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»¯²úÎïÓ뻹ԭ²úÎïµÄÎïÖʵÄÁ¿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

½âÎöÊÔÌâ·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»¯²úÎïÓ뻹ԭ²úÎïµÄÎïÖʵÄÁ¿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¡£
¿¼µã£º¿¼²éÎÞ»ú¿òͼÌâµÄÍƶϡ¢°±ÆøÖƱ¸¡¢NH4£«¼ìÑé¡¢ÑÎÀàË®½â¡¢Íâ½çÌõ¼þ¶ÔË®µçÀëƽºâµÄÓ°ÏìÒÔ¼°Ñõ»¯»¹Ô­·´Ó¦µÄÓйؼÆËãµÈ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ÏÂͼÊÇÎÞ»úÎïA¡«MÔÚÒ»¶¨Ìõ¼þϵÄת»¯¹Øϵ£¨²¿·Ö²úÎï¼°·´Ó¦Ìõ¼þδÁгö£©¡£ÆäÖУ¬IÊǵؿÇÖк¬Á¿×î¸ßµÄ½ðÊô£¬KÊÇÒ»ÖÖºì×ØÉ«ÆøÌå¡£

ÇëÌîдÏÂÁпհףº
£¨1£©×é³Éµ¥ÖÊIµÄÔªËØÔÚÖÜÆÚ±íÖеÄλÖÃΪ                        ¡£
£¨2£©Óû¼ìÑéFÈÜÒºÖÐÊÇ·ñº¬ÓÐÉÙÁ¿M£¬¿ÉÑ¡ÔñµÄÊÔ¼ÁΪ            £¨Ìѧʽ£©¡£
£¨3£©ÔÚ·´Ó¦¢ßÖл¹Ô­¼ÁÓëÑõ»¯¼ÁµÄÎïÖʵÄÁ¿Ö®±ÈΪ___________________¡£
£¨4£©Ä³Í¬Ñ§È¡FµÄÈÜÒº£¬Ëữºó¼ÓÈëKI¡¢µí·ÛÈÜÒº£¬±äΪÀ¶É«¡£Ð´³öÓëÉÏÊö±ä»¯¹ý³ÌÏà¹ØµÄÀë×Ó·½³Ìʽ£º                             ¡£
£¨5£©½«»¯ºÏÎïD ÓëKNO3¡¢KOH ¹²ÈÚ£¬¿ÉÖƵÃÒ»ÖÖ¡°ÂÌÉ«¡±»·±£¸ßЧ¾»Ë®¼ÁK2FeO4£¨¸ßÌúËá¼Ø£©.ͬʱ»¹Éú³ÉKNO2ºÍH2O ¡£¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ__________________________¡£
£¨6£©Ã¾ÓëIµÄºÏ½ðÊÇÒ»ÖÖDZÔÚµÄÖüÇâ²ÄÁÏ£¬¿ÉÔÚë²Æø±£»¤Ï£¬½«Ò»¶¨»¯Ñ§¼ÆÁ¿±ÈµÄMg¡¢Iµ¥ÖÊÔÚÒ»¶¨Î¶ÈÏÂÈÛÁ¶»ñµÃ¡£
¢ÙÈÛÁ¶ÖƱ¸Ã¾IºÏ½ðʱͨÈëë²ÆøµÄÄ¿µÄÊÇ                       ¡£
¢ÚIµç³ØÐÔÄÜÓÅÔ½£¬I-Ag2O µç³Ø¿ÉÓÃ×÷ˮ϶¯Á¦µçÔ´£¬ÆäÔ­ÀíÈçͼËùʾ¡£¸Ãµç³Ø·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                  ¡£

ϱíΪԪËØÖÜÆÚ±íµÄÒ»²¿·Ö£º

×å
ÖÜÆÚ
 
 
 
1
¢Ù
 
 
 
 
 
 
 
2
 
 
 
 
 
¢Ú
 
 
3
¢Û
 
 
¢Ü
 
¢Ý
¢Þ
 
 
£¨1£©Ð´³öÔªËØ¢ÜÔÚÖÜÆÚ±íÖеÄλÖÃ______________¡£
£¨2£©¢Ú¡¢¢Û¡¢¢ÝµÄÔ­×Ӱ뾶ÓÉ´óµ½Ð¡µÄ˳ÐòΪ_________________________¡£
£¨3£©¢Ü¡¢¢Ý¡¢¢ÞµÄÆø̬Ç⻯ÎïµÄÎȶ¨ÐÔÓÉÇ¿µ½ÈõµÄ˳ÐòÊÇ_________________________¡£
£¨4£©¢Ù¡¢¢Ú¡¢¢Û¡¢¢ÞÖеÄijЩԪËØ¿ÉÐγɼȺ¬Àë×Ó¼üÓÖº¬¼«ÐÔ¹²¼Û¼üµÄ»¯ºÏÎд³öÆäÖÐÒ»ÖÖ»¯ºÏÎïµÄµç×Óʽ£º____________________¡£
¢ò.ÓÉÉÏÊö²¿·ÖÔªËØ×é³ÉµÄÎïÖʼ䣬ÔÚÒ»¶¨Ìõ¼þÏ£¬¿ÉÒÔ·¢ÉúÏÂͼËùʾµÄ±ä»¯£¬ÆäÖÐAÊÇÒ»ÖÖµ­»ÆÉ«¹ÌÌå¡£Çë»Ø´ð£º

£¨1£©Ð´³ö¹ÌÌåAÓëÒºÌåX·´Ó¦µÄÀë×Ó·½³Ìʽ                                     ¡£
£¨2£©ÆøÌåYÊÇÒ»ÖÖ´óÆøÎÛȾÎֱ½ÓÅÅ·Å»áÐγÉËáÓê¡£¿ÉÓÃÈÜÒºBÎüÊÕ£¬µ±BÓëYÎïÖʵÄÁ¿Ö®±ÈΪ1:1ÇÒÇ¡ºÃÍêÈ«·´Ó¦Ê±£¬ËùµÃÈÜÒºDµÄÈÜÖÊΪ          £¨Ìѧʽ£©¡£ÒÑÖªÈÜÒºDÏÔËáÐÔ£¬ÔòDÈÜÒºÖи÷ÖÖÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ                              
£¨3£©Ð´³öÆøÌåCÓëÆøÌåY·´Ó¦µÄ»¯Ñ§·½³Ìʽ                                    ¡£

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

¾«Ó¢¼Ò½ÌÍø