ʳÑÎÊÇÈÕ³£Éú»îµÄ±ØÐèÆ·£¬Ò²ÊÇÖØÒªµÄ»¯¹¤Ô­ÁÏ¡£

£¨l£©´ÖʳÑγ£º¬ÓÐÉÙÁ¿K£«¡¢Ca2£«¡¢Mg2£«¡¢Fe3£«¡¢SO42£­µÈÔÓÖÊÀë×Ó£¬ÊµÑéÊÒÌá´¿NaCl

µÄÁ÷³ÌÈçÏ£º

ÌṩµÄÊÔ¼Á£º±¥ºÍNa2CO3ÈÜÒº  ±¥ºÍK2CO3ÈÜÒº   NaOHÈÜÒº   BaCl2ÈÜÒº  Ba(NO3)        ÈÜÒº    75£¥ÒÒ´¼     ËÄÂÈ»¯Ì¼

¢ÙÓû³ýÈ¥ÈÜÒºIÖеÄCa2£«¡¢Mg2£«¡¢Fe3£«¡¢SO42£­ Àë×Ó£¬Ñ¡³öaËù´ú±íµÄÊÔ¼Á£¬°´µÎ¼Ó˳ÐòÒÀ´ÎΪ                                  £¨Ö»Ìѧʽ£©¡£

¢Ú Ï´µÓ³ýÈ¥NaCl¾§Ìå±íÃ渽´øµÄÉÙÁ¿KCl£¬Ñ¡ÓõÄÊÔ¼ÁΪ                     

£¨2£©ÓÃÌá´¿µÄNaClÅäÖÆ500mL4.00mol?L-1NaClÈÜÒº£¬ËùÓÃÒÇÆ÷³ýÒ©³×¡¢²£Á§°ôÍ⻹ÓÐ

                                             £¨ÌîÒÇÆ÷Ãû³Æ£©¡£

£¨3£©µç½â±¥ºÍʳÑÎË®µÄ×°ÖÃÈçͼËùʾ£¬ÈôÊÕ¼¯µÄH2 Ϊ2L£¬ÔòͬÑùÌõ¼þÏÂÊÕ¼¯µÄCl2     £¨Ìî¡°>¡±¡¢¡°=¡±»ò¡°£¼¡±£½2L£¬Ô­ÒòÊÇ                                           ¡£

×°ÖøĽøºó£¬¿ÉÓÃÓÚÖƱ¸NaOHÈÜÒº£¬Èô²â¶¨ÈÜÒºÖÐNaOHµÄŨ¶È£¬³£Óõķ½·¨Îª    

                                    ¡£

£¨4£©ÊµÑéÊÒÖƱ¸H2ºÍCl2ͨ³£²ÉÓÃÏÂÁз´Ó¦£ºZn£«H2SO4ZnSO4£«H2¡ü

MnO2+4HCl£¨Å¨£©MnCl2+Cl2¡ü+2H2O

¾Ý´Ë£¬´ÓÏÂÁÐËù¸øÒÇÆ÷×°ÖÃÖÐÑ¡ÔñÖƱ¸²¢ÊÕ¼¯H2µÄ×°Öà     £¨Ìî´úºÅ£©ºÍÖƱ¸²¢ÊÕ¼¯¸ÉÔï¡¢´¿¾»Cl2µÄ×°Öà              £¨Ìî´úºÅ£©¡£

¿ÉÑ¡ÓÃÖƱ¸ÆøÌåµÄ×°Öãº

£¨1£©¢ÙBaCl2¡¢NaOH¡¢Na2CO3£»¢Ú75%ÒÒ´¼£»

£¨2£©Ììƽ¡¢ÉÕ±­¡¢500mLÈÝÁ¿Æ¿¡¢½ºÍ·µÎ¹Ü£»

£¨3£©£¼  µç½âÉú³ÉµÄÂÈÆøÓëµç½âÉú³ÉµÄNaOH·¢ÉúÁË·´Ó¦     Ëá¼îÖк͵樣»

£¨4£©e   d

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÉúÃüÆðÔ´ÓÚԭʼº£Ñ󡣿Æѧ¼ÒÈÏΪ£¬ÓÉÓÚº£ÑóÖÐAl3+µÄŨ¶È¼«µÍ£¬ËùÒÔÂÁÔªËØδ³ÉΪÈËÌåµÄ±ØÐèÔªËØ£¬Ëü·´µ¹¶ÔÈËÌåÓк¦¡£ÂÁÖ÷ÒªËðº¦ÄÔϸ°û£¬ÊÇÀÏÄêÐÔ³Õ´ôµÄ²¡ÒòÖ®Ò»¡£1989ÄêÊÀ½çÎÀÉú×éÖ¯°ÑÂÁÁÐΪʳƷÎÛȾԴ֮һ¡£Ã¿ÈÕÉãÈëÁ¿Ó¦¿ØÖÆÔÚ4 mgÒÔÏ¡£?

£¨1£©Õ¨ÓÍÌõʱ1 kgÃæ·ÛÐè¼ÓÈë0.5 kgË®£¬4 gÃ÷·¯ºÍ10 gСËÕ´ò¼°ÉÙÁ¿Ê³Ñεȸ¨ÁÏ£¬¾­ÅëÕ¨£¬³ÉÆ·ÓÍÌõµÄ²úÂÊÒ»°ãΪ80%£¬Í¨¹ý¼ÆËã˵Ã÷ÈôÿÌìʳÓÃ100 gÓÍÌõ£¬ÔòÉãÈëÂÁµÄÁ¿ÊÇ__________________¡£?

£¨2£©ÇëÁоÙÎÒ¹ú¹úÃñÔÚÈÕ³£Éú»îÖÐÉãÈëÂÁ£¨³ýʳƷÌí¼Ó¼ÁÍ⣩µÄÈýÖÖ¿ÉÄÜ;¾¶¡£______________________________________________________________________?

 

 

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º0115 Ô¿¼Ìâ ÌâÐÍ£ºÌî¿ÕÌâ

ʳÑÎÊÇÈÕ³£Éú»îµÄ±ØÐèÆ·£¬Ò²ÊÇÖØÒªµÄ»¯¹¤Ô­ÁÏ¡£
£¨1£©´ÖʳÑγ£º¬ÓÐÉÙÁ¿K+¡¢Ca2+¡¢Mg2+¡¢Fe3+¡¢SO42-µÈÔÓÖÊÀë×Ó£¬ÊµÑéÊÒÌá´¿NaClµÄÁ÷³ÌÈçÏÂ
ÌṩµÄÊÔ¼Á£º±¥ºÍNa2CO3ÈÜÒº ±¥ºÍK2CO3ÈÜÒº NaOHÈÜÒº BaCl2ÈÜÒº Ba(NO3)2ÈÜÒº 75%ÒÒ´¼ ËÄÂÈ»¯Ì¼
¢Ù Óû³ýÈ¥ÈÜÒºIÖеÄCa2+¡¢Mg2+¡¢Fe3+¡¢SO42-Àë×Ó£¬Ñ¡³öaËù´ú±íµÄÊÔ¼Á£¬°´µÎ¼Ó˳ÐòÒÀ´ÎΪ__________
£¨Ö»Ìѧʽ£©¡£
¢ÚÏ´µÓ³ýÈ¥NaCl¾§Ìå±íÃ渽´øµÄÉÙÁ¿KCl£¬Ñ¡ÓõÄÊÔ¼ÁΪ__________________¡£
£¨2£©ÓÃÌá´¿µÄNaClÅäÖÆ500mL4.00 mol¡¤L-1NaClÈÜÒº
¢Ù ±¾´ÎʵÑéËùÓÃÒÇÆ÷³ýÒ©³×¡¢²£Á§°ôÍ⻹ÓÐ_____________£¨ÌîÒÇÆ÷³Æ£©¡£
¢ÚÓÃÍÐÅÌÌìƽ³ÆÈ¡µÄNaCl¾§ÌåµÄÖÊÁ¿Îª£º_____________£»
¢ÛÔÚÅäÖÆÈÜÒºµÄ¹ý³ÌÖжà´ÎÓõ½²£Á§°ô£¬ÔÚÈܽâʱ²£Á§°ôµÄ×÷ÓÃÊÇ£º_____________£¬ ÔÚÒÆҺʱ²£Á§°ôµÄ×÷ÓÃÊÇ£º__________________¡£
¢Ü¹Û²ìÒºÃæʱ£¬Èô¸©Êӿ̶ÈÏߣ¬»áʹËùÅäÖƵÄÈÜÒºµÄŨ¶È_________£¨Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡± »ò¡°ÎÞÓ°Ï족£¬ÏÂͬ)£»¼ÓÕôÁóˮʱ²»É÷³¬¹ýÁ˿̶ÈÏߺ󵹳ö²¿·ÖÈÜÒº£¬Ê¹ÒºÃæÓë¿Ì¶ÈÏßÏàÇУ¬»á__________£»
£¨3£©ÓæÑ=1.84g¡¤mL-1£¬ÖÊÁ¿·ÖÊýΪ98%µÄŨÁòËáÅäÖÆ200mL1mol¡¤L-1µÄÏ¡ÁòËáÓëÉÏÊöÅäÖÆÈÜÒºµÄ²½ÖèÉϵIJî±ðÖ÷ÒªÓÐÈýµã£º
¢Ù¼ÆË㣺ÀíÂÛÉÏӦȡŨÁòËáµÄÌå»ýV=___________mL(¾«È·µ½Ð¡ÊýµãºóÁ½Î»)£»
¢ÚÁ¿È¡£ºÓÉÓÚÁ¿Í²ÊÇÒ»ÖÖ´ÖÂÔµÄÁ¿¾ß£¬ÈçÏ뾫ȷÁ¿È¡£¬±ØÐèÑ¡ÓÃ_____________£¨ÌîÒÇÆ÷Ãû³Æ£©¡£
¢ÛÈܽ⣺ϡÊÍŨÁòËáµÄ·½·¨_________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸