ÌâÄ¿ÄÚÈÝ

£¨10·Ö£©ÉսᷨÖÆÑõ»¯ÂÁÉú²ú¹ý³ÌÈçÏ£º

ÒÑÖª£º¢ÙÂÁÍÁ¿óÖ÷Òª³É·ÖΪ£ºAl2O3¡¢SiO2¡¢Fe2O3ºÍTiO2¡£  ¢Ú¸ßÎÂÉÕ½áʱ£¬Al2O3¡¢Fe2O3¡¢TiO2¶¼ÄÜ·¢Éú·´Ó¦·Ö±ðÉú³ÉNaAlO2¡¢Na2Fe2O4ºÍÄÑÈÜÓÚË®µÄCaTiO3¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Na2Fe2O4ÓöË®±ã·¢ÉúË®½â·´Ó¦Éú³ÉFe(OH)3£¬Ð´³öNa2Fe2O4Ë®½â»¯Ñ§·´Ó¦·½³Ìʽ  ¡ø ¡£
£¨2£©½þ³öʱÔÙ¼ÓÈëCaOµÄÄ¿µÄÊÇ         ¡ø           ¡£
£¨3£©½þ³öÒºÖз¢ÉúµÄÉú³ÉAl(OH)3µÄÀë×Ó·´Ó¦·½³Ìʽ      ¡ø      ¡£
£¨4£©ÂËÒºµÄÖ÷Òª³É·ÖÊÇ  ¡ø £¨Ð´»¯Ñ§Ê½£©£»ÂËҺѭ»·Ê¹ÓõÄÓÅµã  ¡ø ¡££¨ÈδðÒ»µã£©
£¨¹²10·Ö£©(1) Na2Fe2O4 + 4H2O ="=" 2NaOH + 2Fe(OH)3¡ý £¨2·Ö£©
(2)³ýÈ¥¿ÉÄÜÈÜÓÚË®µÄ¹èËáÑΣ¬Ìá¸ßÑõ»¯ÂÁ´¿¶È£¨2·Ö£©
(3) AlO2- + CO2 + 2H2O ="=" Al(OH)3¡ý + HCO3- £¨2·Ö£©
(4) NaHCO3¡¢Ca(HCO3)2£¨2·Ö£¬¸÷1·Ö£©£»¼õÉÙ¹¤Òµ·ÏË®ÅÅ·Å»ò½ÚÔ¼Ô­ÁÏ£¬½µµÍ³É±¾£¨2·Ö£©£¨ÆäËüºÏÀí´ð°¸Ò²¸ø·Ö£©
ÂÔ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
¹¤ÒµÉϲÉÓÃÁò»¯ÄÆ£­Ê¯»ÒÌúÑη¨´¦Àí¸ßÉé·ÏË®£¨ÉéµÄÖ÷Òª´æÔÚÐÎʽΪH3AsO3£©È¡µÃÁ˺ܺõÄЧ¹û¡£ÊµÏÖÁË·ÏË®´¦Àí¹ý³ÌµÄ¡°Èý·Ï¡±ÁãÅÅ·Å¡£ÆäÖ÷Òª¹¤ÒÕÁ÷³ÌÈçÏ£º

ÒÑÖª£º¢ÙÑÇÉéËᣨH3AsO3£©»¹Ô­ÐÔ½ÏÇ¿£¬Ò×±»Ñõ»¯ÎªÉéËᣨH3AsO4£©
¢ÚÑÇÉéËáÑεÄÈܽâÐÔ´óÓÚÏàÓ¦µÄÉéËáÑΣÛÈçKsp(FeAsO3)£¾Ksp(FeAsO4)£Ý
Çë»Ø´ðÒÔÏÂÎÊÌ⣺
£¨1£©·ÏÆøµÄÖ÷Òª³É·ÖΪ        £¬Í¨¹ýÓëʯ»Ò·¢Éú        ·´Ó¦£¨Ìî»ù±¾·´Ó¦ÀàÐÍ£©±»ÎüÊÕ¡£
£¨2£©Ð´³öÒ»¼¶³ÁÉé¹ý³ÌÖÐÉú³É´Æ»ÆµÄÀë×Ó·½³Ìʽ£º                               ¡£
£¨3£©ÂËÒºAÖУ¬³ýÁËÓÐNa2SO4¡¢H2SO4ÒÔÍ⣬»¹ÓÐÈÜÖÊ                         ¡£
£¨4£©¶þ¼¶³ÁÉé¹ý³ÌÖÐʹÓÃË«ÑõË®µÄÄ¿µÄÓР             ¡£
A£®½«Èý¼ÛÉéÑõ»¯ÎªÎå¼ÛÉ飬ÒÔÌá¸ß³ýÉéЧ¹û
B£®½«Fe2+Ñõ»¯¿ÉÉú³ÉFe(OH) 3³Áµí£¬ÒÔ¼ÓËÙÐü¸¡ÎïµÄ³Á½µ
C£®×÷ÂÌÉ«Ñõ»¯¼Á£¬²»Òý½øеÄÔÓÖÊ
£¨5£©¹ýÂ˲Ù×÷³£ÓõIJ£Á§ÒÇÆ÷ÓУº                              
£¨6£©ÂËÔüBµÄÖ÷Òª³É·ÖÓР                         £¨Ð´Á½ÖÖ£¬Óû¯Ñ§Ê½±íʾ£©¡£
£¨1£©ºÏ³É°±¹¤Òµ¶Ô»¯Ñ§¹¤ÒµºÍ¹ú·À¹¤Òµ¾ßÓÐÖØÒªÒâÒå¡£¹¤ÒµºÏ³É°±Éú²úʾÒâͼÈçͼ¼×Ëùʾ¡£

¢ÙXµÄ»¯Ñ§Ê½Îª__________£¬ÊôÓÚ________£¨Ìî¡°¼«ÐÔ¡±»ò¡°·Ç¼«ÐÔ¡±£©·Ö×Ó¡£
¢Úͼ¼×ÖÐÌõ¼þÑ¡¶¨µÄÖ÷ÒªÔ­ÒòÊÇ£¨Ñ¡Ìî×ÖĸÐòºÅ£¬ÏÂͬ£©________¡£
A£®Î¶ȡ¢Ñ¹Ç¿¶Ô»¯Ñ§Æ½ºâµÄÓ°Ïì
B£®Ìú´¥Ã½ÔÚ¸ÃζÈʱ»îÐÔ´ó
C£®¹¤ÒµÉú²úÊܶ¯Á¦¡¢²ÄÁÏ¡¢É豸µÈÌõ¼þµÄÏÞÖÆ
¢Û¸Ä±ä·´Ó¦Ìõ¼þ£¬»áʹƽºâ·¢ÉúÒƶ¯¡£Í¼ÒÒ±íʾËæÌõ¼þ¸Ä±ä£¬°±ÆøµÄ°Ù·Öº¬Á¿µÄ±ä»¯Ç÷ÊÆ¡£µ±ºá×ø±êΪѹǿʱ£¬±ä»¯Ç÷ÊÆÕýÈ·µÄÊÇ________£¬µ±ºá×ø±êΪζÈʱ£¬±ä»¯Ç÷ÊÆÕýÈ·µÄÊÇ__________¡£
£¨2£©³£ÎÂÏ°±Æø¼«Ò×ÈÜÓÚË®£¬ÆäË®ÈÜÒº¿ÉÒÔµ¼µç¡£
¢ÙÓ÷½³Ìʽ±íʾ°±ÆøÈÜÓÚË®µÄ¹ý³ÌÖдæÔڵĿÉÄæ·´Ó¦£º
___________________________________________________________________¡£
¢Ú°±Ë®ÖÐË®µçÀë³öµÄc(H+)___________10£­7 mol/L£¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©¡£
¢Û½«°±Ë®ºÍÑÎËá»ìºÏºó£¬Ä³Í¬Ñ§ÍƲâ¸ÃÈÜÒºÖи÷Àë×ÓŨ¶È´óС˳Ðò¿ÉÄÜÓÐÈçÏÂËÄÖÖ¹Øϵ£º
A£®c(Cl£­)£¾c(NH4+)£¾c(H+)£¾c(OH£­)           B£®c(Cl£­)£¾c(NH4+)£¾c(OH£­)£¾c(H+)
C£®c(Cl£­)£¾c(H+)£¾c(NH4+)£¾c(OH£­)           D£®c(NH4+)£¾c(Cl£­)£¾c(OH£­)£¾c(H+)
¢ñ¡¢ÈôÈÜÒºÖÐÖ»ÈܽâÁËÒ»ÖÖÈÜÖÊ£¬¸ÃÈÜÖʵÄÃû³ÆÊÇ             £¬ÉÏÊöÀë×ÓŨ¶È´óС˳Ðò¹ØϵÖÐÕýÈ·µÄÊÇ£¨Ñ¡ÌîÐòºÅ£©                        ¡£
¢ò¡¢ÈôÉÏÊö¹ØϵÖÐCÊÇÕýÈ·µÄ£¬ÔòÈÜÒºÖÐÈÜÖʵĻ¯Ñ§Ê½ÊÇ                   ¡£
¢ó¡¢Èô¸ÃÈÜÒºÖÐÓÉÌå»ýÏàµÈµÄÏ¡ÑÎËáºÍ°±Ë®»ìºÏ¶ø³É£¬ÇÒÇ¡ºÃ³ÊÖÐÐÔ£¬Ôò»ìºÏÇ°
c£¨HCl£©            c£¨NH3¡¤H2O£©£¨Ìî¡°>¡±¡¢¡°<¡±¡¢»ò¡°=¡±£¬ÏÂͬ£©£¬»ìºÏºóÈÜÒºÖÐc£¨NH4+£©Óëc£¨Cl£­£©µÄ¹Øϵc£¨NH4+£©           c£¨Cl£­£©¡£
£¨3£©°±Æø¾ßÓл¹Ô­ÐÔ£¬ÔÚÍ­µÄ´ß»¯×÷ÓÃÏ£¬°±ÆøºÍ·úÆø·´Ó¦Éú³ÉXºÍYÁ½ÖÖÎïÖÊ¡£XΪï§ÑΣ¬YÔÚ±ê×¼×´¿öÏÂΪÆø̬¡£ÔÚ´Ë·´Ó¦ÖУ¬Èôÿ·´Ó¦1Ìå»ý°±Æø£¬Í¬Ê±·´Ó¦0.75Ìå»ý·úÆø£»Èôÿ·´Ó¦8.96 L°±Æø(±ê×¼×´¿ö)£¬Í¬Ê±Éú³É0.3 mol X¡£
¢Ùд³ö°±ÆøºÍ·úÆø·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º ___________________________________¡£
¢ÚÔÚ±ê×¼×´¿öÏ£¬Ã¿Éú³É1 mol Y£¬×ªÒƵç×ÓµÄÎïÖʵÄÁ¿Îª___________mol¡£
£¨4£©ÒÑ֪Һ̬NH3ÓëH2OÏàËÆ£¬Ò²¿ÉÒÔ·¢Éú΢ÈõµÄµçÀ룬µçÀë³öº¬ÓÐÏàͬµç×ÓÊýµÄ΢Á££¬ÔòҺ̬NH3µÄµçÀë·½³ÌʽΪ£º                                         
¹¤ÒµÉÏÓÃÁòÌú¿óΪÖ÷ÒªÔ­ÁϳéÈ¡ÁòËᣬÖ÷ÒªÉ豸ÓзÐÌÚÃ×£¬½Ó´¥ÊÒºÍÎüÒýËþ¡£
£¨1£©ÁòÌú¿óÔÚ½øÈë·ÐÌÚ¯ǰÐèÒª·ÛË飬ÆäÄ¿µÄÊÇ                      ¡£
£¨2£©ÎªÁ˳ä·ÖÀûÓ÷´Ó¦·Å³öµÄÈÈÁ¿£¬½Ó´¥ÊÒÖÐÓ¦°²×°       £¨ÌîÉ豸Ãû³Æ£©£»ÎüÒýËþÖÐÌî³äÐí¶à´É¹Ü£¬Æä×÷ÓÃÊÇ                                               ¡£
£¨3£©ÎüÊÕËþÅŷŵÄβÆøÖк¬ÓÐÉÙÁ¿µÄSO£¬·ÀÖ¹ÎÛȾ´óÆø¡¢³ä·ÖÀûÓÃÔ­ÁÏ£¬ÔÚÅÅ·ÅÇ°±ØÐë½øÐÐβÆø´¦Àí²¢Éè·¨½øÐÐ×ÛºÏÀûÓá£
´«Í³µÄ·½·¨ÊÇ£ºÎ²ÆøÖеÄSOͨ³£ÓÃ×ãÁ¿°±Ë®ÎüÊÕ£¬È»ºóÔÙÓÃÏ¡ÁòËá´¦Àí£¬Ð´³öÉÏÊö¹ý³ÌÖеĻ¯Ñ§·´Ó¦·½³Ìʽ£º                                £¬ÆäÓŵãÊÇ                ¡£
´´Ð·½·¨ÊÇ£º½«Î²ÆøÖеÄSOÓÃNaSOÈÜÒºÎüÊÕ£¬È»ºóÔÙ¼ÓÈÈËùµÃÈÜÒº£¬Ð´³öÉÏÊö¹ý³ÌÖеĻ¯Ñ§·´Ó¦·½³Ìʽ£º                                £¬´´Ð·½·¨Ó봫ͳ·½·¨Ïà±È£¬ÆäÓŵãÊÇ                 ¡£
£¨4£©ÔÚÁòËáµÄ¹¤ÒµÖÆ·¨ÖУ¬ÏÂÁÐÉú²ú²Ù×÷¼°Ëµ·¨Éú²ú²Ù×÷µÄÖ÷ÒªÔ­Òò¶þÕ߶¼ÕýÈ·µÄÊÇ  £¨ÌîÐòºÅ£©
A£®´Ó·ÐÌÚ¯³öÀ´µÄ¯ÆøÐè¾»»¯£¬ÒòΪ¯ÆøÖеÄSOÓëÔÓÖÊ·´Ó¦
B£®ÁòËáÉú²úÖг£²ÉÓøßѹÌõ¼þ£¬Ä¿µÄÊÇÌá¸ßSOµÄת»¯ÂÊ
C£®SO±»Ñõ»¯ÎªSOʱÐèҪʹÓô߻¯¼Á£¬ÕâÑù¿ÉÒÔÌá¸ßSOµÄת»¯ÂÊ
D£®SOÓÃ98.3%ŨÁòËáÎüÊÕ£¬Ä¿µÄÊÇ·ÀÖ¹ÐγÉËáÎí£¬ÓÐÀûÓÚSOÎüÊÕÍêÈ«
£¨5£©Ä³ÁòË᳧ÈôÒªÉú²ú8¶Ö98%µÄŨÁòËáÖÁÉÙÐèÒª±ê×¼×´¿öϵĿÕÆø     m( O¿ÕÆøÖеÄÌå»ý·ÖÊý°´20%¼ÆËã)¡£

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

¾«Ó¢¼Ò½ÌÍø