£¨9·Ö£©£¨1£©Ä³Ò»·´Ó¦ÌåϵÖеÄÎïÖÊÓУºHCl¡¢SnCl2¡¢H2SnCl6¡¢As¡¢H3AsO3¡¢H2O£¬ÒÑÖª£ºHClÊÇ·´Ó¦ÎïÖ®Ò»¡£

¢Ùд³öÅäƽµÄ¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º______________________ ¡ú_______________________

¢Ú±»Ñõ»¯µÄÔªËØÊÇ             ¡£

£¨2£©Ç⻯ÑÇÍ­(CuH)ÊÇÒ»ÖÖÄÑÈÜÎïÖÊ£¬ÓÃCuSO4ÈÜÒººÍ¡°ÁíÒ»ÎïÖÊ¡±ÔÚ40¡«50¡æʱ·´Ó¦¿ÉÉú³ÉËü£®CuH¾ßÓеÄÐÔÖÊÓУº²»Îȶ¨£¬Ò׷ֽ⣻ÔÚÂÈÆøÖÐÄÜȼÉÕ£»ÓëÏ¡ÑÎËá·´Ó¦ÄÜÉú³ÉÆøÌ壻Cu£«ÔÚËáÐÔÌõ¼þÏ·¢ÉúµÄ·´Ó¦ÊÇ£º2Cu£«===Cu2£«£«Cu.

¸ù¾ÝÒÔÉÏÐÅÏ¢£¬½áºÏ×Ô¼ºËùÕÆÎյĻ¯Ñ§ÖªÊ¶£¬»Ø´ðÏÂÁÐÎÊÌ⣺

¢Ùд³öCuHÔÚÂÈÆøÖÐȼÉյĻ¯Ñ§·´Ó¦·½³Ìʽ£º___________________________________.

¢ÚÈç¹û°ÑCuHÈܽâÔÚ×ãÁ¿µÄÏ¡ÏõËáÖÐÉú³ÉµÄÆøÌåÖ»ÓÐNO£¬Çëд³öCuHÈܽâÔÚ×ãÁ¿Ï¡ÏõËáÖз´Ó¦µÄÀë×Ó·½³Ìʽ£º__________________________________________________.

£¨3£©  ÔÚ25¡æ¡¢101kPaÏ£¬1g¼×´¼È¼ÉÕÉú³ÉCO2ºÍҺ̬ˮʱ·ÅÈÈ22£®68kJ¡£Ôò±íʾ¼×´¼È¼ÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽΪ________________________________________________

 

¡¾´ð°¸¡¿

1£©¢Ù12HCl + 3SnCl2 + 2H3AsO3 = 3 H2SnCl6 +2 As + 6H2O  £¨2·Ö£©  ¢Ú Sn £¨1·Ö£©     

(2)¢Ù2CuH£«3Cl22CuCl2£«2HCl £¨2·Ö£©   ¢Ú CuH£«3H£«£«NO===Cu2£«£«2H2O£«NO¡ü£¨2·Ö£©   

£¨3£©CH3OH£¨l£©+O2£¨g£©¡úCO2£¨g£©+2H2O£¨l£©  ¦¤H£½¨C725£®76kJ¡¤mol£­1   £¨2·Ö£©

¡¾½âÎö¡¿ÂÔ

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2013?ÑĮ̀ģÄ⣩Áò¡¢µª¡¢µâ¶¼ÊÇÖØÒªµÄ·Ç½ðÊôÔªËØ£®
£¨1£©Ä³Ò»·´Ó¦ÌåϵÖÐÓз´Ó¦ÎïºÍÉú³ÉÎï¹²5ÖÖÎïÖÊ£ºS¡¢H2S¡¢HNO3¡¢NO¡¢H2O£®¸Ã·´Ó¦Öл¹Ô­¼ÁÊÇ
H2S
H2S
£®Èô·´Ó¦¹ý³ÌÖÐתÒÆÁË0.3molµç×Ó£¬Ôò·´Ó¦µÄHNO3µÄÖÊÁ¿ÊÇ
6.3
6.3
g£®
£¨2£©ÒÑÖª£º2I£¨g£©¨TI2£¨g£©¡÷H=-151 kJ?mol-1£»2H£¨g£©¨TH2£¨g£©¡÷H=-436 kJ?mol-1£»HI£¨g£©=H£¨g£©+I£¨g£©¡÷H=+298 kJ?mol-1£®ÏàͬÌõ¼þÏ£¬H2ÓëI2·´Ó¦Éú³ÉHIµÄÈÈ»¯Ñ§·½³ÌʽΪ£ºH2£¨g£©+I2£¨g£©=2HI£¨g£©¡÷H=
-9
-9
kJ?mol-1£®
£¨3£©½«1molI2£¨g£©ºÍ2molH2£¨g£©ÖÃÓÚij2LÃܱÕÈÝÆ÷ÖУ¬ÔÚÒ»¶¨Î¶ÈÏ·¢Éú·´Ó¦£ºH2£¨g£©+I2£¨g£©?2HI£¨g£©²¢´ïƽºâ£®HIµÄÌå»ý·ÖÊý¦Õ£¨HI£©Ëæʱ¼ä±ä»¯ÈçͼÇúÏßIIËùʾ£º
¢Ù´ïµ½Æ½ºâʱ£¬I2£¨g£©µÄÎïÖʵÄÁ¿Å¨¶ÈΪ
0.05
0.05
mol?L-1£®
¢Ú±£³Ö¼ÓÈëµÄ·´Ó¦ÎïµÄÎïÖʵÄÁ¿²»±ä£¬Èô¸Ä±ä·´Ó¦Ìõ
¼þ£¬ÔÚijһÌõ¼þϦգ¨HI£©µÄ±ä»¯ÈçÇúÏßIËùʾ£¬Ôò¸ÃÌõ¼þ¿ÉÄÜÊÇ
AB
AB
£¨Ìî±àºÅ£©£¬ÔÚÕâÖÖÌõ¼þÏÂƽºâ³£ÊýKÖµ
²»±ä
²»±ä
£¨Ìî¡°Ôö´ó¡±¡¢¡°±äС¡±»ò¡°²»±ä¡¯£¬£©£®
A£®ºãÎÂÌõ¼þÏ£¬ËõС·´Ó¦µÄÈÝ»ý  B£®ºãÈÝÌõ¼þÏ£®¼ÓÈëÊʵ±´ß»¯¼Á
C£®ÔÚºãÈÝÏ£¬Éý¸ßζȠ   D£®ÔÚºãÈÝÏ£¬³äÈëN2ʹÈÝÆ÷ѹǿÔö´ó
¢ÛÈô±£³ÖζȲ»±ä£¬ÔÚÁíÒ»ÏàͬµÄ2LÃܱÕÈÝÆ÷ÖмÓÈë1mol H2£¨g£©ºÍ2mol Hl£¨g£©£¬·¢Éú·´Ó¦´ïµ½Æ½ºâʱ£¬H2µÄÌå»ý·ÖÊýΪ
37%
37%
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

£¨1£©Ä³Ò»·´Ó¦ÌåϵÖÐÓз´Ó¦ÎïºÍÉú³ÉÎï¹²5ÖÖÎïÖÊ£ºS¡¢H2S¡¢HNO3¡¢NO¡¢H2O£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ²¢Åäƽ
 
£®Èô·´Ó¦¹ý³ÌÖÐתÒÆÁË0.3molµç×Ó£¬ÔòÑõ»¯²úÎïµÄÖÊÁ¿ÊÇ
 
g£®
¾«Ó¢¼Ò½ÌÍø
£¨2£©Í¬Ò»ÎïÖʳÊÆø̬µÄìØÖµ×î´ó£¬ÒºÌ¬µÄìØÖµ´ÎÖ®£¬¹Ì̬µÄìØÖµ×îС£®ÈôͬÎÂͬѹÏÂÒ»¸ö»¯Ñ§·´Ó¦Éú³ÉÎïÆøÌåµÄÌå»ýµÈÓÚ·´Ó¦ÎïÆøÌåµÄÌå»ý¾Í¿ÉÒÔ´ÖÂÔÈÏΪ¸Ã·´Ó¦µÄìرäΪ0£®Ä³»¯Ñ§ÐËȤС×飬רÃÅÑо¿ÁËÑõ×åÔªËؼ°ÆäijЩ»¯ºÏÎïµÄ²¿·ÖÐÔÖÊ£®Ëù²é×ÊÁÏÈçÏ£º
a£®íÚ£¨Te£©Îª¹ÌÌ壬H2TeΪÆøÌ壬TeºÍH2²»ÄÜÖ±½Ó»¯ºÏ³ÉH2Te
b£®µÈÎïÖʵÄÁ¿ÑõÆø¡¢Áò¡¢Îø¡¢íÚÓëH2·´Ó¦µÄìʱäÇé¿öÈçͼ1Ëùʾ£º
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
H2ÓëÁò»¯ºÏµÄ·´Ó¦
 
ÈÈÁ¿£¨Ìî¡°·Å³ö¡±»ò¡°ÎüÊÕ¡±£©£®¸ù¾ÝÌâÄ¿Ëù¸øÐÅÏ¢£¬Çë½âÊÍΪʲôTeºÍH2²»ÄÜÖ±½Ó»¯ºÏ
 
£®
£¨3£©ÔÚ¸´Ôӵķ´Ó¦ÖУ¬Òª¿¼ÂÇ·´Ó¦µÄÏȺó˳Ðò£®ÒÑÖªN
H
+
4
+Al
O
-
2
+2H2O¨TAl£¨OH£©3¡ý+NH3?H2O£¬Ïòº¬ÓеÈÎïÖʵÄÁ¿µÄN
H
+
4
¡¢Al3+¡¢H+»ìºÏÈÜÒºÖУ¬ÂýÂýµÎ¼ÓNaOHÈÜÒº£¬Ö±ÖÁ¹ýÁ¿£¬²¢²»¶Ï½Á°è£¬ÒÀ´Î·¢ÉúÁËÊý¸öÀë×Ó·´Ó¦£»ÆäÖУº
µÚ¶þ¸öÀë×Ó·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ
 
£®
×îºóÒ»¸öÀë×Ó·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ
 
£®
£¨4£©½«1mol I2£¨g£©ºÍ2mol H2£¨g£©ÖÃÓÚij2LÃܱÕÈÝÆ÷ÖУ¬ÔÚÒ»¶¨Î¶ÈÏ·¢Éú·´Ó¦£º
H2£¨g£©+I2£¨g£©?2HI£¨g£©£¬¡÷H£¼0£®²¢´ïƽºâ£®HIµÄÌå»ý·ÖÊýHI%ËæʱÎʱ仯ÇúÏßÈçͼ2Ëùʾ£º
¢Ù´ïµ½Æ½ºâʱ£¬I2£¨g£©µÄÎïÖʵÄÁ¿Å¨¶ÈΪ
 
mol/L£®
¢Ú±£³Ö¼ÓÈëµÄ·´Ó¦ÎïµÄÎïÖʵÄÁ¿²»±ä£¬Èô¸Ä±ä·´Ó¦Ìõ¼þ£¬ÔÚijһÌõ¼þÏÂHI%µÄ±ä»¯ÈçÇúÏß1Ëùʾ£¬Ôò¸ÃÌõ¼þ¿ÉÄÜÊÇ£¨Ð´³öËùÓеĿÉÄÜÐÔ£©
 
ÔÚÕâÖÖÌõ¼þÏ£¬Æ½ºâ³£ÊýKÖµ
 
£¨Ìî¡°Ôö´ó¡±¡¢¡°±äС¡±¡¢¡°²»±ä¡±»ò¡°¿ÉÄܱä´óÒ²¿ÉÄܱäС¡±£©
¢ÛÈô±£³ÖζȲ»±ä£¬ÔÚÁíÒ»ÏàͬµÄ2LÃܱÕÈÝÆ÷ÖмÓÈë1mol H2£¨g£©ºÍ2mol HI£¨g£©£¬·¢Éú·´Ó¦´ïµ½Æ½ºâʱ£¬H2µÄÌå»ý·ÖÊýΪ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨9·Ö£©£¨1£©Ä³Ò»·´Ó¦ÌåϵÖеÄÎïÖÊÓУºHCl¡¢SnCl2¡¢H2SnCl6¡¢As¡¢H3AsO3¡¢H2O£¬ÒÑÖª£ºHClÊÇ·´Ó¦ÎïÖ®Ò»¡£

¢Ùд³öÅäƽµÄ¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º______________________ ¡ú_______________________

¢Ú±»Ñõ»¯µÄÔªËØÊÇ            ¡£

£¨2£©Ç⻯ÑÇÍ­(CuH)ÊÇÒ»ÖÖÄÑÈÜÎïÖÊ£¬ÓÃCuSO4ÈÜÒººÍ¡°ÁíÒ»ÎïÖÊ¡±ÔÚ40¡«50¡æʱ·´Ó¦¿ÉÉú³ÉËü£®CuH¾ßÓеÄÐÔÖÊÓУº²»Îȶ¨£¬Ò׷ֽ⣻ÔÚÂÈÆøÖÐÄÜȼÉÕ£»ÓëÏ¡ÑÎËá·´Ó¦ÄÜÉú³ÉÆøÌ壻Cu£«ÔÚËáÐÔÌõ¼þÏ·¢ÉúµÄ·´Ó¦ÊÇ£º2Cu£«===Cu2£«£«Cu.

¸ù¾ÝÒÔÉÏÐÅÏ¢£¬½áºÏ×Ô¼ºËùÕÆÎյĻ¯Ñ§ÖªÊ¶£¬»Ø´ðÏÂÁÐÎÊÌ⣺

¢Ùд³öCuHÔÚÂÈÆøÖÐȼÉյĻ¯Ñ§·´Ó¦·½³Ìʽ£º___________________________________.

¢ÚÈç¹û°ÑCuHÈܽâÔÚ×ãÁ¿µÄÏ¡ÏõËáÖÐÉú³ÉµÄÆøÌåÖ»ÓÐNO£¬Çëд³öCuHÈܽâÔÚ×ãÁ¿Ï¡ÏõËáÖз´Ó¦µÄÀë×Ó·½³Ìʽ£º__________________________________________________.

£¨3£© ÔÚ25¡æ¡¢101kPaÏ£¬1g¼×´¼È¼ÉÕÉú³ÉCO2ºÍҺ̬ˮʱ·ÅÈÈ22£®68kJ¡£Ôò±íʾ¼×´¼È¼ÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽΪ________________________________________________

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨1£©Ä³Ò»·´Ó¦ÌåϵÖÐÓз´Ó¦ÎïºÍÉú³ÉÎï¹²5ÖÖÎïÖÊ£ºS¡¢H2S¡¢HNO3¡¢NO¡¢H2O¡£ 

¸Ã·´Ó¦Öл¹Ô­²úÎïÊÇ______________________£»Èô·´Ó¦¹ý³ÌÖÐתÒÆÁË0£®2µç

×Ó£¬ÔòÑõ»¯²úÎïµÄÖÊÁ¿ÊÇ________________________¡£

    £¨2£©½«¹ýÁ¿Cl2 ͨÈë FeBr2ÈÜÒºÖУ¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º________________________;

    £¨3£©ÏòNH4HCO3ÈÜÒºÖеμÓÉÙÁ¿µÄNaOHÈÜÒº£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ______________;

    £¨4£©¹Û²ìÈçÏ·´Ó¦£¬×ܽá¹æÂÉ£¬È»ºóÍê³ÉÏÂÁÐÎÊÌ⣺

Al£¨OH£©3 £«H2O Al£¨OH£©4- + H+        NH3+H2ONH4+ OH-

¢ÙÒÑÖªB£¨OH£©3ÊÇÒ»ÔªÈõËᣬд³öÆäµçÀë·½³Ìʽ_____________________________

¢ÚN2H4ÊǶþÔªÈõ¼î£¬Ð´³öÆäµÚ¶þ²½µçÀë·½³Ìʽ___________________     ____

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010ÄêÖØÇìÊиßÈýÉÏѧÆÚµÚ¶þ´ÎÔ¿¼Àí×Û»¯Ñ§¾í ÌâÐÍ£ºÌî¿ÕÌâ

£¨1£©Ä³Ò»·´Ó¦ÌåϵÖÐÓз´Ó¦ÎïºÍÉú³ÉÎï¹²5ÖÖÎïÖÊ£ºS¡¢H2S¡¢HNO3¡¢NO¡¢H2O¡£ 

¸Ã·´Ó¦Öл¹Ô­²úÎïÊÇ______________________£»Èô·´Ó¦¹ý³ÌÖÐתÒÆÁË0£®2µç

×Ó£¬ÔòÑõ»¯²úÎïµÄÖÊÁ¿ÊÇ________________________¡£

    £¨2£©½«¹ýÁ¿Cl2 ͨÈë FeBr2ÈÜÒºÖУ¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º________________________;

    £¨3£©ÏòNH4HCO3ÈÜÒºÖеμÓÉÙÁ¿µÄNaOHÈÜÒº£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ______________;

    £¨4£©¹Û²ìÈçÏ·´Ó¦£¬×ܽá¹æÂÉ£¬È»ºóÍê³ÉÏÂÁÐÎÊÌ⣺

Al£¨OH£©3 £«H2O Al£¨OH£©4- + H+        NH3+H2ONH4+ OH-

¢ÙÒÑÖªB£¨OH£©3ÊÇÒ»ÔªÈõËᣬд³öÆäµçÀë·½³Ìʽ_____________________________

¢ÚN2H4ÊǶþÔªÈõ¼î£¬Ð´³öÆäµÚ¶þ²½µçÀë·½³Ìʽ___________________      ____

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸