ÌâÄ¿ÄÚÈÝ

3£®ÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®µÈεÈѹÏ£¬aLO2ÓëbLO3»ìºÏºóµÃµ½»ìºÏÆøÌåµÄÌå»ýΪ£¨a+b£©L
B£®ÅäÖÆ100mL 0.2mol/LCuSO4ÈÜÒº£¬Ðè³ÆÈ¡CuSO4•5H2OµÄÖÊÁ¿Îª5.0 g
C£®ÔÚK2SO4ºÍNaClµÄÖÐÐÔ»ìºÏË®ÈÜÒºÖУ¬Èç¹ûNa+ºÍSO42-µÄÎïÖʵÄÁ¿ÏàµÈ£¬ÔòK+ºÍCl-µÄÎïÖʵÄÁ¿Å¨¶ÈÒ»¶¨Ïàͬ
D£®ÏòijÈÜÒºÖеμÓBaCl2ÈÜÒº£¬²úÉú°×É«³Áµí£¬¼ÌÐøµÎ¼ÓÏ¡ÑÎËá³Áµí²»Èܽ⣬˵Ã÷Ô­ÈÜÒºÖÐÒ»¶¨º¬ÓÐSO42-

·ÖÎö A£®µÈεÈѹÏ£¬»ìºÏÆøÌåµÄÌå»ýµÈÓÚ¸÷×é·ÖÌå»ýÖ®ºÍ£»
B£®¸ù¾Ýn=cV¼ÆËãÁòËáÎïÖʵÄÁ¿£¬ÁòËáÍ­¾§ÌåÎïÖʵÄÁ¿µÈÓÚÁòËáÍ­µÄÎïÖʵÄÁ¿£¬ÔÙ¸ù¾Ým=nM¼ÆËãÁòËáÍ­¾§ÌåÖÊÁ¿£»
C£®ÔÚK2SO4ºÍNaClµÄÖÐÐÔ»ìºÏË®ÈÜÒºÖУ¬¸ù¾ÝµçºÉÊغã¿ÉÖªc£¨Na+£©+c£¨K+£©=2c£¨SO42-£©+c£¨Cl-£©£¬¾Ý´Ë¼ÆËãÅжϣ»
D£®ÈÜÒºÖпÉÄܺ¬ÓÐSO42-»òAg+£®

½â´ð ½â£ºA£®µÈεÈѹÏ£¬»ìºÏÆøÌåµÄÌå»ýµÈÓÚ¸÷×é·ÖÌå»ýÖ®ºÍ£¬¹ÊaLO2ÓëbLO3»ìºÏºóµÃµ½»ìºÏÆøÌåµÄÌå»ýΪ£¨a+b£©L£¬¹ÊAÕýÈ·£»
B£®ÁòËáÎïÖʵÄÁ¿Îª0.1L¡Á0.2mol/L=0.02mol£¬ÁòËáÍ­¾§ÌåÎïÖʵÄÁ¿µÈÓÚÁòËáÍ­µÄÎïÖʵÄÁ¿£¬ÐèÒªÁòËáÍ­¾§ÌåµÄÖÊÁ¿Îª0.02mol¡Á250g/mol=5.0g£¬¹ÊBÕýÈ·£»
C£®ÔÚK2SO4ºÍNaClµÄÖÐÐÔ»ìºÏË®ÈÜÒºÖУ¬¸ù¾ÝµçºÉÊغã¿ÉÖªc£¨Na+£©+c£¨K+£©=2c£¨SO42-£©+c£¨Cl-£©£¬ÈôNa+ºÍSO42-µÄÎïÖʵÄÁ¿ÏàµÈ£¬Ôòc£¨K+£©=c£¨SO42-£©+c£¨Cl-£©£¬¹ÊC´íÎó£»
D£®ÏòijÈÜÒºÖеμÓBaCl2ÈÜÒº£¬²úÉú°×É«³Áµí£¬¼ÌÐøµÎ¼ÓÏ¡ÑÎËá³Áµí²»Èܽ⣬ԭÈÜÒºÖпÉÄܺ¬ÓÐSO42-»òAg+£¬Ó¦ÏȼÓÈë×ãÁ¿ÑÎËᣬûÓа×É«³ÁµíÉú³É£¬ÔÙ¼ÓÈë¼ÓBaCl2ÈÜÒº£¬²úÉú°×É«³Áµí£¬ËµÃ÷Ô­ÈÜÒºÖÐÒ»¶¨º¬ÓÐSO42-£¬¹ÊD´íÎó£¬
¹ÊÑ¡£ºAB£®

µãÆÀ ±¾Ì⿼²éÎïÖʵÄÁ¿ÓйؼÆË㡢ʵÑé·½°¸ÆÀ¼ÛµÈ£¬ÄѶÈÖеȣ¬CÑ¡ÏîÖÐ×¢ÒâÀûÓõçºÉÊغã½â´ð£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø