·Ö×Óɸ¾ßÓоùÔȵÄ΢¿×½á¹¹£¬·Ö×Óɸɸ·Ö×÷ÓüûÏÂͼ¡£ÓÉÓÚ·Ö×Óɸ¾ßÓÐÎü¸½ÄÜÁ¦¸ß£¬ÈÈÎȶ¨ÐÔÇ¿µÈÆäËüÎü¸½¼ÁËùûÓеÄÓŵ㣬ʹµÃ·Ö×Óɸ»ñµÃ¹ã·ºµÄÓ¦Óá£Ä³ÖÖÐͺŵķÖ×ÓɸµÄ¹¤ÒµÉú²úÁ÷³Ì¿É¼òµ¥±íʾÈçÏ£º


ÔÚ¼ÓNH3¡¤H2Oµ÷½ÚpHµÄ¹ý³ÌÖУ¬ÈôpH¿ØÖƲ»µ±»áÓÐAl(OH)3Éú³É£¬¼ÙÉèÉú²úÁ÷³ÌÖÐÂÁÔªËغ͹èÔªËؾùûÓÐËðºÄ£¬ÄÆÔ­×ÓµÄÀûÓÃÂÊΪ10£¥¡£
£¨1£©·Ö×ÓɸµÄ¿×µÀÖ±¾¶Îª4A(1 A=10-10m)³ÆΪ4AÐÍ·Ö×Óɸ£¬µ±Na+±»Ca2+È¡´úʱ¾ÍÖƵÃ5AÐÍ·Ö×Óɸ£¬µ±Na+±»K+È¡´úʱ¾ÍÖƵÃ3AÐÍ·Ö×Óɸ¡£Òª¸ßЧ·ÖÀëÕý¶¡Íé(·Ö×ÓÖ±¾¶Îª4.65A)ºÍÒ춡Íé(·Ö×ÓÖ±¾¶Îª5.6A)Ó¦¸ÃÑ¡Óà         ÐÍ·Ö×Óɸ¡£
£¨2£©A12(SO4)3ÈÜÒºÓëNa2SiO3ÈÜÒº·´Ó¦Éú³É½ºÌåµÄÀë×Ó·½³ÌʽΪ                    
£¨3£©¸ÃÉú²úÁ÷³ÌÖÐËùµÃÂËÒºÀﺬÓеÄÀë×Ó³ýH+¡¢OH-Í⣬Ö÷ҪΪ                       £»¼ìÑéÆäÖнðÊôÑôÀë×ӵIJÙ×÷·½·¨ÊÇ                                       
£¨4£©¼ÓNH3¡¤H2Oµ÷½ÚpHºó£¬¼ÓÈȵ½90¡æ²¢³ÃÈȹýÂ˵ÄÔ­Òò¿ÉÄÜÊÇ                      
£¨5£©¸ÃÉú²úÁ÷³ÌÖÐËùµÃ·Ö×ÓɸµÄ»¯Ñ§Ê½Îª                             
£¨12·Ö£©
£¨1£©5A£¨1·Ö£©
£¨2£© 2A13++3SiO32-+ 6H2O=2Al(OH)3+3H2SiO3     (2·Ö)
£¨3£©Na+¡¢NH4+¡¢SO42-£¨2·Ö£©   ½«²¬Ë¿£¨»òÌúË¿£©·ÅÔھƾ«µÆÍâÑæÉÏ×ÆÉÕÖÁÎÞɫʱ£¬ÕºÈ¡´ý²âÈÜÒºÔÙ·ÅÔھƾ«µÆÍâÑæÉÏ×ÆÉÕ£¬Èô»ðÑæ³Ê»ÆÉ«£¬ËµÃ÷´ý²âÒºÖк¬ÓÐNa+£¨2·Ö£©
£¨4£©¼ÓÈÈÄÜ´Ù½ø½ºÌåÄý¾Û£¬³ÃÈȹýÂË¿É·ÀÖ¹ÆäËüÔÓÖʽᾧÎö³ö £¨2·Ö£©
£¨5£©Na(AlSi3O12)¡¤3H2O»ò ( Na2O¡¤Al2O3¡¤10SiO2¡¤6H2O)£¨3·Ö£¬ÆäËüºÏÀí´ð°¸¾ù¸ø·Ö£©

ÊÔÌâ·ÖÎö£º¸Ã¹¤ÒÕÊÇÀûÓÃÂÁÀë×Ӻ͹èËá¸ùÀë×Ó·¢ÉúË«Ë®½â·´Ó¦£¬µÃµ½½ºÌ壬Ȼºóͨ¹ý±ºÉÕºóµÃµ½·Ö×Óɸ¡£
£¨1£©¸ù¾ÝÖ±¾¶´óС£¬4.65AºÍ5.6A£¬¹ÊʹÓÃ5A·Ö×Óɸ¿ÉÒÔ·ÖÀ룻
£¨2£©¸Ã·´Ó¦ÎªË«Ë®½â·´Ó¦
£¨3£©¸ù¾Ý¼ÓÈëÎïÖÊΪAl2(SO4)3¡¢Na2SiO3¡¢NH3¡¤H2O£¬¹ÊAl¡¢Si¡¢Naת»¯µ½¾§ÌåÖУ¬¹ÊÂËÒºÖÐÓÐNa+¡¢NH4+¡¢SO42-¡£¼ìÑéNa+ÓÃÑæÉ«·´Ó¦£¬¼ìÑéNH4+ÓÃNaOH£¬ÔÙ¼ìÑéNH3£»
£¨5£©¸ù¾ÝÔ­×ÓÀûÓÃÂʼ°Êغã¹Øϵ¿ÉÒÔÈ·¶¨·Ö×Óʽ¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÔÚÒ±½ð¹¤ÒµÖУ¬¾ù²»ÄÜÓÃͨ³£»¯Ñ§»¹Ô­¼ÁÖƵõĽðÊô×éÊÇ£¨ £©
A£®Na¡¢Mg¡¢AlB£®Na¡¢K¡¢Zn¡¢Fe
C£®Zn¡¢Fe¡¢Cu¡¢AgD£®Mg¡¢Al¡¢Zn¡¢Fe

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

¹¤ÒµÉÏÀûÓ÷ÏÌúм(º¬ÉÙÁ¿Ñõ»¯ÂÁ¡¢Ñõ»¯ÌúµÈ)Éú²ú¼îʽÁòËáÌú[Fe(OH)SO4]µÄ¹¤ÒÕÁ÷³ÌÈçÏ£º

ÒÑÖª£º²¿·ÖÑôÀë×ÓÒÔÇâÑõ»¯ÎïÐÎʽ³ÁµíʱÈÜÒºµÄpH¼ûÏÂ±í£º
³ÁµíÎï
Fe(OH)3
Fe(OH)2
Al(OH)3
¿ªÊ¼³Áµí
2.3
7.5
3.4
ÍêÈ«³Áµí
3.2
9.7
4.4
 
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¼ÓÈë¹ýÁ¿·ÏÌúмµÄÄ¿µÄÊÇ                         £¬´ËʱÈÜÒºÖдæÔÚµÄÑôÀë×ÓÖ÷ÒªÓР               £¬ÓÃNaHCO3µ÷ÕûÈÜÒºpHʱµÄÀë×Ó·½³ÌʽÊÇ                  ¡£
£¨2£©ÔÚʵ¼ÊÉú²úÖУ¬·´Ó¦¢òÖг£Í¬Ê±Í¨ÈëO2ÒÔ¼õÉÙNaNO2µÄÓÃÁ¿£¬Í¬Ê±Í¨ÈëO2µÄ×÷ÓÃÊÇ                                   ¡£
£¨3£©¼îʽÁòËáÌúÈÜÓÚË®ºóÄܵçÀë²úÉú[Fe(OH)]2+Àë×Ó£¬Ð´³ö[Fe(OH)]2+·¢ÉúË®½â·´Ó¦Éú³ÉFe(OH)3µÄÀë×Ó·½³Ìʽ                                          ¡£
£¨4£©ÒÑÖª·ÏÌúмÖÐÌúÔªËصÄÖÊÁ¿·ÖÊýΪ84.0%£¬Èô²»¿¼ÂÇÿ²½·´Ó¦ÖÐÌúÔªËصÄËðºÄ£¬ÏÖÓÐ100¶Ö·ÏÌúмÀíÂÛÉÏ×î¶àÄÜÉú²ú             ¶Ö¼îʽÁòËáÌú¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

¹¤ÒµÉÏÓûÆÍ­¿óÒ±Á¶Í­¼°¶Ô¯Ôü×ÛºÏÀûÓõÄÒ»ÖÖ¹¤ÒÕÁ÷³ÌÈçÏ£º

£¨1£©Ò±Á¶¹ý³ÌÖеõ½Cu2OºÍCuµÄ»ìºÏÎï³ÆΪ¡°ÅÝÍ­¡±£¬ÆäÓë½ðÊôA1ÔÚ¸ßÎÂÌõ¼þÏ»ìºÏ·´Ó¦¿ÉµÃ´ÖÍ­£¬·´Ó¦»¯Ñ§·½³ÌʽΪ________¡£´ÖÍ­¾«Á¶Ê±Ó¦½«´ÖÍ­Á¬½ÓÔÚÖ±Á÷µçÔ´µÄ____¼«£¬¿ÉÔÚ____¼«µÃµ½´¿¶È½Ï¸ßµÄ¾«Í­¡£
£¨2£©´«Í³Á¶Í­µÄ·½·¨Ö÷ÒªÊÇ»ð·¨Á¶Í­£¬ÆäÖ÷Òª·´Ó¦Îª£º

ÿÉú³É1 mol Cu£¬¹²ÏûºÄ____mol O2¡£·´Ó¦¢ÛÖеÄÑõ»¯¼ÁÊÇ____¡£
£¨3£©Á¶Í­²úÉúµÄ¯Ôü(º¬)¿ÉÖƱ¸Fe2O3¡£¸ù¾ÝÁ÷³Ì»Ø´ðÏÂÁÐÎÊÌ⣺
¢Ù¼ÓÈëÊÊÁ¿NaClOÈÜÒºµÄÄ¿µÄÊÇ_______ £¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©¡£
¢Ú³ýÈ¥Al3£«µÄÀë×Ó·½³ÌʽÊÇ____¡£
¢ÛÑ¡ÓÃÌṩµÄÊÔ¼Á£¬Éè¼ÆʵÑéÑé֤¯ÔüÖк¬ÓÐFeO¡£ÌṩµÄÊÔ¼ÁÓУºÏ¡ÑÎËᡢϡÁòËá¡¢KSCNÈÜÒº¡¢KMnO4ÈÜÒº¡¢NaOHÈÜÒº¡¢µâË®¡£ËùÑ¡ÊÔ¼ÁÊÇ____¡£ÊµÑéÉè¼Æ£º________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

¹¤ÒµÉÏÉú²ú¸ßÂÈËᣨ·Ðµã£º90oC£©Ê±»¹Í¬Ê±Éú²úÁËÑÇÂÈËáÄÆ£¬Æ乤ÒÕÁ÷³ÌÈçÏ£º

£¨1£©ÀäÈ´¹ýÂ˵ÄÄ¿µÄÊǽµµÍNaHSO4µÄ          £¬²¢·ÖÀë³öNaHSO4¾§Ìå¡£
£¨2£©·´Ó¦Æ÷2Öз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ                                           £¬SO2µÄ
×÷ÓÃÊÇ×÷      ¼Á¡£
£¨3£©ÉÏÊö¹¤ÒµÉú²ú¸ßÂÈËáµÄ»¯Ñ§·´Ó¦Îª:3NaClO3+3H2SO4£¨Å¨£©£½3NaHSO4+HClO4+2ClO2+H2O£¬Ñõ»¯²úÎïÓ뻹ԭ²úÎïµÄÎïÖʵÄÁ¿Ö®±ÈΪ       ¡£
£¨4£©¿ÉÒÔͨ¹ýÕôÁóÂËÒºµÄ·½·¨µÃµ½¸ßÂÈËáµÄÔ­Òò¿ÉÄÜÊǸßÂÈËáµÄ·Ðµã±È½Ï    £¨Ìî¡°¸ß¡±»ò¡°µÍ¡±£©£¬ÈÝÒ×´ÓÈÜÒºÖÐÒݳö£¬Ñ­»·Ê¹ÓõÄÎïÖÊÊÇ        ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

ÀûÓÃËá½â·¨ÖÆîÑ°×·Û²úÉúµÄ·ÏÒº[º¬ÓдóÁ¿FeSO4¡¢H2SO4ºÍÉÙÁ¿Fe2(SO4)3¡¢TiOSO4]£¬Éú²úÌúºìºÍ²¹Ñª¼ÁÈéËáÑÇÌú¡£ÆäÉú²ú²½ÖèÈçÏ£º

ÒÑÖª£ºTiOSO4¿ÉÈÜÓÚË®£¬ÔÚË®ÖпÉÒÔµçÀëΪTiO2+ºÍSO42¡ª¡£Çë»Ø´ð£º
£¨1£©²½Öè¢ÙÖзÖÀëÁòËáÑÇÌúÈÜÒººÍÂËÔüµÄ²Ù×÷ÖÐËùÓõIJ£Á§ÒÇÆ÷ÊÇ                     ¡£
²½Öè¢ÚµÃµ½ÁòËáÑÇÌú¾§ÌåµÄ²Ù×÷ΪÕô·¢Å¨Ëõ¡¢                                          ¡£
£¨1£©²½Öè¢ÜµÄÀë×Ó·½³ÌʽÊÇ                                                           ¡£
£¨1£©²½Öè¢Þ±ØÐë¿ØÖÆÒ»¶¨µÄÕæ¿Õ¶È£¬Ô­ÒòÊÇÓÐÀûÓÚÕô·¢Ë®ÒÔ¼°                           ¡£
£¨1£©ÁòËáÑÇÌúÔÚ¿ÕÆøÖÐìÑÉÕÉú³ÉÌúºìºÍÈýÑõ»¯Áò£¬¸Ã·´Ó¦ÖÐÑõ»¯¼ÁºÍ»¹Ô­¼ÁµÄÎïÖʵÄÁ¿Ö®±ÈΪ             ¡£
£¨1£©ÓÃƽºâÒƶ¯µÄÔ­Àí½âÊͲ½Öè¢ÝÖмÓÈéËáÄܵõ½ÈéËáÑÇÌúµÄÔ­Òò                            ¡£
£¨1£©Îª²â¶¨²½Öè¢ÚÖÐËùµÃ¾§ÌåÖÐFeSO4¡¤7H2OµÄÖÊÁ¿·ÖÊý£¬È¡¾§ÌåÑùÆ·a g£¬ÈÜÓÚÏ¡ÁòËáÅä³É100£®00 mLÈÜÒº£¬È¡³ö20£®00 mLÈÜÒº£¬ÓÃKMnO4ÈÜÒºµÎ¶¨£¨ÔÓÖÊÓëKMnO4²»·´Ó¦£©¡£ÈôÏûºÄ0£®1000 mol?L-1 KMnO4ÈÜÒº20£®00 mL£¬ËùµÃ¾§ÌåÖÐFeSO4¡¤7H2OµÄÖÊÁ¿·ÖÊýΪ£¨ÓÃa±íʾ£©                  ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

ÏÂͼÊǹ¤ÒµÉÏÉú²ú̼Ëá﮵IJ¿·Ö¹¤ÒÕÁ÷³Ì£¬Çë¸ù¾ÝÁ÷³Ìͼ¼°ÒÑÖªÐÅÏ¢»Ø´ðÎÊÌâ¡£

¢Ú
¢Û¼¸ÖÖÎïÖʲ»Í¬Î¶ÈϵÄÈܽâ¶È¡£

£¨1£©´ÓÂËÔü1ÖзÖÀë³öAl2O3µÄ²¿·ÖÁ÷³ÌÈçÏÂͼËùʾ£¬À¨ºÅ±íʾ¼ÓÈëµÄÊÔ¼Á£¬·½¿ò±íʾËùµÃµ½µÄÎïÖÊ¡£Ð´³öͼÖТ١¢¢Ú¡¢¢Û±íʾµÄ¸÷ÎïÖÊ£¬²½ÖèIIÖз´Ó¦µÄÀë×Ó·½³ÌʽÊÇ       ¡£

£¨2£©ÒÑÖªÂËÔü2µÄÖ÷Òª³É·ÖÓÐMg£¨OH£©2ºÍCaCO3£¬Ð´³öÉú³ÉÂËÔü2·´Ó¦µÄÀë×Ó·½³Ìʽ£º
                      ¡£
£¨3£©ÏòÂËÒº2ÖмÓÈë±¥ºÍNa2CO£¬ÈÜÒº£¬¹ýÂ˺ó£¬Óá°ÈÈˮϴµÓ¡±µÄÔ­ÒòÊÇ        ¡£
£¨4£©¹¤ÒµÉÏ£¬½«Li2CO3´ÖÆ·ÖƱ¸³É¸ß´¿Li2CO3µÄ²¿·Ö¹¤ÒÕÈçÏ¡£
¢Ù½«´Ö²úÆ·Li2CO3ÈÜÓÚÑÎËá×÷Óýâ²ÛµÄÑô¼«Òº£¬LiOHÈÜÒº×÷Òõ¼«Òº£¬Á½ÕßÓÃÀë×ÓÑ¡Ôñ°ë͸Ĥ¸ô¿ª£¬ÓöèÐԵ缫µç½â¡£Ñô¼«µÄµç¼«·´Ó¦Ê½ÊÇ          ¡£
¢Úµç½âºóÏò²úÆ·LiOHÈÜÒºÖмÓÈë¹ýÁ¿NH4HCO£¬ÈÜÒºÉú³ÉLi2CO3·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ    ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

¸ßÌúËá¼Ø(K2Fe04)ÊÇÒ»ÖÖ¼¯Ñõ»¯¡¢Îü¸½¡¢ÐõÄýÓÚÒ»ÌåµÄÐÂÐͶ๦ÄÜË®´¦Àí¼Á£¬ÆäÉú²ú¹¤ÒÕÈçÏ£º

ÒÑÖª£º2Fe(NO3)3+3KClO+10KOH=2K2FeO4+6KNO3+3KCl+5H2O
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©½«Cl2ͨÈëKOHÈÜÒºÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ____________¡£
£¨2£©Ð´³ö¹¤ÒµÉÏÖÆÈ¡Cl2µÄ»¯Ñ§·½³Ìʽ____________¡£
£¨3£©ÔÚ¡°·´Ó¦ÒºI¡±ÖмÓÈëKOH¹ÌÌåµÄÄ¿µÄÊÇ____________¡£
£¨4£©K2FeO4¿É×÷ΪÐÂÐͶ๦ÄÜË®´¦Àí¼ÁµÄÔ­ÒòÊÇ____________¡£
£¨5£©ÅäÖÆKOHÈÜҺʱ£¬½«61.6g KOH¹ÌÌåÈܽâÔÚ100 mLË®ÖУ¬ËùµÃÈÜÒºµÄÃܶÈΪ1.47 g ? mL-1£¬Ôò¸ÃÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ____________¡£
£¨6£©´Ó¡°·´Ó¦ÒºII¡±ÖзÖÀë³öK2Fe04ºó£¬¸±²úÆ·ÊÇ___________ (д»¯ Ñ§ Ê½£©¡£
£¨7£©¸Ã¹¤ÒÕÿµÃµ½1.98kgK2FeO4£¬ÀíÂÛÉÏÏûºÄCl2µÄÎïÖʵÄÁ¿Îª______mol¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁдëÊ©Öв»ÄÜÆðµ½½ÚÄܽµºÄºÍ±£»¤»·¾³Ä¿µÄµÄÊÇ
A£®¼Ó´óÌ«ÑôÄÜ¡¢Ë®ÄÜ¡¢·çÄÜ¡¢µØÈÈÄܵÈÄÜÔ´¿ª·¢Á¦¶È£¬¼õÉÙ»¯Ê¯È¼ÁϵÄʹÓÃ
B£®ÑÐÖÆÒÒ´¼ÆûÓͼ¼Êõ£¬½µµÍ»ú¶¯³µÁ¾Î²ÆøÖÐÓк¦ÆøÌåÅÅ·Å
C£®Éú²ú¡¢Éú»îÖУ¬Å¬Á¦ÊµÏÖ×ÊÔ´µÄÑ­»·ÀûÓÃ
D£®ÀûÓøßм¼Êõ£¬Ìá¸ßʯÓÍ¡¢Ãº¡¢ÌìÈ»Æø²úÁ¿£¬ÒÔÂú×㹤ҵÉú²ú¿ìËÙ·¢Õ¹µÄÐèÇó

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸