ÔÚt¡æʱ½«a g NH3ÍêÈ«ÈÜÓÚË®£¬µÃµ½V mLÈÜÒº£¬¼ÙÈç¸ÃÈÜÒºµÄÃܶÈΪd g/cm3£¬ÖÊÁ¿·ÖÊýΪ¦Ø£¬ÆäÖк¬NH4+µÄÎïÖʵÄÁ¿Îªb mol¡£ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ


  1. A.
    ÈÜÖʵÄÖÊÁ¿·ÖÊýΪ¦Ø£½¡Á100%
  2. B.
    °±Ë®µÄÎïÖʵÄÁ¿Å¨¶ÈΪ1000a/(35V)mol¡¤L-1
  3. C.
    ÈÜÒºÖÐc(OH-)£½1000b/V mol¡¤L-1+c(H+)
  4. D.
    ÉÏÊöÈÜÒºÖÐÔÙ¼ÓÈëV mLË®ºó£¬ËùµÃÈÜÒºµÄÖÊÁ¿·ÖÊý´óÓÚ0.5¦Ø
C
A¡¢°±Ë®ÈÜÒºÈÜÖÊΪ°±Æø£¬¸ÃÈÜÒºµÄÃܶÈΪ¦Ñ g?cm-3£¬Ìå»ýΪVmL£¬ËùÒÔÈÜÒºÖÊÁ¿Îª¦ÑVg£¬ÈÜÖÊ°±ÆøµÄÖÊÁ¿Îªag£¬ÈÜÖʵÄÖÊÁ¿·ÖÊýΪag /(¦ÑVg) ¡Á100%£¬¹ÊA´íÎó£»
B¡¢Ë®µÄÃܶȱȰ±Ë®µÄÃܶȴó£¬ÏàµÈÌå»ýµÄ°±Ë®ÓëË®£¬Ë®µÄÖÊÁ¿´ó£¬µÈÌå»ý»ìºÏºóÈÜÒºµÄÖÊÁ¿´óÓÚÔ­°±Ë®µÄ2±¶£¬ÈÜÒºÖа±ÆøµÄÖÊÁ¿Ïàͬ£¬µÈÌå»ý»ìºÏËùµÃÈÜÒºÈÜÖʵÄÖÊÁ¿·ÖÊýСÓÚ0.5w£¬
¹ÊB´íÎó£»
C¡¢ÈÜÒºOH-ÖÐÀ´Ô´ÓëһˮºÏ°±¡¢Ë®µÄµçÀ룬NH4+µÄŨ¶ÈΪbmol /(V¡Á10-3L) ="1000b" /V  (mol/L)£¬Ò»Ë®ºÏ°±µçÀëNH3?H2O?NH4++OH-£¬Ò»Ë®ºÏ°±µçÀë³öµÄOH-Ϊ1000b /V  (mol/L)£¬ËùÒÔÈÜÒºÖÐOH-Ũ¶È´óÓÚ1000b /V  (mol/L)£¬¹ÊC´íÎó£»
D¡¢a g NH3µÄÎïÖʵÄÁ¿Îªag /(17g/mol) ="a" /17 (mol)£¬ÈÜÒºÌå»ýΪVmL£¬ËùÒÔÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ= mol/L£¬¹ÊDÕýÈ·£®
¹ÊÑ¡£ºD£®
µãÆÀ£º¿¼²éÎïÖʵÄÁ¿Å¨¶È¡¢ÖÊÁ¿·ÖÊýµÄ¼ÆËãÓëÏ໥¹ØϵµÈ£¬ÄѶÈÖеȣ¬Àí½â¸ÅÄîÊǽâÌâµÄ¹Ø¼ü£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

£¨2012?½­ËÕÈýÄ££©ºÏÀíµØ¹Ì¶¨ºÍÀûÓÃCO2ÄÜÓÐЧ¼õÉÙ¿ÕÆøÖеÄÎÂÊÒÆøÌ壬²úÎﻹÄÜÔ츣ÈËÀ࣮
£¨1£©11kmÉµ×µÄÎÞ¼¹×µ¶¯ÎïÒÀ¿¿»¯Ñ§×ÔÑø¾ú£¬ÒÔº£µ×ÈÈȪÅç³öÒºÖеÄH2SºÍCO2ºÏ³É£¨C6H10O5£©nºÍÒ»ÖÖµ­»ÆÉ«¹ÌÌ壬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
12nH2S+6nCO2=C6H10O5£©n+12nS¡ý+7nH2O
12nH2S+6nCO2=C6H10O5£©n+12nS¡ý+7nH2O
£®
£¨2£©CO2¿ÉÓÃÓںϳɶþ¼×ÃÑ£¨CH3OCH3£©£¬Óйط´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ£º
CO2£¨g£©+3H2£¨g£©?CH3OH£¨g£©+H2O£¨g£©¡÷H=-49.0kJ?mol-1
2CH3OH£¨g£©?H3OCH3£¨g£©+H2O£¨g£©¡÷H=-23.5kJ?mol-1
¢Ù·´Ó¦2CO2£¨g£©+6H2£¨g£©?CH3OCH3£¨g£©+3H2O£¨g£©µÄ¡÷H=
-121.5kJ/mol
-121.5kJ/mol
£®
¢ÚÒ»¶¨Ìõ¼þÏÂÓÃCO2ºÍH2ºÏ³É¶þ¼×ÃÑ£¬·´Ó¦ÎïÆøÁ÷Á¿¶ÔCO2µÄת»¯ÂÊ¡¢¶þ¼×ÃÑÑ¡ÔñÐÔµÄÓ°Ïì½á¹ûÈçÏÂͼËùʾ£®¿ØÖÆÆøÁ÷Á¿Îª28mL?min-1£¬Éú²ú0.3mol¶þ¼×ÃÑÐèͨÈëCO2µÄÎïÖʵÄÁ¿Îª
20mol
20mol
£®

¢Û·´Ó¦2CH3OH£¨g£©?CH3OCH3£¨g£©+H2O£¨g£©ÔÚT¡æʱµÄƽºâ³£ÊýΪ400£®¸ÃζÈÏ£¬ÔÚÃܱÕÈÝÆ÷ÖмÓÈëCH3OH£¬tʱ¿Ìʱ£¬²âµÃc£¨CH3OH£©=0.03mol?L -1£¬c£¨CH3OCH3£©=0.6mol?L -1£¬´ËʱvÕý
=
=
vÄ棨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°µÈÓÚ¡±£©
¢Ü¶þ¼×ÃÑȼÁϵç³ØµÄ¹¤×÷Ô­ÀíÈçͼ1Ëùʾ£®¸Ãµç³Ø¹¤×÷ʱ£¬aµç¼«µÄ·´Ó¦Ê½Îª
CH3OCH3+3H2O-12e-=2CO2+12H+
CH3OCH3+3H2O-12e-=2CO2+12H+
£®
£¨3£©Ò»ÖÖ¡°Ì¼²¶×½¡±¼¼ÊõΪ£º½«º¬CO2µÄ¹¤ÒµÎ²Æøͨ¹ýNaOHÈÜÒº£¬ÔÚËùµÃÈÜÒºÖмÓCaO£¬³ä·Ö·´Ó¦ºó¹ýÂË£¬ÂËÔü¸ßηֽâµÃµ½µÄ¸ßŨ¶ÈCO2¿ÉÓÃÓÚÖƱ¸¼×´¼µÈ£®¸Ã¼¼Êõµç¿ÉÒÔÑ­»·Ê¹ÓõÄÎïÖÊΪ
Ñõ»¯¸ÆºÍÇâÑõ»¯ÄÆ
Ñõ»¯¸ÆºÍÇâÑõ»¯ÄÆ
£®
£¨4£©ÀûÓùâÄܺ͹â´ß»¯¼Á£¬¿É½«CO2ºÍH2O£¨g£©×ª»¯ÎªCH4ºÍO2£®×ÏÍâ¹âÕÕÉäʱ£¬µÈÁ¿µÄCO2ºÍH2O£¨g£©ÔÚ²»Í¬´ß»¯¼Á£¨¢ñ¡¢¢ò¡¢¢ó£©×÷ÓÃÏ£¬CH4²úÁ¿Ëæ¹âÕÕʱ¼äµÄ±ä»¯Èçͼ2Ëùʾ£®ÔÚ0¡«30СʱÄÚ£¬CH4µÄ°ë¾ùÉú³ÉËÙÂÊv£¨¢ñ£©¡¢v£¨¢ò£©¡¢ºÍv£¨¢ó£©¡¢´Ó´óµ½Ð¡µÄ˳ÐòΪ
v£¨¢ó£©£¾v£¨¢ò£©£¾v£¨¢ñ£©
v£¨¢ó£©£¾v£¨¢ò£©£¾v£¨¢ñ£©
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2009?Õ¿½­¶þÄ££©ÔÚij¸öÈÝ»ýΪ2.0LµÄÃܱÕÈÝÆ÷ÖУ¬ÔÚT¡æʱ°´ÏÂͼ1Ëùʾ·¢Éú·´Ó¦£º
mA£¨g£©+nB£¨g£©pD£¨g£©+qE£¨s£©£¬¡÷H£¼0£¨m¡¢n¡¢p¡¢qΪ×î¼òÕûÊý±È£©£®

£¨1£©Í¼1Ëùʾ£¬·´Ó¦¿ªÊ¼ÖÁ´ïµ½Æ½ºâʱ£¬ÓÃD±íʾµÄƽ¾ù·´Ó¦ËÙÂÊΪ
0.1mol/£¨L?min£©
0.1mol/£¨L?min£©
£®
£¨2£©T¡æʱ¸Ã·´Ó¦µÄ»¯Ñ§Æ½ºâ³£ÊýKµÄÊýֵΪ
0.75
0.75
£®
£¨3£©·´Ó¦´ïµ½Æ½ºâºó£¬µÚ6minʱ£º
¢ÙÈôÉý¸ßζȣ¬DµÄÎïÖʵÄÁ¿±ä»¯ÇúÏß×î¿ÉÄܵÄÊÇ
c
c
£¨ÓÃͼ2ÖеÄa¡«cµÄ±àºÅ×÷´ð£©£»
¢ÚÈôÔÚ6minʱÈÔΪԭƽºâ£¬´Ëʱ½«ÈÝÆ÷µÄÈÝ»ýѹËõΪԭÀ´µÄÒ»°ë£®ÇëÔÚͼ3Öл­³ö6minºóBŨ¶ÈµÄ±ä»¯ÇúÏß
£®
4£©ÔÚT¡æʱ£¬ÏàͬÈÝÆ÷ÖУ¬Èô¿ªÊ¼Ê±¼ÓÈë0.4molA¡¢0.8molB¡¢0.9molDºÍ0.5molE0.5mol£¬·´Ó¦´ïµ½Æ½ºâºó£¬AµÄŨ¶È·¶Î§Îª
0.2mol/L£¼c£¨A£©£¼0.3mol/L
0.2mol/L£¼c£¨A£©£¼0.3mol/L
£®
£¨5£©¸ù¾Ý»¯Ñ§·´Ó¦ËÙÂÊÓ뻯ѧƽºâÀíÂÛ£¬ÁªÏµ»¯¹¤Éú²úʵ¼Ê£¬ÄãÈÏΪÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ
BD
BD
£®
A£®»¯Ñ§·´Ó¦ËÙÂÊÀíÂÛ¿ÉÖ¸µ¼ÔõÑùÔÚÒ»¶¨Ê±¼äÄÚ¿ì³ö²úÆ·
B£®ÓÐЧÅöײÀíÂÛ¿ÉÖ¸µ¼ÔõÑùÌá¸ßÔ­ÁϵÄת»¯ÂÊ
C£®ÀÕÏÄÌØÁÐÔ­Àí¿ÉÖ¸µ¼ÔõÑùʹÓÃÓÐÏÞÔ­Á϶à³ö²úÆ·
D£®´ß»¯¼ÁµÄʹÓÃÊÇÌá¸ß²úÆ·²úÂʵÄÓÐЧ·½·¨
E£®ÕýÈ·ÀûÓû¯Ñ§·´Ó¦ËÙÂʺͻ¯Ñ§·´Ó¦Ï޶ȶ¼¿ÉÒÔÌá¸ß»¯¹¤Éú²úµÄ×ۺϾ­¼ÃЧÒ森

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

t¡æʱ£¬½«3mol A ºÍ1mol BÆøÌ壬ͨÈëÌå»ýΪ2LµÄÃܱÕÈÝÆ÷£¨ÈÝ»ý²»±ä£©£¬·¢Éú·´Ó¦£º3A£¨g£©+B£¨g£©?4C£¨g£© 2minʱ·´Ó¦´ïµ½Æ½ºâ״̬£¨Î¶Ȳ»±ä£©£¬²¢²âµÃC µÄŨ¶ÈΪ0.4mol/L£¬ÇëÌîдÏÂÁпհףº
£¨1£©´Ó·´Ó¦¿ªÊ¼µ½´ïµ½Æ½ºâ״̬£¬Éú³ÉCµÄƽ¾ù·´Ó¦ËÙÂÊΪ
0.2mol/£¨L?min£©
0.2mol/£¨L?min£©
£®
£¨2£©´ïµ½Æ½ºâ״̬ʱ£¬B ÎïÖʵÄת»¯ÂʦÁ£¨B£©=
20%
20%
£¬Æ½ºâ³£ÊýK=
1
27
1
27
£¨Ìî¼ÆËã½á¹û£©£®
£¨3£©Èô¼ÌÐøÏòÔ­»ìºÍÎïÖÊͨÈëÉÙÁ¿º¤Æøºó£¨É躤Æø²»ÓëA¡¢B¡¢C·´Ó¦£©£¬»¯Ñ§Æ½ºâ
C
C
£¨Ìî×Öĸ£©
A¡¢ÏòÕý·´Ó¦·½ÏòÒƶ¯   B¡¢ÏòÄæ·´Ó¦·½ÏòÒƶ¯  C¡¢Æ½ºâ²»Òƶ¯£®
£¨4£©ÈôÏòԭƽºâÖÐÔÙ³äÈëa mol C£¬ÔÚt¡æʱ´ïµ½ÐÂƽºâ£¬´ËʱBµÄÎïÖʵÄÁ¿Îª£ºn £¨B£©=
£¨0.8+0.2a£©
£¨0.8+0.2a£©
 mol£®
£¨5£©Èç¹ûÉÏÊö·´Ó¦ÔÚÏàͬζȺÍÈÝÆ÷ÖнøÐУ¬Óûʹ·´Ó¦´ïµ½Æ½ºâʱCµÄÎïÖʵÄÁ¿·ÖÊýÓëԭƽºâÏàµÈ£¬Æðʼ¼ÓÈëµÄÈýÖÖÎïÖʵÄÎïÖʵÄÁ¿n£¨A£©¡¢n£¨B£©¡¢n£¨C£©Ö®¼ä¸ÃÂú×ãµÄ¹ØϵΪ£º
n£¨A£©=3n£¨B£©£¬n£¨C£©£¾0
n£¨A£©=3n£¨B£©£¬n£¨C£©£¾0
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

»·¾³ÎÊÌⱸÊÜÊÀ½ç¹Ø×¢£®»¯¹¤³§ÒÔ¼°Æû³µÎ²ÆøÅŷŵÄÒ»Ñõ»¯Ì¼£¨CO£©¡¢µªÑõ»¯ÎNOx£©¡¢Áò»¯ÎïµÈÆøÌ强³ÉΪ´óÆøÎÛȾµÄÖ÷ÒªÒòËØ£®
£¨1£©COµÄÖж¾ÊÇÓÉÓÚCOÓëѪҺÖÐѪºìµ°°×µÄѪºìËز¿·Ö·´Ó¦Éú³É̼ÑõѪºìµ°°×£¬·´Ó¦·½³Ìʽ¿É±íʾΪ£ºCO+Hb?O2   O2+Hb?CO
ʵÑé±íÃ÷£¬c£¨Hb?CO£©¼´Ê¹Ö»ÓÐc£¨Hb?O2£©µÄ 1/50£¬Ò²¿ÉÔì³ÉÈ˵ÄÖÇÁ¦ËðÉË£®
ÒÑÖªt¡æʱÉÏÊö·´Ó¦µÄƽºâ³£ÊýK=200£¬ÎüÈë·Î²¿O2µÄŨ¶ÈԼΪ1.0¡Á10-2mol?L-1£¬Èôʹc£¨Hb?CO£©Ð¡ÓÚc£¨Hb?O2£©µÄ 1/50£¬ÔòÎüÈë·Î²¿COµÄŨ¶È²»Äܳ¬¹ý
1.0¡Á10-6
1.0¡Á10-6
mol?L-1£®
£¨2£©Æû³µÎ²ÆøÖÎÀíµÄ·½·¨Ö®Ò»ÊÇÔÚÆû³µµÄÅÅÆø¹ÜÉÏ°²×°Ò»¸ö¡°´ß»¯×ª»¯Æ÷¡±£®
ÒÑÖª·´Ó¦2NO£¨g£©+2CO£¨g£©   N2£¨g£©+2CO2£¨g£©¡÷H=-113kJ?mol-1
¢ÙΪÁËÄ£Äâ´ß»¯×ª»¯Æ÷µÄ¹¤×÷Ô­Àí£¬ÔÚt¡æʱ£¬½«2molNOÓë1mol CO³äÈëlL·´Ó¦ÈÝÆ÷ÖУ¬·´Ó¦¹ý³ÌÖÐNO£¨g£©¡¢CO£¨g£©¡¢N2£¨g£©ÎïÖʵÄÁ¿Å¨¶È±ä»¯ÈçÓÒÏÂͼËùʾ£®·´Ó¦½øÐе½15minʱ£¬NOµÄƽ¾ù·´Ó¦ËÙÂÊΪ
0.4
15
mol?L-1?min-1£¨»ò0.027mol?L-1?min-1£©
0.4
15
mol?L-1?min-1£¨»ò0.027mol?L-1?min-1£©
£¬
¢Ú¹Û²ìÔÚ20min¡«30minÄÚN2£¨g£©·¢Éú±ä»¯µÄÇúÏßͼ£¬ÅжÏÒýÆð¸Ã±ä»¯µÄÌõ¼þ¿ÉÄÜÊÇ
b
b
£»
a£®¼ÓÈë´ß»¯¼Á
b£®½µµÍÌåϵζÈ
c£®ËõСÈÝÆ÷Ìå»ýΪ0.5L
d£®Ôö¼ÓCO2µÄÎïÖʵÄÁ¿
£¨3£©Ãº»¯¹¤Öг£ÐèÑо¿²»Í¬Î¶ÈÏÂƽºâ³£Êý¡¢Í¶Áϱȼ°²úÂʵÈÎÊÌ⣮
ÒÑÖª£ºCO£¨g£©+H2O£¨g£© H2£¨g£©+CO2£¨g£©µÄƽºâ³£ÊýËæζȵı仯ÈçÏÂ±í£º
ζÈ/¡æ 400 500 830 1000
ƽºâ³£ÊýK 10 9 1 0.6
ÊԻشðÏÂÁÐÎÊÌâ
¢ÙÉÏÊöÕýÏò·´Ó¦ÊÇ£º
·ÅÈÈ
·ÅÈÈ
·´Ó¦£¨Ìî¡°·ÅÈÈ¡±¡¢¡°ÎüÈÈ¡±£©£®
¢ÚijζÈÏÂÉÏÊö·´Ó¦Æ½ºâʱ£¬ºãÈÝ¡¢Éý¸ßζȣ¬Õý·´Ó¦ËÙÂÊ
Ôö´ó
Ôö´ó
£¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±£©£¬ÈÝÆ÷ÄÚ»ìºÏÆøÌåµÄѹǿ
Ôö´ó
Ôö´ó
£¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±£©£®
¢ÛÔÚ830¡æ·¢ÉúÉÏÊö·´Ó¦£¬ÒÔϱíÖеÄÎïÖʵÄÁ¿Í¶ÈëºãÈÝ·´Ó¦Æ÷£¬ÆäÖÐÏòÕý·´Ó¦·½ÏòÒƶ¯µÄÓÐ
B¡¢C
B¡¢C
£¨Ñ¡ÌîA¡¢B¡¢C¡¢D£©£®
A B C D
n£¨CO2£© 3 1 0 1
n£¨H2£© 2 1 0 1
n£¨CO£© 1 2 3 0.5
n£¨H2O£© 5 2 3 2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÔÚt¡æʱ£¬½«a gNH3ÍêÈ«ÈܽâÓÚË®£¬µÃV mL±¥ºÍÈÜÒº£¬²âµÃ¸ÃÈÜÒºµÄÃܶÈΪ¦Ñg/cm3£¬ÖÊÁ¿·ÖÊýΪ¦Ø£¬ÆäÖÐn£¨NH4+£© Îªb mol£®ÔòÏÂÁÐÐðÊöÖдíÎóµÄÊÇ£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸