ÌâÄ¿ÄÚÈÝ

£¨15·Ö£©À¶Í­¿óÖ÷Òªº¬2CuCO3¡¤Cu(OH)2£¬»¹º¬ÓÐÉÙÁ¿Fe¡¢SiµÄ»¯ºÏÎ¹¤ÒµÉÏÒÔÀ¶Í­¿óΪԭÁÏÖƱ¸Cu¼°CaCO3£¬·½·¨ÓжàÖÖ¡£
£¨1£©À¶Í­¿óÓ뽹̿¼ÓÈÈ¿ÉÒÔÉú³ÉÍ­¡¢¶þÑõ»¯Ì¼ºÍË®£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ¡£______________________________________________________________¡£
¾ßÌå·½·¨²½ÖèÈçÏ£º 

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨2£©ÈÜÒºAµÄ½ðÊôÀë×ÓÓÐCu2+¡¢Fe2+¡¢Fe3+¡£´ÓÏÂÁÐËù¸øÊÔ¼ÁÖÐÑ¡Ôñ£º¼ìÑéÈÜÒºAÖк¬ÓÐFe2+µÄ×î¼ÑÊÔ¼ÁΪ   £¨Ìî´úºÅ£©£¬ÊµÑé²½ÖèÖÐÊÔ¼Á¢ÙΪ   £¨Ìî´úºÅ£©¡£
a£®KMnO4                      b£®(NH4) 2S              c£®H2O2                   d£®KSCN
£¨3£©ÓÉÈÜÒºC»ñµÃCuSO4¡¤5H2O£¬ÐèÒª¾­¹ý¼ÓÈÈÕô·¢£¬ÀäÈ´½á¾§¡¢¹ýÂ˵ȲÙ×÷£¬¼ÓÈÈÕô·¢ÖÁ            ___________________ʱֹͣ¼ÓÈÈ¡£²£Á§°ôÔÚÕû¸ö²Ù×÷ÖеÄ×÷ÓÃÊÇ         ________________ ¡£
£¨4£©ÖƱ¸CaCO3ʱ£¬Ó¦ÏòCaCl2ÈÜÒºÖÐÏÈͨÈ루»òÏȼÓÈ룩    £¨Ìѧʽ£©¡£ÈôʵÑé¹ý³ÌÖÐÓа±ÆøÒݳö¡¢Ó¦Ñ¡ÓÃÏÂÁР      ×°ÖÃÎüÊÕβÆø£¨Ìî´úºÅ£©¡£

£¨5£©´ÓÈÜÒºCÖлñµÃCuµÄ·½·¨ÓР                   £¨Ð´³öÁ½ÖÖ²»Í¬·½·¨£©¡£
£¨6£©ÓöèÐԵ缫µç½âÁòËáÍ­ÈÜÒºÒ»¶Îʱ¼äºó£¬¼ÓÈë2molCu(OH)2¹ÌÌåʹÁòËáÍ­ÈÜÒº¸´Ô­£¨¸´Ô­ÊÇÖ¸ÈÜÒºÈÜÖʳɷּ°Å¨¶ÈÓëÔ­À´ÍêÈ«Ïàͬ£©£¬Ôò´Ëµç½â¹ý³ÌÖй²×ªÒƵç×Ó        mol¡£
£¨1£©2[2CuCO3¡¤Cu(OH)2]+3C6Cu+7CO2¡ü+2 H2O £¨3·Ö£©                                            
£¨2£©a  c £¨2·Ö£¬¸÷1·Ö£© £¨3£©ÓÐÉÙÁ¿¾§Ä¤³öÏÖʱ£¨ÓÐÉÙÁ¿¾§Ìå³öÏÖÒ²¿É£©£¨2·Ö£©ÒýÁ÷¡¢½Á°è£¨2·Ö£© £¨4£©NH3£¨1·Ö£©    b £¨1·Ö£©     £¨5£©¼ÓÈë»îÆýðÊôFe£¨»òÕßд¾ßÌåÄÜÖƳöCuµÄÆäËü½ðÊôÒ²¿É£©¡¢ÓöèÐԵ缫µç½âÁòËáÍ­ÈÜÒº¡£ £¨2·Ö£©    £¨6£© 8  £¨2·Ö£©
ÂÔ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨10·Ö£©ÏÂÃæÊÇij»¯Ñ§ÐËȤС×éµÄͬѧ×öͬÖÜÆÚÔªËØÐÔÖʵݱä¹æÂÉʵÑéʱ£¬Éè¼ÆÁËÒ»Ì×ʵÑé·½°¸¡£ÆäÖÐʵÑé²½ÖèÈçÏ£º
ʵÑéÐòºÅ
ʵÑé²½Öè
¢Ù
½«Ã¾ÌõÓÃÉ°Ö½´òÄ¥ºó£¬·ÅÈë·ÐË®ÖУ¬ÔÙÏòÈÜÒºÖеμӷÓ̪ÈÜÒº
¢Ú
ÏòÐÂÖƵõ½µÄNa2SÈÜÒºÖеμÓÐÂÖƵÄÂÈË®
¢Û
½«Ò»Ð¡¿é½ðÊôÄÆ·ÅÈëµÎÓзÓ̪ÈÜÒºµÄÀäË®ÖÐ
¢Ü
½«Ã¾ÌõͶÈëÏ¡ÑÎËáÖÐ
¢Ý
½«ÂÁÌõͶÈëÏ¡ÑÎËáÖÐ
 
¢ÅʵÑéÄ¿µÄ                                                                                                            
¢Æд³öʵÑé¢ÙºÍ¢ÚÖеÄʵÑéÏÖÏóºÍ»¯Ñ§·½³Ìʽ
ʵÑé¢Ù£ºÏÖÏó                                          »¯Ñ§·½³Ìʽ                                             
ʵÑé¢Ú£ºÏÖÏó                                          »¯Ñ§·½³Ìʽ                                             
¢ÇʵÑé½áÂÛ                                                                                                                   
(15·Ö)ÒÑÖªCaSO4ÊÜÈȷֽ⣬ÓÉÓÚÊÜÈÈζȲ»Í¬£¬ÆøÌå³É·ÖÒ²²»Í¬¡£ÆøÌå³É·Ö¿ÉÄÜΪSO2¡¢SO3ºÍO2ÖеÄÒ»ÖÖ¡¢¶þÖÖ»òÈýÖÖ¡£Ä³»¯Ñ§¿ÎÍâ»î¶¯Ð¡×é×¼±¸Í¨¹ýϵÁÐʵÑé̽¾¿CaSO4·Ö½âÉú³ÉµÄÆøÌ壬½ø¶øÈ·¶¨CaSO4·Ö½âµÄ»¯Ñ§·½³Ìʽ¡£
¡¾Ìá³ö²ÂÏë¡¿
¢ñ£®ËùµÃÆøÌåµÄ³É·Ö¿ÉÄÜÖ»º¬ SO3Ò»ÖÖ£»
¢ò£®ËùµÃÆøÌåµÄ³É·Ö¿ÉÄܺ¬ÓР              ¶þÖÖ£»£¨Ìî·Ö×Óʽ£©
¢ó£®ËùµÃÆøÌåµÄ³É·Ö¿ÉÄܺ¬ÓÐ SO2¡¢SO3¡¢O2ÈýÖÖ¡£
¡¾Éè¼ÆʵÑé¡¿
¸Ã»¯Ñ§¿ÎÍâ»î¶¯Ð¡×é×¼±¸Í¨¹ý²â¶¨D×°ÖõÄÔöÖØÒÔ¼°Á¿Í²ÖÐË®µÄÌå»ý£¬À´Ì½¾¿CaSO4·Ö½âÉú³ÉµÄÆøÌå³É·Ö£¬½ø¶øÈ·¶¨CaSO4·Ö½âµÄ»¯Ñ§·½³Ìʽ¡£

¡¾ÊµÑé¹ý³Ì¡¿
¸ù¾ÝÉÏÊöʵÑé·½°¸½øÐÐÊÔÑé¡£ÒÑ֪ʵÑé½áÊøʱ£¬CaSO4ÍêÈ«·Ö½â¡£
Çë½áºÏÒÔÏÂʵÑéÏÖÏóºÍ¼Ç¼µÄʵÑéÊý¾Ý½øÐзÖÎö£º
£¨1£©ÈôʵÑé½áÊøʱ£¬GÖÐÁ¿Í²Ã»ÓÐÊÕ¼¯µ½Ë®£¬ÔòÖ¤Ã÷²ÂÏë       ÕýÈ·¡£(Ìî¢ñ»ò¢ò»ò¢ó)
£¨2£©ÈôʵÑé½áÊøʱ£¬×°ÖÃDµÄ×ÜÖÊÁ¿Ôö¼Ó£¬ÄÜ·ñ¶Ï¶¨ÆøÌå²úÎïÖÐÒ»¶¨º¬ÓÐSO2¶ø²»º¬SO3£¿Çë˵Ã÷ÀíÓÉ£º                                                    ¡£
£¨3£©¸ÃʵÑéÉè¼ÆÖУ¬ÈÝÒ׸ø²â¶¨´øÀ´½Ï´óÎó²îµÄÒòËØÓР                        ¡£
(д³öÒ»ÖÖ¼´¿É)
£¨4£©¾­¸Ä½øºó£¬ÓÐÁ½×éͬѧ½øÐиÃʵÑ飬ÓÉÓÚ¼ÓÈÈʱµÄζȲ»Í¬,ʵÑé²âµÃÊý¾ÝÒ²²»Í¬,
Ïà¹ØÊý¾ÝÈçÏ£º
ʵÑéС×é
³ÆÈ¡CaSO4
µÄÖÊÁ¿(g)
×°ÖÃDÔö¼Ó
µÄÖÊÁ¿(g)
Á¿È¡ÆøÌåÌå»ýµÄ×°ÖòâÁ¿µÄÆøÌåÌå»ý (ÕÛËã³É±ê×¼×´¿öÏÂÆøÌåµÄÌå»ý) (mL)
Ò»
4.08
2.56
224
¶þ
5.44
2.56
448
Çëͨ¹ý¼ÆË㣬Íƶϵڶþ×éͬѧµÃ³öµÄCaSO4·Ö½âµÄ»¯Ñ§·´Ó¦·½³Ìʽ£º
µÚ¶þ×飺                                              ¡£
£¨16·Ö£©»ÆÌú¿óµÄÖ÷Òª³É·ÖÊÇFeS2¡£²â¶¨»ÆÌú¿óÖÐFeS2º¬Á¿µÄÁ½ÖÖ·½·¨ÈçÏÂͼËùʾ£º

ÒÑÖª£º¢ÙÍõË®ÊÇÓÉ1Ìå»ýµÄŨÏõËá(¦Ñ=1.42g¡¤cm-3)ºÍ3Ìå»ýµÄŨÑÎËá(¦Ñ=1.19g¡¤cm-3)»ìºÏ¶ø³ÉµÄ¡£
¢Ú»ÆÌú¿óºÍÍõË®·´Ó¦µÄ·½³ÌʽΪFeS2+5HNO3+3HCl=FeCl3+2H2SO4+5NO¡ü+2H2O
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¼òÊöʵÑéÊÒÅäÖÆÍõË®µÄ²Ù×÷¹ý³Ì_____________________________________
________________________________________________________________¡£
£¨2£©·½·¨Ò»ÖйýÂË¡¢Ï´µÓ¡¢×ÆÉÕ¶¼Óõ½µÄÒÇÆ÷ÊÇ_______________¡£
£¨3£©·½·¨¶þÖÐÒªÅжÏBaCl2ÈÜÒºÊÇ·ñ¹ýÁ¿£¬¿ÉÏòÂËÒºÖмÓÈëXÈÜÒº£¬X¿ÉÒÔÊÇ_________(Ìî´úºÅ)
A£®BaCl2B£®NaOHC£®Na2SO4D£®HCl
£¨4£©ÊµÑéÊÒÀûÓÃÏÂÁÐ×°ÖúÍÊÔ¼ÁÖÆÈ¡ÉÙÁ¿ÂÈ»¯ÇâÆøÌåÊÔ¼Á£º¢ÙŨÁòËá¡¡¢ÚŨÑÎËá¡¡¢ÛʳÑιÌÌå

ÈôÑ¡ÓÃÊÔ¼Á¢Ù¢Û£¬ÔòӦѡÔñµÄ×°ÖÃÊÇ___________(Ìî´úºÅ£¬ÏÂͬ)£»ÍƲⷢÉú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ__________________________________________________£»
ÈôÑ¡ÓÃÊÔ¼Á¢Ù¢Ú£¬ÔòÒËÑ¡ÔñµÄ×°ÖÃÊÇ____________¡£
£¨5£©·½·¨Ò»ÖУ¬×ÆÉÕʱ·¢Éú·´Ó¦µÄ»¯Ñ§·´Ó¦·½³ÌʽΪ________________________ £»ÒÑÖª³ÆÈ¡»ÆÌú¿óÑùÆ·µÄÖÊÁ¿Îª1.50g£¬³ÆµÃ×ÆÉÕºó¹ÌÌåµÄÖÊÁ¿Îª0.8g£¬²»¿¼ÂDzÙ×÷Îó²î£¬Ôò¸Ã¿óʯÖÐFeS2µÄÖÊÁ¿·ÖÊýÊÇ________________¡£
£¨12·Ö£©Ä³»¯Ñ§Ñо¿ÐÔѧϰС×éΪ̽¾¿Ä³Æ·ÅÆ»¨ÉúÓÍÖв»±¥ºÍÖ¬·¾ËáµÄº¬Á¿£¬½øÐÐÁËÈçÏÂʵÑ飺

²½Öè¢ñ£º³ÆÈ¡0.4g»¨ÉúÓÍÑùÆ·£¬ÖÃÓÚÁ½¸ö¸ÉÔïµÄµâÆ¿£¨Èçͼ£©ÄÚ£¬¼ÓÈë10mLËÄÂÈ»¯Ì¼£¬ÇáÇáÒ¡¶¯Ê¹ÓÍÈ«²¿Èܽ⡣ÏòµâÆ¿ÖмÓÈë25.00mLº¬0.01mol IBrµÄÎÞË®ÒÒËáÈÜÒº£¬¸ÇºÃÆ¿Èû£¬ÔÚ²£Á§ÈûÓëÆ¿¿ÚÖ®¼äµÎ¼ÓÊýµÎ10%µâ»¯¼ØÈÜÒº·â±Õ·ì϶£¬ÒÔÃâIBrµÄ»Ó·¢Ëðʧ¡£
²½Öè¢ò£ºÔÚ°µ´¦·ÅÖÃ30min£¬²¢²»Ê±ÇáÇáÒ¡¶¯¡£30minºó£¬Ð¡Ðĵشò¿ª²£Á§Èû£¬ÓÃÐÂÅäÖƵÄ10%µâ»¯¼Ø10mLºÍÕôÁóË®50mL°Ñ²£Á§ÈûºÍÆ¿¾±ÉϵÄÒºÌå³åÏ´ÈëÆ¿ÄÚ¡£
²½Öè¢ó£º¼ÓÈëָʾ¼Á£¬ÓÃ0.1mol¡¤L£­1Áò´úÁòËáÄÆÈÜÒºµÎ¶¨£¬ÓÃÁ¦Õñµ´µâÆ¿£¬Ö±ÖÁÖյ㡣
²â¶¨¹ý³ÌÖз¢ÉúµÄÏà¹Ø·´Ó¦ÈçÏ£º
¢Ù£«IBr ¡ú
¢ÚIBr£«KI£½I2£«KBr
¢ÛI2£«2S2O32£­£½2I£­£«S4O62£­
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÅÒÑ֪±ËØ»¥»¯ÎïIBrµÄÐÔÖÊÓë±Ëص¥ÖÊÀàËÆ£¬ÊµÑéÖÐ׼ȷÁ¿È¡IBrÈÜÒºÓ¦Óà ¡ø  £¬Ó÷½³Ìʽ±íʾµâÆ¿±ØÐë¸ÉÔïµÄÔ­Òò  ¡ø  ¡£
¢Æ²½Öè¢òÖеâÆ¿ÔÚ°µ´¦·ÅÖÃ30min£¬²¢²»Ê±ÇáÇáÒ¡¶¯µÄÔ­ÒòÊÇ  ¡ø  ¡£
¢Ç²½Öè¢óÖÐËù¼Óָʾ¼ÁΪ  ¡ø  £¬µÎ¶¨ÖÕµãµÄÏÖÏó  ¡ø  ¡£
¢È·´Ó¦½áÊøºó´ÓÒºÌå»ìºÏÎïÖлØÊÕËÄÂÈ»¯Ì¼£¬ÔòËùÐè²Ù×÷ÓР ¡ø  ¡£

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

¾«Ó¢¼Ò½ÌÍø