ÌâÄ¿ÄÚÈÝ

СÇàͬѧΪÁËÖÆ×÷Ò¶ÂöÊéÇ©£¬´òËãÅäÖÆ125g10£¥µÄNaOHÈÜÒº¡£

£¨1£©¼ÆËãСÇàÐèÒª³ÆÈ¡NaOHµÄÖÊÁ¿¡£

£¨2£©Ð¡Çཫϴ¾»µÄÊ÷Ò¶·ÅÔÚÅäÖÆºÃµÄÈÜÒºÖÐÖó·ÐºóÈ¡³ö£¬ÈÜÒºµÄÖÊÁ¿¼õÉÙÁË5g£¬ÈÜÖÊÖÊÁ¿·ÖÊýÒ²ÓÐËù¼õС¡£ÎªÁ˲ⶨʣÏÂÈÜÒºÖÐNaOHµÄÖÊÁ¿·ÖÊý£¬Ð¡ÇàÏòÈÜÒºÖÐÖð½¥¼ÓÈë7.3£¥µÄÏ¡ÑÎËᣬµ±ÈÜÒºPH=7ʱ£¬ÏûºÄÑÎËá50g¡£¼ÆË㣺

¢ÙÉú³ÉNaClµÄÖÊÁ¿¡£

¢ÚÊ£ÏÂÈÜÒºÖÐNaOHµÄÖÊÁ¿·ÖÊý¡£

 

¡¾´ð°¸¡¿

½â£º£¨1£©m£¨NaOH£©=125g¡Á10£¥=12.5g£¨2·Ö£©

£¨2£©NaOH    £«  HCl   =    NaCl  £«  H2O

      40          36.5        58.5

   m£¨NaOH£©  50g¡Á7.3£¥   m£¨NaCl£©

m£¨NaOH£©=4g      m£¨NaCl£©=5.85g£¨2·Ö£©

¦Ø£¨NaOH£©=¡Á100£¥=3.3£¥£¨2·Ö£©

´ð£ºÂÔ

¡¾½âÎö¡¿ÂÔ

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø