͹èËá¸ùÀë×Ó£¬ÔÙ¸ù¾ÝÈÜÒº³ÊµçÖÐÐÔÅжÏÔ­ÈÜÒºÖÐÒ»¶¨´æÔÚΨһÑôÀë×Ó¼ØÀë×Ó£»
¢ó£®ÔÚ¢òËùµÃÈÜÒºÖÐÔÙ¼ÓÈë¹ýÁ¿NH4HCO3ÈÜÒº£¬ÓÐÆøÌåÉú³É£¬Í¬Ê±Îö³ö°×É«³ÁµíA£¬ËµÃ÷ÈÜÒºÖÐÒ»¶¨´æÔÚÂÁÀë×Ó£¬Ô­ÈÜÒºÖÐÒ»¶¨´æÔÚÆ«ÂÁËá¸ùÀë×Ó£»
IV£ºÔÚ¢óËùµÃÈÜÒºÖк¬ÓйýÁ¿µÄ̼ËáÇâ泥¬¼ÓÈë¹ýÁ¿Ba£¨OH£©2ÈÜÒºÖÁ¹ýÁ¿£¬¼ÓÈÈ»áÓа±ÆøÉú³É£¬°×É«³Áµí¿ÉÄÜΪ̼Ëá±µ»ò̼Ëá±µºÍÁòËá±µµÄ»ìºÏÎ
¸ù¾ÝÒÔÉÏ·ÖÎö½øÐнâ´ð£®
½â´ð£º ½â£º£¨1£©ÎÞÉ«ÈÜÒºÖв»ÄÜ´æÔÚµÄÓÐÉ«µÄÀë×Ó£¬ËùÒÔÒ»¶¨²»´æÔÚMnO4-¡¢Fe2+£¬
¹Ê´ð°¸Îª£ºMnO4-¡¢Fe2+£»
£¨2£©ÏõËá¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬¼ÓÈëÏõËá²úÉúÆøÌ壬ÔòÒ»¶¨´æÔÚCO32-£¬Ò»¶¨²»´æÔÚÓë̼Ëá¸ùÀë×Ó·´Ó¦µÄÀë×Ó£ºAg+¡¢Ba2+¡¢Al3+£»µÃµ½ÎÞÉ«ÈÜÒº£¬ÔòÒ»¶¨²»´æÔÚÉú³ÉÄÑÈÜÎïµÄSiO32-£¬Ò²²»´æÔÚ±»Ñõ»¯Éú³ÉÓÐÉ«ÎïÖʵÄI-£¬ÓÉÓÚÈÜÒº³ÊµçÖÐÐÔ¿ÉÖª£¬ÔòÔ­ÈÜÒºÖÐÒ»¶¨´æÔÚΨһµÄÑôÀë×ÓK+£»
¹Ê´ð°¸Îª£ºAg+¡¢Ba2+¡¢Al3+¡¢I-¡¢SiO32-£»K+¡¢CO32-£» 
£¨3£©ÓÉʵÑé¢ó¿ÉÖª£¬Äܹ»Óë̼ËáÇâ¸ùÀë×Ó·´Ó¦Éú³ÉÆøÌåºÍ³Áµí£¬ÔòÓëÏõËá·´Ó¦ºóµÄÈÜÒºÖдæÔÚAl3+£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£ºAl3++3HCO3-=Al£¨OH£©3¡ý+3CO2¡ü£¬Ô­ÈÜÒºÖÐÒ»¶¨´æÔÚAlO2-£¬
¹Ê´ð°¸Îª£ºAlO2-£»Al3++3HCO3-=Al£¨OH£©3¡ý+3CO2¡ü£»
£¨4£©ÊµÑé¢ôÖпªÊ¼½×¶Î·¢Éú·´Ó¦Îª¹ýÁ¿µÄ̼ËáÇâï§ÓëÇâÑõ»¯±µ·´Ó¦£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º2HCO3-+Ba2++2OH=BaCO3¡ý+2H2O+CO32-£¬
¹Ê´ð°¸Îª£º2HCO3-+Ba2++2OH=BaCO3¡ý+2H2O+CO32-£»
£¨5£©ÓÉÓÚÔ­ÈÜÒºÖпÉÄܺ¬ÓÐÁòËá¸ùÀë×Ó£¬ËùÒÔ°×É«³ÁµíÒ»¶¨º¬ÓÐ̼Ëá±µ£¬¿ÉÄÜΪº¬ÓÐÁòËá±µ£»¼ìÑé°×É«³ÁµíBµÄ·½·¨Îª£ºÈ¡ÉÙÁ¿°×É«³ÁµíB¼ÓÈë×ãÁ¿Ï¡ÑÎËᣬÈô³ÁµíÍêÈ«Èܽ⣬ÔòΪBaCO3£¬Èô³Áµí²¿·ÖÈܽ⣬ÔòΪBaCO3ºÍBaSO4£¬
¹Ê´ð°¸Îª£ºBaCO3»òBaCO3ºÍBaSO4 £»È¡ÉÙÁ¿°×É«³ÁµíB¼ÓÈë×ãÁ¿Ï¡ÑÎËᣬÈô³ÁµíÍêÈ«Èܽ⣬ÔòΪBaCO3£¬Èô³Áµí²¿·ÖÈܽ⣬ÔòΪBaCO3ºÍBaSO4£®
µãÆÀ£º±¾Ì⿼²éÁËÀë×Ó¹²´æ¡¢³£¼ûÀë×ӵļìÑ飬ÌâÄ¿ÄѶÈÖеȣ¬ÊÔÌâµÄÌâÁ¿½Ï´ó£¬×¢ÒâÕÆÎÕ³£¼ûÀë×ÓµÄÐÔÖʼ°¼ìÑé·½·¨£¬Ã÷È·Àë×Ó·´Ó¦·¢ÉúµÄÌõ¼þ¼°Àë×Ó¹²´æµÄÌõ¼þ£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐÎïÖÊ·ÖÀàÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢SO2¡¢SiO2¡¢CO¡¢P2O5¾ùΪËáÐÔÑõ»¯Îï
B¡¢Ï¡¶¹½¬¡¢¹èËá¡¢ÂÈ»¯ÌúÈÜÒº¾ùΪ½ºÌå
C¡¢¿ÕÆø¡¢¸£¶ûÂíÁÖ¡¢Ë®²£Á§¡¢°±Ë®¾ùΪ»ìºÏÎï
D¡¢Éռ±ù´×Ëá¡¢ËÄÂÈ»¯Ì¼¡¢Ê¯Ä«¾ùΪµç½âÖÊ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐÎïÖʵÄÖÆÈ¡£¬ÊµÑé²Ù×÷ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢ÏòµçʯÖлºÂýµÎÈë±¥ºÍʳÑÎË®£¬¿ÉÖÆÈ¡C2H2
B¡¢½«NH4HCO3±¥ºÍÈÜÒºÖÃÓÚÕô·¢ÃóÖмÓÈÈÕô¸É£¬¿ÉÖÆÈ¡NH4HCO3¹ÌÌå
C¡¢ÏòFeCl3±¥ºÍÈÜÒº»ºÂýµÎÈë¹ýÁ¿°±Ë®¼ÓÈÈ£¬¿ÉÖÆÈ¡Fe£¨OH£©3½ºÌå
D¡¢½«CuCl2ÈÜÒºÖÃÓÚÕô·¢ÃóÖмÓÈÈÕô¸É£¬¿ÉÖÆÈ¡ÎÞË®CuCl2¹ÌÌå

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÈçͼËùʾµÄʵÑ飬ÄܴﵽʵÑéÄ¿µÄÊÇ£¨¡¡¡¡£©
A B C D
ÑéÖ¤»¯Ñ§ÄÜת»¯Îªµ×µçÄÜ Ñé֤ζȶÔƽºâÒƶ¯µÄÓ°Ïì ÑéÖ¤Ìú·¢ÉúÇⸯʴ ÑéÖ¤·Ç½ðÊôCl£¾C£¾Si
A¡¢AB¡¢BC¡¢CD¡¢D

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

A¡¢B¡¢C¡¢D¡¢EÊǺ¬ÓÐͬһԪËصij£¼ûÎïÖÊ£®CÎïÖÊÖ»ÓÉÒ»ÖÖÔªËØ×é³É£¬ÔÚ1¸öC·Ö×ÓÖÐÐγɹ²¼Û¼üµÄµç×ÓÊýÓë·Ö×ÓÖÐËùº¬µç×ÓÊýÖ®±ÈΪ3£º7£®C¡¢E¾ù¿ÉÓëÑõÆøÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦Éú³ÉA£®Çë»Ø´ðÒÔÏÂÎÊÌ⣺
£¨1£©³£ÎÂϽ«ÆøÌåBͨÈëË®Öз¢Éú·´Ó¦£¬Éú³ÉAºÍD£¬ÔòDµÄ»¯Ñ§Ê½Îª
 
£®
£¨2£©EµÄ¿Õ¼ä¹¹ÐÍÊÇ
 
£»Ð´³öEÓëÑõÆø·´Ó¦Éú³ÉA»¯Ñ§·½³Ìʽ
 
£®
£¨3£©DÓëE·´Ó¦µÄÉú³ÉÎïÔÚ¹ÌÌåʱµÄ¾§ÌåÀàÐÍÊÇ
 
£»ÆäË®ÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ
 
£®
£¨4£©Ð´³ö¹¤ÒµÉϺϳÉE·´Ó¦µÄ»¯Ñ§·½³Ìʽ
 
£»ÈôʹƽºâÏòÉú³ÉEµÄ·½ÏòÒƶ¯£¬¿É²ÉÈ¡µÄ´ëÊ©ÓÐ
 
£¨Ìîд×Öĸ£©£®
A£®Éý¸ßζȠ    B£®Ôö´óѹǿ    C£®Ê¹Óô߻¯¼Á     D£®Òº»¯·ÖÀë
£¨5£©ÔÚ±ê×¼×´¿öÏ£¬2.24L EÆøÌåÔÚÑõÆøÖÐÍêȫȼÉÕÉú³ÉÆøÌåµ¥ÖÊCºÍҺ̬ˮ£¬·Å³öÈÈÁ¿Q kJ£¬Ôò¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¶Ï¿ª1mol AB£¨g£©·Ö×ÓÖеĻ¯Ñ§¼üʹÆä·Ö±ðÉú³ÉÆø̬AÔ­×ÓºÍÆø̬BÔ­×ÓËùÎüÊÕµÄÄÜÁ¿³ÆΪA-B¼üµÄ¼üÄÜ£®Ï±íÁгöÁËһЩ»¯Ñ§¼üµÄ¼üÄÜE£º
»¯Ñ§¼ü H-H Cl-Cl O¨TO C-Cl C-H O-H H-Cl
E/kJ?mol-1 436 247 x 330 413 463 431
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Èçͼ±íʾij·´Ó¦µÄÄÜÁ¿±ä»¯¹Øϵ£¬Ôò´Ë·´Ó¦Îª
 
£¨Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±£©·´Ó¦£¬ÆäÖС÷H=
 
£¨Óú¬ÓÐa¡¢bµÄ¹Øϵʽ±íʾ£©£®
£¨2£©ÈôͼʾÖбíʾ·´Ó¦H2£¨g£©+
1
2
O2£¨g£©¨TH2O£¨g£©¡÷H=-241.8kJ?mol-1£¬Ôòb=
 
kJ?mol-1£¬
x=
 
£®
£¨3£©ÀúÊ·ÉÏÔøÓᰵؿµ·¨¡±ÖÆÂÈÆø£¬ÕâÒ»·½·¨ÊÇÓÃCuCl2×÷´ß»¯¼Á£¬ÔÚ450¡æÀûÓÿÕÆøÖеÄÑõÆø¸úÂÈ»¯Çâ·´Ó¦ÖÆÂÈÆø£®·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
 
£®
ÈôºöÂÔζȺÍѹǿ¶Ô·´Ó¦ÈȵÄÓ°Ï죬¸ù¾ÝÉÏÌâÖеÄÓйØÊý¾Ý£¬¼ÆËãµ±·´Ó¦ÖÐÓÐ1molµç×ÓתÒÆʱ£¬·´Ó¦µÄÄÜÁ¿±ä»¯Îª
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨1£©ÓöèÐԵ缫µç½âCuSO4ÈÜÒº£¨ÈçͼµÄ×°Öã©£¬¸Ã¹ý³ÌÖеç½â³Øµç¼«·´Ó¦Ê½ÎªÊÇÑô¼«£º
 
£¬Òõ¼«£º
 
£®
£¨2£©Èôµç½â³ØÖÐ×°Èë×ãÁ¿ÈÜÒº£¬µ±Òõ¼«ÔöÖØ3.2gʱ£¬Í£Ö¹Í¨µç£¬´ËʱÑô¼«²úÉúÆøÌåµÄÌå»ý£¨±ê×¼×´¿ö£©Îª
 
£¨¼ÙÉèÆøÌåÈ«²¿Òݳö£©£®
£¨3£©Óûʹµç½âÒº»Ö¸´µ½Æðʼ״̬£¬Ó¦ÏòÈÜÒºÖмÓÈëÊÊÁ¿µÄ
 

A£®CuSO4B£®H2O    C£®CuO      D£®CuSO4?5H2O£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨1£©·´Ó¦£º8CuFeS2+21O2
 ¸ßΠ
.
 
8Cu+4FeO+2Fe2O3+16SO2£¬ÈôCuFeS2ÖÐFeµÄ»¯ºÏ¼ÛΪ+2£¬·´Ó¦Öб»»¹Ô­µÄÔªËØÊÇ
 
£¨ÌîÔªËØ·ûºÅ£©£®
£¨2£©ÊµÑéÊÒ±£´æFeCl2ÈÜҺʱ£¬ÐèÒª¼ÓÈëFe·ÛºÍÑÎËᣬÇëÓÃÀë×Ó·½³Ìʽ±íʾ¼ÓÈëÑÎËáµÄÔ­Òò£º
 
£®
£¨3£©Á¶Í­²úÉúµÄ¯Ôü£¨Ö÷Òª³É·Öº¬Fe2O3¡¢FeO¡¢SiO2¡¢Al2O3£©¿ÉÖƱ¸Fe2O3£®·½·¨Îª£º
¢ÙÓÃÏ¡ÑÎËá½þȡ¯Ôü£¬¹ýÂË£®
¢Ú½«¢ÙÖÐËùµÃÂËÒºÏÈÑõ»¯£¬ÔÙ¼ÓÈë¹ýÁ¿NaOHÈÜÒº£¬¹ýÂË£¬½«³ÁµíÏ´µÓ¡¢¸ÉÔï¡¢ìÑÉյòúÆ·£®
¾ÝÒÔÉÏÐÅÏ¢»Ø´ðÏÂÁÐÎÊÌ⣺
a£®²½Öè¢Ù¡¢¢ÚÖÐÉæ¼°AlÔªËصÄÀë×Ó·½³ÌʽÊÇ
 
¡¢
 
£®
b£®Ñ¡ÓÃÌṩµÄÊÔ¼Á£¬Éè¼ÆʵÑéÑéÖ¤¹ÌÌå¯ÔüÖк¬ÓÐFeO£®
ÌṩµÄÊÔ¼Á£ºÏ¡ÑÎËá  Ï¡ÁòËá  KSCNÈÜÒº  KMnO4ÈÜÒº  NaOHÈÜÒº  µâË®
ËùÑ¡ÊÔ¼ÁΪ
 
£®
¼ø¶¨¹ý³ÌÉæ¼°µÄÑõ»¯»¹Ô­µÄÀë×Ó·½³ÌʽΪ£º
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
Ñ¡Ïî ʵÑé ½âÊÍ»ò½áÂÛ
A ÓýྻµÄPtպȡijÈÜÒº½øÐÐÑæÉ«·´Ó¦£¬»ðÑæ³Ê»ÆÉ« ¸ÃÈÜÒºÖÐÒ»¶¨º¬ÓÐNa+£¬ÎÞk+
B ÓýྻµÄ²£Á§¹ÜÏò°üÓÐNa2O2µÄÍÑÖ¬ÃÞ´µÆø£¬ÍÑÖ¬ÃÞȼÉÕ CO2¡¢H2OÓëNa2O2·´Ó¦ÊÇ·ÅÈÈ·´Ó¦
C ÏòäåË®ÖеÎÈëÖ²ÎïÓÍ£¬Õñµ´ºó£¬ÓͲãÏÔÎÞÉ« äå²»ÈÜÓÚÓÍÖ¬
D ½«ÁòËáËữµÄH2O2µÎÈëFe£¨NO3£©2ÈÜÒº£¬ÈÜÒº±ä»ÆÉ« H2O2µÄÑõ»¯ÐÔ±ÈFe3+Ç¿
A¡¢AB¡¢BC¡¢CD¡¢D

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸