ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿NH4Al(SO4)2¡¢NH4HSO4ÓÃ;¹ã·º¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©³£ÎÂʱ£¬0.1 mol¡¤L-1 NH4Al(SO4)2ÈÜÒºµÄpH=3¡£ÔòÈÜÒºÖÐc(NH4£«)+c(NH3¡¤H2O)_______c (Al3+) + c£ÛAl(OH)3£Ý£¨Ìî¡°©ƒ¡±¡¢¡°©‚¡±»ò¡°=¡±£©£»2c(SO42£­)- c(NH4£«)-3c(Al3+)=________mol¡¤L-1£¨ÌîÊýÖµ£©¡£

£¨2£©80¡æʱ£¬0.1 mol¡¤L-1 NH4Al(SO4)2ÈÜÒºµÄpHСÓÚ3£¬·ÖÎöµ¼ÖÂpHËæζȱ仯µÄÔ­ÒòÊÇ________________________ £¨ÓÃÀë×Ó·½³Ìʽ²¢½áºÏÎÄ×ÖÐðÊö»Ø´ð£©¡£

£¨3£©³£ÎÂʱ£¬Ïò100 mL 0.1 mol¡¤L-1 NH4HSO4ÈÜÒºÖÐµÎ¼Ó 0.1 mol¡¤L-1 NaOHÈÜÒº£¬µÃµ½ÈÜÒºpHÓëNaOHÈÜÒºÌå»ýµÄ¹ØϵÇúÏßÈçͼËùʾ¡£ÏòNH4HSO4ÈÜÒºÖеμÓNaOHÈÜÒºµ½aµãµÄ¹ý³ÌÖУ¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ__________________¡£

£¨4£©Å¨¶È¾ùΪ0.1 mol¡¤L-1 NH4Al(SO4)2ÈÜÒººÍNH4HSO4ÈÜÒº£¬ÆäÖÐ______ÈÜÒºc(NH4£«)´ó¡£

¡¾´ð°¸¡¿ £½ l.0¡Ál0-3£¨»òl.0¡Ál0-3-l.0¡Ál0-11£© Al3+¡¢NH4£«´æÔÚË®½âƽºâ£ºAl3++3H2OAl(OH)3+3H+¡¢NH4£«+H2O NH3¡¤H2O+H+£¬Éý¸ßζÈÆäË®½â³Ì¶ÈÔö´ó£¬c(H+)Ôö´ó£¬pH¼õС H++OH-=== H2O NH4HSO4

¡¾½âÎö¡¿£¨1£©±¾Ì⿼²éÎïÁÏÊغãºÍµçºÉÊغ㣬¸ù¾ÝÎïÁÏÊغ㣬ÒÔ¼°NH4Al(SO4)2£¬ÍƳöc(NH4£«)£«c(NH3¡¤H2O)=c(Al3£«)£«c[Al(OH)3]£¬¸ù¾ÝµçºÉÊغ㣬c(NH4£«)£«3c(Al3£«)£«c(H£«)=c(OH£­)£«2c((SO42£­)£¬Òò´ËÓÐ2c(SO42£­)£­c(NH4£«)£­3c(Al3£«)=c(H£«)£­c(OH£­)=10£­3£­10£­11£»£¨2£©±¾Ì⿼²éÓ°ÏìÑÎÀàË®½âµÄÒòËØ£¬NH4Al(SO4)2ÈÜÒºÏÔËáÐÔ£¬ÊÇÒòΪ´æÔÚNH4£«£«H2ONH3¡¤H2O£«H£«¡¢Al3£«£«3H2OAl(OH)3£«3H£«£¬ÑÎÀàË®½âÊÇÎüÈÈ·´Ó¦£¬Éý¸ßζȣ¬´Ù½øË®½â£¬c(H£«)Ôö´ó£¬pH¼õС£»£¨3£©±¾Ì⿼²éµÎ¶¨ÊµÑ飬NH4HSO4µÄµçÀë·½³ÌʽΪNH4HSO4=NH4£«£«H£«£«SO42£­£¬H£«½áºÏOH£­ÄÜÁ¦Ç¿ÓÚNH4£«£¬ÔÚaµãʱͨÈëNaOHµÄÌå»ýΪ100mL£¬Ç¡ºÃÖкÍNH4HSO4µçÀë³öH£«£¬Òò´ËÀë×Ó·´Ó¦·½³ÌʽΪH£«£«OH£­=H2O£»£¨4£©±¾Ì⿼²éÓ°ÏìÑÎÀàË®½âµÄÒòËØ£¬Al3£«¡¢H£«¶¼ÒÖÖÆNH4£«µÄË®½â£¬H£«ÒÖÖÆNH4£«Ë®½âÄÜÁ¦Ç¿£¬¼´NH4HSO4ÈÜÒºÖÐc(NH4£«)´ó¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿µªµÄ¹Ì¶¨Ò»Ö±ÊÇ¿Æѧ¼ÒÑо¿µÄÖØÒª¿ÎÌ⣬ºÏ³É°±ÔòÊÇÈ˹¤¹Ìµª±È½Ï³ÉÊìµÄ¼¼Êõ£¬ÆäÔ­ÀíΪ:N2(g)+3H2(g)2NH3(g) ¡÷H

£¨1£©ÒÑÖªÆÆ»µ1mol¹²¼Û¼üÐèÒªµÄÄÜÁ¿Èç±íËùʾ

H-H

N-H

N-N

N¡ÔN

435.5kJ

390.8kJ

163kJ

945.8kJ

Ôò¡÷H=__________¡£

£¨2£©ÔÚºãΡ¢ºãѹÈÝÆ÷ÖУ¬°´Ìå»ý±È1:3¼ÓÈëN2ºÍH2½øÐкϳɰ±·´Ó¦£¬´ïµ½Æ½ºâºó£¬ÔÙÏòÈÝÆ÷ÖгäÈëÊÊÁ¿°±Æø£¬´ïµ½ÐÂƽºâʱ£¬c(H2)½«__________£¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±¡¢»ò¡°²»±ä¡±£¬ºóͬ£©£»ÈôÔÚºãΡ¢ºãÈÝÌõ¼þÏÂc(N2)/c(NH3)½«________¡£

£¨3£©ÔÚ²»Í¬Î¶ȡ¢Ñ¹Ç¿ºÍʹÓÃÏàͬ´ß»¯¼ÁÌõ¼þÏ£¬³õʼʱN2¡¢H2·Ö±ðΪ0.1mol¡¢0.3mol ʱ£¬Æ½ºâ»ìºÏÎïÖа±µÄÌå»ý·ÖÊý£¨¦Õ£©ÈçͼËùʾ¡£

¢ÙÆäÖУ¬p1¡¢p2ºÍp3ÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ_______£¬Ô­ÒòÊÇ___________________¡£

¢ÚÈôÔÚ250¡æ¡¢p1Ìõ¼þÏ£¬·´Ó¦´ïµ½Æ½ºâʱµÄÈÝÆ÷Ìå»ýΪ1L£¬Ôò¸ÃÌõ¼þϺϳɰ±µÄƽºâ³£ÊýK=____(½á¹û±£ÁôÁ½Î»Ð¡Êý)¡£

£¨4£©H2NCOONH4Êǹ¤ÒµÓÉ°±ÆøºÏ³ÉÄòËصÄÖмä²úÎï¡£ÔÚÒ»¶¨Î¶ÈÏ¡¢Ìå»ý²»±äµÄÃܱÕÈÝÆ÷Öз¢Éú·´Ó¦H2NCOONH4(s)2NH3(g)+CO2(g),ÄÜ˵Ã÷¸Ã·´Ó¦´ïµ½Æ½ºâ״̬µÄÊÇ_____£¨ÌîÐòºÅ£©

¢ÙÿÉú³É34g NH3µÄͬʱÏûºÄ44g CO2 ¢Ú»ìºÏÆøÌåµÄÃܶȱ£³Ö²»±ä

¢ÛNH3µÄÌå»ý·ÖÊý±£³Ö²»±ä ¢Ü»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿²»±ä ¢Ýc(NH3):c(CO2)=2:1

£¨5£©¿Æѧ¼Ò·¢ÏÖ£¬N2ºÍH2×é³ÉµÄÔ­µç³ØºÏ³É°±Ó빤ҵºÏ³É°±Ïà±È¾ßÓÐЧÂʸߣ¬Ìõ¼þÒ×´ïµ½µÈÓŵ㡣Æä×°ÖÃÈçͼËùʾ¡¢Ð´³ö¸ÃÔ­µç³ØµÄµç¼«·´Ó¦£º________________¡¢_____________£¬ÈôN2À´×ÔÓÚ¿ÕÆø£¬µ±µç¼«Bµ½A¼äͨ¹ý2molH+ʱÀíÂÛÉÏÐèÒª±ê¿öÏ¿ÕÆøµÄÌå»ýΪ_________£¨½á¹û±£ÁôÁ½Î»Ð¡Êý£©¡£

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

¾«Ó¢¼Ò½ÌÍø