ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿¾ÛºÏÁòËáÌúÊÇÒ»ÖÖÐÂÐ͸ßЧµÄÎÞ»ú¸ß·Ö×ÓÐõÄý¼Á¡£ÓÃÁòËáÑÇÌú¾§Ìå¼°ÁòËáΪԭÁÏ´ß»¯Ñõ»¯·¨Éú³ÉÁòËáÌú£¬ÔÙË®½â¡¢¾ÛºÏ³É²úÆ·¡£ÊµÑéÊÒÄ£ÄâÉú²ú¹ý³ÌÈçÏ£º

(1)¢ÙÓÃÔ­ÁÏÅäÖÆ2.50 mol/LµÄFeSO4ÈÜҺʱÓõ½µÄ¶¨Á¿ÒÇÆ÷ÓУº____________________

¢Úд³öÑõ»¯¹ý³ÌÖеÄÀë×Ó·½³Ìʽ£º_________________________________________

(2)×ۺϿ¼ÂÇʵ¼ÊͶÁÏÁòËáÑÇÌúÓëÁòËáµÄÎïÖʵÄÁ¿Ö®±ÈΪ×óÓÒ×î¼Ñ¡£ ¼ÓÈëµÄÁòËá±ÈÀíÂÛÖµÉԶ࣬µ«²»Äܹý¶àµÄÔ­ÒòÊÇ____________________________________________________________________¡£

(3)ÁòËáÌúÈÜҺˮ½â¿ÉÒԵõ½Ò»ÏµÁоßÓо»Ë®×÷ÓõļîʽÁòËáÌú(x Fe2O3¡¤y SO3¡¤z H2O)£¬ÏÖ²ÉÓÃÖØÁ¿·¨²â¶¨x¡¢y¡¢zµÄÖµ¡£

¢Ù²â¶¨Ê±ËùÐèµÄÊÔ¼Á____________¡£

(a) NaOH (b) Ba(OH)2 (c) BaCl2 (d) FeSO4

¢ÚÐèÒª²â¶¨____________ºÍ____________µÄÖÊÁ¿(Ìîд»¯ºÏÎïµÄ»¯Ñ§Ê½)¡£

(4)Ñ¡³ö²â¶¨¹ý³ÌÖÐËùÐèµÄ»ù±¾²Ù×÷____________(°´²Ù×÷ÏȺó˳ÐòÁгö)¡£

a.¹ýÂË¡¢Ï´µÓ

b.Õô·¢¡¢½á¾§

c.ÝÍÈ¡¡¢·ÖÒº

d.ÀäÈ´¡¢³ÆÁ¿

e.ºæ¸É»ò×ÆÉÕ

¡¾´ð°¸¡¿µç×ÓÌìƽ¡¢ÈÝÁ¿Æ¿ 4Fe2+ + O2 + 4H+ 4Fe3+ + 2H2O ÁòËáÉÔ¹ýÁ¿ÊÇΪÁË·ÀÖ¹Fe3+Ë®½â³ÉÇâÑõ»¯Ìú£¬µ±ÁòËá¹ý¶àʱ£¬ºóÐøÖкÍÐèÒªµÄ¼îµÄÓÃÁ¿´ó£¬Ôì³ÉÀË·Ñ ac Fe2O3 BaSO4 aed

¡¾½âÎö¡¿

(1)¢ÙÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÈÜÒºÒª³ÆÁ¿ÈÜÖʵÄÖÊÁ¿£¬ÓÃÈÝÁ¿Æ¿¶¨ÈÝ£»

¢ÚÁòËáÑÇÌúÔÚËáÐÔÌõ¼þϱ»ÑõÆøÑõ»¯³ÉÁòËáÌú£»

(2)ÁòËá¹ý¶à£¬ÔÚºóÃæµ÷½ÚpHʱҪÏûºÄµÄÇâÑõ»¯ÄƵÄÁ¿¾Í¶à£¬Ôì³ÉÀË·Ñ£»

(3)¢Ù²ÉÓÃÖØÁ¿·¨²â¶¨¼îʽÁòËáÌú(x Fe2O3y SO3z H2O)ÖÐx¡¢y¡¢zµÄֵʱ£¬¿ÉÒÔ½«ÑùÆ·ÈÜÓÚÇâÑõ»¯ÄÆÈÜÒº£¬¸ù¾ÝµÃµ½µÄ¹ÌÌåÑõ»¯ÌúµÄÖÊÁ¿È·¶¨xµÄÖµ£¬¹ýÂË£¬ÔÚËùµÃÂËÒºÖмÓÂÈ»¯±µ£¬¸ù¾Ý²úÉúµÄÁòËá±µ³ÁµíµÄÖÊÁ¿¿ÉÈ·¶¨yµÄÖµ£¬¸ù¾ÝÑùÆ·µÄ×ÜÖÊÁ¿½áºÏFe2O3ºÍ¼ÆËãµÃµÄSO3µÄÖÊÁ¿¿ÉÈ·¶¨zµÄÖµ£»

¢Ú¸ù¾Ý¢ÙµÄ·ÖÎö¿ÉÖª£¬Òª²â¶¨Fe2O3¡¢BaSO4µÄÖÊÁ¿£»

(4)²â¶¨¹ý³ÌÖн«ÑùÆ·ÈÜÓÚÇâÑõ»¯ÄÆÈÜÒº£¬¾­¹ý¹ýÂË¡¢Ï´µÓ¡¢ºæ¸É»ò×ÆÉÕ¡¢ÀäÈ´¡¢³ÆÁ¿µÃÑõ»¯ÌúµÄÖÊÁ¿£¬ÔÚËùµÃÂËÒºÖмÓÂÈ»¯±µ£¬¾­¹ý¹ýÂË¡¢Ï´µÓ¡¢ºæ¸É»ò×ÆÉÕ¡¢ÀäÈ´¡¢³ÆÁ¿µÃÁòËá±µµÄÖÊÁ¿£¬¾Ý´Ë´ðÌâ¡£

(1)¢ÙÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÈÜÒºÒª³ÆÁ¿ÈÜÖʵÄÖÊÁ¿£¬ÒªÓõç×ÓÌìƽ£¬ÓÃÈÝÁ¿Æ¿¶¨ÈÝ£¬¹Ê´ð°¸Îª£ºµç×ÓÌìƽ¡¢ÈÝÁ¿Æ¿£»

¢ÚÁòËáÑÇÌúÔÚËáÐÔÌõ¼þϱ»ÑõÆøÑõ»¯³ÉÁòËáÌú£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ4Fe2++O2+4H+=4Fe3++2H2O£¬¹Ê´ð°¸Îª£º4Fe2++O2+4H+=4Fe3++2H2O£»

(2)ÁòËáÉÔ¹ýÁ¿ÊÇΪÁË·ÀÖ¹Fe3+Ë®½â³ÉÇâÑõ»¯Ìú£¬µ±ÁòËá¹ý¶àʱ£¬ºóÐøÖкÍÐèÒªµÄ¼îµÄÓÃÁ¿Á¦´ó£¬Ôì³ÉÀË·Ñ£¬ËùÒÔ¼ÓÈëµÄÁòËá±ÈÀíÂÛÖµÉԶ࣬µ«²»Äܹý¶à£¬¹Ê´ð°¸Îª£ºÁòËáÉÔ¹ýÁ¿ÊÇΪÁË·ÀÖ¹Fe3+Ë®½â³ÉÇâÑõ»¯Ìú£¬µ±ÁòËá¹ý¶àʱ£¬ºóÐøÖкÍÐèÒªµÄ¼îµÄÓÃÁ¿Á¦´ó£¬Ôì³ÉÀË·Ñ£»

(3)¢Ù²ÉÓÃÖØÁ¿·¨²â¶¨¼îʽÁòËáÌú(x Fe2O3y SO3z H2O)ÖÐx¡¢y¡¢zµÄֵʱ£¬¿ÉÒÔ½«ÑùÆ·ÈÜÓÚÇâÑõ»¯ÄÆÈÜÒº£¬¸ù¾ÝµÃµ½µÄ¹ÌÌåÑõ»¯ÌúµÄÖÊÁ¿È·¶¨xµÄÖµ£¬¹ýÂË£¬ÔÚËùµÃÂËÒºÖмÓÂÈ»¯±µ£¬¸ù¾Ý²úÉúµÄÁòËá±µ³ÁµíµÄÖÊÁ¿¿ÉÈ·¶¨yµÄÖµ£¬¸ù¾ÝÑùÆ·µÄ×ÜÖÊÁ¿½áºÏFe2O3ºÍ¼ÆËãµÃµÄSO3µÄÖÊÁ¿¿ÉÈ·¶¨zµÄÖµ£¬¹ÊÑ¡ac£»

¢Ú¸ù¾Ý¢ÙµÄ·ÖÎö¿ÉÖª£¬Òª²â¶¨Fe2O3¡¢BaSO4µÄÖÊÁ¿£¬¹Ê´ð°¸Îª£ºFe2O3¡¢BaSO4£»

(4)²â¶¨¹ý³ÌÖн«ÑùÆ·ÈÜÓÚÇâÑõ»¯ÄÆÈÜÒº£¬¾­¹ý¹ýÂË¡¢Ï´µÓ¡¢ºæ¸É»ò×ÆÉÕ¡¢ÀäÈ´¡¢³ÆÁ¿µÃÑõ»¯ÌúµÄÖÊÁ¿£¬ÔÚËùµÃÂËÒºÖмÓÂÈ»¯±µ£¬¾­¹ý¹ýÂË¡¢Ï´µÓ¡¢ºæ¸É»ò×ÆÉÕ¡¢ÀäÈ´¡¢³ÆÁ¿µÃÁòËá±µµÄÖÊÁ¿£¬¹ÊÑ¡aed¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿È¼ÃºÑÌÆøµÄÍÑÁòÍÑÏõÊÇÄ¿Ç°Ñо¿µÄÈȵ㡣

£¨1£©ÓÃCH4´ß»¯»¹Ô­µªÑõ»¯Îï¿ÉÒÔÏû³ýµªÑõ»¯ÎïµÄÎÛȾ¡£ÒÑÖª£º

¢ÙCH4(g)+4NO2(g)= 4NO(g)+CO2(g)+2H2O(g) ¡÷H= -574 kJmol-1

¢ÚCH4(g)+4NO(g)= 2N2(g)+CO2(g)+2H2O(g) ¡÷H= -1160 kJmol-1

¢ÛH2O(g) = H2O(l) ¡÷H= -44 kJmol-1

д³öCH4(g)ÓëNO2(g)·´Ó¦Éú³ÉN2(g)¡¢CO2(g)ºÍH2O( l ) µÄÈÈ»¯Ñ§·½³Ìʽ_____________¡£

£¨2£©Ä³¿ÆÑÐС×éÑо¿³ôÑõÑõ»¯--¼îÎüÊÕ·¨Í¬Ê±ÍѳýSO2ºÍNO¹¤ÒÕ£¬Ñõ»¯¹ý³Ì·´Ó¦Ô­Àí¼°·´Ó¦ÈÈ¡¢»î»¯ÄÜÊý¾ÝÈçÏ£º

·´Ó¦¢ñ£ºNO(g)+ O3(g) NO2(g)+O2(g) ¡÷H1 = -200.9 kJmol-1 Ea1 = 3.2 kJmol-1

·´Ó¦¢ò£ºSO2(g)+ O3(g) SO3(g)+O2(g) ¡÷H2 = -241.6 kJmol-1 Ea2 = 58 kJmol-1

ÒÑÖª¸ÃÌåϵÖгôÑõ·¢Éú·Ö½â·´Ó¦£º2O3(g) 3O2(g)¡£Çë»Ø´ð£º

ÆäËüÌõ¼þ²»±ä£¬Ã¿´ÎÏòÈÝ»ýΪ2LµÄ·´Ó¦Æ÷ÖгäÈ뺬1.0 mol NO¡¢1.0 mol SO2µÄÄ£ÄâÑÌÆøºÍ2.0 mol O3£¬¸Ä±äζȣ¬·´Ó¦Ïàͬʱ¼ätºóÌåϵÖÐNOºÍSO2µÄת»¯ÂÊÈçͼËùʾ£º

¢ÙÓÉͼ¿ÉÖªÏàͬζÈÏÂNOµÄת»¯ÂÊÔ¶¸ßÓÚSO2£¬½áºÏÌâÖÐÊý¾Ý·ÖÎöÆä¿ÉÄÜÔ­Òò_______¡£

¢ÚÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ________¡£

A£®PµãÒ»¶¨ÎªÆ½ºâ״̬µã

B£®Î¶ȸßÓÚ200¡æºó£¬NOºÍSO2µÄת»¯ÂÊËæζÈÉý¸ßÏÔÖøϽµ¡¢×îºó¼¸ºõΪÁã

C£®ÆäËüÌõ¼þ²»±ä£¬ÈôËõС·´Ó¦Æ÷µÄÈÝ»ý¿ÉÌá¸ßNOºÍSO2µÄת»¯ÂÊ

¢Û¼ÙÉè100¡æʱP¡¢Q¾ùΪƽºâµã£¬´Ëʱ·´Ó¦Ê±¼äΪ10·ÖÖÓ£¬·¢Éú·Ö½â·´Ó¦µÄ³ôÑõÕ¼³äÈë³ôÑõ×ÜÁ¿µÄ10%£¬ÔòÌåϵÖÐÊ£ÓàO3µÄÎïÖʵÄÁ¿ÊÇ________mol£»NOµÄƽ¾ù·´Ó¦ËÙÂÊΪ________£»·´Ó¦¢òÔÚ´ËʱµÄƽºâ³£ÊýΪ_______________¡£

£¨3£©Óõ绯ѧ·¨Ä£Ä⹤ҵ´¦ÀíSO2¡£½«ÁòËṤҵβÆøÖеÄSO2ͨÈëÈçͼװÖ㨵缫¾ùΪ¶èÐÔ²ÄÁÏ£©½øÐÐʵÑ飬¿ÉÓÃÓÚÖƱ¸ÁòËᣬͬʱ»ñµÃµçÄÜ£º

¢ÙM¼«·¢ÉúµÄµç¼«·´Ó¦Ê½Îª___________________¡£

¢Úµ±Íâµç·ͨ¹ý0.2 molµç×Óʱ£¬ÖÊ×Ó½»»»Ä¤×ó²àµÄÈÜÒºÖÊÁ¿_____£¨Ìî¡°Ôö´ó¡±»ò¡°¼õС¡±£©____¿Ë¡£

¡¾ÌâÄ¿¡¿½ðÊô²ÄÁÏ¡¢ÎÞ»ú·Ç½ðÊô²ÄÁÏ¡¢Óлú¸ß·Ö×Ó²ÄÁÏÊÇÈËÀàʹÓõÄÈý´óÀà»ù´¡²ÄÁÏ£¬ËüÃÇÒÔ¸÷×ÔµÄÌصãÂú×ã×ÅÈËÀà¶à·½ÃæµÄÐèÒª¡£

(1)½ðÊô²ÄÁÏÖУ¬ÓÐÒ»ÀàÖüÇâºÏ½ðÄܹ»½áºÏÇâÆøÐγɽðÊô»¯ºÏÎ²¢ÔÚÒ»¶¨Ìõ¼þÏ·ֽâÊͷųöÇâÆø£¬¸ÃÖüÔËÔ­ÀíÊôÓÚ_______±ä»¯£¬¸ÖÌúÊÇÖÆÔìÂÖ´¬µÄÖ÷Òª½ðÊô²ÄÁÏ£¬´¬ÉíÍâͨ³£×°ÉÏÒ»¶¨ÊýÄ¿±ÈÌú¸ü»îÆõĽðÊô¿éÒÔ·ÀÖ¹¸¯Ê´£¬¸Ã½ðÊô¿é¿ÉÒÔÑ¡Ôñ_______£¨Ñ¡Ìî¡°Í­¿é¡±¡¢¡°Ð¿¿é¡±¡¢¡°Ç¦¿é¡±£©£»

(2)ÎÞ»ú·Ç½ðÊô²ÄÁÏÖУ¬ÓÃÓÚµç×Ó¹¤ÒµµÄ¸ß´¿Ì¼Ëá¸Æ¡¢¸ß´¿Ñõ»¯¸ÆÉú²úÁ÷³ÌÈçÏ£º

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

¢ÙÏõËá¸ÆÓë̼Ëáï§ÔÚÈÜÒºÖз´Ó¦£¬Æä»ù±¾·´Ó¦ÀàÐÍΪ_______£»

¢ÚʵÑéÊÒ³£²ÉÓÃ_______²Ù×÷½øÐгÁµí·ÖÀ룻

¢ÛÉú²ú¸ß´¿Ì¼Ëá¸Æʱ£¬Ñ¡Ôñ¡°220¡æºãθÉÔ¶ø²»Ñ¡Ôñ¡°×ÆÉÕ¡±µÄÔ­ÒòÊÇ_______£»

¢Ü¸ß´¿Ñõ»¯¸ÆÉú²ú¹ý³ÌÖУ¬¡°¸ÉÔµÄÖ÷ҪĿµÄÊÇΪÁË·ÀÖ¹______£¨Óû¯Ñ§·½³Ìʽ±íʾ£©£»

(3)Óлú¸ß·Ö×Ó²ÄÁÏ¡°ÓñÃ×ËÜÁÏ¡±£¬ÒòÆä¿É½µ½â±»¹ã·ºÓÃÀ´Ìæ´úÒ»´ÎÐÔÅÝÄ­ËÜÁÏ£¬¡°ÓñÃ×ËÜÁÏ¡±µÄʹÓÿɼõÉÙ_______ÎÛȾ£¬20ÊÀ¼Í30Äê´ú£¬ÄáÁúÒòÆä³é³Éϸ˿¼«Ïñ²ÏË¿¶ø±»ÍÆÏòÊÀ½ç£¬Éú»îÖпɲÉÓÃ______·½·¨Çø·ÖÄáÁúºÍ²ÏË¿¡£

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

¾«Ó¢¼Ò½ÌÍø