ÌâÄ¿ÄÚÈÝ

ÔÚ100mLNaOHÈÜÒºÖмÓÈëNH4NO3ºÍ(NH4)2SO4µÄ¹ÌÌå»ìºÏÎ¼ÓÈȳä·Ö·´Ó¦£¬ÏÂͼ±íʾ¼ÓÈëµÄ»ìºÏÎïµÄÖÊÁ¿ºÍ²úÉúÆøÌåµÄÌå»ý(±ê×¼×´¿ö)¹ØÏµ¡£

                                             

(1)ÊÔ¼ÆËãNaOHÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È¡£

(2)µ±NaOHÈÜÒºµÄÌå»ýΪ140mL£¬¹ÌÌå»ìºÏÎïµÄÖÊÁ¿Îª51£®6g£¬³ä·Ö·´Ó¦ºó£¬Éú³ÉÆøÌåµÄÌå»ý(±ê×¼×´¿ö)Ϊ¶àÉÙÉý?

(3)µ±NaOHÈÜÒºµÄÌå»ýΪ180mL£¬¹ÌÌå»ìºÏÎïµÄÖÊÁ¿ÈÔΪ51£®6g£¬³ä·Ö·´Ó¦ºó£¬Éú³ÉÆøÌåµÄÌå»ý(±ê×¼×´¿ö)Ϊ¶àÉÙÉý?

 

½â£º£¨1£©5mol/L £¨3·Ö£©   £¨2£©15.68L£¨3·Ö£© £¨3£©16.8L£¨3·Ö£©

½âÎö:ÂÔ

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø