ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿ºÏ³ÉÆøµÄÖ÷Òª³É·ÖÊÇÒ»Ñõ»¯Ì¼ºÍÇâÆø£¬¿ÉÓÃÓںϳɶþ¼×ÃѵÈÇå½àȼÁÏ¡£´ÓÌìÈ»Æø»ñµÃºÏ³ÉÆø¹ý³ÌÖпÉÄÜ·¢ÉúµÄ·´Ó¦ÓУº
CH4(g)+H2O(g)CO(g)+3H2(g) ¦¤H1=+206.1kJ/mol ¢Ù
CH4(g)+CO2(g)2CO(g)+2H2(g) ¦¤H2=+247.3kJ/mol ¢Ú
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)ÔÚÒ»ÃܱÕÈÝÆ÷ÖнøÐз´Ó¦¢Ù£¬²âµÃCH4µÄÎïÖʵÄÁ¿Å¨¶ÈË淴Ӧʱ¼äµÄ±ä»¯Èçͼ1Ëùʾ¡£·´Ó¦½øÐеÄÇ°5minÄÚ£¬v(H2)=____£»10minʱ£¬¸Ä±äµÄÍâ½çÌõ¼þ¿ÉÄÜÊÇ_____¡£
(2)Èçͼ2Ëùʾ£¬Ôڼס¢ÒÒÁ½ÈÝÆ÷Öзֱð³äÈëµÈÎïÖʵÄÁ¿µÄCH4ºÍCO2£¬Ê¹¼×¡¢ÒÒÁ½ÈÝÆ÷³õʼÈÝ»ýÏàµÈ¡£ÔÚÏàͬζÈÏ·¢Éú·´Ó¦¢Ú£¬²¢Î¬³Ö·´Ó¦¹ý³ÌÖÐζȲ»±ä¡£ÒÑÖª¼×ÈÝÆ÷ÖÐCH4µÄת»¯ÂÊËæʱ¼ä±ä»¯µÄͼÏñÈçͼ3Ëùʾ£¬ÇëÔÚͼ3Öл³öÒÒÈÝÆ÷ÖÐCH4µÄת»¯ÂÊËæʱ¼ä±ä»¯µÄͼÏñ_____¡£
(3)800¡æʱ£¬·´Ó¦CO(g)+H2O(g)CO2(g)+H2(g)µÄ»¯Ñ§Æ½ºâ³£ÊýK=1.0£¬Ä³Ê±¿Ì²âµÃ¸ÃζÈϵÄÃܱÕÈÝÆ÷Öи÷ÎïÖʵÄÎïÖʵÄÁ¿¼ûÏÂ±í£¬´Ëʱ·´Ó¦ÖÐÕý¡¢Äæ·´Ó¦ËÙÂʵĹØϵʽÊÇ____£¨ÌîÐòºÅ£©¡£
a.v(Õý)£¾v(Äæ)
b.v(Õý)£¼v(Äæ)
c.v(Õý)=v(Äæ)
d.ÎÞ·¨ÅжÏ
CO | H2O | CO2 | H2 |
0.5mol | 8.5mol | 2.0mol | 2.0mol |
(4)¶þ¼×ÃÑ£¨CH3OCH3£©¿ÉÓɺϳÉÆø£¨COºÍH2£©ÔÚÒ»¶¨Ìõ¼þÏÂÖƵã¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º____¡£
(5)ÒÔ¶þ¼×ÃÑ¡¢¿ÕÆø¡¢ÇâÑõ»¯¼ØÈÜҺΪÔÁÏ£¬Ê¯Ä«Îªµç¼«¹¹³ÉȼÁϵç³Ø¡£¸Ãµç³ØµÄ¸º¼«µç¼«·´Ó¦·½³ÌʽÊÇ£º____¡£Óöþ¼×ÃÑȼÁϵç³ØΪµçÔ´£¬Óò¬µç¼«µç½âKClºÍCuSO4µÄ»ìºÏÈÜÒº£¬µ±µç·ÖÐͨ¹ý0.3molµç×ӵĵçÁ¿Ê±£¬ÒõÑôÁ½¼«¶¼²úÉú2.8LµÄÆøÌå(±ê×¼×´¿ö)£¬Èôµç½âºóÈÜÒºÌå»ýΪ1L£¬ÔòÑô¼«ÆøÌåµÄ³É·Ö¼°ÎïÖʵÄÁ¿Îª£º____¡£
¡¾´ð°¸¡¿0.3mol/(L¡¤min) Éý¸ßζȻò³åÈëË®ÕôÆø a 2CO+4H2¡úCH3OCH3+H2O£¨Ò»¶¨Ìõ¼þ£© CH3OCH3+16OH--12e-=2CO32-+11H2O O2£º0.025mol¡¢Cl2£º0.10mol
¡¾½âÎö¡¿
(1)·´Ó¦½øÐеÄÇ°5minÄÚ£¬CH4µÄŨ¶È±ä»¯Á¿Îª0.50mol/L£¬ÓÉ´Ë¿ÉÇó³öv(H2)£»10minʱ£¬¸Ä±äµÄÍâ½çÌõ¼þ£¬2minÄÚCH4µÄŨ¶È±ä»¯Á¿Îª0.25mol/L£¬Óëǰһʱ¼ä¶Î½øÐжԱȣ¬´Ó¶øµÃ³ö·´Ó¦ËÙÂʼӿ죬ÓÉ´ËÔ¤²â¿ÉÄÜ·¢Éú¸Ä±äµÄÍâ½çÌõ¼þ¡£
(2)ÒòΪ·´Ó¦CH4(g)+CO2(g)2CO(g)+2H2(g)·¢Éúºó£¬ÆøÌå·Ö×ÓÊýÔö´ó£¬ËùÒÔÒÒÈÝÆ÷Ï൱ÓÚ¼×ÈÝÆ÷¼õСѹǿ£¬ÓÉ´Ë¿ÉÈ·¶¨Æ½ºâÒƶ¯µÄ·½Ïò¡¢CH4µÄת»¯Âʱ仯¼°ËùÐèµÄʱ¼ä¡£
(3)800¡æʱ£¬·´Ó¦CO(g)+H2O(g)CO2(g)+H2(g)µÄ»¯Ñ§Æ½ºâ³£ÊýK=1.0£¬ÀûÓñíÖÐÊý¾ÝÇóŨ¶ÈÉÌ£¬²¢ÓëK½øÐбȽϣ¬´Ó¶øÈ·¶¨Æ½ºâÒƶ¯µÄ·½Ïò£¬Óɴ˵óöÕý¡¢Äæ·´Ó¦ËÙÂʵÄÏà¶Ô´óС¡£
(4) COºÍH2ÔÚÒ»¶¨Ìõ¼þÏÂÖƵÃCH3OCH3£¬ÓÉ´Ë¿Éд³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ¡£
(5)ÒÔ¶þ¼×ÃÑ¡¢¿ÕÆø¡¢ÇâÑõ»¯¼ØÈÜҺΪÔÁÏ£¬Ê¯Ä«Îªµç¼«¹¹³ÉȼÁϵç³Ø¡£¸Ãµç³ØµÄ¸º¼«Îª¶þ¼×ÃÑÔÚ¼îÐÔÌõ¼þÏÂʧµç×Ó£¬×ª»¯ÎªCO32-µÈ¡£Óöþ¼×ÃÑȼÁϵç³ØΪµçÔ´£¬Óò¬µç¼«µç½âKClºÍCuSO4µÄ»ìºÏÈÜÒº£¬Ñô¼«ÏÈ·¢ÉúCl-ʧµç×ÓÉú³ÉCl2µÄ·´Ó¦£¬ºó·¢ÉúH2Oʧµç×ÓÉú³ÉO2µÄ·´Ó¦£»Òõ¼«ÏÈ·¢ÉúCu2+µÃµç×ӵķ´Ó¦£¬ºó·¢ÉúH2OµÃµç×ÓÉú³ÉH2µÄ·´Ó¦£¬´úÈëÊý¾Ý½¨Á¢·½³Ìʽ£¬¼´¿ÉÇó³ö½á¹û¡£
(1)´ÓͼÖпÉÌáÈ¡ÒÔÏÂÐÅÏ¢£¬·´Ó¦½øÐеÄÇ°5minÄÚ£¬CH4µÄŨ¶È±ä»¯Á¿Îª0.50mol/L£»10minºó£¬¸Ä±äijÌõ¼þ£¬2minÄÚCH4µÄŨ¶È±ä»¯Á¿Îª0.25mol/L¡£Ôò·´Ó¦½øÐеÄÇ°5minÄÚ£¬v(CH4)==0.10mol/(L¡¤min)£¬ÒÀ¾Ý»¯Ñ§·½³ÌʽCH4(g)+H2O(g)CO(g)+3H2(g)£¬¿ÉµÃ³öv(H2)=3v(CH4)= 0.30mol/(L¡¤min)£»Óë10minÇ°½øÐжԱȣ¬·´Ó¦ËÙÂʼӿìÇÒƽºâÕýÏòÒƶ¯£¬ËùÒԸıäµÄÍâ½çÌõ¼þ¿ÉÄÜÊÇÉý¸ßζȻò³åÈëË®ÕôÆø¡£´ð°¸Îª£º0.30mol/(L¡¤min)£»Éý¸ßζȻò³åÈëË®ÕôÆø£»
(2)ÎÒÃÇÒÔ¼×Ϊ²ÎÕÕ£¬Ëæ×Å·´Ó¦µÄ½øÐУ¬ÆøÌå·Ö×ÓÊýÔö´ó£¬Ñ¹Ç¿Ôö´ó£¬¶ÔÓÚºãѹÈÝÆ÷ÒÒ£¬Ï൱ÓÚ¼×Ôö´óÈÝ»ý¼õСѹǿ£¬Ôò·´Ó¦ËÙÂʼõÂý£¬µ«Æ½ºâÕýÏòÒƶ¯£¬CH4µÄת»¯ÂÊÔö´ó£¬ÓÉ´Ë¿ÉÔÚͼ3Öл³öÒÒÈÝÆ÷ÖÐCH4µÄת»¯ÂÊËæʱ¼ä±ä»¯µÄͼÏóΪ¡£´ð°¸Îª£º£»
(3)ÓɱíÖÐÊý¾Ý¼°·´Ó¦CO(g)+H2O(g)CO2(g)+H2(g)£¬¿ÉÇó³öŨ¶ÈÉÌΪQ==0.94<K£¬ËùÒÔƽºâÕýÏòÒƶ¯£¬v(Õý)£¾v(Äæ)£¬¹ÊÑ¡a¡£´ð°¸Îª£ºa£»
(4) COºÍH2ÔÚÒ»¶¨Ìõ¼þÏÂÖƵÃCH3OCH3£¬»¯Ñ§·½³ÌʽΪ2CO+4H2CH3OCH3+H2O¡£´ð°¸Îª£º2CO+4H2CH3OCH3+H2O£»
(5)ÒÔ¶þ¼×ÃÑ¡¢¿ÕÆø¡¢ÇâÑõ»¯¼ØÈÜҺΪÔÁÏ£¬Ê¯Ä«Îªµç¼«¹¹³ÉȼÁϵç³Ø¡£¸Ãµç³ØµÄ¸º¼«Îª¶þ¼×ÃÑÔÚ¼îÐÔÌõ¼þÏÂʧµç×Ó£¬×ª»¯ÎªCO32-µÈ£¬µç¼«·´Ó¦·½³ÌʽÊÇ£ºCH3OCH3+16OH--12e-=2CO32-+11H2O¡£
Ñô¼«·¢ÉúµÄ·´Ó¦Îª£º2Cl--2e-span>==Cl2¡ü£¬2H2O-4e-==O2¡ü+4H+¡£n(ÆøÌå)==0.125mol£¬Éèn(Cl2)=x£¬Ôòn(O2)=0.125-x£¬´Ó¶øµÃ³ö2x+4¡Á(0.125-x)=0.3mol£¬´Ó¶øÇó³öx=0.10mol£¬0.125-x=0.025mol£¬¼´O2£º0.025mol¡¢Cl2£º0.10mol¡£´ð°¸Îª£ºO2£º0.025mol¡¢Cl2£º0.10mol¡£
¡¾ÌâÄ¿¡¿Ïò1Lº¬0.01molNaAlO2ºÍµÄÈÜÒºÖлºÂýͨÈë¶þÑõ»¯Ì¼£¬ËæÔö´ó£¬ÏȺó·¢ÉúÈý¸ö²»Í¬µÄ·´Ó¦£¬µ±0.01mol<n(CO2)¡Ü0.015molʱ·¢ÉúµÄ·´Ó¦ÊÇ:2NaAlO2+CO2+3H2O=2Al(OH)3¡ý+Na2CO3£¬ÏÂÁжÔÓ¦¹ØϵÕýÈ·µÄÊÇ£¨ £©
Ñ¡Ïî | ÈÜÒºÖÐÀë×ÓµÄÎïÖʵÄÁ¿Å¨¶È | |
A | 0 | |
B | 0.01 | |
C | 0.015 | |
D | 0.03 |
A.AB.BC.CD.D
¡¾ÌâÄ¿¡¿ÀûÓÃË®îÜ¿ó(Ö÷Òª³É·ÖΪCo2O3£¬º¬ÉÙÁ¿Fe2O3¡¢A12O3¡¢MnO¡¢MgO¡¢CaOµÈ)ÖÆÈ¡²ÝËáîܵŤÒÕÁ÷³ÌÈçͼ£º
ÒÑÖª£º¢Ù½þ³öÒºº¬ÓеÄÑôÀë×ÓÖ÷ÒªÓÐH+¡¢Co2+¡¢Fe2+¡¢Mn2+¡¢Ca2+¡¢Mg2+¡¢Al3+µÈ£»
¢Ú²¿·ÖÑôÀë×ÓÒÔÇâÑõ»¯ÎïÐÎʽ³ÁµíʱÈÜÒºµÄpH¼ûÏÂ±í£º
³ÁµíÎï | Fe(OH)3 | Al(OH)3 | Co(OH)2 | Fe(OH)2 | Mn(OH)2 |
ÍêÈ«³ÁµíµÄpH | 3.7 | 5.2 | 9.2 | 9.6 | 9.8 |
(1)½þ³ö¹ý³ÌÖмÓÈëNa2SO3µÄÄ¿µÄÊÇ_____¡£
(2)NaClO3ÔÚ·´Ó¦ÖÐÂÈÔªËر»»¹ÔΪ×îµÍ¼Û£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ_____¡£
(3)¼ÓNa2CO3ÄÜʹ½þ³öÒºÖÐijЩ½ðÊôÀë×Óת»¯³ÉÇâÑõ»¯Îï³Áµí¡£ÊÔÓÃÀë×Ó·½³ÌʽºÍ±ØÒªµÄÎÄ×Ö¼òÊöÆäÔÀí£º_____¡£
(4)ÝÍÈ¡¼Á¶Ô½ðÊôÀë×ÓµÄÝÍÈ¡ÂÊÓëpHµÄ¹ØϵÈçͼËùʾ£¬ÔÚÂËÒºIIÖÐÊÊÒËÝÍÈ¡µÄpHΪ_____×óÓÒ¡£
a.2.0~2.5 b.3.0~3.5 c.4.0~4.5
(5)ÂËÒºI¡°³ý¸Æ¡¢Ã¾¡±Êǽ«Æäת»¯ÎªMgF2¡¢CaF2³Áµí¡£ÒÑÖªKsp(MgF2)=7.35¡Á10-11¡¢Ksp(CaF2)=1.05¡Á10-10£¬µ±¼ÓÈë¹ýÁ¿NaFºó£¬ËùµÃÂËÒº=____¡£
(6)¹¤ÒµÉÏÓð±Ë®ÎüÊÕ·ÏÆøÖеÄSO2¡£ÒÑÖªNH3¡¤H2OµÄµçÀëƽºâ³£ÊýK=1.8¡Á10-5£¬H2SO3µÄµçÀëƽºâ³£ÊýK1=1.2¡Á10-2£¬K2=1.3¡Á10-8¡£ÔÚͨ·ÏÆøµÄ¹ý³ÌÖУ¬µ±Ç¡ºÃÐγÉÕýÑÎʱ£¬ÈÜÒºÖÐÀë×ÓŨ¶ÈµÄ´óС¹ØϵΪ____¡£
¡¾ÌâÄ¿¡¿½«Ò»¶¨Á¿´¿¾»µÄ°±»ù¼×Ëá粒ÌÌåÖÃÓÚºãÈݵÄÃܱÕÕæ¿ÕÈÝÆ÷ÖУ¬Ôں㶨ζÈÏÂʹÆä´ïµ½·Ö½âƽºâ£ºNH2COONH4£¨s£© 2NH3£¨g£©£«CO2£¨g£©£¬ÊµÑé²âµÃ²»Í¬Î¶ÈϵÄƽºâÊý¾ÝÁÐÓÚÏÂ±í£º
ζÈ/¡æ | 15 | 20 | 25 | 30 | 35 |
ƽºâ×Üѹǿ/kPa | 5.7 | 8.3 | 12.0 | 17.1 | 24.0 |
ƽºâÆøÌå×ÜŨ¶È/ 10-3mol¡¤L£1 | 2.4 | 3.4 | 4.8 | 6.8 | 9.4 |
ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
A. ¸Ã·´Ó¦ÔÚµÍÎÂÏ¿ÉÒÔ×Ô·¢½øÐÐ
B. µ±ÌåϵÖÐÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿²»±äʱ£¬ËµÃ÷¸Ã·´Ó¦´ïµ½ÁËƽºâ״̬
C. ºãÎÂÌõ¼þÏ£¬ÏòÈÝÆ÷ÖÐÔÙ³äÈë2mol NH3ºÍ1molCO2£¬Æ½ºâÏò×óÒƶ¯£¬Æ½ºâºó£¬NH3µÄŨ¶È¼õС
D. 15¡æʱ£¬¸Ã·´Ó¦µÄƽºâ³£ÊýԼΪ2.05¡Á10-9
¡¾ÌâÄ¿¡¿°±ÊÇ»¯Ñ§ÊµÑéÊÒ¼°»¯¹¤Éú²úÖеÄÖØÒªÎïÖÊ£¬Ó¦Óù㷺¡£
£¨1£©ÒÑÖª25¡æʱ£ºN2(g)£«O2(g)2NO(g) ¦¤H=+183 kJ/mol
2H2(g)£«O2(g)£½2H2O(l) ¦¤H=£571.6 kJ/mol
4NH3(g)£«5O2(g)£½4NO(g)£«6H2O(l) ¦¤H=£1164.4 kJ/mol
ÔòN2(g)£«3H2(g)2NH3(g) ¦¤H=______kJ/mol
£¨2£©ÔÚºãκãÈÝÃܱÕÈÝÆ÷ÖнøÐкϳɰ±·´Ó¦£¬ÆðʼͶÁÏʱ¸÷ÎïÖÊŨ¶ÈÈçÏÂ±í£º
N2 | H2 | NH3 | |
ͶÁÏ¢ñ | 1.0 mol/L | 3.0 mol /L | 0 |
ͶÁÏ¢ò | 0.5 mol/L | 1.5 mol/L | 1.0 mol/L |
¢Ù°´Í¶ÁÏ¢ñ½øÐз´Ó¦£¬²âµÃ´ïµ½»¯Ñ§Æ½ºâ״̬ʱH2µÄת»¯ÂÊΪ40%£¬Ôò¸ÃζÈϺϳɰ±·´Ó¦µÄƽºâ³£Êý±í´ïʽΪ_______¡£
¢Ú°´Í¶ÁÏ¢ò½øÐз´Ó¦£¬Æðʼʱ·´Ó¦½øÐеķ½ÏòΪ________£¨Ìî¡°ÕýÏò¡±»ò¡°ÄæÏò¡±£©¡£
¢ÛÈôÉý¸ßζȣ¬ÔòºÏ³É°±·´Ó¦µÄ»¯Ñ§Æ½ºâ³£Êý________£¨Ìî¡°±ä´ó¡±¡¢¡°±äС¡±»ò¡°²»±ä¡±£©¡£
¢ÜL£¨L1¡¢L2£©¡¢X¿É·Ö±ð´ú±íѹǿ»òζȡ£ÏÂͼ±íʾLÒ»¶¨Ê±£¬ºÏ³É°±·´Ó¦ÖÐH2(g)µÄƽºâת»¯ÂÊËæXµÄ±ä»¯¹Øϵ¡£
¢¡ X´ú±íµÄÎïÀíÁ¿ÊÇ______¡£
¢¢ ÅжÏL1¡¢L2µÄ´óС¹Øϵ£¬²¢¼òÊöÀíÓÉ______¡£
£¨3£©µç»¯Ñ§ÆøÃô´«¸ÐÆ÷¿ÉÓÃÓÚ¼à²â»·¾³ÖÐNH3µÄº¬Á¿£¬Æ乤×÷ÔÀíʾÒâͼÈçÏ£º
¢Ùµç¼«bÉÏ·¢ÉúµÄÊÇ______·´Ó¦£¨Ìî¡°Ñõ»¯¡±»ò¡°»¹Ô¡±£©¡£
¢Úд³öµç¼«aµÄµç¼«·´Ó¦Ê½_________¡£