ÌâÄ¿ÄÚÈÝ

¹¤ÒµÉϺϳɰ±ÔÚÒ»¶¨Ìõ¼þϽøÐÐÈçÏ·´Ó¦£ºN2£¨g£©+3H2£¨g£©2NH3£¨g£© ¡÷H=-92£®44 kJ£¯mol¡£Æ䲿·Ö¹¤ÒÕÁ÷³ÌÈçÏÂͼËùʾ£º

£¨1£©ºÏ³É°±ËùÐèÒªµÄÔ­ÁÏÆøÖУ¬µªÆøÈ¡×Ô           £¬ÇâÆøÀ´Ô´ÓÚ             ¡£
£¨2£©¶ÔÔ­ÁÏÆø½øÐо»»¯´¦ÀíµÄÄ¿µÄÊÇ                    ¡£
£¨3£©É豸AµÄÃû³ÆÊÇ          £¬É豸BµÄÃû³ÆÊÇ           ¡£
£¨4£©ÔÚ20¡«50 Mpaʱ£¬¹¤ÒµºÏ³É°±Ñ¡ÔñÔÚ400¡ª500¡æµÄζȽøÐз´Ó¦£¬
Ö÷ÒªÔ­ÒòÊÇ          ¡£
£¨5£©¾Ý¡¶¿Æѧ¡·ÔÓÖ¾±¨µÀ£¬Ï£À°»¯Ñ§¼ÒÔÚ³£Ñ¹Ï½«ÇâÆøºÍÓÃÇâÆøÏ¡Ê͵ĵªÆø·Ö±ðͨÈËÒ»¸ö¼ÓÈȵ½570¡æµÄµç½â³Ø£¨Èçͼ£©ÖУ¬ÇâºÍµªÔڵ缫ÉϺϳÉÁË°±£¬ÇÒת»¯ÂÊ´ïµ½ÁË78£¥¡£Ôò£ºÑô¼«·´Ó¦Îª               £¬Òõ¼«·´Ó¦Îª                      ¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
 ¡¾»¯Ñ§¨D¨DÑ¡ÐÞ»¯Ñ§Óë¼¼Êõ¡¿(15·Ö£©
°±ÔÚ¹úÃñ¾­¼ÃÖÐÕ¼ÓÐÖØÒªµØ룬ÏÂͼÊǺϳɰ±µÄ¼òÒªÁ÷³Ì£º

I.Ô­ÁÏÆøµÄÖƱ¸£º
(1) ºÏ³É°±ËùÐ赪ÆøÀ´×Ô¿ÕÆø£¬·½·¨Ö®Ò»Êǽ«¿ÕÆøÒº»¯ºóÔÙ¼ÓÈȷֹݣ»ÇëÁíÉè¼ÆÒ»ÖÖ´Ó¿ÕÆøÖзÖÀë³öµªÆøµÄ·½·¨£º_________________________________________________________£»
(2) Çëд³ö¹¤ÒµÉÏ»ñµÃÇâÆøµÄÒ»ÖÖ·½·¨£¨Óû¯Ñ§·½³Ìʽ±íʾ£©____________
II.Ô­ÁÏÆøµÄ¾»»¯£º¡¢
Ϊ·ÀÖ¹´ß»¯¼Á¡°Öж¾¡±£¬Ô­ÁÏÆøÔÚ½øÈËѹËõ»ú֮ǰ±ØÐë¾­¹ý¾»»¯¡¢¾«ÖÆ´¦Àí£¬¡°¾«ÖÆ¡±¹ý³Ìͨ³£Êǽ«º¬ÓÐÉÙÁ¿CO¡¢CO2¡¢O2ºÍH2SµÈÔÓÖʵÄÔ­ÁÏÆøÌåͨÈ뺬Óа±Ë®µÄ´×ËáÑÇÍ­¶þ°±£¨)ÈÜÒº£¬ÒÔ»ñµÃ´¿¾»Ô­ÁÏÆø¡£ÆäÖУ¬ÎüÊÕCOµÄ·´Ó¦Îª£º

(3) ΪÌáó{COÎüÊÕÂÊ£¬¿É²ÉÈ¡µÄÓÐЧ´ëÊ©ÊÇ__________________
(4) ³ýÈ¥ÑõÆøʱ£¬ÑõÆø½«Ñõ»¯Îª£¬Ôò·´Ó¦Öл¹Ô­¼ÁÓëÑõ»¯¼ÁµÄÎïÖʵÄÁ¿Ö®±ÈÊÇ____________£»

III.°±µÄºÏ³É£º
(5)¾Ý±¨µÀ£¬¿Æѧ¼Ò²ÉÓøßÖÊ×Óµ¼µçÐÔµÄSCYÌÕ´É(ÄÜ´«µÝH+)Ϊ½éÖÊ£¬ÓÃÎü¸½ÔÚËüÄÚÍâ±íÃæÉϵĽðÊôîٶྦྷ±¡Ä¤×öµç¼«£¬ÊµÏÖÁËó{γ£Ñ¹Ï¸ßת»¯Âʵĵ绯ѧºÏ³É°±¡£ÆäʵÑé×°ÖÃÈçͼ¡£Çëд³öîٵ缫AÉϵĵ缫·´Ó¦Ê½________________________
£¨I£©¶àÏîÑ¡ÔñÌ⣨6·Ö£©
»¯Ñ§Óë¿Æѧ¡¢¼¼Êõ¡¢Éç»á¡¢»·¾³ÃÜÇÐÏà¹Ø£¬ÏÂÁÐ×ö·¨ÖÐÕýÈ·µÄÊÇ__________¡£
A£®ÑÐÖÆÒÒ´¼ÆûÓÍ(ÆûÓÍÖÐÌí¼ÓÒ»¶¨±ÈÀýÒÒ´¼)¼¼Êõ£¬²»ÄܽµµÍ»ú¶¯³µÎ²ÆøÖÐÓк¦ÆøÌåÅÅ·Å
B£®¹¤ÒµÉÏÓÃʯ»ÒÈé¶ÔúȼÉÕºóÐγɵÄÑÌÆø½øÐÐÍÑÁò£¬×îÖÕÄÜÖƵÃʯ¸à
C£®ÎªÁËÓÐЧµÄ·¢Õ¹Çå½àÄÜÔ´£¬²ÉÓõç½âË®µÄ·½·¨´óÁ¿ÖƱ¸H2
D£®ÊÀ²©Í£³µ³¡°²×°´ß»¯¹â½âÉèÊ©£¬¿É½«Æû³µÎ²ÆøÖÐCOºÍNOx·´Ó¦Éú³ÉÎÞ¶¾ÆøÌå
E£®ÀûÓû¯Ñ§·´Ó¦Ô­Àí£¬Éè¼ÆºÍÖÆÔìеÄÒ©Îï
£¨II£©£¨14·Ö£©
ºÆ嫵ĺ£ÑóÊÇÒ»¸ö¾Þ´óµÄ×ÊÔ´±¦¿â£¬Ô̲Ø×Å·áÈĵĿó²ú£¬ÊDZ¦¹óµÄ»¯Ñ§×ÊÔ´£¬ÏÂͼÊǺ£Ë®¼Ó¹¤µÄʾÒâͼ£¬¸ù¾ÝÏÂͼ»Ø´ðÎÊÌâ¡£

£¨1£©º£Ë®µ­»¯¹¤³§Í¨³£²ÉÓõÄÖƱ¸µ­Ë®µÄ·½·¨ÓР                £¨Ð´³öÁ½ÖÖ£©¡£
£¨2£©ÏÂͼÊÇ´ÓŨËõº£Ë®ÖÐÌáÈ¡äåµÄÁ÷³Ìͼ¡£Ð´³öÏÂͼÖТ٢ڵĻ¯Ñ§Ê½£º¢Ù      £¬¢Ú      £¬ÎüÊÕËþÖз¢ÉúµÄ»¯Ñ§·´Ó¦·½³ÌʽΪ                        ¡£

£¨3£©ÖƱ¸½ðÊôþÊÇͨ¹ýµç½âÈÛÈÚµÄMgC12£¬¶ø²»ÓÃMgO£¬ÆäÔ­ÒòÊÇ             ¡£
£¨4£©Ê³ÑÎÒ²ÊÇÒ»ÖÖÖØÒªµÄ»¯¹¤Ô­ÁÏ£¬Âȼҵ¾ÍÊÇͨ¹ýµç½â±¥ºÍʳÑÎË®À´ÖƱ¸NaOH¡¢H2ºÍC12¡£º£Ë®Öеõ½µÄ´ÖÑÎÖÐÍùÍùº¬ÓÐһЩÔÓÖÊ£¬±ØÐë¼ÓÈëһЩ»¯Ñ§ÊÔ¼Á£¬Ê¹ÔÓÖʳÁµí£¬´¦ÀíºóµÄÑÎË®»¹Ðè½øÈëÑôÀë×Ó½»»»Ëþ£¬ÆäÔ­ÒòÊÇ                 ¡£µç½âʳÑÎË®ÔÚÀë×Ó½»»»Ä¤µç½â²ÛÖнøÐУ¬Àë×Ó½»»»Ä¤µÄ×÷ÓÃÊÇ              ¡£
£¨5£©¶à¾§¹èÖ÷Òª²ÉÓÃSiHCl3»¹Ô­¹¤ÒÕÉú²ú£¬Æ丱²úÎïSiCl4¿Éת»¯ÎªSiHCl3¶øÑ­»·Ê¹Óá£Ò»¶¨Ìõ¼þÏ£¬ÔÚ20LºãÈÝÃܱÕÈÝÆ÷Öеķ´Ó¦£º3 SiCl4£¨g£©+2 H2£¨g£©+Si£¨g£© 4 SiHCl3£¨g£©¡£´ïƽºâºó£¬H2ÓëSiHCl3ÎïÖʵÄÁ¿Å¨¶È·Ö±ðΪ0.140mol/LºÍ0.020mol/L£¬ÈôH2È«²¿À´Ô´ÓÚÂȼҵ£¬ÀíÂÛÉÏÐèÏûºÄ´¿NaClµÄÖÊÁ¿Îª       kg¡£
£¨12·Ö£©¸ßÌúËá¼ØÊÇÒ»ÖÖ¸ßЧµÄ¶à¹¦ÄܵÄË®´¦Àí¼Á¡£¹¤ÒµÉϳ£²ÉÓÃNaClOÑõ»¯·¨Éú²ú£¬Ô­ÀíΪ£º
3NaClO + 2Fe(NO3)3 + 10NaOH£½2Na2FeO4¡ý+ 3NaCl + 6NaNO3 + 5H2O
Na2FeO4£«2KOH£½K2FeO4£«2NaOH
Ö÷ÒªµÄÉú²úÁ÷³ÌÈçÏ£º

£¨1£©Ð´³ö·´Ó¦¢ÙµÄÀë×Ó·½³Ìʽ_____________________________¡£
£¨2£©Á÷³ÌͼÖС°×ª»¯¡±ÊÇÔÚijµÍÎÂϽøÐеģ¬ËµÃ÷´ËζÈÏÂKsp(K2FeO4) _________ Ksp(Na2FeO4)£¨Ìî¡°£¾¡±»ò¡°£¼¡±»ò¡°£½¡±£©¡£
£¨3£©·´Ó¦µÄζȡ¢Ô­ÁϵÄŨ¶ÈºÍÅä±È¶Ô¸ßÌúËá¼ØµÄ²úÂʶ¼ÓÐÓ°Ïì¡£
ͼ1Ϊ²»Í¬µÄζÈÏ£¬Fe(NO3)3²»Í¬ÖÊÁ¿Å¨¶È¶ÔK2FeO4Éú³ÉÂʵÄÓ°Ï죻
ͼ2Ϊһ¶¨Î¶ÈÏ£¬Fe(NO3)3ÖÊÁ¿Å¨¶È×î¼Ñʱ£¬NaClOŨ¶È¶ÔK2FeO4Éú³ÉÂʵÄÓ°Ïì¡£

¢Ù¹¤ÒµÉú²úÖÐ×î¼ÑζÈΪ_______¡æ£¬´ËʱFe(NO3)3ÓëNaClOÁ½ÖÖÈÜÒº×î¼ÑÖÊÁ¿Å¨¶ÈÖ®±ÈΪ_______¡£
¢ÚÈôNaClO¼ÓÈë¹ýÁ¿£¬Ñõ»¯¹ý³ÌÖлáÉú³ÉFe(OH)3£¬Ð´³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ£º
____________________________________________________¡£
ÈôFe(NO3)3¼ÓÈë¹ýÁ¿£¬ÔÚ¼îÐÔ½éÖÊÖÐK2FeO4ÓëFe3+·¢ÉúÑõ»¯»¹Ô­·´Ó¦Éú³ÉK3FeO4£¬´Ë·´Ó¦µÄÀë×Ó·½³Ìʽ£º____________________________________¡£
£¨4£©K2FeO4 ÔÚË®ÈÜÒºÖÐÒ×Ë®½â£º4FeO42¡ª+10H2O4Fe(OH)3+8OH¡ª+3O2¡ü¡£ÔÚ¡°Ìá´¿¡±K2FeO4ÖвÉÓÃÖؽᾧ¡¢Ï´µÓ¡¢µÍκæ¸ÉµÄ·½·¨£¬ÔòÏ´µÓ¼Á×îºÃÑ¡ÓÃ_______ÈÜÒº£¨ÌîÐòºÅ£©¡£
A£®H2OB£®CH3COONa¡¢Òì±û´¼C£®NH4Cl¡¢Òì±û´¼D£®Fe(NO3)3¡¢Òì±û´¼

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

¾«Ó¢¼Ò½ÌÍø