ÌâÄ¿ÄÚÈÝ
ÒÑ֪ͨ³£×´¿öϼס¢ÒÒ¡¢±û¡¢¶¡µÈΪÆøÌåµ¥ÖÊ£¬A¡¢B¡¢C¡¢D¡¢E¡¢F¡¢G¡¢HµÈΪ»¯ºÏÎÆäÖÐA¡¢B¡¢E¡¢G¾ùΪÆøÌå,CΪ³£¼ûÒºÌå¡£·´Ó¦¢Ù¡¢¢Ú¡¢¢Û¶¼ÊÇÖØÒªµÄ»¯¹¤·´Ó¦£¬·´Ó¦¢ÜÊÇÖØÒªµÄʵÑéÊÒÖÆÈ¡ÆøÌåµÄ·´Ó¦¡£ÓйصÄת»¯¹ØϵÈçÏÂͼËùʾ(·´Ó¦Ìõ¼þ¾ùÒÑÂÔÈ¥£©¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)·´Ó¦¢ÜµÄ»¯Ñ§·½³ÌʽΪ: ________________________¡£
(2)BºÍEÔÚÒ»¶¨Ìõ¼þÏ¿ɷ¢Éú·´Ó¦,ÕâÊÇÒ»¸ö¾ßÓÐʵ¼ÊÒâÒåµÄ·´Ó¦,¿ÉÏû³ýE¶Ô»·¾³µÄÎÛȾ£¬¸Ã·´Ó¦Ñõ»¯²úÎïÓ뻹ԲúÎïµÄÎïÖʵÄÁ¿Ö®±ÈΪ________¡£
(3)0.1mol lL-1AÈÜÒººÍ0.1mol ?L-1BÈÜÒºµÈÌå»ý»ìºÏ£¬ÈÜÒº³Ê________ÐÔ,ÔÒòÊÇ(ÓÃÀë×Ó·½³Ìʽ˵Ã÷) ________________¡£
(4)ÇëÉè¼ÆʵÑé¼ìÑéD¾§ÌåÖеÄÑôÀë×Ó(¼òÊöʵÑé²Ù×÷¡¢ÏÖÏóºÍ½áÂÛ£©£º________________¡£
(5)pHÏàͬµÄA¡¢D¡¢HÈýÖÖÈÜÒº£¬ÓÉË®µçÀë³öµÄc(OH£)µÄ´óС¹ØϵÊÇ(ÓÃA¡¢D¡¢H±íʾ) : ____________¡£
(6)ÏòÒ»¶¨Á¿µÄFe¡¢FeO¡¢Fe3O4µÄ»ìºÏÎïÖУ¬¼ÓÈë1mol? L£1 AµÄÈÜÒº100 mL£¬Ç¡ºÃʹ»ìºÏÎïÈ«²¿Èܽ⣬Çҷųö336mL(±ê×¼×´¿öÏÂ)µÄÆøÌ壬ÏòËùµÃÈÜÒºÖмÓÈëKSCNÈÜÒº£¬ÈÜÒºÎÞºìÉ«³öÏÖ£»ÈôȡͬÖÊÁ¿µÄFe¡¢FeO¡¢Fe3O4»ìºÏÎ¼ÓÈë1 mol ? L£1 HµÄÈÜÒº,ҲǡºÃʹ»ìºÏÎïÈ«²¿Èܽ⣬ÇÒÏòËùµÃÈÜÒºÖмÓÈëKSCNÈÜÒº,ÈÜÒºÒ²ÎÞºìÉ«³öÏÖ,ÔòËù¼ÓÈëµÄHÈÜÒºµÄÌå»ýÊÇ________¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)·´Ó¦¢ÜµÄ»¯Ñ§·½³ÌʽΪ: ________________________¡£
(2)BºÍEÔÚÒ»¶¨Ìõ¼þÏ¿ɷ¢Éú·´Ó¦,ÕâÊÇÒ»¸ö¾ßÓÐʵ¼ÊÒâÒåµÄ·´Ó¦,¿ÉÏû³ýE¶Ô»·¾³µÄÎÛȾ£¬¸Ã·´Ó¦Ñõ»¯²úÎïÓ뻹ԲúÎïµÄÎïÖʵÄÁ¿Ö®±ÈΪ________¡£
(3)0.1mol lL-1AÈÜÒººÍ0.1mol ?L-1BÈÜÒºµÈÌå»ý»ìºÏ£¬ÈÜÒº³Ê________ÐÔ,ÔÒòÊÇ(ÓÃÀë×Ó·½³Ìʽ˵Ã÷) ________________¡£
(4)ÇëÉè¼ÆʵÑé¼ìÑéD¾§ÌåÖеÄÑôÀë×Ó(¼òÊöʵÑé²Ù×÷¡¢ÏÖÏóºÍ½áÂÛ£©£º________________¡£
(5)pHÏàͬµÄA¡¢D¡¢HÈýÖÖÈÜÒº£¬ÓÉË®µçÀë³öµÄc(OH£)µÄ´óС¹ØϵÊÇ(ÓÃA¡¢D¡¢H±íʾ) : ____________¡£
(6)ÏòÒ»¶¨Á¿µÄFe¡¢FeO¡¢Fe3O4µÄ»ìºÏÎïÖУ¬¼ÓÈë1mol? L£1 AµÄÈÜÒº100 mL£¬Ç¡ºÃʹ»ìºÏÎïÈ«²¿Èܽ⣬Çҷųö336mL(±ê×¼×´¿öÏÂ)µÄÆøÌ壬ÏòËùµÃÈÜÒºÖмÓÈëKSCNÈÜÒº£¬ÈÜÒºÎÞºìÉ«³öÏÖ£»ÈôȡͬÖÊÁ¿µÄFe¡¢FeO¡¢Fe3O4»ìºÏÎ¼ÓÈë1 mol ? L£1 HµÄÈÜÒº,ҲǡºÃʹ»ìºÏÎïÈ«²¿Èܽ⣬ÇÒÏòËùµÃÈÜÒºÖмÓÈëKSCNÈÜÒº,ÈÜÒºÒ²ÎÞºìÉ«³öÏÖ,ÔòËù¼ÓÈëµÄHÈÜÒºµÄÌå»ýÊÇ________¡£
£¨Ã¿¿Õ2·Ö£¬×îºóÒ»¿Õÿ¿Õ3·Ö£¬¹²¼Æ15·Ö£©
£¨1£©2NH4Cl£«Ca(OH)2 CaCl2£«2H2O£«2NH3¡ü £¨2£©2:3
£¨3£©ËáÐÔ£»NH4£«£«H2ONH3¡¤H2O£«H£«
£¨4£©È¡Ò»Ö§ÊԹܼÓÈëÊÊÁ¿µÄD¾§Ì壬Ȼºó¼ÓÈëÊÊÁ¿µÄNaOHÈÜÒº£¬¼ÓÈÈ£¬ÔÚÊԹܿڷÅÖÃʪÈóµÄºìɫʯÈïÊÔÖ½£¬ÈôÊÔÖ½±äÀ¶£¬ÔòÖ¤Ã÷D¾§ÌåÖк¬ÓÐNH4£«
£¨5£©D£¾A£½H £¨6£©110ml
£¨1£©2NH4Cl£«Ca(OH)2 CaCl2£«2H2O£«2NH3¡ü £¨2£©2:3
£¨3£©ËáÐÔ£»NH4£«£«H2ONH3¡¤H2O£«H£«
£¨4£©È¡Ò»Ö§ÊԹܼÓÈëÊÊÁ¿µÄD¾§Ì壬Ȼºó¼ÓÈëÊÊÁ¿µÄNaOHÈÜÒº£¬¼ÓÈÈ£¬ÔÚÊԹܿڷÅÖÃʪÈóµÄºìɫʯÈïÊÔÖ½£¬ÈôÊÔÖ½±äÀ¶£¬ÔòÖ¤Ã÷D¾§ÌåÖк¬ÓÐNH4£«
£¨5£©D£¾A£½H £¨6£©110ml
ÊÔÌâ·ÖÎö£ºCΪ³£¼ûÒºÌ壬ÔòC¿ÉÄÜÊÇË®¡£Í¨³£×´¿öϼס¢ÒÒ¡¢±û¡¢¶¡µÈΪÆøÌåµ¥ÖÊ£¬Ôò±ûºÍ¶¡Ó¦¸ÃÊÇÇâÆøºÍÑõÆø¡£AºÍB¶¼ÊÇ»¯ºÏÎÇÒ¶¼ÊÇÆøÌå¡£ÓÖÒòΪAºÍB¿ÉÒÔ·´Ó¦Éú³ÉD£¬DºÍF·´Ó¦Éú³ÉB¡¢C¡¢I£¬ÇҸ÷´Ó¦ÊÇÖØÒªµÄʵÑéÊÒÖÆÈ¡ÆøÌåµÄ·´Ó¦£¬ÓÉ´Ë¿ÉÒÔÍƲ⣬¸Ã·´Ó¦ÊÇʵÑéÊÒÖÆÈ¡°±ÆøµÄ·´Ó¦£¬ËùÒÔ¶¡ÊÇÇâÆø£¬±ûÊÇÑõÆø£¬BÄܺÍÑõÆø·´Ó¦Éú³ÉË®ºÍE£¬ËùÒÔBÊÇ°±Æø£¬ÔòÒÒÊǵªÆø£¬AÊÇÂÈ»¯Ç⣬¼×ÊÇÂÈÆø£¬DÊÇÂÈ»¯ï§£¬FÊÇÇâÑõ»¯¸Æ£¬IÊÇÂÈ»¯¸Æ¡£EÊÇNO£¬GÊÇNO2£¬NO2ÈÜÓÚË®Éú³ÉÏõËáºÍNO£¬ÔòHÊÇÏõËá¡£
£¨1£©·´Ó¦¢ÜµÄ»¯Ñ§·½³ÌʽΪ2NH4Cl£«Ca(OH)2 CaCl2£«2H2O£«2NH3¡ü¡£
£¨2£©°±ÆøÓëNO·´Ó¦Éú³É°±ÆøºÍË®£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ4NH3£«6NO£½5N2£«6H2O£¬ÆäÖл¹Ô¼ÁÊÇ°±Æø£¬Ñõ»¯¼ÁÊÇNO£¬µªÆø¼ÈÊÇÑõ»¯²úÎҲÊÇ»¹Ô²úÎï¡£ÓÉÓÚ»¹Ô¼ÁÓëÑõ»¯¼ÁµÄÎïÖʵÄÁ¿Ö®±ÈÊÇ2:3£¬ËùÒԸ÷´Ó¦Ñõ»¯²úÎïÓ뻹ԲúÎïµÄÎïÖʵÄÁ¿Ö®±ÈҲΪ2:3¡£
£¨3£©0.1mol lL-1AÈÜÒººÍ0.1mol ?L-1BÈÜÒºµÈÌå»ý»ìºÏÇ¡ºÃ·´Ó¦Éú³ÉÂÈ»¯ï§£¬ÈÜÒºÖÐNH4£«Ë®½â£¬ËùÒÔÈÜÒºÏÔËáÐÔ£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪNH4£«£«H2ONH3¡¤H2O£«H£«¡£
£¨4£©DÊÇÂÈ»¯ï§£¬ÑôÀë×ÓÊÇNH4£«£¬NH4£«ÄܺͼӦÉú³É°±Æø£¬Òò´Ë¿ÉÒÔͨ¹ý¼ìÑé°±ÆøÀ´Ö¤Ã÷¾§ÌåÖк¬ÓÐNH4£«£¬ËùÒÔÕýÈ·µÄ²Ù×÷ÊÇÈ¡Ò»Ö§ÊԹܼÓÈëÊÊÁ¿µÄD¾§Ì壬Ȼºó¼ÓÈëÊÊÁ¿µÄNaOHÈÜÒº£¬¼ÓÈÈ£¬ÔÚÊԹܿڷÅÖÃʪÈóµÄºìɫʯÈïÊÔÖ½£¬ÈôÊÔÖ½±äÀ¶£¬ÔòÖ¤Ã÷D¾§ÌåÖк¬ÓÐNH4£«¡£
£¨5£©ÑÎËáºÍÏõËᶼÊÇËáÒÖÖÆË®µÄµçÀ룬ÔÚ¶þÕßŨ¶ÈÏàͬµÄÌõ¼þ϶ÔË®µÄÒÖÖƳ̶ÈÏàͬ¡£ÂÈ»¯ï§ÊÇÇ¿ËáÈõ¼îÑΣ¬ÈÜÓÚË®NH4£«Ë®½â£¬´Ù½øË®µÄµçÀ룬ËùÒÔpHÏàͬµÄA¡¢D¡¢HÈýÖÖÈÜÒº£¬ÓÉË®µçÀë³öµÄc(OH£)µÄ´óС¹ØϵÊÇD£¾A£½H¡£
£¨6£©ÈÜÒºÖоùûÓгöÏÖºìÉ«£¬ÔòÈÜÒºÖеÄÈÜÖÊ·Ö±ðÊÇÂÈ»¯ÑÇÌúºÍÏõËáÑÇÌú£¬ÇÒ¶þÕßµÄÎïÖʵÄÁ¿¾ùÊÇ£½0.05mol¡£336mlÆøÌåÊÇÇâÆø£¬ÆäÎïÖʵÄÁ¿ÊÇ0.336L¡Â22.4L/mol£½0.015mol£¬Ôò¸ù¾Ýµç×ÓתÒÆÏàµÈ¿ÉÖª£¬±»»¹ÔµÄÏõËáµÄÎïÖʵÄÁ¿ÊÇ£½0.01mol£¬ËùÒÔ¸ù¾ÝµªÔ×ÓÊغã¿ÉÖª£¬ÏõËáµÄÎïÖʵÄÁ¿ÊÇ0.05mol¡Á2£«0.01mol£½0.11mol£¬Òò´ËÏõËáµÄÌå»ýÊÇ0.11mol¡Â1mol/L£½0.11L£½110ml¡£4£«¼ìÑé¡¢ÑÎÀàË®½â¡¢Íâ½çÌõ¼þ¶ÔË®µçÀëƽºâµÄÓ°ÏìÒÔ¼°Ñõ»¯»¹Ô·´Ó¦µÄÓйؼÆËãµÈ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿