(14·Ö) (1) 2009Ä꣬³¤´ºÓ¦Óû¯Ñ§Ñо¿ËùÔÚ¼×´¼È¼Áϵç³Ø¼¼ÊõÉÏ»ñµÃÐÂÍ»ÆÆ£¬Ô­ÀíÈçÏÂͼËùʾ¡£
¢ÙÇëд³ö´ÓC¿ÚͨÈëO2·¢ÉúµÄµç¼«·´Ó¦Ê½___________________¡£
¢ÚÒÔʯīµç¼«µç½â±¥ºÍʳÑÎË®£¬µç½â¿ªÊ¼ºóÔÚ______________µÄÖÜΧ£¨Ìî¡°Òõ¼«¡±»ò¡°Ñô¼«¡±£©ÏȳöÏÖºìÉ«¡£¼ÙÉèµç³ØµÄÀíÂÛЧÂÊΪ80%£¨µç³ØµÄÀíÂÛЧÂÊÊÇÖ¸µç³Ø²úÉúµÄ×î´óµçÄÜÓëµç³Ø·´Ó¦ËùÊͷŵÄÈ«²¿ÄÜÁ¿Ö®±È£©£¬ÈôÏûºÄ6.4g¼×´¼ÆøÌ壬Íâµç·ͨ¹ýµÄµç×Ó¸öÊýΪ__________________(±£ÁôÁ½Î»ÓÐЧÊý×Ö)¡£

(2)¹¤Òµ·ÏË®Öг£º¬ÓÐCu2+µÈÖؽðÊôÀë×Ó£¬Ö±½ÓÅÅ·Å»áÔì³ÉÎÛȾ£¬Ä¿Ç°ÔÚ¹¤Òµ·ÏË®´¦Àí¹ý³ÌÖУ¬ÒÀ¾Ý³Áµíת»¯µÄÔ­Àí£¬³£ÓÃFeSµÈÄÑÈÜÎïÖÊ×÷Ϊ³Áµí¼Á³ýÈ¥ÕâЩÀë×Ó¡£ÒÑÖªÊÒÎÂÏÂKsp(FeS)=6.3¡Á10-18mol2?L-2£¬Ksp(CuS)=1.3¡Á10-36mol2?L-2¡£
ÇëÓÃÀë×Ó·½³Ìʽ˵Ã÷ÉÏÊö³ýÔÓµÄÔ­Àí___________________________________________¡£
(3)¹¤ÒµÉÏΪÁË´¦Àíº¬ÓÐCr2O72£­µÄËáÐÔ¹¤Òµ·ÏË®£¬ÓÃÂÌ·¯£¨FeSO4¡¤7H2O£©°Ñ·ÏË®ÖеÄÁù¼Û¸õÀë×Ó»¹Ô­³ÉÈý¼Û¸õÀë×Ó£¬ÔÙ¼ÓÈë¹ýÁ¿µÄʯ»ÒË®£¬Ê¹¸õÀë×Óת±äΪCr(OH)3³Áµí¡£
¢ÙÑõ»¯»¹Ô­¹ý³ÌµÄÀë×Ó·½³ÌʽΪ______________________________________________¡£
¢Ú³£ÎÂÏ£¬Cr(OH)3µÄÈܶȻýKsp =1¡Á10¡ª32 mol4?L-4£¬ÈÜÒºµÄpHÖÁÉÙΪ____£¬²ÅÄÜʹCr3+³ÁµíÍêÈ«¡£
¢ÛÏÖÓÃÉÏÊö·½·¨´¦Àí100m3º¬¸õ£¨£«6¼Û£©78mg?L¡ª1µÄ·ÏË®£¬ÐèÓÃÂÌ·¯µÄÖÊÁ¿Îª        kg¡££¨±£ÁôÖ÷Òª¼ÆËã¹ý³Ì£©

(14·Ö) (1) ¢ÙO2+4e¡ª+4H+=2H2O(2·Ö)
¢ÚÒõ¼«£¨1·Ö£©   5.8¡Á1023£¨2·Ö£©
(2) FeS(s)+ Cu2+(aq)= CuS(s)+Fe2+(aq) £¨2·Ö£©
(3)¢ÙCr2O72¡ª£«6Fe2+£«14H+£½2Cr3+£«6Fe3+£«7H2O£¨2·Ö£©
¢Ú5 £¨2·Ö£©
¢ÛCr3+      ¡«  3 FeSO4¡¤7H2O    (3·Ö)

n(FeSO4¡¤7H2O)=450mol     m(FeSO4¡¤7H2O)= 450mol¡Á278g?mol¡ª1=125.1kg

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨14·Ö)£¨1£©ÔÚ2LµÄÃܱÕÈÝÆ÷ÖзÅÈë4molN2O5£¬·¢ÉúÈçÏ·´Ó¦£º2N2O5(g)  4NO2(g)+ O2(g)¡£·´Ó¦5minºó£¬²âµÃN2O5ת»¯ÁË20%£¬Ôò£º¦Ô£¨NO2)Ϊ                        ¡¢5minʱ£¬N2O5Õ¼»ìºÏÆøÌåÌå»ý·ÖÊýÊÇ         ¡£

ÔĶÁ×ÊÁÏ£¬»Ø´ð(2)¡¢(3)СÌâ

пͭԭµç³ØÓû­Í¼µÄ·½Ê½£¨Èçͼ£©±íʾºÜ²»·½±ã£¬³£³£²ÉÓõç³Øͼʽ±í´ïʽ£¬ÈçZn|ZnSO4(1mol/L)||CuSO4£¨1mol/L£©|Cu   ÉÏʽÖУ¬·¢ÉúÑõ»¯·´Ó¦µÄ¸º¼«Ð´ÔÚ×ó±ß£¬·¢Éú»¹Ô­·´Ó¦µÄÕý¼«Ð´ÔÚÓұߡ£ÓÃʵ´¹Ïß¡°|¡±±íʾµç¼«ÓëÈÜÒºÖ®¼äµÄ½çÃ棬ÓÃ˫ʵ´¹Ïß¡°||¡±±íʾÑÎÇÅ¡£

£¨2£©ÉÏÊö×ÊÁÏÁоٵĵç³ØÖУ¬Ð¿Æ¬ÉÏ·¢ÉúµÄµç¼«·´Ó¦Ê½ÊÇ                         £¬

£¨3£©ÏÖÓÐÒ»µç³Ø£¬Æäͼʽ±í´ïʽΪCu|CuSO4(1mol/L)||Fe2(SO4)3£¨0.5mol/L£©|C¡£¸Ãµç³ØÖУ¬Õý¼«µÄµç¼«·´Ó¦Ê½ÊÇ                £¬¸º¼«µÄµç¼«·´Ó¦Ê½ÊÇ                ¡£

£¨4£©Ð´³öÖ§Á´Ö»ÓÐÒ»¸öÒÒ»ùÇÒʽÁ¿×îСµÄÍéÌþµÄ½á¹¹¼òʽ                       

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010Äê¹óÖÝÊ¡×ñÒåËÄÖиßÒ»ÉÏѧÆÚÆÚÖп¼ÊÔ»¯Ñ§¾í ÌâÐÍ£ºÌî¿ÕÌâ

(14·Ö)£¨1£©448mLijÆøÌåÔÚ±ê×¼×´¿öϵÄÖÊÁ¿Îª1.28g£¬¸ÃÆøÌåµÄĦ¶ûÖÊÁ¿Ô¼Îª         
£¨2£©12.4gNa2RÖк¬Na+0.4mol£¬ÔòNa2RµÄĦ¶ûÖÊÁ¿Îª      ,RµÄÏà¶ÔÔ­×ÓÖÊÁ¿Îª       
£¨3£©5molCO2µÄÖÊÁ¿ÊÇ_________£»ÔÚ±ê×¼×´¿öÏÂËùÕ¼µÄÌå»ýԼΪ__________£»Ëùº¬µÄ·Ö×ÓÊýĿԼΪ____________________£»Ëùº¬ÑõÔ­×ÓµÄÊýĿԼΪ___________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012-2013ѧÄê¹ã¶«Ê¡Â½·áÊÐíÙʯÖÐѧ¸ßÒ»µÚÒ»´ÎÔ¿¼»¯Ñ§ÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£ºÌî¿ÕÌâ

( 14·Ö)£¨1£©3.6¿ËH2OµÄÎïÖʵÄÁ¿ÊÇ     £¬º¬ÓР     ¸öH2O£¬º¬ÓР      molH¡£
£¨2£©3.01¡Á1023¸öOH¡ªµÄÎïÖʵÄÁ¿Îª        £¬ÖÊÁ¿Îª        £¬º¬ÓÐÖÊ×ÓµÄÎïÖʵÄÁ¿Îª         £¬º¬Óеç×ÓµÄÎïÖʵÄÁ¿Îª         ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012-2013ѧÄ꼪ÁÖÊ¡³¤´ºÊÐʮһ¸ßÖиßÒ»ÉÏѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£ºÌî¿ÕÌâ

(14·Ö)£¨1£©Ïò1 L AlCl3ºÍFeCl3»ìºÏÈÜÒºÖмÓÈ뺬a mol NaOHµÄÈÜҺʱ£¬²úÉúµÄ³ÁµíÁ¿¿É´ï×î´óÖµ£»¼ÌÐø¼ÓÈëNaOHÈÜÒº£¬³Áµí¿ªÊ¼Èܽ⣬µ±Ç°ºó¼ÓÈëµÄNaOH×ÜÁ¿´ïµ½b molʱ£¬³Áµí²»ÔÙ¼õÉÙ£¬ÔòÔ­ÈÜÒºÖÐFe3+µÄÎïÖʵÄÁ¿Å¨¶ÈΪ       £¬AlCl3µÄÎïÖʵÄÁ¿         ¡£
£¨2£©ÒÑÖª£º2Fe3++2I£­¡¡=¡¡2Fe2++ I2£»¡¡¡¡2Fe2++Br=¡¡2Fe3++2Br£­
¢ÙÏòº¬ÓÐ1mol FeI2ºÍ1.5mol FeBr2µÄÈÜÒºÖÐͨÈë2mol Cl2£¬´Ëʱ±»Ñõ»¯µÄÀë×ÓÊÇ¡¡¡¡¡£
¢ÚÈç¹ûÏò¢ÙµÄÈÜÒºÖÐͨÈë3mol Cl2£¬Ôò±»Ñõ»¯µÄÀë×Ó¶ÔÓ¦µÄÑõ»¯²úÎï·Ö±ðÊÇ¡¡¡¡¡¡¡¡ ¡£
(3£©¢ÙÈôm gÌúмÓ뺬ÓÐ n gHNO3µÄÏõËáÈÜҺǡºÃÍêÈ«·´Ó¦£¬Èô m : n =" 1" : 2.7£¬ ¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ __________________________________________________£¨¼ÙÉ軹ԭ²úÎïÖ»ÓÐÒ»ÖÖ£¬ÇÒÖ»Éú³ÉÒ»ÖÖÑΣ©
¢ÚÈôº¬ n g HNO3µÄÏ¡ÏõËáÈÜҺǡºÃʹ5.6gÌú·ÛÍêÈ«Èܽ⣬ÈôÓÐ n/4 gHNO3±»»¹Ô­³ÉNO£¨ÎÞÆäËü»¹Ô­²úÎÔò n µÄ·¶Î§Îª_________________________           
¢ÛijÌõ¼þÏÂпºÍÏõËᷴӦʱµÄÎïÖʵÄÁ¿Ö®±ÈΪ2:5£¬´ËʱÏõËáµÄ»¹Ô­²úÎïÊÇ____________

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄêɽ¶«Ê¡¸ßÈý9ÔÂÖÊÁ¿¼ì²â»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ

(14·Ö)£¨1£©ÒÑÖª:

Fe£¨s£©+1/2O2£¨g£©=FeO£¨s£©  

2Al£¨s£©+3/2O2£¨g£©= Al2O3£¨s£© 

AlºÍFeO·¢ÉúÂÁÈÈ·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ________________________________________¡£

£¨2£©·´Ó¦ÎïÓëÉú³ÉÎï¾ùΪÆø̬µÄij¿ÉÄæ·´Ó¦ÔÚ²»Í¬Ìõ¼þϵķ´Ó¦Àú³Ì·Ö±ðΪA¡¢B£¬ÈçͼËùʾ¡£

¢Ù¾ÝͼÅжϸ÷´Ó¦ÊÇ_____(Ìî¡°Îü¡±»ò¡°·Å¡±)  ÈÈ·´Ó¦£¬µ±·´Ó¦´ïµ½Æ½ºâºó£¬ÆäËûÌõ¼þ²»±ä£¬Éý¸ßζȣ¬·´Ó¦ÎïµÄת»¯ÂÊ__   _ (Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±)

¢ÚÆäÖÐBÀú³Ì±íÃ÷´Ë·´Ó¦²ÉÓõÄÌõ¼þΪ______ (Ñ¡ÌîÏÂÁÐÐòºÅ×Öĸ)

A¡¢Éý¸ßζȠ           B¡¢Ôö´ó·´Ó¦ÎïµÄŨ¶È

C¡¢½µµÍζȠ           D¡¢Ê¹ÓÃÁË´ß»¯¼Á

£¨3£© 1000¡æʱ£¬ÁòËáÄÆÓëÇâÆø·¢ÉúÏÂÁз´Ó¦£ºNa2SO4(s) + 4H2(g)  Na2S(s) + 4H2O(g) ¡£

¢Ù¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽΪ____________________¡£ÒÑÖªK1000¡æ£¼K1200¡æ£¬Ôò¸Ã·´Ó¦ÊÇ________·´Ó¦£¨Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±£©¡£

¢ÚÓÃÓйØÀë×Ó·½³Ìʽ˵Ã÷ÉÏÊö·´Ó¦ËùµÃ¹ÌÌå²úÎïµÄË®ÈÜÒºµÄËá¼îÐÔ_______ _____

                                                                    

£¨4£©³£ÎÂÏ£¬Èç¹ûÈ¡0.1mol¡¤L£­1HAÈÜÒºÓë0.1mol¡¤L£­1NaOHÈÜÒºµÈÌå»ý»ìºÏ£¨»ìºÏºóÈÜÒºÌå»ýµÄ±ä»¯ºöÂÔ²»¼Æ£©£¬²âµÃ»ìºÏÒºµÄpH£½8£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺

¢Ù»ìºÏÈÜÒºÖÐË®µçÀë³öµÄc(H£«)Óë0.1mol¡¤L£­1NaOHÈÜÒºÖÐË®µçÀë³öµÄc(H£«)±È½Ï

                  £¨Ì¡¢£¾¡¢£½£©¡£

¢ÚÒÑÖªNH4AÈÜҺΪÖÐÐÔ£¬ÓÖÖª½«HAÈÜÒº¼Óµ½Na2CO3ÈÜÒºÖÐÓÐÆøÌå·Å³ö£¬ÊÔÍƶÏ(NH4)2CO3ÈÜÒºµÄpH             7£¨Ì¡¢£¾¡¢£½£©£»ÏàͬζÈÏ£¬µÈÎïÖʵÄÁ¿Å¨¶ÈµÄÏÂÁÐËÄÖÖÑÎÈÜÒº°´pHÓÉ´óµ½Ð¡µÄÅÅÁÐ˳ÐòΪ                          ¡££¨ÌîÐòºÅ£©

a.NH4HCO3          b.NH4A          c.(NH4)2CO3         d.NH4Cl

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸