ÌâÄ¿ÄÚÈÝ

ÒÑÖªÎïÖÊMÓÉͬһ¶ÌÖÜÆÚµÄX¡¢YÁ½ÖÖÔªËØ×é³É£¬X Ô­×ÓµÄ×îÍâ²ãµç×ÓÊýÊÇÆä×îÄÚ²ãµç×ÓÊýµÄÒ»°ë£¬YÔªËØµÄ×î¸ßÕý¼ÛÓëÆä×îµÍ¸º¼ÛµÄ´úÊýºÍΪ6¡£MÓëÆäËûÎïÖʵÄת»¯¹ØÏµÈçÏÂËùʾ£¨²¿·Ö²úÎïÒÑÂÔÈ¥£©
(1)¹¤Òµµç½âMÈÜÒºµÄ»¯Ñ§·½³ÌʽΪ____________________
(2)ÈôAÊÇÓëX¡¢YͬÖÜÆÚµÄÒ»ÖÖ³£¼û½ðÊô£¬ÔòAÔªËØÔÚÖÜÆÚ±íÖеÄλÖÃÊǵÚ____ÖÜÆÚµÚ____×壬д³öAÓëBÈÜÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ____________________¡£
(3)ÈôAÊÇÄ³ÔªËØµÄÒ»ÖÖ³£¼ûËáÐÔÑõ»¯Î¿ÉÓÃÓÚÖÆÔì¹âµ¼ÏËά¡£Ôò¸ÃÔªËØµÄÔ­×ӽṹʾÒâͼΪ
________£¬Ð´³öEÓëF·´Ó¦µÄÀë×Ó·½³Ìʽ____________________¡£
(4)BµÄµç×ÓʽΪ____________£¬ÆäÖк¬ÓеĻ¯Ñ§¼üΪ____________¡£
(5)д³öMÎïÖʵÄÒ»ÖÖÓÃ;£º________________¡£
(1)2NaCl+2H2O2NaOH+H2¡ü+Cl2¡ü
(2)Èý£»¢óA£»2Al+2NaOH+2H2O=2NaAlO2+3H2¡ü
(3)£»2H++SiO32-=H2SiO3¡ý
(4)£»Àë×Ó¼üºÍ¹²¼Û¼ü
(5)Âȼҵ£¨´ð°¸ºÏÀí¾ù¿É£©
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
A¡¢B¡¢C¡¢D¡¢EÊÇλÓÚ¶ÌÖÜÆÚµÄÖ÷×åÔªËØ£¬ÆäÖÐA¡¢BΪ½ðÊôÔªËØ£¬DµÄ×îÍâ²ãµç×ÓÊýÊÇÆäÄÚ²ãµç×ÓÊýµÄÈý±¶£®ÒÑÖª£º¢ÙÈÈÎȶ¨ÐÔ£ºHmD£¾HmC£»¢ÚCm-¡¢E£¨m-1£©-¾ßÓÐÏàͬµÄµç×Ó²ã½á¹¹£»¢ÛBÓëEÔÚͬһÖÜÆÚ£¬ÔÚ¸ÃÖÜÆÚÖУ¬EµÄÔ­×Ó°ë¾¶×îС£¬BµÄÀë×Ó°ë¾¶×îС£»¢ÜA¡¢BËùÐγɵĵ¥ºËÀë×Ó¾ßÓÐÏàͬµÄµç×Ó²ã½á¹¹£¬BµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÓëA¡¢EµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯Îï¶¼ÄÜ·´Ó¦£®ÒÀ¾ÝÉÏÊöÐÅÏ¢»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©HmDµÄµç×Óʽ£º
£®AºÍEÐγɵϝºÏÎïµÄ¾§ÌåÀàÐÍÊô
Àë×Ó¾§Ìå
Àë×Ó¾§Ìå
£»¸Ã¾§Ìå½á¹¹ÖÐÀëA×î½üµÄAÓÐ
12
12
¸ö£®
£¨2£©ÄÜÖ¤Ã÷Cm-¡¢E£¨m-1£©-»¹Ô­ÐÔÇ¿ÈõµÄ·´Ó¦µÄÀë×Ó·½³ÌʽΪ
Cl2+S2-¨T2Cl-+S¡ý
Cl2+S2-¨T2Cl-+S¡ý
£®
£¨3£©Óû¯Ñ§ÓÃÓï±íʾBµÄ×î¸ßÕý¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÔÚË®ÈÜÒºÖеĵçÀë¹ý³Ì
H2O+H++AlO2-?Al£¨OH£©3?Al3++3OH-
H2O+H++AlO2-?Al£¨OH£©3?Al3++3OH-
£®
£¨4£©ÒÑÖª17g HmCÔÚÒ»¶¨Ìõ¼þÏÂÍêȫȼÉÕÉú³ÉÎȶ¨»¯ºÏÎïʱ·Å³öÈÈÁ¿Îªa kJ£¬Ð´³öÄܱíʾ¸ÃÎïÖÊȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ
2H2S£¨g£©+3O2£¨g£©?2SO2£¨g£©+2H2O£¨l£©¡÷H=-4akJ/mol
2H2S£¨g£©+3O2£¨g£©?2SO2£¨g£©+2H2O£¨l£©¡÷H=-4akJ/mol
£®
£¨5£©³£ÎÂÏ£¬½«µÈÎïÖʵÄÁ¿Å¨¶ÈµÄHmCÈÜÒººÍAµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÈÜÒºµÈÌå»ý»ìºÏ£¬ËùµÃÈÜÒºÏÔ¼îÐÔ£¬ÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ
c£¨Na+£©£¾c£¨HS-£©£¾c£¨OH-£©£¾c£¨H+£©£¾c£¨S2-£©
c£¨Na+£©£¾c£¨HS-£©£¾c£¨OH-£©£¾c£¨H+£©£¾c£¨S2-£©
£®
A¡¢B¡¢C¡¢D¡¢EÊÇλÓÚ¶ÌÖÜÆÚµÄÖ÷×åÔªËØ£¬ÆäÖÐA¡¢BΪ½ðÊôÔªËØ£®ÒÑÖª£º¢ÙÈÈÎȶ¨ÐÔ£ºHmD£¾HmC£»¢ÚCm-¡¢E£¨m-1£©-¾ßÓÐÏàͬµÄµç×Ó²ã½á¹¹£»¢ÛBÓëEÔÚͬһÖÜÆÚ£¬ÔÚ¸ÃÖÜÆÚÖУ¬EµÄÔ­×Ó°ë¾¶×îС£¬BµÄÀë×Ó°ë¾¶×îС£»¢ÜA¡¢BËùÐγɵĵ¥ºËÀë×Ó¾ßÓÐÏàͬµÄµç×Ó²ã½á¹¹£¬BµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÓëA¡¢EµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯Îï¶¼ÄÜ·´Ó¦£®ÒÀ¾ÝÉÏÊöÐÅÏ¢ÓÃÏàÓ¦µÄ»¯Ñ§ÓÃÓï»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©HmDµÄµç×Óʽ£º
£®
£¨2£©Cm-¡¢E£¨m-1£©-µÄ»¹Ô­ÐÔÇ¿ÈõΪ
S2-
S2-
£¾
Cl-
Cl-
£¬ÄÜÖ¤Ã÷Æä»¹Ô­ÐÔÇ¿ÈõµÄÀë×Ó·½³ÌʽΪ
Cl2+S2-=2Cl-+S¡ý
Cl2+S2-=2Cl-+S¡ý
£®
£¨3£©Ð´³öBµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïºÍAµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯Îï·´Ó¦µÄÀë×Ó·½³Ìʽ
Al£¨OH£©3+OH-=AlO2-+2H2O
Al£¨OH£©3+OH-=AlO2-+2H2O
£®
£¨4£©³£ÎÂÏ£¬½«µÈÎïÖʵÄÁ¿Å¨¶ÈµÄHmCÈÜÒººÍAµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÈÜÒºµÈÌå»ý»ìºÏ£¬ËùµÃÈÜÒºÏÔ¼îÐÔ£¬ÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ
c£¨Na+£©£¾c£¨HS-£©£¾c£¨OH-£©£¾c£¨H+£©£¾c£¨S2-£©
c£¨Na+£©£¾c£¨HS-£©£¾c£¨OH-£©£¾c£¨H+£©£¾c£¨S2-£©
£®

£¨5£©ÔÚB¡¢C¡¢Eµ¥ÖÊÖУ¬·ûºÏÏÂÁÐת»¯¹ØÏµµÄÊÇ
S
S
£®
A¡¢B¡¢C¡¢D¡¢EÊÇλÓÚ¶ÌÖÜÆÚµÄÖ÷×åÔªËØ£¬ÆäÖÐA¡¢BΪ½ðÊôÔªËØ£¬DµÄ×îÍâ²ãµç×ÓÊýÊÇÆäÄÚ²ãµç×ÓÊýµÄÈý±¶£®ÒÑÖª£º¢ÙÈÈÎȶ¨ÐÔ£ºHmD£¾HmC£»¢ÚCm-¡¢E£¨m-1£©-¾ßÓÐÏàͬµÄµç×Ó²ã½á¹¹£»¢ÛBÓëEÔÚͬһÖÜÆÚ£¬ÔÚ¸ÃÖÜÆÚÖУ¬EµÄÔ­×Ó°ë¾¶×îС£¬BµÄÀë×Ó°ë¾¶×îС£»¢ÜA¡¢BËùÐγɵļòµ¥Àë×Ó¾ßÓÐÏàͬµÄµç×Ó²ã½á¹¹£¬BµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÓëA¡¢EµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯Îï¶¼ÄÜ·´Ó¦£®ÒÀ¾ÝÉÏÊöÐÅÏ¢Óû¯Ñ§ÓÃÓï»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öAÓëCÐγɻ¯ºÏÎïµÄµç×Óʽ
£®
£¨2£©Cm-¡¢E£¨m-1£©-µÄ»¹Ô­ÐÔÇ¿ÈõΪ
S2-
S2-
£¾
Cl-
Cl-
£¬ÄÜÖ¤Ã÷Æä»¹Ô­ÐÔÇ¿ÈõµÄÀë×Ó·½³ÌʽΪ
Cl2+S2-=2Cl-+S¡ý
Cl2+S2-=2Cl-+S¡ý
£®
£¨3£©Ð´³öBµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïºÍAµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯Îï·´Ó¦µÄÀë×Ó·½³Ìʽ
Al£¨OH£©3+OH-=AlO2-+2H2O
Al£¨OH£©3+OH-=AlO2-+2H2O
£®
£¨4£©ÔªËØBÐγɵĵ¥ÖÊÓëFe3O4·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
8Al+3Fe3O4
 ¸ßΠ
.
 
4Al2O3+9Fe
8Al+3Fe3O4
 ¸ßΠ
.
 
4Al2O3+9Fe
£®
£¨5£©ÉÏÊöÎåÖÖÔªËØ£¬ÓÉÆäÖÐÈýÖÖÔªËØ×é³ÉµÄÒ×ÈÜÓÚË®µÄÎïÖÊÖУ¬ÄÜ´Ù½øË®µçÀëµÄÎïÖÊÊÇ£º
Na2SO3
Na2SO3
£¨Ð´³öÒ»ÖÖÎïÖʵĻ¯Ñ§Ê½¼´¿É£¬ÏÂͬ£©£¬¸ÃÎïÖʵÄË®ÈÜÒºÏÔ
¼î
¼î
ÐÔ£¬ÓÃÀë×Ó·½³Ìʽ±íʾÆäÔ­Òò£º
SO32-+H2OHSO3-+OH-
SO32-+H2OHSO3-+OH-
£®
A¡¢B¡¢C¡¢D¡¢EÊÇλÓÚ¶ÌÖÜÆÚµÄÖ÷×åÔªËØ£¬ÆäÖÐA¡¢BΪ½ðÊôÔªËØ£¬DµÄ×îÍâ²ãµç×ÓÊýÊÇÆäÄÚ²ãµç×ÓÊýµÄÈý±¶£®ÒÑÖª£º¢ÙÈÈÎȶ¨ÐÔ£ºHmD£¾HmC£»¢ÚCm-¡¢E£¨m-1£©-¾ßÓÐÏàͬµÄµç×Ó²ã½á¹¹£»¢ÛBÓëEÔÚͬһÖÜÆÚ£¬ÔÚ¸ÃÖÜÆÚÖУ¬EµÄÔ­×Ó°ë¾¶×îС£¬BµÄÀë×Ó°ë¾¶×îС£»¢ÜA¡¢BËùÐγɵĵ¥ºËÀë×Ó¾ßÓÐÏàͬµÄµç×Ó²ã½á¹¹£¬BµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÓëA¡¢EµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯Îï¶¼ÄÜ·´Ó¦£®ÒÀ¾ÝÉÏÊöÐÅÏ¢»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©HmDµÄµç×Óʽ£º______£®AºÍEÐγɵϝºÏÎïµÄ¾§ÌåÀàÐÍÊô______£»¸Ã¾§Ìå½á¹¹ÖÐÀëA×î½üµÄAÓÐ______¸ö£®
£¨2£©ÄÜÖ¤Ã÷Cm-¡¢E£¨m-1£©-»¹Ô­ÐÔÇ¿ÈõµÄ·´Ó¦µÄÀë×Ó·½³ÌʽΪ______£®
£¨3£©Óû¯Ñ§ÓÃÓï±íʾBµÄ×î¸ßÕý¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÔÚË®ÈÜÒºÖеĵçÀë¹ý³Ì______£®
£¨4£©ÒÑÖª17g HmCÔÚÒ»¶¨Ìõ¼þÏÂÍêȫȼÉÕÉú³ÉÎȶ¨»¯ºÏÎïʱ·Å³öÈÈÁ¿Îªa kJ£¬Ð´³öÄܱíʾ¸ÃÎïÖÊȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ______£®
£¨5£©³£ÎÂÏ£¬½«µÈÎïÖʵÄÁ¿Å¨¶ÈµÄHmCÈÜÒººÍAµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÈÜÒºµÈÌå»ý»ìºÏ£¬ËùµÃÈÜÒºÏÔ¼îÐÔ£¬ÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ______£®

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

¾«Ó¢¼Ò½ÌÍø