ÌâÄ¿ÄÚÈÝ
(14·Ö)X¡¢Y¡¢Z¡¢WÊÇÔªËØÖÜÆÚ±íÖÐÇ°ËÄÖÜÆڵij£¼ûÔªËØ£¬ÆäÏà¹ØÐÅÏ¢ÈçÏÂ±í£»
¢Å XλÓÚÔªËØÖÜÆÚ±íµÚ ×å¡£XµÄÒ»ÖÖµ¥ÖÊÈÛµãºÜ¸ß£¬Ó²¶ÈºÜ´ó£¬ÔòÕâÖÖµ¥Öʵľ§ÌåÊôÓÚ ¾§Ìå¡£
¢Æ XÓëYÖе縺ÐÔ½ÏÇ¿µÄÊÇ(ÌîÔªËØ·ûºÅ£© £»XY2µÄµç×ÓʽÊÇ £¬·Ö×ÓÖдæÔÚ ¸ö¦Ò¼ü¡£
¢ÇZ2Y2Öк¬ÓеĻ¯Ñ§¼üÀàÐÍÓÐ ¡£Òõ¡¢ÑôÀë×ӵĸöÊý±ÈΪ ¡£
¢ÈWµÄ»ù̬Ô×ÓºËÍâµç×ÓÅŲ¼Ê½ÊÇ ¡£
¢É·Ï¾ÉÓ¡Ë¢µç·°åÉÏÓÐWµÄµ¥ÖÊA¡£ÓÃH2O2ºÍH2SO4µÄ»ìºÏÈÜÒº¿ÉÈܳöÓ¡Ë¢µç·°åÉϵÄA¡£ÒÑÖª£º
A(s)+H2SO4(aq) ="=" ASO4(aq) + H2(g)¡¡ ¦¤H=+64.4kJ¡¤mol-1
2H2O2(l) ="=" 2H2O(l) + O2(g)¡¡ ¦¤H= -196.4kJ¡¤mol-1
H2(g)+O2(g) ="=" H2O(l)¡¡ ¦¤H= -285.8kJ¡¤mol-1
Çëд³öAÓëH2SO4¡¢H2O2·´Ó¦Éú³ÉASO4(aq)ºÍH2O(l)µÄÈÈ»¯Ñ§·½³Ìʽ(AÓû¯Ñ§Ê½±íʾ)£º
¡¡ ¡¡¡£
ÔªËØ | Ïà¹ØÐÅÏ¢ |
X | XµÄ»ù̬Ô×ÓºËÍâ3¸öÄܼ¶ÉÏÓеç×Ó£¬ÇÒÿ¸öÄܼ¶Éϵĵç×ÓÊýÏàµÈ |
Y | Ô×Ó×îÍâ²ãµç×ÓÊýÊÇ´ÎÍâ²ãµÄÈý±¶ |
Z | µ¥Öʼ°Æ仯ºÏÎïµÄÑæÉ«·´Ó¦Îª»ÆÉ« |
W | WÔªËØ»ù̬Ô×ÓµÄM²ãÈ«³äÂú£¬N²ãÖ»ÓÐÒ»¸öµç×Ó |
¢Æ XÓëYÖе縺ÐÔ½ÏÇ¿µÄÊÇ(ÌîÔªËØ·ûºÅ£© £»XY2µÄµç×ÓʽÊÇ £¬·Ö×ÓÖдæÔÚ ¸ö¦Ò¼ü¡£
¢ÇZ2Y2Öк¬ÓеĻ¯Ñ§¼üÀàÐÍÓÐ ¡£Òõ¡¢ÑôÀë×ӵĸöÊý±ÈΪ ¡£
¢ÈWµÄ»ù̬Ô×ÓºËÍâµç×ÓÅŲ¼Ê½ÊÇ ¡£
¢É·Ï¾ÉÓ¡Ë¢µç·°åÉÏÓÐWµÄµ¥ÖÊA¡£ÓÃH2O2ºÍH2SO4µÄ»ìºÏÈÜÒº¿ÉÈܳöÓ¡Ë¢µç·°åÉϵÄA¡£ÒÑÖª£º
A(s)+H2SO4(aq) ="=" ASO4(aq) + H2(g)¡¡ ¦¤H=+64.4kJ¡¤mol-1
2H2O2(l) ="=" 2H2O(l) + O2(g)¡¡ ¦¤H= -196.4kJ¡¤mol-1
H2(g)+O2(g) ="=" H2O(l)¡¡ ¦¤H= -285.8kJ¡¤mol-1
Çëд³öAÓëH2SO4¡¢H2O2·´Ó¦Éú³ÉASO4(aq)ºÍH2O(l)µÄÈÈ»¯Ñ§·½³Ìʽ(AÓû¯Ñ§Ê½±íʾ)£º
¡¡ ¡¡¡£
¢Å IV A (1·Ö) Ô×Ó(1·Ö)
¢Æ O (1·Ö) (2·Ö) 2 (1·Ö)
¢Ç Àë×Ó¼ü¡¢·Ç¼«ÐÔ¼ü(»ò¹²¼Û¼ü) (2·Ö) 1¡Ã2 (1·Ö)
¢È [Ar]3d104s1 (2·Ö)
¢É Cu (s) + H2O2(l) + H2SO4(aq) = CuSO4(aq) + 2H2O(l) ?H =" -" 319.6 kJ / mol (3·Ö)
¢Æ O (1·Ö) (2·Ö) 2 (1·Ö)
¢Ç Àë×Ó¼ü¡¢·Ç¼«ÐÔ¼ü(»ò¹²¼Û¼ü) (2·Ö) 1¡Ã2 (1·Ö)
¢È [Ar]3d104s1 (2·Ö)
¢É Cu (s) + H2O2(l) + H2SO4(aq) = CuSO4(aq) + 2H2O(l) ?H =" -" 319.6 kJ / mol (3·Ö)
ÂÔ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿