£¨8·Ö£©ÔÚ»ð¼ýÍƽøÆ÷ÖÐ×°ÓÐÇ¿»¹Ô­¼Á루N2H4£©ºÍÇ¿Ñõ»¯¼Á£¨H2O2£©£¬µ±ËüÃÇ»ìºÏʱ£¬¼´²úÉú´óÁ¿µÄN2ºÍË®ÕôÆø£¬²¢·Å³ö´óÁ¿ÈÈ¡£ÒÑÖª0.4molҺ̬ëºÍ×ãÁ¿H2O2·´Ó¦£¬Éú³ÉµªÆøºÍË®ÕôÆø£¬·Å³ö256.65kJµÄÈÈÁ¿¡£

£¨1£©Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ__________________________________________¡£

£¨2£©ÒÑÖªH2O(l)=H2O(g) ¡÷H=+44kJ¡¤mol-1£¬Ôò16 gҺ̬ëºÍ×ãÁ¿H2O2·´Ó¦£¬Éú³ÉµªÆøºÍҺ̬ˮʱ£¬·Å³öµÄÈÈÁ¿ÊÇ________kJ¡£

£¨3£©ÉÏÊö·´Ó¦Ó¦ÓÃÓÚ»ð¼ýÍƽø¼Á£¬³ýÊÍ·Å´óÁ¿µÄÈȺͿìËÙ²úÉú´óÁ¿ÆøÌåÍ⣬»¹ÓÐÒ»¸öºÜÍ»³öµÄÓŵãÊÇ________________________¡£

£¨4£©ÒÑÖªN2(g)+2O2(g)=2 NO2(g)   ¡÷H=+67.7 kJ¡¤mol-1£¬ N2H4(g)+O2(g)= N2(g)+2H2O (g) ¡÷H=-534 kJ¡¤mol-1£¬¸ù¾Ý¸Ç˹¶¨ÂÉд³öëÂÓëNO2ÍêÈ«·´Ó¦Éú³ÉµªÆøºÍÆø̬ˮµÄÈÈ»¯Ñ§·½³Ìʽ______________________                ____¡£

 

¡¾´ð°¸¡¿

£¨1£©N2H4(l)+2 H2O2 (l) = N2(g)+ 4 H2O (g)  ¡÷H= - 641.625 kJ¡¤mol-1£¨2·Ö£©

£¨2£©408.8   £¨2·Ö£©£¨3£©Éú³ÉN2ºÍH2O£¬¶Ô»·¾³ÎÞÎÛȾ£¨2·Ö£©

£¨4£©2N2H4(g)+2 NO2 (g) =3N2(g)+4 H2O(g)   ¡÷H= - 1135.7kJ¡¤mol-1£¨2·Ö£©

¡¾½âÎö¡¿ÂÔ

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¶ÌÖÜÆÚÔªËØA¡¢B¡¢C¡¢D¡¢E¡¢F£¬Ô­×ÓÐòÊýÒÀ´ÎÔö´ó£®ÒÑÖª£º¢ÙAÓëEͬÖ÷×壬EµÄµ¥ÖÊÓëD2·´Ó¦¿ÉÉú³ÉE2DºÍE2D2Á½ÖÖ¹ÌÌ壻¢ÚFµÄµ¥ÖÊÔÚD2ÖÐȼÉյIJúÎï¿ÉʹƷºìÈÜÒºÍÊÉ«£»BµÄµ¥ÖÊÔÚD2ÖÐȼÉÕ¿ÉÉú³ÉBDºÍBD2Á½ÖÖÆøÌ壻¢ÛCA4++DA-=CA3¡ü+A2D£¬¸÷ÖÖ·´Ó¦ºÍÉú³ÉÎïµÄµç×ÓÊý¶¼ÓëE+ÏàµÈ£®Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©BD2µÄ½á¹¹Ê½Îª
O=C=O
O=C=O
£¬FµÄÀë×ӽṹʾÒâͼΪ
£¬CA3·Ö×ӵĿռ乹ÐÍΪ
Èý½Ç׶ÐÍ
Èý½Ç׶ÐÍ
£®
£¨2£©·Ï¾ÉÓ¡Ë¢µç·°åÉϺ¬ÓÐÍ­£¬ÒÔÍù»ØÊյķ½·¨Êǽ«Æä×ÆÉÕʹÓÃͭת»¯ÎªÑõ»¯Í­£¬ÔÙÓÃÏ¡ÁòËáÀ´Èܽ⣮ÏÖ¸ÄÓÃA2D2ºÍÏ¡ÁòËá½þÅݼ´´ïÄ¿µÄ£¬ÓÖ±£»¤ÁË»·¾³£¬Æä·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
Cu+H2O2+H2SO4=CuSO4+2H2O
Cu+H2O2+H2SO4=CuSO4+2H2O
£»
£¨3£©ÔÚ»ð¼ýÍƽøÆ÷ÖÐ×°ÓÐÇ¿»¹Ô­¼ÁC2A4ºÍÇ¿Ñõ»¯¼ÁA2D2£¬ÒÑÖª0.5molҺ̬C2A4ºÍ×ãÁ¿ÒºÌ¬A2D2·´Ó¦£¬Éú³ÉÒ»ÖÖÎÞÉ«ÎÞζÎÞ¶¾µÄÆøÌåºÍË®ÕôÆø£¬·Å³ö320kJÈÈÁ¿£¬Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ
N2H4£¨1£©+2H2O2£¨1£©=N2£¨g£©+4H2O£¨g£©£»¡÷H=-640kJ?mol-1
N2H4£¨1£©+2H2O2£¨1£©=N2£¨g£©+4H2O£¨g£©£»¡÷H=-640kJ?mol-1
£®
£¨4£©Àë×Ó»¯ºÏÎïCA5µÄµç×ÓʽΪ
£¬Æ侧ÌåµÄ×îС½á¹¹µ¥ÔªÈçͼËùʾ¾­ÊµÑé²âµÃCA5¾§ÌåµÄÃܶÈΪdg/cm3£¬°¢·ü¼ÓµÂÂÞ³£ÊýÓÃNA±íʾ£¬Ôò¸Ã¾§ÌåÖÐÒõÀë×ÓÓëÑôÀë×ÓÖ®¼äµÄ×î¶Ì¾àÀëΪ
3
2
3
19
NAd
3
2
3
19
NAd
cm£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÔÚ»ð¼ýÍƽøÆ÷ÖÐ×°ÓÐÇ¿»¹Ô­¼Á루N2H4£©ºÍÇ¿Ñõ»¯¼Á£¨H2O2£©£¬µ±ËüÃÇ»ìºÏʱ£¬¼´²úÉú´óÁ¿µÄN2ºÍË®ÕôÆø£¬²¢·Å³ö´óÁ¿ÈÈ£®ÒÑÖª0.4molҺ̬ëºÍ×ãÁ¿H2O2·´Ó¦£¬Éú³ÉµªÆøºÍË®ÕôÆø£¬·Å³ö256.65kJµÄÈÈÁ¿£®
£¨1£©Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ
N2H4£¨l£©+2H2O2£¨l£©¨TN2£¨g£©+4H2O£¨g£©¡÷H=-641.625 kJ?mol-1
N2H4£¨l£©+2H2O2£¨l£©¨TN2£¨g£©+4H2O£¨g£©¡÷H=-641.625 kJ?mol-1
£®
£¨2£©ÒÑÖªH2O£¨l£©¨TH2O£¨g£©¡÷H=+44kJ?mol-1£¬Ôò16gҺ̬ëÂÓë×ãÁ¿ÒºÌ¬¹ýÑõ»¯Çâ·´Ó¦Éú³ÉµªÆøºÍҺ̬ˮʱ£¬·Å³öµÄÈÈÁ¿ÊÇ
408.8
408.8
kJ£®
£¨3£©ÉÏÊö·´Ó¦Ó¦ÓÃÓÚ»ð¼ýÍƽø¼Á£¬³ýÊÍ·Å´óÁ¿µÄÈȺͿìËÙ²úÉú´óÁ¿ÆøÌåÍ⣬»¹ÓÐÒ»¸öºÜÍ»³öµÄÓŵãÊÇ
Éú³ÉÎïΪµªÆøºÍË®£¬²»ÎÛȾ¿ÕÆø
Éú³ÉÎïΪµªÆøºÍË®£¬²»ÎÛȾ¿ÕÆø
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÔÚ»ð¼ýÍƽøÆ÷ÖÐ×°ÓÐÇ¿»¹Ô­¼Á루N2H4£©ºÍÇ¿Ñõ»¯¼ÁH2O2£¬µ±ËüÃÇ»ìºÏʱ£¬¼´²úÉú´óÁ¿N2ºÍË®ÕôÆø£¬²¢·Å³ö´óÁ¿ÈÈ£¬ÒÑÖª0.4molҺ̬ëÂÓë×ãÁ¿ÒºÌ¬H2O2·´Ó¦£¬Éú³ÉµªÆøºÍË®ÕôÆø£¬·Å³ö256.65KJÈÈÁ¿£®
£¨1£©Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ
N2H4£¨l£©+2H2O2£¨l£©¨TN2£¨g£©+4H2O£¨g£©¡÷H=-641.625 kJ?mol-1
N2H4£¨l£©+2H2O2£¨l£©¨TN2£¨g£©+4H2O£¨g£©¡÷H=-641.625 kJ?mol-1
£®
£¨2£©ÒÑÖªH2O£¨l£©=H2O£¨g£©-44KJ£¬Ôò16gҺ̬ëÂÓë×ãÁ¿ÒºÌ¬¹ýÑõ»¯Çâ·´Ó¦Éú³ÉµªÆøºÍҺ̬ˮʱ£¬·Å³öµÄÈÈÁ¿ÊÇ
408.8
408.8
KJ£®
£¨3£©ÉÏÊö·´Ó¦ÓÃÓÚ»ð¼ýÍƽøÆ÷£¬³ýÊͷųö´óÁ¿ÈȺͿìËÙ²úÉú´óÁ¿ÆøÌåÍ⣬»¹ÓÐÒ»¸öºÜÍ»³öµÄÓŵãÊÇ
Éú³ÉÎïΪµªÆøºÍË®£¬²»ÎÛȾ¿ÕÆø
Éú³ÉÎïΪµªÆøºÍË®£¬²»ÎÛȾ¿ÕÆø
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2009?Ïæ̶һģ£©¸ù¾ÝÏÂÁÐÊÂʵÍê³ÉÏÂÁз´Ó¦·½³Ìʽ£º
£¨1£©AsH3ÊÇÒ»ÖÖºÜÇ¿µÄ»¹Ô­¼Á£®ÊÒÎÂÏ£¬ËüÄÜÔÚ¿ÕÆøÖÐ×Ôȼ£¬ÆäÑõ»¯²úÎïΪAs2O3£¬Ð´³ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
2AsH3+3O2=As2O3+3H2O
2AsH3+3O2=As2O3+3H2O
£®
£¨2£©°×Á×ÔÚÒ»¶¨Ìõ¼þÏ¿ɱ»HClO3µÄË®ÈÜÒºÑõ»¯£¬Éú³ÉÁ×µÄ×î¸ß¼Û̬»¯ºÏÎ¶øÂÈÔªËر»»¹Ô­Îª×îµÍ¼Û̬£¬Ð´³ö·´Ó¦»¯Ñ§·½³Ìʽ
3P4+10HClO3+18H2O=10HCl+12H3PO4
3P4+10HClO3+18H2O=10HCl+12H3PO4
£®
£¨3£©Ï¡ÁòËáÓëÍ­²»·´Ó¦£¬µ«ÔÚÏ¡ÁòËáÖмÓH2O2ºó£¬Ôò¿Éʹͭ˳ÀûÈܽ⣬д³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ£º
H2O2+Cu+2H+=Cu2++2H2O
H2O2+Cu+2H+=Cu2++2H2O
£®
£¨4£©ÔÚ»ð¼ýÍƽøÆ÷ÖÐ×°ÓÐÇ¿»¹Ô­¼Á£¨N2H4£©ºÍÇ¿Ñõ»¯¼ÁH2O2£®µ±ËüÃÇ»ìºÏʱ£¬¼´²úÉú´óÁ¿µªÆøºÍË®ÕôÆø£¬²¢·Å³ö´óÁ¿ÈÈ£®ÒÑÖª0.4molҺ̬ëºÍ×ãÁ¿ÒºÌ¬H2O2·´Ó¦£¬Éú³ÉµªÆøºÍË®ÕôÆø£¬·Å³ö256.65KJµÄÈÈÁ¿£®Ð´³ö´Ë·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º
N2H4£¨l£©+2H2O2£¨l£©¨TN2£¨g£©+4H2O£¨g£©¡÷H=-641.625 kJ?mol-1
N2H4£¨l£©+2H2O2£¨l£©¨TN2£¨g£©+4H2O£¨g£©¡÷H=-641.625 kJ?mol-1
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

ÓÐM¡¢A¡¢B¡¢D¡¢N¡¢EÎåÖÖ¶ÌÖÜÆÚÔªËØ£¬Ô­×ÓÐòÊýÒÀ´ÎÔö´ó£®MÔªËصĵ¥ÖÊÊÇ×ÔÈ»½ç×îÇáµÄÆøÌ壬NÔªËصÄÔ­×Ӱ뾶ÊÇËùÔÚÖÜÆÚÔ­×Ӱ뾶×î´óµÄ£®A¡¢B¡¢D¡¢E·Ö±ðÔÚ±í£¨ÖÜÆÚ±íµÄÒ»²¿·Ö£©Õ¼ÓÐÏàÓ¦µÄλÖã¬ËüÃǵÄÔ­×ÓÐòÊýÖ®ºÍΪ37£®ÊԻشð£º
£¨1£©ÔÚB¡¢D¡¢EÈýÖÖÔªËصÄ×î¼òµ¥Ç⻯ÎïÖУ¬×îÎȶ¨µÄÊÇ
H2O
H2O
£¨Ìѧʽ£©£®M¡¢D¡¢E¡¢NÐγɵļòµ¥Àë×ӵİ뾶ÓÉ´óµ½Ð¡µÄ¹ØϵÊÇ£¨ÓÃÀë×Ó·ûºÅ±íʾ£©
S2-£¾O2-£¾Na+£¾H+
S2-£¾O2-£¾Na+£¾H+
£®
£¨2£©A¡¢B¡¢D¡¢M¿É×é³É¶àÖÖ18µç×Ó·Ö×Ó£¬Çëд³ö2ÖÖ¾ßÓÐ18µç×ÓµÄÓлúÎïµÄ½á¹¹¼òʽ£¨Óýṹ¼òʽ±íʾ£©
CH3OH
CH3OH
¡¢
CH3CH3
CH3CH3
£®
£¨3£©ÓÉA¡¢D¡¢NÈýÖÖÔªËØ×é³ÉµÄÎÞ»úÎÏà¶Ô·Ö×ÓÖÊÁ¿106£©£¬ÆäË®ÈÜÒº³Ê¼îÐÔ£¬ÓÃÀë×Ó·½³Ìʽ±íʾÆäÔ­Òò
CO32-+H2O?HCO3-+OH-
CO32-+H2O?HCO3-+OH-
£®
£¨4£©ÔªËØBµÄÇ⻯ÎïÓëÔªËØDµ¥ÖÊÒ»¶¨Ìõ¼þÏ·¢ÉúÖû»·´Ó¦£¬Ôڸ÷´Ó¦ÖÐÑõ»¯¼ÁÓ뻹ԭ¼ÁµÄÎïÖʵÄÁ¿Ö®±ÈΪ
3£º4
3£º4
£®
£¨5£©ÔÚ»ð¼ýÍƽøÆ÷ÖÐ×°ÓÐÇ¿»¹Ô­¼ÁB2M4ºÍÇ¿Ñõ»¯¼ÁM2D2£¬ÒÑÖª0.5molҺ̬B2M4ºÍ×ãÁ¿ÒºÌ¬M2D2·´Ó¦£¬Éú³ÉÒ»ÖÖÎÞÉ«ÎÞζÎÞ¶¾µÄÆøÌåºÍÒ»ÖÖÕôÆø£¬·Å³ö320.8kJÈÈÁ¿£¬
H2O£¨1£©=H2O£¨g£©¡÷H=+44kJ?mol-1
2M2D2£¨1£©=2M2D£¨1£©+D2£¨g£©¡÷H=-196.4kJ?mol-1
д³öB2M4ÓëÑõÆø·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ
N2H4£¨1£©+O2£¨g£©=N2£¨g£©+2H2O£¨1£©¡÷H=-621.2kJ?mol-1£»
N2H4£¨1£©+O2£¨g£©=N2£¨g£©+2H2O£¨1£©¡÷H=-621.2kJ?mol-1£»
£®
Çëд³öB2M4¡¢D2ÓëKOHÈÜÒº×é³ÉµÄ¼îÐÔµç³ØµÄ¸º¼«·´Ó¦Ê½£¨ÒÑÖª£ºB2M4¿ÉÒԺ͠D2·´Ó¦Éú³ÉB2ºÍM2D£©
N2H4-4e©¤+4OH©¤=N2+4H2O
N2H4-4e©¤+4OH©¤=N2+4H2O
£»¸Ãµç³ØÖÐOH-ÒÆÏò
¸º
¸º
¼«£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸