ÌâÄ¿ÄÚÈÝ

A¡¢B¡¢C¡¢D¾ùΪÖÐѧ»¯Ñ§Öг£¼ûµÄÎïÖÊ£¬ËüÃÇÖ®¼äµÄת»¯¹ØϵÈçÏÂͼ£¨²¿·Ö²úÎïÒÑÂÔÈ¥£©£º

ÊԻشð£º
£¨1£©ÈôDÊǾßÓÐÑõ»¯ÐԵĵ¥ÖÊ£¬ÔòÊôÓÚ¶ÌÖÜÆÚµÄÖ÷×å½ðÊôÔªËØAΪ      £¨ÌîÔªËØ·ûºÅ£©¡£
£¨2£©ÈôDÊǽðÊôµ¥ÖÊ£¬DÔÚ³±ÊªµÄ¿ÕÆøÖÐÒ×·¢ÉúÎüÑõ¸¯Ê´£¬CÈÜÒºÔÚ±£´æʱӦ¼ÓÈëËáºÍÉÙÁ¿D·ÀÖ¹Æä±äÖÊ£¬Èô²»¼ÓDÔòCÈÜÒºÔÚ¿ÕÆøÖбäÖʵÄÀë×Ó·½³ÌʽΪ                    £»½«DµÄÂÈ»¯ÎïµÄË®ÈÜÒºÕô¸É²¢×ÆÉÕ²úÎïÊÇ            ¡£
£¨3£©ÈôA¡¢B¡¢C¾ùΪÎÞ»ú»¯ºÏÎÇÒ¾ùº¬µØ¿ÇÖк¬Á¿×î¸ßµÄ½ðÊôÔªËØE£¬ÔÚÈÜÒºÖÐAºÍC·´Ó¦Éú³ÉB¡£Çëд³öBת»¯ÎªCµÄËùÓпÉÄܵÄÀë×Ó·½³Ìʽ                      ¡£
£¨4£©½«µÚ£¨1£©ÌâÍƳöµÄAµ¥ÖÊÓëµÚ£¨3£©ÌâEµ¥ÖʵĻìºÏÎï11.9gͶÈëÒ»¶¨Á¿µÄË®Öгä·Ö·´Ó¦£¬AÓëE¾ùûÓÐÊ£Ó࣬¹²ÊÕ¼¯µ½±ê×¼×´¿öϵÄÆøÌåvL¡£ÏòËùµÃÈÜÒºÖÐÖðµÎ¼ÓÈëŨ¶ÈΪ2mol?L-1µÄH2SO4ÈÜÒº£¬ÖÁ100mLʱ°×É«³Áµí´ïµ½×î´óÁ¿¡£Ôòv=       ¡£

£¨1£©Na
£¨2£©4Fe2£«£«O2+4H+£½4Fe3£«+2H2O£»    Fe2 O3
£¨3£©Al(OH)3£«3H£«£½Al3£«£«3H2O¡¢Al(OH)3£«OH£­£½AlO2£­£«2H2O
£¨4£©7.84L

½âÎöÊÔÌâ·ÖÎö£º£¨1£©³£¼ûÓÐÑõ»¯ÐԵĵ¥ÖÊΪO2¡¢Cl2£¬AΪ¶ÌÖÜÆÚ½ðÊôÔªËØ£¬ÔòÓÉת»¯¹Øϵ¿ÉÒÔAÊÇNa£¬BÊÇNa2O£¬CÊÇNa2O2£¬DÊÇO2¡£
£¨2£©DÊÇÔÚ³±ÊªµÄ¿ÕÆøÖÐÒ×·¢ÉúÎüÑõ¸¯Ê´µÄ½ðÊôµ¥ÖÊ£¬ÊÇÌúÔªËØ£¬ÔòCÖк¬ÓÐFe2+£¬½«FeCl3µÄË®ÈÜÒºÕô¸É²¢×ÆÉÕ£¬ÔòÓÚFe3+µÄË®½âºÍHClµÄ»Ó·¢£¬×îÖյõ½µÄ²úÎïÊÇFe2O3¡£
£¨3£©µØ¿ÇÖк¬Á¿×î¸ßµÄ½ðÊôÔªËØÊÇÂÁ£¬ÓÉA¡¢B¡¢CÏ໥ת»¯¹Øϵ¿ÉÖªBÊÇAl(OH)3£¬AºÍCÖзֱðº¬ÓÐAl3+¡¢AlO2£­£¬µ«ÎÞ·¨È·ÈϾßÌ庬ÓÐÄĸöÀë×Ó£¬¹ÊBת»¯ÎªC¿ÉÄÜÊÇÓÉAl(OH)3ת»¯ÎªAl3£«»òÓÉAl(OH)3ת»¯ÎªAlO2£­¡£
£¨4£©¼ÓÈë100 mL 2mol?L-1µÄH2SO4ÈÜҺʱ³Áµí´ï×î´óÁ¿£¬´ËʱÈÜÒºÖнöº¬Na2SO4£¬ÓÉÔ­×ÓÊغã¿ÉÖª£¬n(Na)=2n(Na2SO4)=2n(H2SO4)="2" ¡Á0.1 L¡Á2mol?L-1="0.4" mol£¬ÔòÔ­»ìºÏÎïÖÐÓÐÄƵÄÖÊÁ¿Îª0.4 mol¡Á23 mol·L£­1=9.2g£¬ÂÁµÄÖÊÁ¿Îª11.9g£­9.2g=2.7g£¬¹Ê·´Ó¦²úÉúH2µÄÌå»ýΪ(¡Á0.4mol+¡Á)¡Á22.4L/mol="7.84" L¡£
¿¼µã£ºÄƵĻ¯Ñ§ÐÔÖÊ£»»¯Ñ§·½³ÌʽµÄÓйؼÆË㣻ÂÁµÄ»¯Ñ§ÐÔÖÊ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¶ÌÖÜÆÚµÄÎåÖÖÔªËØA¡¢B¡¢C¡¢D¡¢E,Ô­×ÓÐòÊýÒÀ´ÎÔö´ó¡£A¡¢B¡¢CÈýÖÖÔªËصç×Ó²ãÊýÖ®ºÍÊÇ5¡£A¡¢BÁ½ÔªËØÔ­×Ó×îÍâ²ãµç×ÓÊýÖ®ºÍµÈÓÚCÔªËØÔ­×Ó×îÍâ²ãµç×ÓÊý;BÔªËØÔ­×Ó×îÍâµç×Ó²ãÉϵĵç×ÓÊýÊÇËüµÄµç×Ó²ãÊýµÄ2±¶,AÓëD¿ÉÒÔÐγÉÔ­×Ó¸öÊý±È·Ö±ðΪ1¡Ã1ºÍ2¡Ã1µÄÁ½ÖÖҺ̬»¯ºÏÎï;Eµ¥ÖÊÓÃÓÚ¾»»¯Ë®ÖÊ¡£
Çë»Ø´ð:
(1)д³öDÔÚÔªËØÖÜÆÚ±íÖеÄλÖá¡                    
EµÄÔ­×ӽṹʾÒâͼÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£ 
ÏÂÁпÉÒÔÑéÖ¤CÓëDÁ½ÔªËØÔ­×ӵõç×ÓÄÜÁ¦Ç¿ÈõµÄʵÑéÊÂʵÊÇ¡¡¡¡¡¡¡¡(Ìîд±àºÅ)¡£ 
A.±È½ÏÕâÁ½ÖÖÔªËصÄÆø̬Ç⻯ÎïµÄ·Ðµã
B.±È½ÏÖ»ÓÐÕâÁ½ÖÖÔªËØËùÐγɵĻ¯ºÏÎïÖеĻ¯ºÏ¼Û
C.±È½ÏÕâÁ½ÖÖÔªËصÄÆø̬Ç⻯ÎïµÄÎȶ¨ÐÔ
D.±È½ÏÕâÁ½ÖÖÔªËصĵ¥ÖÊÓëÇâÆø»¯ºÏµÄÄÑÒ×
(2)ÓÉA¡¢BÁ½ÖÖÔªËØ×é³ÉµÄ×î¼òµ¥µÄ»¯ºÏÎï,д³öÆäµç×Óʽ¡¡¡¡¡¡¡¡¡£ 
(3)¾ùÓÉA¡¢B¡¢C¡¢DËÄÖÖÔªËØ×é³ÉµÄ¼×¡¢ÒÒÁ½ÖÖ»¯ºÏÎï,¶¼¼È¿ÉÒÔÓëÑÎËá·´Ó¦ÓÖ¿ÉÒÔÓëNaOHÈÜÒº·´Ó¦,¼×ΪÎÞ»úÑÎ,Æ仯ѧʽΪ¡¡¡¡¡¡¡¡,ÒÒΪÌìÈ»¸ß·Ö×Ó»¯ºÏÎïµÄË®½â²úÎï,ÇÒÊÇͬÀàÎïÖÊÖÐÏà¶Ô·Ö×ÓÖÊÁ¿×îСµÄ,Æä½á¹¹¼òʽΪ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£ 
(4)½ºÌ¬´ÅÁ÷ÌåÔÚҽѧÉÏÓÐÖØÒªµÄÓÃ;,¶øÄÉÃ×¼¶Fe3O4ÊÇ´ÅÁ÷ÌåÖеÄÖØÒªÁ£×Ó,ÆäÖƱ¸¹ý³Ì¿É¼òµ¥±íʾÈçÏÂ:
¢Ù½«»¯ºÏÎïCA3ͨÈëµÈÎïÖʵÄÁ¿µÄFeSO4¡¢Fe2(SO4)3µÄ»ìºÏÈÜÒºÖÐ,Éú³ÉÁ½ÖÖ¼î,д³ö¸Ã·´Ó¦¹ý³ÌµÄ×ܵÄÀë×Ó·½³Ìʽ¡¡                                  ¡£ 
¢ÚÉÏÊö·´Ó¦Éú³ÉµÄÁ½ÖÖ¼î¼ÌÐø×÷ÓÃ,µÃµ½Fe3O4¡£
(5)ÒÑ֪ϱíÊý¾Ý:

ÎïÖÊ
Fe(OH)2
Fe(OH)3
Ksp/25 ¡æ
2.0¡Á10-16
4.0¡Á10-36
 
Èôʹ»ìºÏÒºÖÐFeSO4¡¢Fe2(SO4)3µÄŨ¶È¾ùΪ2.0 mol¡¤L-1,Ôò»ìºÏÒºÖÐc(OH-)²»µÃ´óÓÚ¡¡¡¡¡¡¡¡mol¡¤L-1¡£ 

ij¿óʯ¿ÉÄÜÊÇÓÉFeCO3¡¢SiO2¡¢Al2O3ÖеÄÒ»ÖÖ»ò¼¸ÖÖ×é³É£¬Ä³Ñо¿ÐÔѧϰС×éÓû·ÖÎöÆä³É·Ö£¬ÊµÑé¼Ç¼ÈçÏÂͼËùʾ¡£

£¨1£©¸Ã¿óʯÖк¬ÓР                                                 (Ìѧʽ)£¬Çëд³öʵÑé¹ý³Ì¢ÙÖз´Ó¦µÄÀë×Ó·½³Ìʽ                                                         ¡£
£¨2£©½«Ñõ»¯Îï¢òÔÚÈÛÈÚ״̬ϵç½â£¬¿ÉÒԵõ½Ä³½ðÊôµ¥ÖÊ¡£µ±Ñô¼«ÉÏÊÕ¼¯µ½ÆøÌå33£®6L(ÒÑÕÛËã³É±ê×¼×´¿ö)ʱ£¬Òõ¼«Éϵõ½¸Ã½ðÊô           g¡£
£¨3£©ÊÔд³ö¸Ã½ðÊôÓëÑõ»¯ÎïI·´Ó¦µÄ»¯Ñ§·½³Ìʽ                                £»½«¸Ã·´Ó¦µÄ²úÎï¼ÓÈëµ½×ãÁ¿ÉÕ¼îÈÜÒºÖУ¬Çëд³ö·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ                                       ¡£
£¨4£©¾­½øÒ»²½·ÖÎö¸Ã¿óʯÖл¹º¬ÓÐ΢Á¿µÄSrCO3(ÉÏÊö·½°¸¼ì²â²»³ö)¡£ïÈ(Sr)ΪµÚÎåÖÜÆÚ¢òA×åÔªËØ¡£ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ               (ÌîÐòºÅ)¡£
¢ÙÂÈ»¯ïÈ(SrCl2)ÈÜÒºÏÔËáÐÔ
¢ÚSrSO4ÄÑÈÜÓÚË®
¢Û¹¤ÒµÉÏ¿ÉÒÔÓõç½âSrCl2µÄË®ÈÜÒºÖÆÈ¡½ðÊôïÈ(Sr)
¢Ü¸ß´¿ÁùË®ÂÈ»¯ïȾ§Ìå(SrCl2¡¤6H2O)±ØÐëÔÚHCl·ÕΧÖмÓÈȲÅÄܵõ½SrCl2

³£¼ûÔªËØA¡¢B¡¢M×é³ÉµÄËÄÖÖÎïÖÊ·¢Éú·´Ó¦:¼×+ÒÒ=±û+¶¡,ÆäÖм×ÓÉAºÍM×é³É,ÒÒÓÉBºÍM×é³É,±ûÖ»º¬ÓÐM¡£
£¨1£©Èô¼×Ϊµ­»ÆÉ«¹ÌÌ壬ÒҺͱû¾ùΪ³£ÎÂϵÄÎÞÉ«ÎÞζÆøÌå¡£ÔòÒҵĵç×ÓʽΪ         ;Éú³É±ê×¼×´¿öÏÂ5.6L±ûתÒƵĵç×ÓÊýΪ         ;³£ÎÂ϶¡ÈÜÒºpH     7,ÓÃÀë×Ó·½³Ìʽ½âÊÍ                                                                  ¡£
£¨2£©Èô¶¡ÎªÄÜʹƷºìÍÊÉ«µÄÎÞÉ«ÆøÌ壬±ûΪ³£¼ûºìÉ«½ðÊô£¬»¯ºÏÎï¼×¡¢ÒÒÖÐÔ­×Ó¸öÊý±È¾ùΪ1:2(M¾ùÏÔ+1¼Û),Ô­×ÓÐòÊýB´óÓÚA¡£Ôò¢ÙAÔÚÖÜÆÚ±íÖÐλÖÃΪ           ¢Ú¶¡ÓëË®·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                 Ïò·´Ó¦ºóÈÜÒºÖеμÓÁ½µÎ×ÏɫʯÈïÊÔÒºµÄÏÖÏóΪ                                 
¢ÛÕýÈ·ÊéдÉÏÊöÉú³É±ûµÄ»¯Ñ§·½³Ìʽ                                           
¢ÜÏòMCl2µÄÈÜÒºÖÐͨÈ붡,¿É¹Û²ìµ½°×É«µÄMCl³Áµí,д³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ                                                      ¡£

X¡¢Y¡¢Z¡¢Q¡¢MΪ³£¼ûµÄ¶ÌÖÜÆÚÔªËØ£¬ÆäÔ­×ÓÐòÊýÒÀ´ÎÔö´ó¡£ÓйØÐÅÏ¢Èçϱí:

X
¶¯Ö²ÎïÉú³¤²»¿ÉȱÉÙµÄÔªËØ£¬Êǵ°°×ÖʵÄÖØÒª³É·Ö
Y
µØ¿ÇÖк¬Á¿¾ÓµÚһλ
Z
¶ÌÖÜÆÚÖÐÆäÔ­×Ӱ뾶×î´ó
Q
Éú»îÖдóÁ¿Ê¹ÓÃÆäºÏ½ðÖÆÆ·£¬¹¤ÒµÉÏ¿ÉÓõç½âÆäÑõ»¯ÎïµÄ·½·¨ÖƱ¸
M
º£Ë®ÖдóÁ¿¸»¼¯µÄÔªËØÖ®Ò»£¬Æä×î¸ßÕý»¯ºÏ¼ÛÓ븺¼ÛµÄ´úÊýºÍΪ6
 
£¨1£©XµÄÆø̬Ç⻯ÎïµÄ´óÁ¿Éú²úÔø¾­½â¾öÁ˵ØÇòÉÏÒòÁ¸Ê³²»×ã¶øµ¼Öµļ¢¶öºÍËÀÍöÎÊÌ⣬Çëд³ö¸ÃÆø̬Ç⻯ÎïµÄµç×Óʽ____________¡£
£¨2£©ÒÑÖª37RbºÍ53I¶¼Î»ÓÚµÚÎåÖÜÆÚ£¬·Ö±ðÓëZºÍMͬһÖ÷×å¡£ÏÂÁÐÓйØ˵·¨ÕýÈ·µÄÊÇ____________£¨ÌîÐòºÅ£©¡£
A£®Ô­×Ӱ뾶£º  Rb>I                   
B£®RbMÖк¬Óй²¼Û¼ü
C£®Æø̬Ç⻯ÎïÈÈÎȶ¨ÐÔ£ºM>I
D£®Rb¡¢Q¡¢MµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯Îï¿ÉÒÔÁ½Á½·¢Éú·´Ó¦
£¨3£©»¯ºÏÎïQXµ¼ÈÈÐԺã¬ÈÈÅòÕÍϵÊýС£¬ÊÇÁ¼ºÃµÄÄÍÈȳå»÷²ÄÁÏ¡£¿¹ÈÛÈÚ½ðÊôÇÖÊ´µÄÄÜÁ¦Ç¿£¬ÊÇÈÛÖý´¿Ìú¡¢ÂÁ»òÂÁºÏ½ðÀíÏëµÄÛáÛö²ÄÁÏ¡£Óйػ¯ºÏÎïQXµÄÖƱ¸¼°»¯Ñ§ÐÔÖÊÈçÏ£¨ËùÓÐÈÈÁ¿Êý¾Ý¾ùÒÑÕÛºÏΪ25¡æ¡¢101.3 kPaÌõ¼þϵÄÊýÖµ£©
¿ÉÓÃQºÍXµÄµ¥ÖÊÔÚ800 ~ 1000¡æÖƵã¬Ã¿Éú³É1 mol QX£¬ÎüÊÕa kJµÄÈÈÁ¿¡£
¿ÉÓÃQµÄÑõ»¯Îï¡¢½¹Ì¿ºÍXµÄµ¥ÖÊÔÚ1600 ~ 1750¡æÉú³ÉQX£¬Ã¿Éú³É1 mol QX£¬ÏûºÄ18 g̼£¬ÎüÊÕb kJµÄÈÈÁ¿¡£
Çë¸ù¾ÝÉÏÊöÐÅϢд³öÔÚÀíÂÛÉÏQµÄÑõ»¯Îï¸ú½¹Ì¿·´Ó¦Éú³ÉQµ¥ÖʺÍCOµÄÈÈ»¯Ñ§·½³Ìʽ________________¡£
£¨4£©X¡¢Y×é³ÉµÄÒ»ÖÖÎÞÉ«ÆøÌåÓö¿ÕÆø±äΪºì×ØÉ«¡£½«±ê×¼×´¿öÏÂ40 £Ì¸ÃÎÞÉ«ÆøÌåÓë15 £ÌÑõÆøͨÈëÒ»¶¨Å¨¶ÈµÄNaOHÈÜÒºÖУ¬Ç¡ºÃ±»ÍêÈ«ÎüÊÕ£¬Í¬Ê±Éú³ÉÁ½ÖÖÑΡ£Çëд³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ                 ¡£

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

¾«Ó¢¼Ò½ÌÍø