ÌâÄ¿ÄÚÈÝ


A¡¢B¡¢C¡¢D¡¢EÊÇÖÐѧ»¯Ñ§Öг£¼ûµÄÎÞÉ«ÆøÌ壬ËüÃǾùÓɶÌÖÜÆÚÔªËØ×é³É¡£A¡¢B¡¢CÏ໥ת»¯µÄ¹ØÏµÈçͼËùʾ(²¿·Ö²úÎïÒÑÂÔÈ¥)¡£

¢ÙAÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶£»C¡¢DΪ¿ÕÆøÖеÄÖ÷Òª³É·Ö£»B¡¢EÊÇÓж¾ÆøÌ壬ËüÃǵķÖ×Ó¾ùÓÉͬÖÜÆÚÔªËØÔ­×Ó¹¹³É¡£

¢ÚBºÍDÏàÓöÉú³Éºì×ØÉ«ÆøÌå¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)CµÄµç×ÓʽÊÇ________¡£

(2)д³öA¡úB·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º____________________________________¡£

(3)BºÍE·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ__________________________ ___________________________________________________¡£

(4)³£ÎÂÏ£¬°ÑÒ»¶¨Á¿µÄAͨÈëË®ÖУ¬²âµÃÈÜÒºµÄpH£½12£¬Ôò¸ÃÈÜÒºÖÐÓÉË®µçÀë³öµÄOH£­µÄŨ¶ÈΪ________£»¸ÃÈÜÒºÖУ¬Ò»¶¨ÄÜ´óÁ¿¹²´æµÄÀë×Ó×éÊÇ________(ÌîÐòºÅ)¡£

a£®Na£«¡¡NO¡¡Fe2£«¡¡Cl£­

b£®Ag£«¡¡Mg2£«¡¡Ca2£«¡¡NO

c£®Al3£«¡¡K£«¡¡AlO¡¡Cl£­

d£®CO¡¡Na£«¡¡K£«¡¡NO


½âÎö¡¡ÒòAÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶£¬ÔòAΪNH3,4NH3£«5O24NO£«6H2O£¬B¡¢EÊÇÓÉͬÖÜÆÚÔªËØÔ­×Ó¹¹³ÉµÄÓж¾ÆøÌ壬¸ù¾ÝÐÅÏ¢¢Ú¿ÉÖª£¬BΪNO£¬EΪCO£¬ÓÖÒò2NO£«2CON2£«2CO2£¬N2£«O22NO£¬ÔòCΪN2£¬DΪO2¡£(1)CΪN2£¬Æä·Ö×ÓÖдæÔÚN¡ÔN£¬ÓÉ´Ë¿ÉÊéдµç×Óʽ¡£(2)A¡úB¼´ÎªNH3¡úNO£¬ÎªNH3µÄ´ß»¯Ñõ»¯£¬ÐèÒª´ß»¯¼ÁºÍ¼ÓÈÈ¡£(3)BΪNO£¬EΪCO£¬¸ù¾ÝÆä²úÎïΪC(N2)£¬ÔòNO±íÏÖÑõ»¯ÐÔ£¬CO±íÏÖ»¹Ô­ÐÔÉú³ÉCO2¡£(4)NH3ÈÜÓÚË®Éú³ÉNH3¡¤H2O£¬ÆäÈÜÒºÖеÄH£«¾ùÓÉH2OµçÀë²úÉú£¬Ôòc(H£«)£½c(H£«)H2O£½c(OH£­)H2O£½1.0¡Á10£­12 mol¡¤L£­1£¬ÓÉÓÚFe2£«£«2NH3¡¤H2O===Fe(OH)2¡ý£«2NH£¬Ag£«£«NH3¡¤H2O===AgOH¡ý£«NH£¬Al3£«£«3AlO£«6H2O===4Al(OH)3¡ý£¬Al3£«£«3NH3¡¤H2O===Al(OH)3¡ý£«3NH£¬È·¶¨dΪÕýÈ·Ñ¡Ïî¡£

´ð°¸¡¡(1)¡ÃN⋮⋮N¡Ã¡¡(2)4NH3£«5O24NO£«6H2O¡¡(3)2NO£«2CON2£«2CO2¡¡(4)1¡Á10£­12 mol¡¤L£­1¡¡d


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¶þÑõ»¯ÁòÊÇÖØÒªµÄ¹¤ÒµÔ­ÁÏ£¬Ì½¾¿ÆäÖÆ±¸·½·¨ºÍÐÔÖʾßÓзdz£ÖØÒªµÄÒâÒå¡£

(1)¹¤ÒµÉÏÓûÆÌú¿ó(FeS2£¬ÆäÖÐÁòÔªËØÎª£­1¼Û)ÔÚ¸ßÎÂϺÍÑõÆø·´Ó¦ÖƱ¸SO2£º4FeS2£«11O28SO2£«2Fe2O3£¬¸Ã·´Ó¦Öб»Ñõ»¯µÄÔªËØÊÇ________(ÌîÔªËØ·ûºÅ)¡£µ±¸Ã·´Ó¦×ªÒÆ2.75 molµç×Óʱ£¬Éú³ÉµÄ¶þÑõ»¯ÁòÔÚ±ê×¼×´¿öϵÄÌå»ýΪ________ L¡£

(2)ʵÑéÊÒÖÐÓÃÏÂÁÐ×°ÖòⶨSO2´ß»¯Ñõ»¯ÎªSO3µÄת»¯ÂÊ¡£(ÒÑÖªSO3µÄÈÛµãΪ16.8 ¡æ£¬¼ÙÉèÆøÌå½øÈë×°ÖÃʱ·Ö±ð±»ÍêÈ«ÎüÊÕ£¬ÇÒºöÂÔ¿ÕÆøÖÐCO2µÄÓ°Ïì)

¢Ù¼òÊöʹÓ÷ÖҺ©¶·ÏòÔ²µ×ÉÕÆ¿ÖеμÓŨÁòËáµÄ²Ù×÷£º________________________________________¡£

¢ÚʵÑé¹ý³ÌÖУ¬ÐèҪͨÈëÑõÆø¡£ÊÔд³öÒ»¸öÓÃÈçÏÂͼËùʾװÖÃÖÆÈ¡ÑõÆøµÄ»¯Ñ§·½³Ìʽ£º______________________________________¡£

¢Ûµ±Í£Ö¹Í¨ÈëSO2£¬Ï¨Ãð¾Æ¾«µÆºó£¬ÐèÒª¼ÌÐøÍ¨Ò»¶Îʱ¼äµÄÑõÆø£¬ÆäÄ¿µÄÊÇ___________________________________________________________¡£

¢ÜʵÑé½áÊøºó£¬Èô×°ÖÃDÔö¼ÓµÄÖÊÁ¿Îªm g£¬×°ÖÃEÖвúÉú°×É«³ÁµíµÄÖÊÁ¿Îªn g£¬Ôò´ËÌõ¼þ϶þÑõ»¯ÁòµÄת»¯ÂÊÊÇ________(Óú¬×ÖĸµÄ´úÊýʽ±íʾ£¬²»Óû¯¼ò)¡£

(3)ijѧϰС×éÉè¼ÆÓÃÈçͼװÖÃÑéÖ¤¶þÑõ»¯ÁòµÄ»¯Ñ§ÐÔÖÊ¡£

¢ÙÄÜ˵Ã÷¶þÑõ»¯Áò¾ßÓÐÑõ»¯ÐÔµÄʵÑéÏÖÏóΪ_______________________ ________________________________________________________¡£

¢ÚΪÑéÖ¤¶þÑõ»¯ÁòµÄ»¹Ô­ÐÔ£¬³ä·Ö·´Ó¦ºó£¬È¡ÊÔ¹ÜbÖеÄÈÜÒº·Ö³ÉÈý·Ý£¬·Ö±ð½øÐÐÈçÏÂʵÑé¡£

·½°¸¢ñ£ºÏòµÚÒ»·ÝÈÜÒºÖмÓÈëAgNO3ÈÜÒº£¬Óа×É«³ÁµíÉú³É

·½°¸¢ò£ºÏòµÚ¶þ·ÝÈÜÒºÖмÓÈëÆ·ºìÈÜÒº£¬ºìÉ«ÍÊÈ¥

·½°¸¢ó£ºÏòµÚÈý·ÝÈÜÒºÖмÓÈëBaCl2ÈÜÒº£¬²úÉú°×É«³Áµí

ÉÏÊö·½°¸ÖкÏÀíµÄÊÇ________(Ìî¡°¢ñ¡±¡¢¡°¢ò¡±»ò¡°¢ó¡±)£»ÊÔ¹ÜbÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ____________________________________________¡£

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

¾«Ó¢¼Ò½ÌÍø