ÌâÄ¿ÄÚÈÝ

¸ù¾ÝÒªÇóÍê³ÉÏÂÁи÷СÌâ
£¨1£©¢ÙʵÑéÊÒÓüÓÈȹÌÌå»ìºÏÎïµÄ·½·¨ÖƱ¸°±ÆøµÄ»¯Ñ§·´Ó¦·½³ÌʽÊÇ             ¡£
¢ÚΪÁ˵õ½¸ÉÔïµÄNH3£¬ÓÃ________×ö¸ÉÔï¼Á¡££¨Ìî±àºÅ£©

A£®¼îʯ»Ò B£®Å¨H2SO4 C£®ÎÞË®CaCl2 D£®P2O5 
£¨2£©H2O²ÎÓëµÄÖû»·´Ó¦£º
·ûºÏX+W¡úY+V£¬ÒÑÖªXºÍY·Ö±ðÊǶÌÖÜÆÚͬÖ÷×åÁ½ÖÖÔªËØÐγɵĵ¥ÖÊ£¬ W¡¢VÊÇ»¯ºÏÎï
¢ÙWÊÇË®ÇÒVµÄÑæÉ«·´Ó¦Îª»ÆÉ«£¬Àë×Ó·½³Ìʽ                               £»
¢ÚVÊÇË®£¬»¯Ñ§·½³ÌʽΪ                                            ¡£

(7·Ö)
£¨1£©¢Ù2NH4Cl + Ca(OH)22NH3¡ü + CaCl2  + 2H2O£¨2·Ö£©
¢ÚA£¨1·Ö£©
(2)¢Ù2Na +2 H2O ="2Na+" +2OH- + H2¡ü£¨2·Ö£©
¢Ú2H2S+O2 ==2S+2H2O£¨2·Ö£©

½âÎöÊÔÌâ·ÖÎö£º£¨1£©¢ÙʵÑéÊÒÓüÓÈȹÌÌå»ìºÏÎïµÄ·½·¨ÖƱ¸°±ÆøµÄ»¯Ñ§·´Ó¦·½³ÌʽÊÇ2NH4Cl + Ca(OH)22NH3¡ü + CaCl2  + 2H2O¢ÚΪÁ˵õ½¸ÉÔïµÄNH3£¬ÓÃA¡¢¼îʯ»Ò×ö¸ÉÔï¼Á¡££¨2£©¢ÙWÊÇË®ÇÒVµÄÑæÉ«·´Ó¦Îª»ÆÉ«£¬ËµÃ÷VÖк¬ÓÐÄÆÔªËØ£¬Àë×Ó·½³Ìʽ2Na +2 H2O ="2Na+" +2OH- + H2¡ü¢ÚVÊÇË®£¬ »¯Ñ§·½³ÌʽΪ2H2S+O2 ==2S+2H2O¡£
¿¼µã£º±¾Ì⿼²é°±ÆøµÄÖƱ¸Óë¸ÉÔÓëË®²ÎÓëµÄ·´Ó¦·½³ÌʽµÄÊéд¡£

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                                                      ¡£

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

¾«Ó¢¼Ò½ÌÍø