ÌâÄ¿ÄÚÈÝ

¡°ÀðæÒº¡±ÊÇÖÆÓ¡Ë¢Ð¿°åʱ£¬ÓÃÏ¡ÏõËḯʴп°åºóµÃµ½µÄ¡°·ÏÒº¡±(º¬ÓÐÉÙÁ¿µÄCl£­¡¢Fe3£«)£¬Ä³»¯Ñ§ÐËȤС×éÄâÓá°ÀðæÒº¡±ÖÆÈ¡Zn(NO3)2¡¤6H2OµÄ¹ý³ÌÈçÏ£º
               
ÒÑÖª£ºZn(NO3)2¡¤6H2OÊÇÒ»ÖÖÎÞÉ«¾§Ì壬ˮÈÜÒº³ÊËáÐÔ£¬Zn(NO3)2ÄÜÓë¼î·´Ó¦£¬µÃµ½µÄ²úÎï¾ßÓÐÁ½ÐÔ¡£
(1)¡°ÀðæÒº¡±ÖÐÈÜÖʵÄÖ÷Òª³É·ÖÊÇ________(Ìѧʽ)£¬ÈôÏ¡ÏõËḯʴп°å²úÉúµÄÆøÌåΪN2O£¬Ð´³öÏ¡ÏõËḯʴп°å·´Ó¦µÄÖ÷Òª»¯Ñ§·½³Ìʽ_____________________________¡£
(2)ÔÚ²Ù×÷¢ÙÖб£³ÖpH£½8µÄÄ¿µÄÊÇ____________________________¡£
(3)³Áµí¢ñµÄÖ÷Òª³É·ÖÊÇ_______________________________________¡£
(4)²Ù×÷¢ÛÖмÓÈÈ¡¢Öó·ÐµÄÄ¿µÄÊÇ________________________________£»
´Ë²½Öè²Ù×÷µÄÀíÂÛÒÀ¾ÝÊÇ____________________________________¡£
(5)²Ù×÷¢Ü±£³ÖpH£½2µÄÄ¿µÄÊÇ__________________________________£»
´Ë²½Öè²Ù×÷ÖÐËùÓõÄÖ÷ÒªÒÇÆ÷ÊÇ________________________________¡£
(1)Zn(NO3)2
4Zn£«10HNO3===4Zn(NO3)2£«N2O¡ü£«5H2O
(2)·ÀÖ¹Éú³ÉµÄZn(OH)2³Áµí±»Èܽâ
(3)Zn(OH)2ºÍFe(OH)3
(4)´ÙʹFe3£«Íêȫˮ½â¡¡Î¶ÈÔ½¸ß£¬Ë®½â³Ì¶ÈÔ½´ó
(5)ÒÖÖÆZn2£«Ë®½âΪZn(OH)2
Õô·¢Ã󡢾ƾ«µÆ¡¢Ìú¼Ų̈¡¢²£Á§°ô
ÂÔ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ʵÑéÊÒ³£ÀûÓü×È©·¨(HCHO)²â¶¨(NH4)2SO4ÑùÆ·ÖеªµÄÖÊÁ¿·ÖÊý£¬Æä·´Ó¦Ô­ÀíΪ£º
4NH4£« £«6HCHO =3H£«£«6H2O£«(CH2)6N4H£« £ÛµÎ¶¨Ê±£¬1 mol (CH2)6N4H£«ÏûºÄNaOHÓë l mol H£«Ï൱£Ý£¬È»ºóÓÃNaOH±ê×¼ÈÜÒºµÎ¶¨·´Ó¦Éú³ÉµÄËᡣijÐËȤС×éÓü×È©·¨½øÐÐÁËÈçÏÂʵÑ飺
²½ÖèI ³ÆÈ¡ÑùÆ·1.500 g¡£
²½ÖèII ½«ÑùÆ·Èܽâºó£¬ÍêÈ«×ªÒÆµ½250 mLÈÝÁ¿Æ¿ÖУ¬¶¨ÈÝ£¬³ä·ÖÒ¡ÔÈ¡£
²½ÖèIII ÒÆÈ¡25.00 mLÑùÆ·ÈÜÒºÓÚ250 mL×¶ÐÎÆ¿ÖУ¬¼ÓÈë10 mL 20£¥µÄÖÐÐÔ¼×È©ÈÜÒº£¬Ò¡ÔÈ¡¢¾²ÖÃ5 minºó£¬¼ÓÈë1~2µÎ·Ó̪ÊÔÒº£¬ÓÃNaOH±ê×¼ÈÜÒºµÎ¶¨ÖÁÖյ㡣°´ÉÏÊö²Ù×÷·½·¨ÔÙÖØ¸´2´Î¡£
(1)¸ù¾Ý²½ÖèIIIÌî¿Õ£º
¢Ù¼îʽµÎ¶¨¹ÜÓÃÕôÁóˮϴµÓºó£¬Ö±½Ó¼ÓÈëNaOH±ê×¼ÈÜÒº½øÐе樣¬Ôò²âµÃÑùÆ·ÖеªµÄÖÊÁ¿·ÖÊý________(Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족)¡£
¢Ú×¶ÐÎÆ¿ÓÃÕôÁóˮϴµÓºó£¬Ë®Î´µ¹¾¡£¬ÔòµÎ¶¨Ê±ÓÃÈ¥NaOH±ê×¼ÈÜÒºµÄÌå»ý________________(Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족)
¢ÛµÎ¶¨Ê±±ßµÎ±ßÒ¡¶¯×¶ÐÎÆ¿£¬ÑÛ¾¦Ó¦¹Û²ì____________
(A)µÎ¶¨¹ÜÄÚÒºÃæµÄ±ä»¯    (B)×¶ÐÎÆ¿ÄÚÈÜÒºÑÕÉ«µÄ±ä»¯
¢ÜµÎ¶¨´ïµ½ÖÕµãʱ£¬·Óָ̪ʾ¼ÁÓÉ_________É«±ä³É_________É«¡£
(2)µÎ¶¨½á¹ûÈçϱíËùʾ£º
µÎ¶¨
´ÎÊý
´ý²âÈÜÒºµÄÌå»ý
/mL
±ê×¼ÈÜÒºµÄÌå»ý/mL
µÎ¶¨Ç°¿Ì¶È
µÎ¶¨ºó¿Ì¶È
1
25.00
1.02
21.03
2
25.00
2.00
21.99
3
25.00
0.20
20.20
µÎ¶¨Ê±ÏûºÄNaOH±ê×¼ÈÜÒºµÄÌå»ýµÄƽ¾ùÖµV="__________mL;" ÈôNaOH±ê×¼ÈÜÒºµÄŨ¶ÈΪ0.1010 mol¡¤L£­1£¬Ôò250 mLÈÜÒºÖеÄn(NH4+)=__________mol,¸ÃÑùÆ·ÖеªµÄÖÊÁ¿·ÖÊýΪ______¡£
ij»¯Ñ§ÐËȤС×éÓú¬ÓÐÂÁ¡¢Ìú¡¢Í­µÄºÏ½ðÖÆÈ¡´¿¾»µÄÂÈ»¯ÂÁÈÜÒº¡¢ÂÌ·¯¾§ÌåºÍµ¨·¯¾§Ì壬ÒÔ̽Ë÷¹¤Òµ·ÏÁϵÄÔÙÀûÓá£ÆäʵÑé·½°¸ÈçÏ£º

»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öºÏ½ðÓëÉÕ¼îÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ                            £¬ÓÐÈËÈÏΪºÏ½ðÓëÉÕ¼îÈÜÒºÐγÉÁËÔ­µç³Ø£¬Ôò×÷Ϊԭµç³Ø¸º¼«µÄÎïÖÊÊÇ           ¡£
£¨2£©ÓÉÂËÒºAÖÆAlCl3ÈÜÒºµÄ;¾¶ÓТٺ͢ÚÁ½ÖÖ£¬ÄãÈÏΪºÏÀíµÄÊÇ                      ¡£ÉÏÊöʵÑé·½°¸¶à´¦²ÉÓÃÁ˹ýÂ˲Ù×÷£¬¹ýÂËËùÓõ½µÄ²£Á§ÒÇÆ÷ÓР                   ºÍ²£Á§°ô£»ÆäÖв£Á§°ôµÄ×÷ÓÃÊÇ                 ¡£
£¨3£©ÓôÖÖÆÑõ»¯Í­Í¨¹ýÁ½ÖÖ;¾¶ÖÆÈ¡µ¨·¯£¬Óë;¾¶¢ÛÏà±È£¬Í¾¾¶¢ÜÃ÷ÏÔ¾ßÓеÄÁ½¸öÓŵãÊÇ:                            ¡¢                                    ¡£
£¨4£©Í¨¹ý;¾¶¢ÜʵÏÖÓôÖÖÆÑõ»¯Í­ÖÆÈ¡µ¨·¯£¬±ØÐë½øÐеÄʵÑé²Ù×÷²½Ö裺ËáÈÜ¡¢¼ÓÈÈͨÑõÆø¡¢¹ýÂË¡¢                     ¡¢ÀäÈ´½á¾§¡¢     ¡¢×ÔÈ»¸ÉÔï¡£ÆäÖС°¼ÓÈÈͨÑõÆø¡±ËùÆðµÄ×÷ÓÃΪ                                             £¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©¡£
£¨5£©ÔڲⶨËùµÃµ¨·¯£¨CuSO4¡¤xH2O£©ÖнᾧˮxÖµµÄʵÑé¹ý³ÌÖУº³ÆÁ¿²Ù×÷ÖÁÉÙ½øÐР   ´Î¡£Èô²â¶¨½á¹ûxֵƫ¸ß£¬¿ÉÄܵÄÔ­ÒòÊÇ         ¡£
a£®¼ÓÈÈζȹý¸ß   b£®µ¨·¯¾§ÌåµÄ¿ÅÁ£½Ï´ó
c£®¼ÓÈȺó·ÅÔÚ¿ÕÆøÖÐÀäÈ´             d£®µ¨·¯¾§Ì岿·Ö·ç»¯            
e£®¼ÓÈÈʱµ¨·¯¾§Ìå·É½¦³öÀ´           f£®ËùÓÃÛáÛöÊÂÏÈδ¸ÉÔ³±Êª£©

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

¾«Ó¢¼Ò½ÌÍø