ÌâÄ¿ÄÚÈÝ

(1)ÈôijҩƷÖÊÁ¿Ô¼Îª32.0g£¬ ÍÐÅÌÌìƽ׼ȷ³ÆÆäÖÊÁ¿£¬ÈôÓáý±íʾÔÚÓÒÅÌ·ÅÉÏíÀÂ룬Óáü±íʾ½«íÀÂëÈ¡Ï£¬ÔÚÏÂÁбí¸ñµÄ¿Õ¸ñÄÚ£¬ÓáýºÍ¡ü±íʾÏàÓ¦íÀÂëµÄ·ÅÉÏ»òÈ¡Ï£®

50g

20g

20g

10g

5g

(2)ÅäÖÆ500mL 0.1mol.L-1 Na2CO3ÈÜÒº£¬Í¼ÖвÙ×÷¢ÚÖÐÓ¦¸ÃÌîдµÄÊý¾ÝΪ__________£¬ÊµÑé²Ù×÷µÄÏȺó˳ÐòΪ________________ (Ìî±àºÅ)¡£

(3)ÔÚÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÈÜҺʱ£¬Óá°Æ«¸ß¡¢Æ«µÍ¡¢ÎÞÓ°Ï족±íʾÏÂÁвÙ×÷¶ÔËùÅäÈÜҺŨ¶ÈµÄÓ°Ïì¡£

¢ÙÓÃÁ¿Í²È¡ÒºÌ¬ÈÜÖÊ£¬¶ÁÊýʱ£¬¸©ÊÓÁ¿Í²£¬ËùÅäÖÆÈÜÒºµÄŨ¶È___________

¢Ú½«Á¿È¡ÒºÌ¬ÈÜÖʵÄÁ¿Í²ÓÃˮϴµÓ£¬Ï´µÓÒºµ¹ÈëÈÝÁ¿Æ¿£¬ËùÅäÖÆÈÜÒºµÄŨ¶È___________

¢Û¶¨ÈÝÒ¡ÔȺó£¬ÓÐÉÙÁ¿ÈÜÒºÍâÁ÷£¬¶ÔËùÅäÖÆÈÜÒºµÄŨ¶È___________

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
9£®Åð¼°Æ仯ºÏÎïÔÚ¹¤ÒµÉÏÓÐÐí¶àÓÃ;£®ÒÔÌúÅð¿ó£¨Ö÷Òª³É·ÖΪMg2B2O5•H2O£¬»¹ÓÐÉÙÁ¿Fe2O3¡¢FeO¡¢CaO¡¢Al2O3ºÍSiO2µÈ£©ÎªÔ­ÁÏÖƱ¸ÅðËᣨH3BO3£©µÄ¹¤ÒÕÁ÷³ÌÈçͼËùʾ£º

ÒÑÖª£ºFe3+¡¢Al3+¡¢Fe2+ºÍMg2+ÒÔÇâÑõ»¯ÎïÐÎʽÍêÈ«³Áµíʱ£¬ÈÜÒºµÄpH·Ö±ðΪ3.2¡¢4.9¡¢9.7ºÍ12.4£®
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÎªÌá¸ß¡°½þ³ö¡±ËÙÂÊ£¬³ýÊʵ±Ôö¼ÓÁòËáŨ¶ÈÍ⣬»¹¿É²ÉÈ¡µÄ´ëÊ©ÓÐÌá¸ß·´Ó¦Î¶Ȼò¼õСÌúÅð¿ó·ÛÁ£¾¶£»
£¨2£©ÊµÑéÊÒÖйýÂ˲Ù×÷ËùÐèÒªµÄ²£Á§ÒÇÆ÷ÓУº²£Á§°ô¡¢ÉÕ±­ºÍ©¶·£»
£¨3£©¡°½þÔü¡±ÖеÄÎïÖÊÊÇSiO2¡¢CaSO4£¨»¯Ñ§Ê½£©£»
£¨4£©¡°¾»»¯³ýÔÓ¡±ÐèÏȼÓH2O2ÈÜÒº£¬×÷ÓÃÊǽ«ÑÇÌúÀë×ÓÑõ»¯ÎªÌúÀë×Ó£¬È»ºóÔÙµ÷½ÚÈÜÒºµÄpHԼΪ5£¬Ä¿µÄÊÇʹÌúÀë×Ó¡¢ÂÁÀë×ÓÐγÉÇâÑõ»¯Îï³Áµí¶ø³ýÈ¥£»
£¨5£©¡°´ÖÅðËᡱÖеÄÖ÷ÒªÔÓÖÊÊÇÆßË®ÁòËáþ£¨ÌîÃû³Æ£©£»
£¨6£©µ¥ÖÊÅð¿ÉÓÃÓÚÉú²ú¾ßÓÐÓÅÁ¼¿¹³å»÷ÐÔÄܵÄÅð¸Ö£®ÒÔÅðËáºÍ½ðÊôþΪԭÁÏÔÚ¼ÓÈÈÌõ¼þÏ¿ÉÖƱ¸µ¥ÖÊÅð£¬
Óû¯Ñ§·½³Ìʽ±íʾÖƱ¸¹ý³Ì2H3BO3$\frac{\underline{\;\;¡÷\;\;}}{\;}$B2O3+3H2O¡¢B2O3+3Mg$\frac{\underline{\;\;¡÷\;\;}}{\;}$2B+3MgO£®

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

¾«Ó¢¼Ò½ÌÍø