ÌâÄ¿ÄÚÈÝ
£¨04ÄêÈ«¹ú¾í£©£¨l6·Ö£© ·Ûĩ״ÊÔÑùAÊÇÓɵÈÎïÖʵÄÁ¿µÄMgOºÍFe2O3×é³ÉµÄ»ìºÏÎï¡£½øÐÐÈçÏÂʵÑ飺
¢ÙÈ¡ÊÊÁ¿A½øÐÐÂÁÈÈ·´Ó¦£¬²úÎïÖÐÓе¥ÖÊBÉú³É£»
¢ÚÁíÈ¡20 g AÈ«²¿ÈÜÓÚ0.15 L 6.0 mol?
ÑÎËáÖУ¬µÃÈÜÒºC£»
¢Û½«¢ÙÖеõ½µÄµ¥ÖÊBºÍÈÜÒºC·´Ó¦£¬·Å³ö l.12 L£¨±ê¿ö£©ÆøÌ壬ͬʱÉú³ÉÈÜÒºD£¬»¹²ÐÁôÓйÌÌåÎïÖÊB£»
¢ÜÓÃKSCNÈÜÒº¼ìÑéʱ£¬ÈÜÒºD²»±äÉ«¡£
ÇëÌî¿Õ£º
£¨1£©¢ÙÖÐÒý·¢ÂÁÈÈ·´Ó¦µÄʵÑé²Ù×÷ÊÇ_____¡£²úÎïÖеĵ¥ÖÊBÊÇ__________________¡£
£¨2£©¢ÚÖÐËù·¢ÉúµÄ¸÷·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ_______________¡£
£¨3£©¢ÛÖÐËù·¢ÉúµÄ¸÷·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ________________¡£
£¨4£©ÈôÈÜDµÄÌå»ýÈÔÊÓΪ0.15 L£¬Ôò¸ÃÈÜÒºÖÐc£¨Mg2+£©Îª_____________________________£¬
c£¨Fe2+£©Îª____________________________________________________¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿