ÌâÄ¿ÄÚÈÝ
£¨16·Ö£©
£¨1£©ÒÑÖª298Kʱ£¬Fe(OH)3µÄÈܶȻý³£Êý=2.6¡Á10-39£¬Mg(OH)2µÄÈܶȻý³£Êý=5.6¡Á£¬È¡ÊÊÁ¿Å¨¶È¾ùΪ0.1mol/LµÄMgCl2ºÍFeCl3»ìºÏÒº£¨º¬HCl£©£¬¼ÓÈëÒ»¶¨Á¿µÄMgCO3´ïµ½³ÁµíÈÜҺƽºâ£¬²âµÃpH=4.00£¬Ôò´ËζÈϲÐÁôÔÚÈÜÒºÖеÄc(Fe3+)=______________£»
ÓÐûÓÐMg(OH)2³ÁµíÉú³É______________Ìî¡°ÓС±»ò¡°ÎÞ¡±£©£¬ÀíÓÉÊÇ________________________________¡£
£¨2£©ËáHnAÓë¼îB(OH)mÍêÈ«·´Ó¦Éú³ÉÕýÑÎ.
¢ÙÈôHnAΪHCl£¬ÇÒ¸ÃÑÎÈÜÒºµÄpH£¼7£¬ÓÃÀë×Ó·½³Ìʽ˵Ã÷ÔÒò£º
¢ÚÈô½«0.4mol¡¤L£1µÄNaOHÈÜÒºÓë0.2mol¡¤L£1µÄHnAÈÜÒºµÈÌå»ý»ìºÏºópH=10£¬
ÔòHnAΪ £¨ÌîÐòºÅ£©.
a.һԪǿËá b. Ò»ÔªÈõËá c. ¶þԪǿËá d. ¶þÔªÈõËá
£¨3£©Ä³Ñ§Ð£¿ÎÍâÐËȤС×é´Óº£Ë®É¹ÑκóµÄÑα(Ö÷Òªº¬Na+¡¢Mg2+¡¢Cl£¡¢Br£µÈ)ÖÐÄ£Ä⹤ҵÉú²úÀ´Ìáȡþ£¬Ö÷Òª¹ý³ÌÈçÏ£º»Ø´ðÏÂÁÐÎÊÌ⣺
¢¡.´Ó¹ý³Ì¢ÙµÃµ½µÄMg(OH)2³ÁµíÖлìÓÐÉÙÁ¿µÄCa(OH)2 £¬³ýÈ¥ÉÙÁ¿Ca(OH)2µÄ·½·¨ÊÇÏȽ«³Áµí¼ÓÈ뵽ʢÓÐ ÈÜÒºµÄÉÕ±ÖУ¬³ä·Ö½Á°èºó¾¹ýÂË¡¢Ï´µÓ¿ÉµÃ´¿¾»µÄMg(OH)2¡£
¢¢.¹ý³Ì¢ÛµÄת»¯ÐèÒªÔÚHClÆøÁ÷ÖмÓÈÈ£¬HClµÄ×÷ÓÃÊÇ
¢£.д³ö¹ý³Ì¢ÜÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ
½âÎö
I.ÈçÏÂͼËùʾ£¬½«2molAÆøÌåºÍ1molBÆøÌå³äÈëÒ»ÈÝ»ý¿É±äµÄÃܱÕÈÝÆ÷ÖС£
·¢Éú·´Ó¦£º2A(g)+B(g)2C(g)¡£·´Ó¦¿ªÊ¼Ê±¿É»¬¶¯µÄ»îÈûµÄλÖÃÈçͼ1Ëùʾ£¬µ±·´Ó¦´ïµ½Æ½ºâʱ£¬»îÈûλÖÃÈçͼ2Ëùʾ.Ôò´ïµ½Æ½ºâʱ£¬AµÄת»¯ÂÊΪ________________;
¸ÃÌõ¼þÏ·´Ó¦µÄƽºâ³£ÊýΪ_______________________________¡£
|
|
|
|
|
|
|
¢ò.£¨1£©ÒÑÖª298Kʱ£¬1molC2H6ÔÚÑõÆøÖÐÍêȫȼÉÕÉú³É¶þÑõ»¯Ì¼ºÍҺ̬ˮ£¬·Å³öÈÈÁ¿1558.3KJ¡£Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ__________________________________________.
£¨20ÀûÓø÷´Ó¦Éè¼ÆÒ»¸öȼÁϵç³Ø:ÓÃÇâÑõ»¯¼ØÈÜÒº×öµç½â ÖÊÈÜÒº£¬Óöà¿×ʯī×öµç¼«£¬Ôڵ缫ÉÏ·Ö±ð³äÈëÒÒÍéºÍÑõÆø¡£Ð´³ö¸º¼«µÄµç¼«·´Ó¦Ê½____________________________
(3)ÓÐÈçÓÒ»¯Ñ§ÊµÑé×°ÖÃͼ£¬
ʯī°ôÉϵĵ缫·´Ó¦Ê½Îª_______________________________;
Èç¹ûÆðʼʱʢÓÐ1000mLpH=5µÄÁòËáÍÈÜÒº£¨25¡æ£©£¨CuSO4 ×ãÁ¿£©£¬Ò»¶Îʱ¼äºóÈÜÒºµÄpH±äΪ1£¬ÈôҪʹÈÜÒº»Ö¸´µ½ÆðʼŨ¶È£¨ºöÂÔÈÜÒºÌå»ýµÄ±ä»¯£©£¬¿ÉÏòÈÜÒºÖмÓÈë________(ÌîÎïÖÊÃû³Æ)£¬ÆäÖÊÁ¿Îª_______________.