ÌâÄ¿ÄÚÈÝ

ij¹¤³§ÅųöµÄÎÛË®Öк¬ÓдóÁ¿µÄFe2£«¡¢Zn2£«¡¢Hg2£«ÈýÖÖ½ðÊôÀë×Ó¡£ÒÔÏÂÊÇij»¯Ñ§Ñо¿ÐÔѧϰС×éµÄͬѧÉè¼ÆµÄ³ýÈ¥ÎÛË®ÖеĽðÊôÀë×Ó£¬»ØÊÕÂÌ·¯¡¢ð©·¯(ZnSO4·7H2O)ºÍ¹¯µÄ·½°¸¡£
[Ò©Æ·]¡¡NaOHÈÜÒº¡¢Áò»¯ÄÆÈÜÒº¡¢Áò»¯ÑÇÌú¡¢Ï¡ÁòËá¡¢Ìú·Û
[ʵÑé·½°¸]

[ÎÊÌâ̽¾¿]
£¨1£©²½Öè¢òËù·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ_________________________________________________¡£
£¨2£©²½Öè¢óÖеijéÂËΪ¼õѹÌõ¼þϵĹýÂË£¬¿ÉÒÔ¼Ó¿ì¹ýÂËËٶȣ»¸Ã²½ÖèÉæ¼°·´Ó¦µÄÀë×Ó·½³ÌʽÓÐZn2£«£«4OH£­=ZnO22-£«2H2OºÍ________________¡£
£¨3£©²½Öè¢öÖеõ½ÁòËáпÈÜÒºµÄÀë×Ó·´Ó¦·½³ÌʽΪ______________________________¡£
£¨4£©ÓûʵÏÖ²½Öè¢õ£¬ËùÐè¼ÓÈëµÄÊÔ¼ÁÓÐ________¡¢________£¬ËùÉæ¼°µÄÖ÷Òª²Ù×÷ÒÀ´ÎΪ______________________¡£
£¨5£©²½Öè¢ô³£Óõķ½·¨ÊǼÓÈÈ£¬¸Ã²½ÖèÊÇ·ñ¶Ô»·¾³ÓÐÓ°Ï죿__________(Ìî¡°ÊÇ¡±»ò¡°·ñ¡±)£¬ÈçÓÐÓ°Ï죬ÇëÄãÉè¼ÆÒ»¸ö»·¾³±£»¤·½°¸À´ÊµÏÖ²½Öè¢ôµÄ·´Ó¦________________________¡£
£¨1£©FeS£«2H£«=Fe2£«£«H2S¡ü£¬ZnS£«2H£«=Zn2£«£«H2S¡ü
£¨2£©4Fe2£«£«O2£«8OH£­£«2H2O=4Fe(OH)3¡ý
£¨3£©ZnO22-£«4H£«=Zn2£«£«2H2O
£¨4£©Ï¡ÁòËá¡¡Ìú·Û¡¡¹ýÂË¡¢Å¨Ëõ½á¾§¡¢¹ýÂË
£¨5£©¡¡ÊÇ¡¡ÔÚÃܱÕÈÝÆ÷ÖмÓÈÈHgS

ÊÔÌâ·ÖÎö£º£¨1£©²½Öè¢ò¼ÓÈëÏ¡ÁòËᣬÓëFeS¡¢ZnS·Ö±ð·¢Éú¸´·Ö½â·´Ó¦£¬¿Éд³ÉÀë×Ó·½³Ìʽ¡£
£¨2£©FeSO4¡¢ZnSO4»ìºÏÈÜÒºÖмÓÈë¹ýÁ¿µÄNaOHʱ£¬Fe2£«ÓëOH£­·´Ó¦Éú³ÉFe(OH)2£¬Fe(OH)2ѸËÙ±»¿ÕÆøÖеÄO2Ñõ»¯ÎªFe(OH)3£¬Æä×Ü·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º4Fe2£«£«O2£«8OH£­£«2H2O=4Fe(OH)3¡ý¡£
£¨3£©¸ù¾Ý¿òͼת»¯¹Øϵ£¬¿ÉÒÔ¿´³öZn(OH)2µÄÐÔÖÊÀàËÆÓÚAl(OH)3£¬ËùÒÔNa2ZnO2ÈÜÒºÖмÓÈë¹ýÁ¿µÄH2SO4Éú³ÉZnSO4ÈÜÒº£¬ÆäÀë×Ó·½³ÌʽΪ£ºZnO22-£«4H£«=Zn2£«£«2H2O¡£
£¨4£©ÇâÑõ»¯ÌúÄÜÓëÏ¡ÁòËá·´Ó¦Éú³ÉÁòËáÌú£¬ÁòËáÌúÄÜÓë¹ýÁ¿Ìúµ¥ÖÊ·´Ó¦Éú³É¶þ¼ÛÌúÀë×Ó£¬È»ºó¹ýÂË¡¢Å¨Ëõ½á¾§¡¢¹ýÂ˵õ½FeSO4?7H2O¡£
£¨5£©HgSÔÚ¿ÕÆøÖмÓÈȿɵÃHg¡¢SO2£¬HgÕôÆøºÍSO2¶¼»á¶Ô»·¾³²úÉúÎÛȾ£¬µ«ÔÚÃܱÕÈÝÆ÷ÖмÓÈÈHgS¿ÉÓÐЧ·ÀÖ¹HgÕôÆøºÍSO2Êͷŵ½´óÆøÖУ¬´Ó¶ø±£»¤ÁË»·¾³¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÑÇÂÈËáÄÆ£¨NaClO2£©ÊÇÒ»ÖÖÖØÒªµÄº¬ÂÈÏû¶¾¼Á£¬Ö÷ÒªÓÃÓÚ¹¤ÒµÉú²úµÄƯ°×¼°Ïû¶¾¡£
ÏÂͼÊǹýÑõ»¯Çâ·¨Éú²úÑÇÂÈËáÄƵŤÒÕÁ÷³Ìͼ£º
Ïà¹ØÎïÖʵÄÐÔÖÊÈçÏ£º
¢Ù ClO2·ÐµãµÍÒ×Æø»¯£»Å¨¶È½Ï¸ßµÄClO2ÆøÌåÒ׷ֽⱬը¡£
¢Ú NaClO2Èܽâ¶ÈËæζÈÉý¸ß¶øÔö´ó£¬Êʵ±Ìõ¼þÏ¿ɽᾧÎö³öNaClO2¡¤3H2O¡£

£¨1£©.ÔÚClO2·¢ÉúÆ÷ÖÐͨÈëSO2µÄͬʱ¹ÄÈë¿ÕÆø£¬Æä×÷ÓÃÊÇ________(ÌîÐòºÅ)¡£
A£®½«SO2Ñõ»¯³ÉSO3£¬ÔöÇ¿ËáÐÔ
B£®Ï¡ÊÍClO2ÒÔ·ÀÖ¹±¬Õ¨
C£®½«NaClO3Ñõ»¯³ÉClO2
£¨2£©ÎªÈ·±£H2O2³ä·Ö²ÎÓë·´Ó¦£¬ÎüÊÕËþÄÚζȲ»Ò˽ϸߣ¬ÆäÔ­ÒòÊÇ______________¡£
£¨3£©160 g¡¤L-1 NaOHÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ______________£¬ÔÚʵÑéÊÒÈôÒªÅäÖÆ450mL¸ÃÎïÖʵÄÁ¿Å¨¶ÈNaOHÈÜÒº£¬Ó¦³ÆÈ¡NaOHµÄÖÊÁ¿Îª__________¿Ë¡£
£¨4£©Ð´³öÎüÊÕËþÄÚ·´Ó¦µÄ»¯Ñ§·½³Ìʽ_____________________________________¡£
£¨5£©´ÓÂËÒºÖеõ½NaClO2¡¤3H2O´Ö¾§ÌåµÄʵÑé²Ù×÷ÒÀ´ÎÊÇ____________(ÌîÐòºÅ)¡£
A£®ÕôÁó      B£®Õô·¢      C£®×ÆÉÕ      D£®¹ýÂË      E£®ÀäÈ´½á¾§
ÒªµÃµ½¸ü´¿µÄNaClO2¡¤3H2O¾§Ìå±ØÐë½øÒ»²½µÄ²Ù×÷ÊÇ________(Ìî²Ù×÷Ãû³Æ)¡£
Ò»ÖÖº¬ÂÁ¡¢ï®¡¢îܵÄÐÂÐ͵ç×Ó²ÄÁÏ£¬Éú²úÖвúÉúµÄ·ÏÁÏÊýÁ¿¿É¹Û£¬·ÏÁÏÖеÄÂÁÒÔ½ðÊôÂÁ²­µÄÐÎʽ´æÔÚ£»îÜÒÔCo2O3·CoOµÄÐÎʽ´æÔÚ£¬Îü¸½ÔÚÂÁ²­µÄµ¥Ãæ»òË«Ã棻﮻ìÔÓÓÚÆäÖС£
´Ó·ÏÁÏÖлØÊÕÑõ»¯îÜ£¨CoO£©µÄ¹¤ÒÕÁ÷³ÌÈçÏ£º

£¨1£©¹ý³ÌIÖвÉÓÃNaOHÈÜÒºÈܳö·ÏÁÏÖеÄAl£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ                  ¡£
£¨2£©¹ý³ÌIIÖмÓÈëÏ¡H2SO4Ëữºó£¬ÔÙ¼ÓÈëNa2S2O3ÈÜÒº½þ³öîÜ¡£Ôò½þ³öîܵĻ¯Ñ§·´Ó¦·½³ÌʽΪ£¨²úÎïÖÐÖ»ÓÐÒ»ÖÖËá¸ù£©                    ¡£ÔÚʵÑéÊÒÄ£Ä⹤ҵÉú²úʱ£¬Ò²¿ÉÓÃÑÎËá½þ³öîÜ£¬µ«Êµ¼Ê¹¤ÒµÉú²úÖв»ÓÃÑÎËᣬÇë´Ó·´Ó¦Ô­Àí·ÖÎö²»ÓÃÑÎËá½þ³öîܵÄÖ÷ÒªÔ­Òò_______________¡£
£¨3£©¹ý³Ì¢óµÃµ½ï®ÂÁÔüµÄÖ÷Òª³É·ÖÊÇLiFºÍAl(OH)3£¬Ì¼ËáÄÆÈÜÒºÔÚ²úÉúAl(OH)3ʱÆðÖØÒª×÷Óã¬Çëд³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ________________________¡£
£¨4£©Ì¼ËáÄÆÈÜÒºÔÚ¹ý³ÌIIIºÍIVÖÐËùÆð×÷ÓÃÓÐËù²»Í¬£¬Çëд³öÔÚ¹ý³ÌIVÖÐÆðµÄ×÷ÓÃÊÇ
____________________________________________________________¡£
£¨5£©ÔÚNa2CO3ÈÜÒºÖдæÔÚ¶àÖÖÁ£×Ó£¬ÏÂÁи÷Á£×ÓŨ¶È¹ØϵÕýÈ·µÄÊÇ______£¨ÌîÐòºÅ£©¡£
A£®c(Na+) = 2c(CO32-)B£®c(Na+) > c(CO32-) > c(HCO3-)
C£®c(OH-) > c(HCO3-) > c(H+)D£®c(OH-) - c(H+)£½c(HCO3-) + 2c(H2CO3)
ʳÑÎÊÇÈÕ³£Éú»îµÄ±ØÐèÆ·£¬Ò²ÊÇÖØÒªµÄ»¯¹¤Ô­ÁÏ¡£
´ÖʳÑγ£º¬ÓÐÉÙÁ¿Ca2£«¡¢Mg2£«¡¢Fe3£«¡¢SO42¡ªµÈÔÓÖÊÀë×Ó£¬ÊµÑéÊÒÌṩµÄÊÔ¼ÁÈçÏ£º±¥ºÍNa2CO3ÈÜÒº¡¢±¥ºÍK2CO3ÈÜÒº¡¢NaOHÈÜÒº¡¢BaCl2ÈÜÒº¡¢Ba(NO3)2ÈÜÒº¡¢75%ÒÒ´¼¡¢ËÄÂÈ»¯Ì¼¡£ÊµÑéÊÒÌá´¿NaClµÄÁ÷³ÌÈçÏ£º

£¨1£©Óû³ýÈ¥ÈÜÒº¢ñÖеÄCa2£«¡¢Mg2£«¡¢Fe3£«¡¢SO42¡ªÀë×Ó£¬Ñ¡³öAËù´ú±íµÄ¶àÖÖÊÔ¼Á£¬°´µÎ¼Ó˳ÐòÒÀ´ÎΪi  NaOH   ii                      iii                 £¨Ìѧʽ£©¡£
£¨2£©Çëд³öÏÂÁÐÊÔ¼Á¼ÓÈëʱ·¢Éú·´Ó¦µÄÀë×Ó·´Ó¦·½³Ìʽ£º
¼ÓÈëÊÔ¼Ái£º                                              £»
¼ÓÈëÊÔ¼Áiii£º                                              ¡£
£¨3£©Ï´µÓ³ýÈ¥NaCl¾§Ìå±íÃ渽´øµÄÉÙÁ¿KCl£¬Ñ¡ÓõÄÊÔ¼Á  Îª                 ¡££¨´ÓÌṩµÄÊÔ¼ÁÖÐÑ¡£©
£¨4£©ÊµÑéÖÐÓõ½µÄÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈΪ0£®400mol/L£¬ÏÖʵÑéÊÒijŨÑÎËáÊÔ¼ÁÆ¿ÉϵÄÓйØÊý¾ÝÈçÏ£º

ÓûÓÃÉÏÊöŨÑÎËáÅäÖÆʵÑéËùÐèŨ¶ÈµÄÏ¡ÑÎËá480mL£¬
¢ÙÅäÖÆÐèÒªµÄ²£Á§ÒÇÆ÷ÓР                       £¨ÌîÒÇÆ÷Ãû³Æ£©
¢ÚÐèÁ¿È¡µÄŨÑÎËáµÄÌå»ýΪ£º                         ¡£
ÓÃ1£­¶¡´¼¡¢ä廯ÄƺͽÏŨH2SO4»ìºÏÎïΪԭÁÏ£¬ÔÚʵÑéÊÒÖƱ¸1£­ä嶡Í飬²¢¼ìÑé·´Ó¦µÄ²¿·Ö¸±²úÎï¡££¨ÒÑÖª£ºNaCl+H2SO4(Ũ)=NaHSO4+HCl¡ü£©ÏÖÉè¼ÆÈçÏÂ×°Öã¬ÆäÖмгÖÒÇÆ÷¡¢¼ÓÈÈÒÇÆ÷¼°ÀäÈ´Ë®¹ÜûÓл­³ö¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÒÇÆ÷DµÄÃû³ÆÊÇ           ¡£
£¨2£©¹Ø±ÕaºÍb¡¢½ÓͨÊúÖ±ÀäÄý¹ÜµÄÀäÄýË®£¬¸øA¼ÓÈÈ30·ÖÖÓ£¬ÖƱ¸1£­ä嶡Í顣д³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ                                                      ¡£
£¨3£©ÀíÂÛÉÏ£¬ÉÏÊö·´Ó¦µÄÉú³ÉÎﻹ¿ÉÄÜÓУº¶¡ÃÑ¡¢1£­¶¡Ï©¡¢ä廯ÇâµÈ¡£Ï¨ÃðA´¦¾Æ¾«µÆ£¬ÔÚÊúÖ±ÀäÄý¹ÜÉÏ·½ÈûÉÏÈû×Ó£¬´ò¿ªa£¬ÀûÓÃÓàÈȼÌÐø·´Ó¦Ö±ÖÁÀäÈ´£¬Í¨¹ýB¡¢C×°ÖüìÑ鲿·Ö¸±²úÎï¡£B¡¢CÖÐӦʢ·ÅµÄÊÔ¼Á·Ö±ðÊÇ            ¡¢           ¡£
£¨4£©ÔÚʵÑé¹ý³ÌÖУ¬·¢ÏÖAÖÐÒºÌåÓÉÎÞÉ«Öð½¥±ä³ÉºÚÉ«£¬¸ÃºÚÉ«ÎïÖÊÓëŨÁòËá·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                    £¬¿ÉÔÚÊúÖ±ÀäÄý¹ÜµÄÉ϶ËÁ¬½ÓÒ»¸öÄÚ×°ÎüÊÕ¼Á¼îʯ»ÒµÄ¸ÉÔï¹Ü£¬ÒÔÃâÎÛȾ¿ÕÆø¡£
£¨5£©Ïà¹ØÓлúÎïµÄÊý¾ÝÈçÏ£º
ÎïÖÊ
ÈÛµã/0C
·Ðµã/0C
1£­¶¡´¼
-89£®5
117£®3
1£­ä嶡Íé
-112£®4
101£®6
¶¡ÃÑ
-95£®3
142£®4
1£­¶¡Ï©
-185£®3
-6£®5
ΪÁ˽øÒ»²½¾«ÖÆ1£­ä嶡Í飬¼ÌÐø½øÐÐÁËÈçÏÂʵÑ飺´ýÉÕÆ¿ÀäÈ´ºó£¬°ÎÈ¥ÊúÖ±µÄÀäÄý¹Ü£¬ÈûÉÏ´øζȼƵÄÏðƤÈû£¬¹Ø±Õa£¬´ò¿ªb£¬½ÓͨÀäÄý¹ÜµÄÀäÄýË®£¬Ê¹ÀäË®´Ó     £¨Ìîc»òd£©´¦Á÷È룬ѸËÙÉý¸ßζÈÖÁ       ¡æ£¬ÊÕ¼¯ËùµÃÁó·Ö¡£
£¨6£©ÈôʵÑéÖÐËùÈ¡1£­¶¡´¼¡¢NaBr·Ö±ðΪ7£®4 g¡¢13£®0 g£¬Õô³öµÄ´Ö²úÎï¾­Ï´µÓ¡¢¸ÉÔïºóÔÙ´ÎÕôÁóµÃµ½9£®6 g 1£­ä嶡Í飬Ôò1£­ä嶡ÍéµÄ²úÂÊÊÇ            ¡£

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

¾«Ó¢¼Ò½ÌÍø