ÌâÄ¿ÄÚÈÝ

£¨13·Ö£©¢ñ£®Í¨³£×´¿öÏ£¬X¡¢YºÍZÊÇÈýÖÖÆø̬µ¥ÖÊ¡£XµÄ×é³ÉÔªËØÊǵÚÈýÖÜÆÚÔ­×Ӱ뾶×îСµÄÔªËØ(Ï¡ÓÐÆøÌåÔªËسýÍâ)£»YºÍZ¾ùÓÉÔªËØR×é³É£¬·´Ó¦Y+2I-+2H+====I2+Z+H2O³£×÷ΪYµÄ¼ø¶¨·´Ó¦¡£WÊǶÌÖÜÆÚÔªËØ£¬×îÍâ²ãµç×ÓÊýÊÇ×îÄÚ²ãµç×ÓÊýµÄÈý±¶£¬ÎüÒýµç×Ó¶ÔµÄÄÜÁ¦±ÈXµ¥ÖʵÄ×é³ÉÔªËØÒªÈõ¡£
£¨1£© ZµÄ»¯Ñ§Ê½__________________
£¨2£©½«YºÍ¶þÑõ»¯Áò·Ö±ðͨÈëÆ·ºìÈÜÒº£¬¶¼ÄÜʹƷºìÍÊÉ«¡£¼òÊöÓÃÍÊÉ«µÄÈÜÒºÇø±ðYºÍ¶þÑõ»¯ÁòµÄʵÑé·½·¨:________________________________________________________¡£
£¨3£©¾Ù³öʵÀý˵Ã÷XµÄÑõ»¯ÐÔ±ÈWµ¥ÖÊÑõ»¯ÐÔÇ¿(½öÓÃÒ»¸ö»¯Ñ§·½³Ìʽ±íʾ):_____________¡£
¢ò£®ÈçͼÊÇ0.1 mol¡¤L£­1ËÄÖÖµç½âÖÊÈÜÒºµÄpHËæζȱ仯µÄͼÏñ¡£

£¨1£©ÆäÖзûºÏ0.1 mol¡¤L£­1NH4Al(SO4)2µÄpHËæζȱ仯µÄÇúÏßÊÇ________(ÌîдÐòºÅ)£¬
£¨2£©20 ¡æʱ£¬0.1 mol¡¤L£­1NH4Al(SO4)2ÖÐ2c(SO42-)£­c(NH4+)£­3c(Al3£«)£½________¡££¨¼ÆË㾫ȷֵ£©
£¨3£©ÊÒÎÂʱ£¬Ïò100 mL 0.1 mol¡¤L£­1 NH4HSO4ÈÜÒºÖеμÓ0.1 mol¡¤L£­1 NaOHÈÜÒº£¬µÃµ½ÈÜÒºpHÓëNaOHÈÜÒºÌå»ýµÄ¹ØϵÇúÏßÈçͼËùʾ£º

ÊÔ·ÖÎöͼÖÐa¡¢b¡¢c¡¢dËĸöµã£¬Ë®µÄµçÀë³Ì¶È×î´óÊÇ________µã£»ÔÚbµã£¬ÈÜÒºÖи÷Àë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄÅÅÁÐ˳ÐòÊÇ______________________

¢ñ£®£¨1£©O2£¨2·Ö£©
£¨2£©¼ÓÈÈÍÊÉ«ºóµÄÈÜÒº£¬ÈôÈÜÒº»Ö¸´ºìÉ«£¬ÔòԭͨÈëÆøÌåΪSO2£»ÈôÈÜÒº²»±äºì£¬ÔòԭͨÈëÆøÌåÊÇO3£¨2·Ö£©
£¨3£© H2S+Cl2====S£«2HCl£¨1·Ö£©(ÆäËûºÏÀí´ð°¸¾ù¿É)
¢ò£®£¨1£©¢ñ£¨2·Ö£©
£¨2£©£¨10£­3-10£­11£©mol¡¤L£­1    £¨2·Ö£© £¨3)a£¨2·Ö£©¡¡c(Na£«)£¾c(SO42-)£¾c(NH4+)£¾c(OH£­)£½c(H£«) £¨2·Ö£©

½âÎöÊÔÌâ·ÖÎö£ºI£ºÊ×Ïȸù¾ÝÐÅϢȷ¶¨XΪCl2£¬RΪÑõÔªËØ£¬YΪO3£¬ZΪO2£¬WΪÁòÔªËØ¡£O3Òò¾ßÓÐÇ¿Ñõ»¯ÐÔʹƷºìÍÊÉ«£¬ÓëSO2²»Í¬£¬¹Ê¶ÔÍÊÉ«ºóµÄÈÜÒº½øÐмÓÈÈ£¬ÈôÈÜÒº»Ö¸´ºìÉ«£¬ÔòԭͨÈëÆøÌåΪSO2£»ÈôÈÜÒº²»±äºì£¬ÔòԭͨÈëÆøÌåÊÇO3£»ÄÜ˵Ã÷Cl2µÄÑõ»¯ÐÔ±ÈSµÄÑõ»¯ÐÔÇ¿µÄ·´Ó¦ÓÐH2S+Cl2====S£«2HCl£¬K2S+Cl2====S£«2KClµÈ¡£
¢ò¢ÅNH4Al(SO4)2ÖÐNH4+¡¢Al3£«·¢ÉúË®½âʹÈÜÒºÏÔËáÐÔ£¨pH£¼7£©£¬ÇÒζÈÉý¸ß£¬´Ù½øË®½â£¬ËáÐÔÔöÇ¿£¬pH¼õС£¬ÇúÏߢñÕýÈ·¡£
¢Æ¾ÝNH4Al(SO4)2ÈÜÒºÖеĵçºÉÊغãʽΪ2c(SO42-)+ c(OH£­)£½c(H+)+c(NH4+)+3c(Al3£«)£¬Ôò2c(SO42-)£­c(NH4+)£­3c(Al3£«)£½c(H+) £­c(OH£­)£½£¨1¡Á10£­3£­1¡Á10£­11£©mol¡¤L£­1£»
¢ÇÔÚaµãÁ½ÎïÖÊÇ¡ºÃ·´Ó¦Éú³É(NH4)2SO4£¬NH4+·¢ÉúË®½â£¬Ë®µÄµçÀë³Ì¶È×î´ó£»ÔÚbµã[ÔÚaµã£¨Ç¡ºÃÉú³É(NH4)2SO4£©Ö®ºóÔÙÉÔ¶à¼ÓÒ»µãNaOH£¬Ê¹ÉÙÁ¿NH4+ÓëOH£­·´Ó¦Éú³ÉNH3¡¤H2O£¬¼´c(Na£«)£¾c(SO42-)£¾c(NH4+)]£¬ÓÖ¸ù¾ÝÈÜҺΪÖÐÐÔ£¬¹ÊÓÐc(Na£«)£¾c(SO42-)£¾c(NH4+)£¾c(OH£­)£½c(H£«)¡£
¿¼µã£º±¾Ì⿼²éÔªËØÖÜÆÚÂÉ¡¢ÑεÄË®½â¡¢µçºÉÊغ㡢Àë×ÓŨ¶È´óС±È½ÏµÈ֪ʶ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

£¨15·Ö£©¡¾»¯Ñ§¡ªÑ¡ÐÞ3£ºÎïÖʽṹÓëÐÔÖÊ¡¿
¡¡a¡¢b¡¢c¡¢d¡¢fÎåÖÖÇ°ËÄÖÜÆÚÔªËØ£¬Ô­×ÓÐòÊýÒÀ´ÎÔö´ó£»a¡¢b¡¢cÈýÖÖÔªËصĻù̬ԭ×Ó¾ßÓÐÏàͬµÄÄܲãºÍÄܼ¶£¬µÚÒ»µçÀëÄÜI1(a)<I1(c)<I1(b)£¬ÇÒÆäÖлù̬bÔ­×ÓµÄ2p¹ìµÀ´¦ÓÚ°ë³äÂú״̬£» dΪÖÜÆÚ±íÇ°ËÄÖÜÆÚÖе縺ÐÔ×îСµÄÔªËØ£»fµÄÔ­×ÓÐòÊýΪ29¡£Çë»Ø´ðÏÂÁÐÎÊÌâ¡£(ÈçÐè±íʾ¾ßÌåÔªËØÇëÓÃÏàÓ¦µÄÔªËØ·ûºÅ)
£¨1£©Ð´³öac2µÄµç×Óʽ__________£»»ù̬fÔ­×ÓµÄÍâΧµç×ÓÅŲ¼Ê½Îª                  ¡£
£¨2£©Ð´³öÒ»ÖÖÓëac2»¥ÎªµÈµç×ÓÌåµÄÎïÖʵĻ¯Ñ§Ê½                    ¡£
£¨3£©bµÄ¼òµ¥Ç⻯ÎïµÄ·Ðµã±Èͬ×åÔªËØÇ⻯ÎïµÄ·Ðµã         ¡££¨Ìî¡°¸ß¡±»ò¡°µÍ¡±£©

£¨4£©»¯ºÏÎïMÓÉc¡¢dÁ½ÖÖÔªËØ×é³É£¬Æ侧°û½á¹¹Èç¼×£¬ÔòMµÄ»¯Ñ§Ê½Îª                ¡£
£¨5£©»¯ºÏÎïNµÄ²¿·Ö½á¹¹ÈçÒÒ£¬NÓÉa¡¢bÁ½ÔªËØ×é³É£¬ÔòÓ²¶È³¬¹ý½ð¸Õʯ¡£ÊԻش𣺢ÙNµÄ¾§ÌåÀàÐÍΪ________________________£¬ÆäÓ²¶È³¬¹ý½ð¸ÕʯµÄÔ­ÒòÊÇ___________________¡£
¢ÚN¾§ÌåÖÐa¡¢bÁ½ÔªËØÔ­×ÓµÄÔÓ»¯·½Ê½¾ùΪ___________________¡£

£¨15·Ö£©ÒÑÖªA¡¢B¡¢C¡¢D¡¢E¶¼ÊÇÖÜÆÚ±íÖÐÇ°ËÄÖÜÆÚµÄÔªËØ£¬ËüÃǵĺ˵çºÉÊýÒÀ´ÎÔö¼Ó¡£Ïà¹ØÐÅÏ¢ÈçϱíËùʾ£¬¸ù¾ÝÍƶϻشðÏÂÁÐÎÊÌ⣺£¨´ðÌâʱA¡¢B¡¢C¡¢D¡¢EÓÃËù¶ÔÓ¦µÄÔªËØ·ûºÅ±íʾ£©

A
AµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎﻯѧʽΪH2AO3
B
BÔªËصĵÚÒ»µçÀëÄܱÈͬÖÜÆÚÏàÁÚÁ½¸öÔªËض¼´ó
C
CÔ­×ÓÔÚͬÖÜÆÚÔ­×ÓÖа뾶×î´ó£¨Ï¡ÓÐÆøÌå³ýÍ⣩£¬Æäµ¥ÖÊÑæɫΪ»ÆÉ«
D
ZµÄ»ù̬ԭ×Ó×îÍâ²ãµç×ÓÅŲ¼Ê½Îª3s23p2
E
EÓëCλÓÚ²»Í¬ÖÜÆÚ£¬EÔ­×ÓºËÍâ×îÍâ²ãµç×ÓÊýÓëCÏàͬ£¬ÆäÓà¸÷²ãµç×Ó¾ù³äÂú
£¨1£©CÔÚÖÜÆÚ±íÖÐλÓÚµÚ¡¡¡¡¡¡ÖÜÆÚµÚ¡¡¡¡×壬E»ù̬ԭ×ÓºËÍâµç×ÓÅŲ¼Ê½ÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡
£¨2£©A¡¢B¡¢DÈýÖÖÔªËص縺ÐÔÓÉ´óµ½Ð¡ÅÅÁÐ˳ÐòΪ¡¡¡¡¡¡¡¡¡¡¡¡¡¡£¬ÆäÖÐAµÄ×î¸ß¼ÛÂÈ»¯Îï¹¹³É¾§ÌåµÄ΢Á£¼ä×÷ÓÃÁ¦Îª¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡
£¨3£©AºÍBµÄ×î¼òµ¥Ç⻯ÎïÖнÏÎȶ¨µÄÊÇ¡¡¡¡¡¡¡¡£¨Ìѧʽ£©¡£BµÄ×î¼òµ¥Ç⻯ÎïºÍEµÄºÚÉ«Ñõ»¯Îï¹ÌÌåÔÚ¼ÓÈÈʱ¿É·´Ó¦£¬Ð´³öÆä·´Ó¦·½³Ìʽ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡
£¨4)EµÄµ¥Öʺ͹ýÑõ»¯ÇâÔÚÏ¡ÁòËáÖпɷ´Ó¦£¬ÓÐÈ˽«Õâ¸ö·´Ó¦Éè¼Æ³ÉÔ­µç³Ø£¬Çëд³öÕý¼«·´Ó¦·½³Ìʽ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡
£¨5£©ÃºÈ¼ÉÕ²úÉúµÄÑÌÆøÖÐÓÐBµÄÑõ»¯Î»áÒýÆðÑÏÖصĻ·¾³ÎÊÌ⣬Òò´Ë£¬³£ÓÃAH4´ß»¯»¹Ô­ÒÔÏû³ýÎÛȾ£¬ÒÑÖª£º 
¢Ù AH4(g)+2 BO2£¨g)£½ B2(g)+AO2(g)+2H2O (g)  ¡÷H1£½£­867kJ£¯mol
¢Ú 2BO2(g) ?B2O4(g)  ¡÷H2=£­56.9 kJ£¯mol
д³öAH4ºÍB2O4·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡

£¨14·Ö£©Ï±íΪԪËØÖÜÆÚ±íµÄÒ»²¿·Ö£º

×å
ÖÜÆÚ
 
 
 
1
¢Ù
 
 
 
 
 
 
 
2
 
 
 
 
 
¢Ú
 
 
3
¢Û
 
 
¢Ü
 
¢Ý
¢Þ
 
¢ñ£®Óû¯Ñ§ÓÃÓï»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öÔªËØ¢ÜÔÚÖÜÆÚ±íÖеÄλÖ㺡¡¡¡¡¡¡¡                                           ¡¡¡¡£»
£¨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д³öÆäÖÐÁ½ÖÖ»¯
ºÏÎïµÄµç×Óʽ£º¡¡¡¡¡¡                                 ¡¡¡¡¡¡¡£
¢ò£®ÓÉÉÏÊö²¿·ÖÔªËØ×é³ÉµÄÎïÖʼ䣬ÔÚÒ»¶¨Ìõ¼þÏ£¬¿ÉÒÔ·¢ÉúÏÂͼÖеı仯£¬ÆäÖÐAÊÇÒ»
ÖÖµ­»ÆÉ«¹ÌÌå¡£Ôò£º

£¨1£©Ð´³ö¹ÌÌåAÓëÒºÌåX·´Ó¦µÄÀë×Ó·½³Ìʽ£º¡¡¡¡¡¡¡¡¡¡¡¡      ¡¡¡¡¡¡¡¡£»
£¨2£©ÆøÌåYÊÇÒ»ÖÖ´óÆøÎÛȾÎֱ½ÓÅÅ·Å»áÐγÉËáÓê¡£¿ÉÓÃÈÜÒºBÎüÊÕ£¬µ±BÓëYÎïÖʵÄÁ¿Ö®±ÈΪ1¡Ã1ÇÒÇ¡ºÃÍêÈ«·´Ó¦Ê±£¬ËùµÃÈÜÒºDµÄÈÜÖÊΪ¡¡¡¡ ¡¡¡¡£¨Ìѧʽ£©£»ÒÑÖªÈÜÒºDÏÔËáÐÔ£¬ÔòDÈÜÒºÖи÷ÖÖÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ                           ¡¡£»
£¨3£©ÔÚ100 mL 18 mol/LµÄFŨÈÜÒºÖмÓÈë¹ýÁ¿Í­Æ¬£¬¼ÓÈÈʹ֮³ä·Ö·´Ó¦£¬²úÉúÆøÌåµÄÌå»ý£¨±ê¿öÏ£©¿ÉÄÜΪ£º¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£
A£®40.32 L    B£®30.24 L     C£®20.16 L     D£®13.44 L

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

¾«Ó¢¼Ò½ÌÍø