ÌâÄ¿ÄÚÈÝ

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£º

A£®°ÑNaHCO3ºÍNa2CO3»ìºÍÎï6.85 gÈÜÓÚË®ÖƳÉ100 mLÈÜÒº£¬ÆäÖÐc(Na£«)£½1 mol¡¤L-1£¬¸ÃÈÜÒºÖмÓÈëÒ»¶¨Á¿ÑÎËáÇ¡ºÃÍêÈ«·´Ó¦£¬½«ÈÜÒºÕô¸Éºó£¬ËùµÃ¹ÌÌåÖÊÁ¿ÎÞ·¨¼ÆËã

B£®½«54.4 gÌú·ÛºÍÑõ»¯ÌúµÄ»ìºÏÎïÖмÓÈë4.0mol/L200 mLµÄÏ¡ÁòËᣬǡºÃÍêÈ«·´Ó¦£¬·Å³öÇâÆø4.48 L(±ê×¼×´¿ö)£¬·´Ó¦ºóµÄÈÜÒºÖеμÓKSCN²»ÏÔºìÉ«£¬ÇÒÎÞ¹ÌÌåÊ£ÓàÎ·´Ó¦ºóµÃµ½FeSO4µÄÎïÖʵÄÁ¿ÊÇ0.8mol

C£®ÏÖÏòÒ»ÃܱÕÈÝÆ÷ÖгäÈë1mol N2ºÍ3mol H2£¬ÔÚÒ»¶¨Ìõ¼þÏÂʹ¸Ã·´Ó¦·¢Éú£¬´ïµ½»¯Ñ§Æ½ºâʱ£¬N2¡¢H2ºÍNH3µÄÎïÖʵÄÁ¿Å¨¶ÈÒ»¶¨ÏàµÈ

D£®Ä³ÈÜÒºÖпÉÄܺ¬ÓÐH+¡¢Na+¡¢NH4+¡¢Mg2+¡¢Fe3+¡¢Al3+¡¢SO42+ µÈÀë×Ó£¬µ±Ïò¸ÃÈÜÒºÖмÓÈëijŨ¶ÈµÄNaOHÈÜҺʱ£¬·¢ÏÖÉú³É³ÁµíµÄÎïÖʵÄÁ¿ËæNaOHÈÜÒºµÄÌå»ý±ä»¯ÈçͼËùʾ£¬ÓÉ´Ë¿ÉÖª£¬¸ÃÈÜÒºÖп϶¨º¬ÓеÄÑôÀë×ÓÊÇH+¡¢NH4+¡¢Mg2+¡¢Al3+

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø