(14·Ö)

ÏÖÓÐA¡¢B¡¢C¡¢D¡¢E¡¢FÁùÖÖ¶ÌÖÜÆÚÖ÷×åÔªËØ£¬ËüÃǵÄÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£¬DÓëEµÄÇ⻯Îï·Ö×Ó¹¹ÐͶ¼ÊÇVÐÍ£®A¡¢BÔ­×ÓµÄ×îÍâ²ãµç×ÓÊýÖ®ºÍÓëCÔ­×ÓµÄ×îÍâ²ãµç×ÓÊýÏàµÈ£¬AÄÜ·Ö±ðÓëB¡¢C¡¢DÐγɵç×Ó×ÜÊýÏàµÈµÄ·Ö×Ó£¬ÇÒAÓëD¿ÉÐγɵĻ¯ºÏÎï³£ÎÂϾùΪҺ̬£®Çë»Ø´ðÏÂÁÐÎÊÌ⣨Ìî¿ÕʱÓÃʵ¼ÊÎïÖʵĻ¯Ñ§Ê½±íʾ£©£º

£¨1£©CÓëFÁ½ÖÖÔªËØÐγÉÒ»ÖÖ»¯ºÏÎï·Ö×Ó£¬¸÷Ô­×Ó×îÍâ²ã´ï8µç×ӽṹ£¬Ôò¸Ã·Ö×ӵĽṹʽ      £¬Æä¿Õ¼ä¹¹ÐÍΪ           £®

£¨2£©¿Æѧ¼ÒÖƳöÁíÒ»ÖÖÖ±ÏßÐÍÆø̬»¯ºÏÎï B2D2·Ö×Ó£¬ÇÒ¸÷Ô­×Ó×îÍâ²ã¶¼Âú×ã8µç×ӽṹ£¬ÔòB2D2µç×ÓʽΪ                        £®

£¨3£©C4·Ö×ӽṹÈçÌâ26ͼËùʾ£¬ÒÑÖª¶ÏÁÑlmol C£­CÎüÊÕ167kJÈÈÁ¿£¬Éú³É1mo1C¡ÔC·Å³ö942kJÈÈÁ¿.ÔòÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ          £®

¢Ù C4ÊôÓÚÒ»ÖÖÐÂÐ͵Ļ¯ºÏÎï                          ¢Ú C4·Ðµã±ÈP4(°×Á×)µÍ

¢Û lmol C4ÆøÌåת±äΪC2ÎüÊÕ882kJÈÈÁ¿        ¢Ü C2ÓëC4»¥ÎªÍ¬ËØÒìÐÎÌå

¢Ý C4ÓëC2»¥ÎªÍ¬·ÖÒì¹¹Ìå

£¨4£©ÒÑÖªA2(g)ÓëC2(g)Ò»¶¨Ìõ¼þÏ·´Ó¦Éú³É0.5mol CA3(g)·Å³ö23kJµÄÈÈÁ¿£¬Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ                                 £®

£¨5£©ÎªÁ˳ýÈ¥»¯ºÏÎïÒÒ(A2ED4)µÄÏ¡ÈÜÒºÖлìÓеÄA2ED3£¬³£¼ÓÈëÑõ»¯¼ÁA2D2£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º                                 £®

 

¡¾´ð°¸¡¿

 

£¨1£©     (2·Ö)    Èý½Ç׶ÐΠ (2·Ö)

£¨2£©              (2·Ö)

£¨3£© ¢Ú¢Ü         (2·Ö)

£¨4£©N2(g)+3H2(g) 2NH3(g)£»  ¡÷H =-92kJ/mol (3·Ö)

£¨5£©H2O2+H2SO3=2H+ Ê®SO42£­+H2O £¨3·Ö£©

¡¾½âÎö¡¿

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

(14·Ö)

ÏÖÓÐA¡¢B¡¢C¡¢D¡¢E¡¢FÁùÖÖ¶ÌÖÜÆÚÔªËØ£¬A¡¢E´¦ÓÚͬһÖ÷×åÇÒEÊǶÌÖÜÆÚÔªËØÖÐÔ­×Ӱ뾶×î´óµÄÔªËØ(²»º¬Ï¡ÓÐÆøÌå)¡£ÒÑÖªCµÄÔ­×ÓÐòÊýµÈÓÚA¡¢BµÄÔ­×ÓÐòÊýÖ®ºÍ£¬DµÄÔ­×ÓÐòÊýµÈÓÚA¡¢CµÄÔ­×ÓÐòÊýÖ®ºÍ¡£ÈËÀàÒÑÖªµÄ»¯ºÏÎïÖУ¬ÓÐÒ»À໯ºÏÎïµÄÖÖÀàÒѳ¬¹ýÈýǧÍò£¬ÕâÀ໯ºÏÎïÖÐÒ»°ã¶¼º¬ÓÐA¡¢BÁ½ÖÖÔªËØ¡£FÊÇÒ»ÖÖÉú»îÖг£ÓõĽðÊô¡£¸ù¾ÝÒÔÉÏÐÅÏ¢£¬»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ÔªËØBÔÚÖÜÆÚ±íÖеÄλÖÃÊÇ                ¡£

(2)FµÄµ¥ÖÊÓëEµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯Îï·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ              ¡£

(3)C¡¢D×îµÍ¼ÛÇ⻯ÎïÎȶ¨ÐÔµÄÇ¿Èõ˳ÐòÊÇ              (Ìѧʽ) £»D¡¢E¡¢FÔ­×Ӱ뾶µÄ´óС˳ÐòÊÇ                   (ÌîÔªËØ·ûºÅ)¡£

(4)ÔÚA¡¢C¡¢D×é³ÉµÄ»¯ºÏÎïÖУ¬¼ÈÓÐÀë×Ó¼üÓÖÓй²¼Û¼üµÄÊÇ          (Ìѧʽ)¡£

(5)ÔÚA¡¢B¡¢DÖУ¬ÓÉÁ½ÖÖ»òÈýÖÖÔªËØ×é³ÉÁ½ÖÖ¿ÉÒÔ·´Ó¦µÄÒõÀë×Ó£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ                                  ¡£

(6)»¯ºÏÎïBA4DÓëO2¡¢Ï¡ÁòËá¿É×é³ÉȼÁϵç³Ø£¬´Ëµç³ØµÄ¸º¼«·´Ó¦Ê½ÊÇ           ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

(14·Ö)

ÏÖÓÐA¡¢B¡¢C¡¢D¡¢E¡¢FÁùÖÖ¶ÌÖÜÆÚÖ÷×åÔªËØ£¬ËüÃǵÄÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£¬DÓëEµÄÇ⻯Îï·Ö×Ó¹¹ÐͶ¼ÊÇVÐÍ£®A¡¢BÔ­×ÓµÄ×îÍâ²ãµç×ÓÊýÖ®ºÍÓëCÔ­×ÓµÄ×îÍâ²ãµç×ÓÊýÏàµÈ£¬AÄÜ·Ö±ðÓëB¡¢C¡¢DÐγɵç×Ó×ÜÊýÏàµÈµÄ·Ö×Ó£¬ÇÒAÓëD¿ÉÐγɵĻ¯ºÏÎï³£ÎÂϾùΪҺ̬£®Çë»Ø´ðÏÂÁÐÎÊÌ⣨Ìî¿ÕʱÓÃʵ¼ÊÎïÖʵĻ¯Ñ§Ê½±íʾ£©£º

£¨1£©CÓëFÁ½ÖÖÔªËØÐγÉÒ»ÖÖ»¯ºÏÎï·Ö×Ó£¬¸÷Ô­×Ó×îÍâ²ã´ï8µç×ӽṹ£¬Ôò¸Ã·Ö×ӵĽṹʽ      £¬Æä¿Õ¼ä¹¹ÐÍΪ          £®

£¨2£©¿Æѧ¼ÒÖƳöÁíÒ»ÖÖÖ±ÏßÐÍÆø̬»¯ºÏÎï B2D2·Ö×Ó£¬ÇÒ¸÷Ô­×Ó×îÍâ²ã¶¼Âú×ã8µç×ӽṹ£¬ÔòB2D2µç×ÓʽΪ                       £®

£¨3£©C4·Ö×ӽṹÈçÌâ26ͼËùʾ£¬ÒÑÖª¶ÏÁÑlmol C£­CÎüÊÕ167kJÈÈÁ¿£¬Éú³É1mo1C¡ÔC·Å³ö942kJÈÈÁ¿.ÔòÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ          £®

¢Ù C4ÊôÓÚÒ»ÖÖÐÂÐ͵Ļ¯ºÏÎï                         ¢Ú C4·Ðµã±ÈP4(°×Á×)µÍ

¢Û lmol C4ÆøÌåת±äΪC2ÎüÊÕ882kJÈÈÁ¿        ¢Ü C2ÓëC4»¥ÎªÍ¬ËØÒìÐÎÌå

¢Ý C4ÓëC2»¥ÎªÍ¬·ÖÒì¹¹Ìå

£¨4£©ÒÑÖªA2(g)ÓëC2(g)Ò»¶¨Ìõ¼þÏ·´Ó¦Éú³É0.5mol CA3(g)·Å³ö23kJµÄÈÈÁ¿£¬Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ                                £®

£¨5£©ÎªÁ˳ýÈ¥»¯ºÏÎïÒÒ(A2ED4)µÄÏ¡ÈÜÒºÖлìÓеÄA2ED3£¬³£¼ÓÈëÑõ»¯¼ÁA2D2£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º                                £®

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012½ì¹óÖÝÊ¡¿­ÀïÒ»ÖиßÈýµÚ¶þ´ÎÔ¿¼Àí×ۺϲâÊÔ£¨»¯Ñ§²¿·Ö£© ÌâÐÍ£ºÌî¿ÕÌâ

(14·Ö)ÏÖÓÐA¡¢B¡¢C¡¢D¡¢EÎåÖÖÇ¿µç½âÖÊ£¬ËüÃÇÔÚË®ÖпɵçÀë²úÉúÏÂÁÐÀë×Ó(¸÷ÖÖÀë×Ó²»Öظ´)¡£

ÑôÀë×Ó
H£«¡¢Na£«¡¢Al3£«¡¢Ag£«¡¢Ba2£«
ÒõÀë×Ó
OH£­¡¢Cl£­¡¢CO32£­¡¢NO3£­¡¢SO42£­
Çë¸ù¾ÝÒÔÏÂʵÑéÊÂʵÖð²½ÍƳöËüÃǵÄÃû³Æ²¢»Ø´ðÎÊÌ⣺
(1)
ÎïÖʼø¶¨ÊµÑé
ÍƵ¼½áÂÛ
¢ÙÓÃpHÊÔÖ½²â³öA¡¢BÈÜÒº³Ê¼îÐÔ£¬C¡¢D¡¢EÈÜÒº³ÊËáÐÔ
A¡¢BÖк¬ÓеÄÒõÀë×ÓΪ________£¬C¡¢D¡¢EÖк¬ÓеÄÑôÀë×ÓΪ________
¢ÚAÈÜÒºÓëEÈÜÒº·´Ó¦¼ÈÓÐÆøÌåÓÖÓгÁµí²úÉú£»AÓëC·´Ó¦Ö»ÓÐÆøÌå²úÉú
AΪ________£¬Cº¬ÓеÄÑôÀë×ÓΪ________
¢ÛDÈÜÒºÓëÁíÍâËÄÖÖÈÜÒº·´Ó¦¶¼ÄܲúÉú³Áµí£»CÖ»ÄÜÓëD·´Ó¦²úÉú³Áµí
DΪ________
(2)д³öEÈÜÒºÓë¹ýÁ¿µÄBÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ£º_____________________________¡£
(3)ÔÚ100 mL 0.1 mol/LµÄEÈÜÒºÖУ¬ÖðµÎ¼ÓÈë35 mL 2 mol/L NaOHÈÜÒº£¬×îÖյõ½³ÁµíµÄÎïÖʵÄÁ¿Îª__________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011½ìÖØÇìÊÐÆßÇø2011½ì¸ßÈýµÚÒ»´Îµ÷ÑвâÊÔ£¨Àí×Û£©»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ

(14·Ö)
ÏÖÓÐA¡¢B¡¢C¡¢D¡¢E¡¢FÁùÖÖ¶ÌÖÜÆÚÔªËØ£¬A¡¢E´¦ÓÚͬһÖ÷×åÇÒEÊǶÌÖÜÆÚÔªËØÖÐÔ­×Ӱ뾶×î´óµÄÔªËØ(²»º¬Ï¡ÓÐÆøÌå)¡£ÒÑÖªCµÄÔ­×ÓÐòÊýµÈÓÚA¡¢BµÄÔ­×ÓÐòÊýÖ®ºÍ£¬DµÄÔ­×ÓÐòÊýµÈÓÚA¡¢CµÄÔ­×ÓÐòÊýÖ®ºÍ¡£ÈËÀàÒÑÖªµÄ»¯ºÏÎïÖУ¬ÓÐÒ»À໯ºÏÎïµÄÖÖÀàÒѳ¬¹ýÈýǧÍò£¬ÕâÀ໯ºÏÎïÖÐÒ»°ã¶¼º¬ÓÐA¡¢BÁ½ÖÖÔªËØ¡£FÊÇÒ»ÖÖÉú»îÖг£ÓõĽðÊô¡£¸ù¾ÝÒÔÉÏÐÅÏ¢£¬»Ø´ðÏÂÁÐÎÊÌ⣺
(1)ÔªËØBÔÚÖÜÆÚ±íÖеÄλÖÃÊÇ                ¡£
(2)FµÄµ¥ÖÊÓëEµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯Îï·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ              ¡£
(3)C¡¢D×îµÍ¼ÛÇ⻯ÎïÎȶ¨ÐÔµÄÇ¿Èõ˳ÐòÊÇ              (Ìѧʽ) £»D¡¢E¡¢FÔ­×Ӱ뾶µÄ´óС˳ÐòÊÇ                   (ÌîÔªËØ·ûºÅ)¡£
(4)ÔÚA¡¢C¡¢D×é³ÉµÄ»¯ºÏÎïÖУ¬¼ÈÓÐÀë×Ó¼üÓÖÓй²¼Û¼üµÄÊÇ          (Ìѧʽ)¡£
(5)ÔÚA¡¢B¡¢DÖУ¬ÓÉÁ½ÖÖ»òÈýÖÖÔªËØ×é³ÉÁ½ÖÖ¿ÉÒÔ·´Ó¦µÄÒõÀë×Ó£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ                                  ¡£
(6)»¯ºÏÎïBA4DÓëO2¡¢Ï¡ÁòËá¿É×é³ÉȼÁϵç³Ø£¬´Ëµç³ØµÄ¸º¼«·´Ó¦Ê½ÊÇ           ¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸