Ô¥ÄϾÅУ2013¡ª2014ѧÄêÉÏѧÆÚ12Ô·ÝÁª¿¼

¸ßÈý»¯Ñ§´ð°¸

Ò»¡¢Ñ¡ÔñÌ⣨ÿÌâÖ»ÓÐÒ»¸öÕýÈ·Ñ¡Ïî,ÿСÌâ3·Ö£¬¹²48·Ö£©

1 A 2 D 3 C 4 B 5 B 6 C 7 A 8 A 9C 10 A

11 C 12 B 13 C 14 D 15 C 16 D

17£¨14·Ö£©£¨1£©¢Ú£¬Éú³ÉÕýÑÎË®½âÏÔ¼îÐÔ £¨¸÷1·Ö£© £¨2£©B£¨1·Ö£©

£¨3£©10-8/£¨a ¨C 0.1£©£¬£¨2·Ö£©£¨4£©10-6 £¨2·Ö£©

¢É¢Ù ¢Û>¢Ú>¢Ü>¢Ù £¨1·Ö£© ¢Ú ¢Û>¢Ú>¢Ù>¢Ü£¨1·Ö£©

¢ÛÔÚÏ¡ÁòËáÖУ¬ÓÉH2SO4µÚÒ»²½µçÀë³öµÄH+£¬ÄÜÒÖÖÆHSO- 4µÄµçÀ룬´Ùʹ

HSO- 4
H++SO42-ªªµÄµçÀëƽºâÏò×óÒƶ¯£¬¶øÔÚNaHSO4ÈÜÒºÖв»´æÔÚÉÏÊöµÄ×÷Óã¨2·Ö£©

£¨6£©ÒÒ£¬£¨1·Ö£© Mn2+ - 5e- + 4H2O = MnO4- + 8H+ £¨2·Ö£©

18£¨15·Ö£©£¨1£©CaCl2+H2O2+2NH3+8H2O = CaO2¡¤8H2O¡ý+2NH4Cl £¨2·Ö£©

£¨2£©È¡×îºóÒ»´ÎÏ´µÓÒºÉÙÐíÓÚÊÔ¹ÜÖУ¬ÔٵμÓÏ¡ÏõËáËữµÄÏõËáÒøÈÜÒº£¬¿´ÊÇ·ñ²úÉú°×É«³Áµí¡£Èô²»²úÉú°×É«³ÁµíÔòˮϴºÏ¸ñ£¨2·Ö£©

£¨3£©¢ÙζȵͿɼõÉÙ¹ýÑõ»¯ÇâµÄ·Ö½â£¬Ìá¸ß¹ýÑõ»¯ÇâµÄÀûÓÃÂÊ

£¨»ò·ÀÖ¹¹ýÑõ»¯ÇâµÄ·Ö½â£©£¨2·Ö£©

¢ÚͨÈë¹ýÁ¿NH3ʹÈÜÒº³Ê¼îÐÔ£¬ÒÖÖÆCaO2¡¤8H2OµÄÈܽ⣨»òʹÈÜÒº³Ê¼îÐÔ£¬¼õÉÙCaO2¡¤8H2OµÄÈܽ⣬»òÌá¸ß²úÆ·µÄ²úÂÊ¡£Òâ˼Ïà½ü¾ù¸ø·Ö£©£¨2·Ö£©

£¨4£©¢Ùµ±µÎÈë×îºóÒ»µÎKMnO4±ê×¼ÈÜÒººóÈÜÒºÓÉÎÞÉ«±ädzºìÉ«£¬ÇÒ30s²»ÍÊÉ«£¨2·Ö£©

¢Ú b £¨2·Ö£© ¢Û 82.91%£¨3·Ö£©

19£¨12·Ö£©£¨Ã¿¿Õ2·Ö£©£¨1£©2Al+2OH- +2H2O=2 AlO2- +3 H2¡ü£¨2£©Öƹ⵼ÏËά

£¨3£©¢Ù2F2+2H2O=4HF+O2¡ü ¢ÚH2SµÄµç×Óʽ(ÂÔ) ¢Û Àë×Ó¼üºÍ¹²¼Û¼ü£¨4£©1£º2

20£¨11·Ö£©

£¨1£©¢ÙEa(Ag)< Ea(Pt)

¢Ú 8.9¡Á10-7 £¨2·Ö£©

(2) C £¨2·Ö£©(Òò¸Ã·´Ó¦ÊDz»¿ÉÄæ·´Ó¦£¬²»¿¼ÂÇζȶÔƽºâÒƶ¯µÄÓ°Ï졣ζÈÉý¸ß·´Ó¦ËÙ

Âʼӿ죬Òò¶ø¾­¹ýÏàͬ·´Ó¦Ê±¼äºó²âµÃµÄO2Ìå»ýËæζÈÉý¸ß¶øÔö´ó )

(3) ¢Ù·Å£¨1·Ö£© ¢Ú<£¨2·Ö£© ¢Û ʹÓô߻¯¼Á£¨2·Ö£©