ÌâÄ¿ÄÚÈÝ


¡¾»¯Ñ§¡ª¡ªÑ¡ÐÞ£º»¯Ñ§Óë¼¼Êõ¡¿£¨15·Ö£©

·Ö×Óɸ¾ßÓÐÎü¸½ÄÜÁ¦Ç¿£¬ÈÈÎȶ¨ÐԸߵÈÓÅÁ¼ÐÔÄÜ£¬Ê¹µÃ·Ö×Óɸ»ñµÃ¹ã·ºÓ¦Óã¬Ä³ÖÖÐͺŷÖ×ÓɸµÄ¹¤ÒµÉú²úÁ÷³Ì¿É±íʾÈçÏ£º

ÔڵμӰ±Ë®µ÷½ÚpH=9µÄ¹ý³ÌÖУ¬ÈôpH¿ØÖƲ»µ±£¬»áÓÐAl(OH)3Éú³É£¬¼ÙÉèÉú²ú¹ý³ÌÖÐÂÁÔªËØ¡¢¹èÔªËØ¾ùûÓÐËðʧ£¬ÄÆÔ­×ÓÀûÓÃÂÊΪ10%¡£

(1)·Ö×ÓɸµÄÖ±¾¶Îª4A(1A=10£­10m)³ÆÎª4AÐÍ·Ö×Óɸ£¬µ±Na+±»Ca2+È¡´úʱ¾ÍÖÆµÃ5AÐÍ·Ö×Óɸ£¬µ±Na+±»K+È¡´úʱ¾ÍÖÆµÃ6AÐÍ·Ö×Óɸ¡£Òª¸ßЧ·ÖÀëÕý¶¡Íé(·Ö×ÓÖ±¾¶Îª4.65A)ºÍÒì¶¡Íé(·Ö×ÓÖ±¾¶Îª5.6A)Ó¦¸ÃÑ¡ÓÃ_______ÐÍ·Ö×Óɸ¡£

(2)Al2(SO4)3ÈÜÒºÓëNa2SiO3ÈÜÒº·´Ó¦Éú³É½ºÌåµÄÀë×Ó·½³ÌʽΪ________________________

(3)¸ÃÉú²úÁ÷³ÌÖÐËùµÃÂËÒºÀﺬÓеÄÀë×Ó³ýH+¡¢OH£­Í⣬Ö÷Òª»¹ÓР         µÈÀë×Ó¡£¼ìÑéÆäÖнðÊôÑôÀë×ӵIJÙ×÷·½·¨ÊÇ             £»

(4)¼ÓNH3•H2Oµ÷½ÚpHºó£¬¼ÓÈȵ½90¡æ²¢³ÃÈȹýÂ˵ÄÔ­Òò¿ÉÄÜÊÇ                   ¡£

(5)¸ÃÉú²úÁ÷³ÌÖÐËùµÃ·Ö×ÓɸµÄ»¯Ñ§Ê½Îª                  ¡££¨ÓÃÑõ»¯ÎïÐÎʽ±íʾ£©


(15·Ö)£¨1£©5A£¨1·Ö£©

£¨2£©2Al3++3SiO32-+6H2O=2Al(OH)3+3H2SiO3£¨3·Ö£©

£¨3£©Na+,NH4+,SO42-;£¨3·Ö£©½«²¬Ë¿£¨»òÌúË¿£©Ôھƾ«µÆÉϼÓÈÈÖÁÎÞÉ«£¬ÕºÈ¡´ý²âÈÜÒº·ÅÔھƾ«µÆÍâÑæÉÏׯÉÕ£¬Èô»ðÑæ³Ê»ÆÉ«£¬Ôò´ý²âÒºÖк¬ÓÐNa+;(2·Ö)

£¨4£©¼ÓÈÈÄÜ´Ùʹ½ºÌåÄý¾Û£¬³ÃÈȹýÂ˿ɷÀÖ¹ÆäËüÔÓÖʾ§ÌåÎö³ö¡££¨2·Ö£©

£¨5£©Na2O¡¤Al2O3¡¤10SiO2¡¤6H2O£¨4·Ö£©


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø