ÌâÄ¿ÄÚÈÝ

A¡¢B¡¢C¡¢D¡¢EΪԭ×ÓÐòÊýÒÀ´ÎÔö´óµÄ¶ÌÖÜÆÚÔªËØ,ÒÑÖªA¡¢B¡¢E 3ÖÖÔ­×Ó×îÍâ²ã¹²ÓÐ11¸öµç×Ó,ÇÒÕâ3ÖÖÔªËصÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïÁ½Á½½ÔÄÜ·¢Éú·´Ó¦Éú³ÉÑκÍË®,CÔªËصÄ×îÍâ²ãµç×ÓÊý±È´ÎÍâ²ãµç×ÓÊýÉÙ4,DÔªËØÔ­×Ó´ÎÍâ²ãµç×ÓÊý±È×îÍâ²ãµç×ÓÊý¶à3¡£
(1)д³öÏÂÁÐÔªËØ·ûºÅ:
A¡¡¡¡¡¡¡¡,B¡¡¡¡¡¡¡¡,C¡¡¡¡¡¡¡¡,D¡¡¡¡¡¡¡¡,E¡£ 
(2)д³öA¡¢BÁ½ÔªËØ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïÏ໥·´Ó¦µÄ»¯Ñ§·½³Ìʽ:¡¡________________¡£ 
(3)±È½ÏC¡¢DµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄËáÐÔ:(Óû¯Ñ§Ê½±íʾ)¡¡____________¡£ 

(1)Na¡¡Al¡¡Si¡¡P¡¡Cl(¸÷1·Ö,¹²5·Ö)
(2)Al(OH)3+NaOHNaAlO2+2H2O(2·Ö)
(3)H3PO4>H2SiO3(1·Ö)

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Ñõ»¯ÂÁ(Al2O3) ºÍµª»¯¹è£¨Si3N4£©ÊÇÓÅÁ¼µÄ¸ßνṹÌÕ´É£¬ÔÚ¹¤ÒµÉú²úºÍ¿Æ¼¼ÁìÓòÓÐÖØÒªÓÃ;¡£
£¨1£©AlÓëNaOHÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪ                                    ¡£
£¨2£©ÏÂÁÐʵÑéÄܱȽÏþºÍÂÁµÄ½ðÊôÐÔÇ¿ÈõµÄÊÇ           £¨ÌîÐòºÅ£©¡£
a£®²â¶¨Ã¾ºÍÂÁµÄµ¼µçÐÔÇ¿Èõ
b£®²â¶¨µÈÎïÖʵÄÁ¿Å¨¶ÈµÄAl2(SO4)3ºÍMgSO4ÈÜÒºµÄpH
c£®Ïò0.1 mol/LAlCl3ºÍ0.1 mol/L MgCl2ÖмӹýÁ¿NaOHÈÜÒº
£¨3£©ÂÁÈÈ·¨Êdz£ÓõĽðÊôÒ±Á¶·½·¨Ö®Ò»¡£
ÒÑÖª£º4Al (s)+3O2(g) =2Al2O3(s)   ¦¤H1 =" -3352" kJ/mol
Mn(s)+ O2(g) =MnO2 (s)    ¦¤H2 =" -521" kJ/mol
AlÓëMnO2·´Ó¦Ò±Á¶½ðÊôMnµÄÈÈ»¯Ñ§·½³ÌʽÊÇ                            ¡£
£¨4£©µª»¯¹è¿¹¸¯Ê´ÄÜÁ¦ºÜÇ¿£¬µ«Ò×±»Çâ·úËḯʴ£¬µª»¯¹èÓëÇâ·úËá·´Ó¦Éú³ÉËÄ·ú»¯¹èºÍÒ»ÖÖï§ÑΣ¬Æä·´Ó¦·½³ÌʽΪ                                          ¡£
£¨5£©¹¤ÒµÉÏÓû¯Ñ§ÆøÏà³Á»ý·¨ÖƱ¸µª»¯¹è£¬Æä·´Ó¦ÈçÏ£º3SiCl4(g) + 2N2(g) + 6H2(g)Si3N4(s) + 12HCl(g)  ¡÷H£¼0  
ijζȺÍѹǿÌõ¼þÏ£¬·Ö±ð½«0.3mol SiCl4(g)¡¢0.2mol N2(g)¡¢0.6mol H2(g)³äÈë2LÃܱÕÈÝÆ÷ÄÚ£¬½øÐÐÉÏÊö·´Ó¦£¬5min´ïµ½Æ½ºâ״̬£¬ËùµÃSi3N4(s)µÄÖÊÁ¿ÊÇ5.60g¡£
¢ÙH2µÄƽ¾ù·´Ó¦ËÙÂÊÊÇ         mol£¯(L¡¤min)¡£
¢ÚÈô°´n(SiCl4) : n(N2) : n(H2) =" 3" : 2 : 6µÄͶÁÏÅä±È£¬ÏòÉÏÊöÈÝÆ÷²»¶ÏÀ©´ó¼ÓÁÏ£¬SiCl4(g)µÄת»¯ÂÊÓ¦     £¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±£©¡£
£¨6£©298Kʱ£¬Ksp[Ce(OH)4]£½1¡Á10¡ª29¡£Ce(OH)4µÄÈܶȻý±í´ïʽΪKsp£½              ¡£
ΪÁËʹÈÜÒºÖÐCe4£«³ÁµíÍêÈ«£¬¼´²ÐÁôÔÚÈÜÒºÖеÄc(Ce4+)СÓÚ1¡Á10£­5mol¡¤L£­1£¬Ðèµ÷½ÚpHΪ     ÒÔÉÏ¡£

¶ÌÖÜÆÚÔªËØA¡¢B¡¢C¡¢DÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£¬ÇÒCÔªËØ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÄܵçÀë³öµç×ÓÊýÏàµÈµÄÒõ¡¢ÑôÀë×Ó¡£A¡¢CλÓÚͬһÖ÷×壬AΪ·Ç½ðÊôÔªËØ£¬BµÄ×îÍâ²ãµç×ÓÊýÊÇ´ÎÍâ²ãµç×ÓÊýµÄ3±¶£¬B¡¢CµÄ×îÍâ²ãµç×ÓÊýÖ®ºÍÓëDµÄ×îÍâ²ãµç×ÓÊýÏàµÈ¡£Eµ¥ÖÊÊÇÉú»îÖг£¼û½ðÊô£¬ÆäÖÆÆ·ÔÚ³±Êª¿ÕÆøÖÐÒ×±»¸¯Ê´»òË𻵡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)CµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄ»¯Ñ§Ê½Îª                          £¬
ÆäÖк¬ÓеĻ¯Ñ§¼üÀàÐÍΪ                                              ¡£
(2)ÓÉÉÏÊöA¡¢B¡¢C¡¢DËÄÖÖÔªËØÖеÄÈýÖÖ×é³ÉijÖÖÑΣ¬Ë®ÈÜÒºÏÔ¼îÐÔ£¬ÊǼÒÓÃÏû¶¾¼ÁµÄÖ÷Òª³É·Ö¡£½«¸ÃÑÎÈÜÒºµÎÈëKIµí·ÛÈÜÒºÖУ¬ÈÜÒº±äΪÀ¶É«£¬Ôò·´Ó¦µÄÀë×Ó·½³ÌʽΪ                                                          ¡£
(3)EÔªËØÓëDÔªËØ¿ÉÐγÉED2ºÍED3Á½ÖÖ»¯ºÏÎÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ       (ÌîÐòºÅ)¡£
¢Ù±£´æED2ÈÜҺʱ£¬ÐèÏòÈÜÒºÖмÓÈëÉÙÁ¿Eµ¥ÖÊ
¢ÚED2Ö»ÄÜͨ¹ýÖû»·´Ó¦Éú³É£¬ED3Ö»ÄÜͨ¹ý»¯ºÏ·´Ó¦Éú³É
¢ÛͭƬ¡¢Ì¼°ôºÍED3ÈÜÒº×é³ÉÔ­µç³Ø£¬µç×ÓÓÉͭƬÑص¼ÏßÁ÷Ïò̼°ô
¢ÜÏòµí·Ûµâ»¯¼ØÈÜÒººÍ±½·ÓÈÜÒºÖзֱðµÎ¼Ó¼¸µÎED3µÄŨÈÜÒº£¬Ô­ÎÞÉ«ÈÜÒº¶¼±ä³É×ÏÉ«

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

¾«Ó¢¼Ò½ÌÍø