ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿Ô×ÓÐòÊýÒÀ´ÎÔö´óµÄËÄÖÖÖ÷×åÔªËØA£¬B£¬C£¬D·Ö±ð´¦ÓÚµÚÒ»ÖÁµÚËÄÖÜÆÚ£¬ÆäÖÐAÔªËØÔ×ÓºËÊÇÒ»¸öÖÊ×Ó£»BÔªËØÔ×ÓºËÍâµç×ÓÓÐ6ÖÖ²»Í¬µÄÔ˶¯×´Ì¬£¬BÓëC¿ÉÐγÉÕýËÄÃæÌåÐηÖ×Ó£¬DÔ×ÓÍâΧµç×ÓÅŲ¼Îª3d104s1¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÕâËÄÖÖÔªËØÖе縺ÐÔ×î´óµÄÊÇ________(ÌîÔªËØ·ûºÅ)£¬µÚÒ»µçÀëÄÜ×îСµÄÊÇ________(ÌîÔªËØ·ûºÅ)¡£
£¨2£©CËùÔÚµÄÖ÷×åÔªËØÆø̬Ç⻯ÎïÖУ¬·Ðµã×îµÍµÄÊÇ________(Ìѧʽ)¡£
£¨3£©BÔªËØ¿ÉÐγɶàÖÖµ¥ÖÊ£¬ÆäÖС°Ö»ÓÐÒ»²ãÔ×Óºñ¡±µÄÎïÖÊ£¬±»¹«ÈÏΪĿǰÊÀ½çÉÏÒÑÖªµÄ×¡¢×î¼áÓ²¡¢´«µ¼µç×ÓËÙ¶È×î¿ìµÄÐÂÐͲÄÁÏ£¬¸Ã²ÄÁϾ§Ìå½á¹¹ÈçͼËùʾ£¬ÆäÔ×ÓµÄÔÓ»¯ÀàÐÍΪ________ÔÓ»¯¡£
£¨4£©DµÄË®ºÏ´×ËáÑξ§Ìå¾Ö²¿½á¹¹ÈçͼËùʾ£¬¸Ã¾§ÌåÖк¬ÓеĻ¯Ñ§¼üÊÇ________(ÌîÑ¡ÏîÐòºÅ)¡£
¢Ù¼«ÐÔ¼ü¡¡¢Ú·Ç¼«ÐÔ¼ü¡¡¢ÛÅäλ¼ü¡¡¢Ü½ðÊô¼ü
¡¾´ð°¸¡¿Cl Cu HCl sp2 ¢Ù¢Ú¢Û
¡¾½âÎö¡¿
Ô×ÓÐòÊýÒÀ´ÎÔö´óµÄËÄÖÖÖ÷×åÔªËØA¡¢B¡¢C¡¢D·Ö±ð´¦ÓÚµÚÒ»ÖÁµÚËÄÖÜÆÚ£¬ÆäÖÐAÔ×ÓºËÊÇÒ»¸öÖÊ×Ó£¬ÔòAΪHÔªËØ£»BÔ×ÓºËÍâµç×ÓÓÐ6ÖÖ²»Í¬µÄÔ˶¯×´Ì¬£¬¼´ºËÍâÓÐ6¸öµç×Ó£¬ÔòBΪCÔªËØ£»BÓëC¿ÉÐγÉÕýËÄÃæÌåÐÍ·Ö×Ó£¬ÇÒC´¦ÓÚµÚÈýÖÜÆÚ£¬ÔòCΪClÔªËØ£»DÔ×ÓÍâΧµç×ÓÅŲ¼Îª3d104s1£¬ÔòDΪCuÔªËØ£»
£¨1£©ËÄÖÖÔªËØÖУ¬ÂÈÔªËصķǽðÊôÐÔ×îÇ¿£¬Ôòµç¸ºÐÔ×î´ó£»ÍΪ½ðÊô£¬ÆäËüΪ·Ç½ðÊô£¬ËùÒÔ͵ÚÒ»µçÀëÄÜ×îС£¬¹Ê´ð°¸Îª£ºCl£»Cu£»
£¨2£©CΪ¢§A×åÔªËØ£¬HFÖдæÔÚÇâ¼ü£¬·Ðµã±ÈHCl¸ß£¬ÆäËüÇ⻯ÎïÏà¶Ô·Ö×ÓÖÊÁ¿Ô½´ó£¬·ÐµãÔ½¸ß£¬ËùÒÔHClµÄ·Ðµã×îµÍ£¬¹Ê´ð°¸Îª£ºHCl£»
£¨3£©ÔÚÆä²ã×´½á¹¹ÖÐ̼̼¼ü¼ü½ÇΪ120¡ã£¬Ã¿¸ö̼Ô×Ó¶¼½áºÏ×Å3¸ö̼Ô×Ó£¬Ì¼Ô×Ó¼äÐγÉÀëÓò´ó¦Ð¼ü£¬¹Ê´ð°¸Îª£ºsp2£»
£¨4£©Óɽṹͼ¿ÉÖª£¬¸Ã¾§ÌåÖк¬ÓÐC-H¼üΪ¼«ÐÔ¼ü¡¢C-C¼üΪ·Ç¼«ÐÔ¼ü¡¢Åäλ¼ü£¬¹ÊÑ¡£º¢Ù¢Ú¢Û¡£
¡¾ÌâÄ¿¡¿KIO3 ÊÇÒ»ÖÖÖØÒªµÄÎÞ»ú»¯ºÏÎ¿É×÷ΪʳÑÎÖеIJ¹µâ¼Á¡£ÀûÓá°KClO3 Ñõ»¯·¨¡±ÖƱ¸ KIO3 ¹¤ÒÕÁ÷³ÌÈçÏÂͼËùʾ£¬»Ø´ðÏÂÁÐÎÊÌ⣺
(1)¡°Ëữ·´Ó¦¡±ËùµÃ²úÎïÓÐ KH(IO3)2¡¢Cl2 ºÍ KCl¡£¡°Öð Cl2¡±²ÉÓõķ½·¨ÊÇ________£»
(2)¡°ÂËÒº¡±ÖеÄÈÜÖÊÖ÷ÒªÊÇ_______£»¡°µ÷ pH¡±Öз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_____________________£»
(3)½á¾§¹ýÂËʱ¹¤ÈË·¢ÏÖ KH(IO3)2 ΪÐü¸¡Ðõ×´³Áµí£¬ÒÔÏ¿ÉÒÔ²ÉÓõķ½·¨ÊÇ________£»
A£®ÇãÎö·¨ B.Öؽᾧ·¨ C.³éÂË·¨ D. ÀëÐÄ·ÖÀë·¨
(4)Éú²úÖУ¬Èç¹ûÊ¡È¥¡°Ëữ¡±¡¢¡°ÖðÂÈ¡±¡¢¡°½á¾§¢Ù¡¢¹ýÂË¡±ÕâÈý²½²Ù×÷£¬Ö±½ÓÓÃÊÔ¼Á X µ÷Õû·´Ó¦ºóÈÜÒºµÄ pH£¬¶ÔÉú²úµâËá¼ØÓÐʲô¾ßÌåÓ°Ïì_________________________________________________________________£»
(5)KIO3 Ò²¿É²ÉÓá°µç½â·¨¡±ÖƱ¸£¬×°ÖÃÈçͼËùʾ¡£Óë¡°µç½â·¨¡±Ïà±È£¬¡°KClO3 Ñõ»¯·¨¡±µÄÖ÷Òª²»×ãÖ®´¦ÓÐ_____________________________________£»(д³öÒ»µã)
(6)ÒÑÖª£ºKIO3£«5KI£«3H2SO4=3K2SO4£«3I2£«3H2O£»I2£«2S2O32-=2I££«S4O62-¡£²â¶¨¼ÓµâʳÑÎÖеâµÄº¬Á¿£¬Ñ§Éú¼×Éè¼ÆµÄʵÑé²½ÖèÈçÏ£º
a£®×¼È·³ÆÈ¡ w g ʳÑΣ¬¼ÓÊÊÁ¿ÕôÁóˮʹÆäÍêÈ«Èܽ⣻
b£®ÓÃÏ¡ÁòËáËữËùµÃÈÜÒº£¬¼ÓÈë¹ýÁ¿ KI ÈÜÒº£¬Ê¹ KIO3 Óë KI ·´Ó¦ÍêÈ«£»
c£®ÒÔµí·ÛΪָʾ¼Á£¬¼ÓÈëÎïÖʵÄÁ¿Å¨¶ÈΪ2.0¡Á10-3mol¡¤L-1 µÄ Na2S2O3 ÈÜÒº 10.0mL Ç¡ºÃ·´Ó¦¡£ Ôò¼ÓµâʳÑÎÑùÆ·ÖеĵâÔªËغ¬Á¿ÊÇ______________mg/kg£»(ÒÔº¬ w µÄ´úÊýʽ±íʾ)
(7)ѧÉúÒÒÓÖ¶Ô´¿¾»µÄ NaCl(²»º¬ KIO3)½øÐÐÁËÏÂÁÐʵÑ飺
²Ù×÷²½Öè | ʵÑéÏÖÏó |
È¡ 1g ´¿¾»µÄ NaCl£¬¼Ó 3mL Ë®Åä³ÉÈÜÒº¡£ | ÈÜÒºÎޱ仯 |
µÎÈë 5 µÎµí·ÛÈÜÒººÍ 1mL 0.1 mol¡¤L£1KI ÈÜÒº£¬Õñµ´¡£ | ÈÜÒºÎޱ仯 |
È»ºóÔÙµÎÈë 1 µÎ 1mol¡¤L£1 µÄ H2SO4£¬Õñµ´¡£ | ÈÜÒº±äÀ¶É« |
¸ù¾ÝѧÉúÒÒµÄʵÑé½á¹û£¬Çë¶ÔѧÉú¼×µÄʵÑé½á¹û×÷³ö¼òÒªÆÀ¼Û___________________________¡£