ÌâÄ¿ÄÚÈÝ

»¯Ñ§·´Ó¦ÔÚ¹¤Å©ÒµÉú²úÖÐÓÐ×ÅÖØÒªµÄÓ¦Óᣰ´ÒªÇó»Ø´ðÏÂÁÐÎÊÌ⣺
(1)ÀûÓû¯Ñ§·´Ó¦¿ÉÒÔÖƱ¸Ðí¶àÎïÖÊ¡£
¢ÙʵÑéÊÒÓÃÍ­ÖƱ¸NOµÄÀë×Ó·½³ÌʽΪ___________________¡£
¢Ú¿ÉÓÃAlºÍFe2O3ÖÆFe£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ___________________¡£
¢Ûº£Ë®Ìáäå¹ý³ÌÖУ¬ÏòŨËõµÄº£Ë®ÖÐͨÈëÂÈÆø£¬ÔÙÓÃÈÈ¿ÕÆø´µ³öÉú³ÉµÄä壬ȻºóÓÃ̼ËáÄÆÈÜÒºÎüÊÕä壬äåÆ绯ΪBr£­ºÍBrO3£­¡£ÕâÁ½²½·´Ó¦µÄÀë×Ó·½³Ìʽ·Ö±ðΪ__________¡¢___________¡£
(2)»¯Ñ§·´Ó¦ÄÜΪÈËÀàÉú²úÉú»îÌṩÄÜÔ´¡£
¢ÙÓÉ·´Ó¦CH4 +2O2CO2 +2H2O£¬¿ÉÒÔÉè¼Æ³öÒÔNaOHÈÜҺΪµç½âÖÊÈÜÒºµÄȼÁϵç³Ø£¬¸Ãµç³Ø¹¤×÷ʱ¸º¼«µÄµç¼«·´Ó¦Ê½Îª£º______________¡£
¢Ú2011Äêɽ¶«¸ß¿¼»¯Ñ§ÖÐÔøÌáµ½ÄÆÁò¸ßÄܵç³Ø£¬ÓÒͼÊǸõç³ØµÄ½á¹¹Ê¾Òâͼ£¬¸Ãµç³ØµÄ¹¤×÷ζÈΪ320¡æ×óÓÒ£¬µç³Ø·´Ó¦Îª2Na+xS£½Na2Sx¡£¸Ãµç³Ø¸º¼«Îª________(Ìѧʽ)£¬Õý¼«µÄµç¼«·´Ó¦Ê½Îª                 ¡£Óøõç³Ø×÷µçÔ´½øÐдÖÍ­¾«Á¶Ê±£¬µ±µÃµ½64g¾«Í­Ê±£¬ÀíÂÛÉϸõç³Ø¸º¼«ÏûºÄµÄÖÊÁ¿Îª_____g¡£

(1)¢Ù3Cu+8H++2NO3£­£½3Cu2++2NO¡ü+4H2O
¢Ú2Al+Fe2O32Fe+Al2O3£»(ȱÉÙÌõ¼þ¼õ1·Ö)
¢Û2Br£­+Cl2£½2Cl£­+Br2£¬3Br2+3CO32£­£½5Br£­+BrO3£­+3CO2¡ü
(2)¢ÙCH4+10OH£­£­8e£­£½CO32£­+7H2O£»¢ÚNa£¬xS+2e£­£½Sx2£­£¬46

½âÎöÊÔÌâ·ÖÎö£º(1)¢ÙÏõËá¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬ÄܺͽðÊôÍ­·¢ÉúÑõ»¯»¹Ô­·´Ó¦£¬Òò´Ë¿ÉÓÃÓÚÏ¡ÏõËáÓëÍ­·´Ó¦ÖƱ¸NO£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ3Cu+8H++2NO3£­£½3Cu2++2NO¡ü+4H2O¡£
¢ÚÂÁÊÇ»îÆõĽðÊô£¬¿ÉÒÔͨ¹ýÂÁÈÈ·´Ó¦Ò±Á¶½ðÊô£¬ËùÒÔ¿ÉÓÃAlºÍFe2O3ÖÆFe£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ2Al+Fe2O32Fe+Al2O3¡£
¢ÛÂÈÆøÑõ»¯ÐÔÇ¿ÓÚµ¥ÖÊäåµÄ£¬ËùÒÔÂÈÆøÄÜ°ÑäåÀë×ÓÑõ»¯Éú³Éµ¥ÖÊä壬·´Ó¦µÄÀë×Ó·½³ÌʽΪ2Br£­+Cl2£½2Cl£­+Br2¡£µ¥ÖÊäåÈÜÓÚË®ÏÔËáÐÔ£¬ÄܺÍ̼ËáÄÆÈÜÒº·¢ÉúÆ绯·´Ó¦Éú³ÉNaBr¡¢NaBrO3ºÍË®£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ3Br2+3CO32£­£½5Br£­+BrO3£­+3CO2¡ü¡£
£¨2£©¢ÙÔ­µç³ØÖиº¼«Ê§È¥µç×Ó£¬·¢ÉúÑõ»¯·´Ó¦¡£Õý¼«µÃµ½µç×Ó·¢Éú»¹Ô­·´Ó¦£¬Ôò¸ù¾Ý·´Ó¦CH4 +2O2CO2 +2H2O¿ÉÖª£¬»¹Ô­¼ÁÊǼ×Í飬ÑõÆøÊÇÑõ»¯¼Á£¬ËùÒÔÈç¹ûÉè¼Æ³ÉÔ­µç³Ø£¬ÔòÑõÆøÔÚÕý¼«Í¨È룬¼×ÍéÔÚ¸º¼«Í¨Èë¡£ÓÉÓÚµç½âÖÊÈÜÒºÊÇÇâÑõ»¯ÄÆÈÜÒº£¬ËùÒÔ¸º¼«µç¼«·´Ó¦Ê½ÎªCH4+10OH£­£­8e£­£½CO32£­+7H2O¡£
¢ÚÔ­µç³ØÖиº¼«Ê§È¥µç×Ó£¬·¢ÉúÑõ»¯·´Ó¦¡£Õý¼«µÃµ½µç×Ó·¢Éú»¹Ô­·´Ó¦£¬Ôò¸ù¾Ýµç³Ø·´Ó¦Îª2Na+xS£½Na2Sx¿ÉÖª£¬ÄÆÊÇ»¹Ô­¼Á£¬ËùÒԸõç³Ø¸º¼«ÎªNa¡£Õý¼«ÊÇSµÃµ½µç×Ó£¬ÔòÕý¼«µç¼«·´Ó¦Ê½ÎªxS+2e£­£½Sx2£­¡£64gÍ­µÄÎïÖʵÄÁ¿£½64g¡Â64g/mol£½1mol£¬¸ù¾Ýµç¼«·´Ó¦Ê½Cu2£«£«2e£­£½Cu¿ÉÖª£¬×ªÒÆ2molµç×Ó£¬ËùÒÔ¸ù¾Ýµç×ÓÊغã¿ÉÖª£¬¸º¼«ÏûºÄ½ðÊôÄƵÄÎïÖʵÄÁ¿ÊÇ2mol£¬ÖÊÁ¿£½2mol¡Á23g/mol£½46g¡£
¿¼µã£º¿¼²éÎïÖʵÄÐÔÖÊ¡¢·½³ÌʽµÄÊéд£»µç»¯Ñ§Ô­ÀíµÄÓйØÅжϡ¢Ó¦ÓÃÓë¼ÆËã

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

A¡¢B¡¢W¡¢D¡¢EΪ¶ÌÖÜÆÚÔªËØ£¬ÇÒÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£¬ÖÊ×ÓÊýÖ®ºÍΪ39£¬
B¡¢WͬÖÜÆÚ£¬A¡¢DͬÖ÷×壬A¡¢WÄÜÐγÉÁ½ÖÖҺ̬»¯ºÏÎïA2WºÍA2W2£¬EÔªËصÄÖÜÆÚÐòÊýÓëÖ÷×åÐòÊýÏàµÈ¡£
ÊԻشð£º
£¨1£©EÔªËØÔÚÖÜÆÚ±íÖеÄλÖÃΪ      £¬Ð´³öEµÄ×î¸ß¼ÛÑõ»¯ÎïÓëDµÄ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯Îï·´Ó¦µÄÀë×Ó·½³Ìʽ                              ¡£
£¨2£© ÓÉA¡¢B¡¢WÈýÖÖÔªËØ×é³ÉµÄ18µç×Ó΢Á£µÄ½á¹¹¼òʽΪ             ¡£
£¨3£©¾­²â¶¨A2W2Ϊ¶þÔªÈõËᣬÆäËáÐÔ±È̼ËáµÄ»¹ÒªÈõ£¬Çëд³öÆäµÚÒ»²½µçÀëµÄµçÀë·½³Ìʽ                   ¡£³£ÓÃÁòËá´¦ÀíBaO2À´ÖƱ¸A2W2£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ               ¡£
£¨4£©·ÏÓ¡Ë¢µç·°åÉϺ¬ÓÐÍ­£¬ÒÔÍùµÄ»ØÊÕ·½·¨Êǽ«Æä×ÆÉÕʹͭת»¯ÎªÑõ»¯Í­£¬ÔÙÓÃÁòËáÈܽ⡣ÏÖ¸ÄÓÃA2W2ºÍÏ¡ÁòËá½þÅÝ·ÏÓ¡Ë¢µç·°å¼È´ïµ½ÉÏÊöÄ¿µÄ£¬ÓÖ±£»¤ÁË»·¾³£¬ÊÔд³ö·´Ó¦µÄÀë×Ó·½³Ìʽ             ¡£
£¨5£© ÔªËØDµÄµ¥ÖÊÔÚÒ»¶¨Ìõ¼þÏ£¬ÄÜÓëAµ¥ÖÊ»¯ºÏÉú³ÉÒ»ÖÖÇ⻯ÎïDA£¬ÈÛµãΪ800¡æ¡£DAÄÜÓëË®·´Ó¦·ÅÇâÆø£¬Èô½«1mol DAºÍ1 mol Eµ¥ÖÊ»ìºÏ¼ÓÈë×ãÁ¿µÄË®£¬³ä·Ö·´Ó¦ºóÉú³ÉÆøÌåµÄÌå»ýÊÇ           L£¨±ê×¼×´¿öÏ£©¡£

³£¼ûÔªËØA¡¢B¡¢M×é³ÉµÄËÄÖÖÎïÖÊ·¢Éú·´Ó¦:¼×+ÒÒ=±û+¶¡,ÆäÖм×ÓÉAºÍM×é³É,ÒÒÓÉBºÍM×é³É,±ûÖ»º¬ÓÐM¡£
£¨1£©Èô¼×Ϊµ­»ÆÉ«¹ÌÌ壬ÒҺͱû¾ùΪ³£ÎÂϵÄÎÞÉ«ÎÞζÆøÌå¡£ÔòÒҵĵç×ÓʽΪ         ;Éú³É±ê×¼×´¿öÏÂ5.6L±ûתÒƵĵç×ÓÊýΪ         ;³£ÎÂ϶¡ÈÜÒºpH     7,ÓÃÀë×Ó·½³Ìʽ½âÊÍ                                                                  ¡£
£¨2£©Èô¶¡ÎªÄÜʹƷºìÍÊÉ«µÄÎÞÉ«ÆøÌ壬±ûΪ³£¼ûºìÉ«½ðÊô£¬»¯ºÏÎï¼×¡¢ÒÒÖÐÔ­×Ó¸öÊý±È¾ùΪ1:2(M¾ùÏÔ+1¼Û),Ô­×ÓÐòÊýB´óÓÚA¡£Ôò¢ÙAÔÚÖÜÆÚ±íÖÐλÖÃΪ           ¢Ú¶¡ÓëË®·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                 Ïò·´Ó¦ºóÈÜÒºÖеμÓÁ½µÎ×ÏɫʯÈïÊÔÒºµÄÏÖÏóΪ                                 
¢ÛÕýÈ·ÊéдÉÏÊöÉú³É±ûµÄ»¯Ñ§·½³Ìʽ                                           
¢ÜÏòMCl2µÄÈÜÒºÖÐͨÈ붡,¿É¹Û²ìµ½°×É«µÄMCl³Áµí,д³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ                                                      ¡£

ÏÖÓÐÎåÖÖ¿ÉÈÜÐÔÎïÖÊA¡¢B¡¢C¡¢D¡¢E£¬ËüÃÇËùº¬µÄÒõ¡¢ÑôÀë×Ó»¥²»Ïàͬ£¬·Ö±ðº¬ÓÐÎåÖÖÑôÀë×ÓNa+¡¢Al3+¡¢Mg2+¡¢Ba2+¡¢Fe3+ºÍÎåÖÖÒõÀë×ÓCl?¡¢OH?¡¢NO3?¡¢CO¡¢XÖеÄÒ»ÖÖ¡£
£¨1£©Ä³Í¬Ñ§Í¨¹ý±È½Ï·ÖÎö£¬ÈÏΪÎÞÐè¼ìÑé¾Í¿ÉÅжÏÆäÖбØÓеÄÁ½ÖÖÎïÖÊÊÇ          
£¨Ìѧʽ£©£»
£¨2£©ÎªÁËÈ·¶¨X£¬ÏÖ½«£¨1£©ÖеÄÁ½ÖÖÎïÖʼÇΪAºÍB£¬µ±CÓëBµÄÈÜÒº»ìºÏʱ£¬²úÉúºìºÖÉ«³ÁµíºÍÎÞÉ«ÎÞζÆøÌ壻µ±CÓëAµÄÈÜÒº»ìºÏʱ²úÉú×Ø»ÒÉ«³Áµí£¬Ïò¸Ã³ÁµíÖеÎÈëÏ¡HNO3£¬³Áµí²¿·ÖÈܽ⣬×îºóÁôÓа×É«³Áµí²»ÔÙÈܽ⡣Ôò£º
¢ÙXΪ           £»
A£®SO       B£®SO      C£®CH3COO     D£®SiO
¢ÚAÖеĻ¯Ñ§¼üÀàÐÍΪ                £»
¢Û½«0.02 molµÄAÓë0.01molµÄCͬʱÈܽâÔÚ×ãÁ¿µÄÕôÁóË®ÖУ¬³ä·Ö·´Ó¦ºó£¬×îÖÕËùµÃ³ÁµíµÄÖÊÁ¿Îª             £¨¾«È·µ½0.1g)£»
¢ÜÀûÓÃÉÏÊöÒѾ­È·¶¨µÄÎïÖÊ£¬¿ÉÒÔ¼ìÑé³öD¡¢EÖеÄÑôÀë×Ó¡£Çë¼òÊöʵÑé²Ù×÷²½Öè¡¢ÏÖÏó¼°½áÂÛ                                                          £»
£¨3£©½«CuͶÈëµ½×°ÓÐDÈÜÒºµÄÊÔ¹ÜÖУ¬Cu²»Èܽ⣻ÔٵμÓÏ¡H2SO4£¬CuÖð½¥Èܽ⣬¹Ü¿Ú¸½½üÓкì×ØÉ«ÆøÌå³öÏÖ£¬Ôò¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º                     ¡£

ÅçȪÊÇÒ»ÖÖ³£¼ûµÄÏÖÏó£¬Æä²úÉúÔ­ÒòÊÇ´æÔÚѹǿ²î¡£

£¨1£©Í¼¢ñΪ»¯Ñ§½ÌѧÖг£ÓõÄÅçȪʵÑé×°Öá£ÔÚÉÕÆ¿ÖгäÂú¸ÉÔïÆøÌ壬½ºÍ·µÎ¹Ü¼°ÉÕ±­ÖзֱðÊ¢ÓÐÒºÌå¡£ÏÂÁÐ×éºÏÖÐÄÜÐγÉÅçȪµÄÊÇ        
A. SO2ºÍH2O                   B. CO2ºÍNaOHÈÜÒº   
C. Cl2ºÍH2O                    D. NOºÍH2O
£¨2£©ÔÚͼ¢òµÄ׶ÐÎÆ¿ÖУ¬·Ö±ð¼ÓÈë×ãÁ¿µÄÏÂÁÐÎïÖʺó£¬ÄܲúÉúÅçȪµÄÊÇ    
A. CuÓëÏ¡ÁòËá     B. CuÓëÏ¡ÏõËá      C. AlÓëŨÏõËá     D. FeÓëŨÏõËá
£¨3£©±È½Ïͼ¢ñºÍͼ¢òÁ½Ì××°Ö㬴ӲúÉúÅçȪµÄÔ­ÀíÀ´·ÖÎö£¬Í¼¢ñÊÇ     Éϲ¿ÉÕÆ¿ÄÚÆøÌåѹǿ£»Í¼¢òÊÇ      Ï²¿×¶ÐÎÆ¿ÄÚÆøÌåѹǿ£¨¾ùÌî¡°Ôö´ó¡±»ò¡°¼õС¡±£©¡£
£¨4£©Ä³Ñ§Éú»ý¼«Ë¼¿¼ÅçȪԭÀíµÄÓ¦Óã¬Éè¼ÆÁËÈçͼ¢óËùʾµÄ×°Öá£
¢Ù Èç¹û¹Ø±Õ»îÈûc£¬´ò¿ª»îÈûa¡¢b£¬ÔÙ¼·Ñ¹½ºÍ·µÎ¹Ü¡£Ôò¿ÉÄܳöÏÖµÄÏÖÏóΪ                                                                    ¡£
¢ÚÔÚ¢Ù²Ù×÷µÄ»ù´¡ÉÏ£¬ÈôÒªÔÚ¸Ã×°ÖÃÖвúÉúË«ÅçȪÏÖÏó£¬Æä²Ù×÷·½·¨ÊÇ       ¡£

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

¾«Ó¢¼Ò½ÌÍø