£¨1£©·¢ÉäÎÀÐÇʱ¿ÉÓÃ루N2H4£©ÎªÈ¼ÁϺÍNO2×÷Ñõ»¯¼Á£¬Á½Õß·´Ó¦Éú³ÉN2ºÍË®ÕôÆø£®ÓÖÒÑÖª£º
¢ÙN2£¨g£©+2O2£¨g£©¨T2NO2£¨g£©¡÷H=+67.7kJ/mol            ¢Ù
¢ÚN2H4£¨g£©+O2£¨Æø£©=N2£¨g£©+2H2O£¨g£©¡÷H=-534kJ/mol     ¢Ú
ÊÔд³öëÂÓëNO2·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ
 
£®
£¨2£©0.3molÆøÌ¬¸ßÄÜȼÁÏÒÒÅðÍ飨·Ö×ÓʽB2H6£©£¬ÔÚÑõÆøÖÐȼÉÕ£¬Éú³É¹Ì̬ÈýÑõ»¯¶þÅðºÍҺ̬ˮ£¬·Å³ö649.5kJµÄÈÈÁ¿£¬ÔòÆäÈÈ»¯Ñ§·½³ÌʽΪ£º
 
£®ÓÖÒÑÖªH2O£¨l£©=H2O£¨g£©£»¡÷H=+44kJ?mol-1£¬Ôò11.2L±ê×¼×´¿öϵÄÒÒÅðÍéÍêȫȼÉÕÉú³ÉÆøÌ¬Ë®Ê±·Å³öµÄÈÈÁ¿ÊÇ
 
kJ£®
¿¼µã£ºÈÈ»¯Ñ§·½³Ìʽ
רÌ⣺»¯Ñ§·´Ó¦ÖеÄÄÜÁ¿±ä»¯
·ÖÎö£º£¨1£©ÒÀ¾ÝÈÈ»¯Ñ§·½³ÌʽµÄÒâÒåºÍ¸Ç˹¶¨ÂɵÄÄÚÈÝͨ¹ýºÏ²¢¼ÆËãµÃµ½ÈÈ»¯Ñ§·½³Ìʽ£»
£¨2£©0.3molÆøÌ¬¸ßÄÜȼÁÏÒÒÅðÍéÔÚÑõÆøÖÐȼÉÕ£¬Éú³É¹Ì̬ÈýÑõ»¯¶þÅðºÍҺ̬ˮ£¬·Å³ö649.5KJµÄÈÈÁ¿£¬Ôò1molÆøÌ¬¸ßÄÜȼÁÏÒÒÅðÍéÔÚÑõÆøÖÐȼÉÕ£¬Éú³É¹Ì̬ÈýÑõ»¯¶þÅðºÍҺ̬ˮ£¬·Å³ö2165KJµÄÈÈÁ¿£¬¸ù¾Ý¸Ç˹¶¨ÂÉд³öÆäÈÈ»¯Ñ§·´Ó¦·½³Ìʽ£¬ºó¸ù¾ÝÎïÖʵÄÎïÖʵÄÁ¿Óë·´Ó¦·Å³öµÄÈÈÁ¿³ÉÕý±ÈÀ´½â´ð£®
½â´ð£º ½â£º£¨1£©¢ÙN2£¨g£©+O2£¨g£©=2NO2£¨g£©¡÷H1=+67.7kJ/mol
¢ÚN2H4£¨g£©+O2£¨g£©=N2£¨g£©+2H2O£¨g£©¡÷H2=-534kJ/mol
ÒÀ¾Ý¸Ç˹¶¨ÂÉ£º¢Ú¡Á2-¢ÙµÃµ½£º2N2H4£¨g£©+2NO2£¨g£©=3N2£¨g£©+4H2O£¨g£©¡÷H=-1135.7KJ/mol£»
¹Ê´ð°¸Îª£º2N2H4£¨g£©+2NO2£¨g£©=3N2£¨g£©+4H2O£¨g£©¡÷H=-1135.7KJ/mol£»
£¨2£©0.3molÆøÌ¬¸ßÄÜȼÁÏÒÒÅðÍéÔÚÑõÆøÖÐȼÉÕ£¬Éú³É¹Ì̬ÈýÑõ»¯¶þÅðºÍҺ̬ˮ£¬·Å³ö649.5KJµÄÈÈÁ¿£¬Ôò1molÆøÌ¬¸ßÄÜȼÁÏÒÒÅðÍéÔÚÑõÆøÖÐȼÉÕ£¬Éú³É¹Ì̬ÈýÑõ»¯¶þÅðºÍҺ̬ˮ£¬·Å³ö2165KJµÄÈÈÁ¿£¬·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ£ºB2H6£¨g£©+3O2£¨g£©¨TB2O3£¨s£©+3H2O£¨l£©¡÷H=-2165kJ/mol£¬
¢ÙB2H6£¨g£©+3O2£¨g£©¨TB2O3£¨s£©+3H2O£¨l£©¡÷H=-2165kJ/mol£¬
¢ÚH2O£¨l£©¡úH2O£¨g£©£»¡÷H=+44kJ/moL£¬
ÓɸÇ˹¶¨ÂÉ¿ÉÖª¢Ù+¢Ú¡Á3µÃ£ºB2H6£¨g£©+3O2£¨g£©¨TB2O3£¨s£©+3H2O£¨g£©¡÷H=-2033kJ/mol£¬
11.2L£¨±ê×¼×´¿ö£©¼´0.5molÒÒÅðÍéÍêȫȼÉÕÉú³ÉÆøÌ¬Ë®Ê±·Å³öµÄÈÈÁ¿ÊÇ2033kJ¡Á0.5=1016.5kJ£¬
¹Ê´ð°¸Îª£ºB2H6£¨g£©+3O2£¨g£©¨TB2O3£¨s£©+3H2O£¨l£©¡÷H=-2165kJ/mol£»1016.5£»
µãÆÀ£º±¾Ì⿼²éÁËÈÈ»¯Ñ§·½³ÌʽµÄÊéд·½·¨ºÍ¸Ç˹¶¨ÂɵļÆËãÓ¦ÓÃÒÔ¼°ÈÈÁ¿µÄ¼ÆË㣬ÐèҪעÒâµÄÓУº·´Ó¦ÈȵÄÊýÖµÓ뻯ѧ·½³ÌÊ½Ç°ÃæµÄϵÊý³ÉÕý±È£®ÌâÄ¿ÄѶÈÖеȣ®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÔÚFe£¨OH£©3½ºÌåÖУ¬ÖðµÎ¼ÓÈëHIÏ¡ÈÜÒº£¬»á³öÏÖһϵÁб仯£®
£¨1£©ÏȳöÏÖºìºÖÉ«³Áµí£¬Ô­ÒòÊÇ
 
£®
£¨2£©Ëæºó³ÁµíÈܽ⣬ÈÜÒº³Ê»ÆÉ«£¬Ð´³ö´Ë·´Ó¦µÄÀë×Ó·½³Ìʽ
 
£®
£¨3£©×îºóÈÜÒºÑÕÉ«¼ÓÉ´ËÔ­ÒòµÄ·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¸ù¾ÝÒªÇ󻨴ðÏà¹ØÎÊÌ⣺
£¨1£©³ýÈ¥ÏÂÁÐÎïÖÊÖÐËù»ìÓеÄÉÙÁ¿ÔÓÖÊ£¨À¨ºÅÄÚΪÔÓÖÊ£©£¬Ð´³öÓйصķ´Ó¦·½³Ìʽ£®
¢ÙÍ­·Û £¨ÂÁ·Û£©
 
£»
¢ÚFeCl3 ÈÜÒº£¨FeCl2 £©
 
£»
¢ÛN2 £¨O2£©
 
£®
£¨2£©ÈçͼËùʾ¸÷Ïî±ä»¯µÄδ֪ÊýÖоùº¬ÓÐÄÆÔªËØ£¬EΪµ­»ÆÉ«·ÛÄ©£¬¾Ý´Ë»Ø´ðÏÂÁÐÎÊÌ⣺
¢Ùд³öA¡úBµÄ»¯Ñ§·½³Ìʽ£¬±ê³öµç×Ó×ªÒÆµÄ·½ÏòºÍÊýÄ¿£»
 

¢ÚÔÚB¡úCµÄ±ä»¯ÖÐËùµÃCµÄÈÜÒºÍùÍù²»´¿£¬ÆäÖеÄÔÓÖÊ£¨²»°üÀ¨Ë®£©¿ÉÄÜÊÇ
 
£¬Ö÷ÒªÔ­ÒòÊÇ
 
£»»¹¿ÉÄÜÊÇ
 
£¬Ö÷ÒªÔ­ÒòÊÇ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÓÐËÄÖÖ³£¼ûÒ©Î¢Ù°¢Ë¾Æ¥ÁÖ¢ÚÇàÃ¹ËØ¢ÛÎ¸ÊæÆ½¢ÜÂ黯¼î£®Çë»Ø´ð£º
£¨1£©Ä³Í¬Ñ§Î¸Ëá¹ý¶à£¬Ó¦Ñ¡ÓõÄÒ©ÎïÊÇ
 
£¨ÌîÐòºÅ£©£¬¿¹ËáÒ©µÄ³É·ÖAl£¨OH£©3ÖкÍθËá¹ý¶àµÄÀë×Ó·½³ÌʽΪ
 
£®
£¨2£©ÓÉÓÚ¾ßÓÐÐË·Ü×÷Ó㬹ú¼Ê°Âί»áÑϽûÔ˶¯Ô±·þÓõÄÒ©ÎïÊÇ
 
£®
£¨3£©´ÓÓÃÒ©°²È«½Ç¶È¿¼ÂÇ£¬Ê¹ÓÃǰҪ½øÐÐÆ¤·ôÃô¸ÐÐÔ²âÊÔµÄÒ©ÎïÊÇ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÔÚ±×åÔªËØÖУ¬Ñõ»¯ÐÔ×îÇ¿µÄÊÇ
 
£¬Ô­×Ó°ë¾¶×îСµÄÊÇ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÓÐÏÂÁÐÎïÖÊ£º¢Ù´×Ëá¢Ú¿ÁÐÔÄÆ¢Û¹èËὺÌå¢ÜÁòËá¢ÝÕáÌÇ¢ÞʳÑÎÈÜÒº¢ß¾Æ¾«¢à¶þÑõ»¯Ì¼¢áÂÈÆø¢âKAl£¨SO4£©2£¬»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÉÏÊöÎïÖÊÖÐÈÜÓÚË®Äܵ¼µçÇÒÊôÓڷǵç½âÖʵÄÊÇ
 
£¨ÌîÐòºÅ£©£¬Çø±ð¢ÛºÍ¢ÞÁ½ÖÖ·Öɢϵ×î¼òµ¥µÄ·½·¨ÊÇÀûÓÃ
 
£®
£¨2£©¢Ù¡¢¢âµÄµçÀë·½³Ìʽ·Ö±ðΪ
 
¡¢
 
£®
£¨3£©ÔÚ0.1mol/L HClÈÜÒºÖмÓÈëÉÙÁ¿¢Ú£¬ºöÂÔÌå»ýºÍζȵı仯£¬ÈÜÒºµÄµ¼µçÐÔ
 
£¨Ìî¡°ÔöÇ¿¡±¡¢¡°¼õÈõ¡±¡¢¡°¼¸ºõ²»±ä¡±£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁл¯Ñ§·½³ÌʽÖУ¬¿ÉÓÃÀë×Ó·½³ÌʽH++OH-=H2O±íʾµÄÊÇ£¨¡¡¡¡£©
A¡¢HCl+Fe£¨OH£©3=FeCl3+3H2O
B¡¢2HCl+Cu£¨OH£©2=CuCl2+2 H2O
C¡¢H2SO4+Ba£¨OH£©2=BaSO4¡ý+2 H2O
D¡¢HNO3+KOH=KNO3+H2O

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

NA´ú±í°¢·ü¼ÓµÂÂÞ³£Êý£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢2.3gÄÆÓÉÔ­×Ó±ä³ÉÀë×Óʱ£¬µÃµ½µÄµç×ÓÊýΪ0.1NA
B¡¢0.2NA¸öÁòËá·Ö×ÓÓë19.6gÁ×ËẬÓÐÏàͬµÄÑõÔ­×ÓÊý
C¡¢28gµªÆøËùº¬ÓеÄÔ­×ÓÊýΪNA
D¡¢NA¸öÑõ·Ö×ÓÓëNA¸öÇâ·Ö×ÓµÄÖÊÁ¿±ÈΪ8£º1

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐËÄÖÖÑÎÈÜÒº³ÊËáÐÔµÄÊÇ£¨¡¡¡¡£©
A¡¢NaCl
B¡¢CH3COONa
C¡¢FeCl3
D¡¢CH3COONH4

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸