25¡æʱ£¬ÔÚ20 mL 0.1 mol?L£­1 NaOHÈÜÒºÖÐÖðµÎ¼ÓÈë0.2 mol?L£­1 ´×ËáÈÜÒº£¬ÇúÏßÈçͼËùʾ£¬ÓйØÁ£×ÓŨ¶È¹ØϵµÄ±È½ÏÖв»ÕýÈ·µÄÊÇ£¨¡¡ £©

A£®ÔÚAµã£ºc(Na£«)£¾c(OH£­)£¾c(CH3COO£­)£¾c(H£«)

B£®ÔÚBµã£ºc(OH£­)£½c(H£«)£¬c(Na£«)£½c(CH3COO£­)

C£®ÔÚCµã£ºc(CH3COO£­)£¾c(Na£«)£¾c(H£«)£¾c(OH£­)

D£®ÔÚCµã£ºc(CH3COO£­)£«c(CH3COOH)£½2c(Na£«)

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

ÁªºÏ¹úÆøºò±ä»¯´ó»á2009Äê12ÔÂ7ÈÕÔÚµ¤ÂóÊ׶¼¸ç±¾¹þ¸ùÀ­¿ªá¡Ä»£¬½µµÍ´óÆøÖÐCO2µÄº¬Á¿¼°ÓÐЧµØ¿ª·¢ÀûÓÃCO2£¬ÒýÆðÁ˸÷¹úµÄÆÕ±éÖØÊÓ£®
£¨1£©¹¤ÒµÉÏÓÐÒ»ÖÖ·½·¨ÊÇÓÃCO2À´Éú²úȼÁϼ״¼£®298.15Kʱ£¬CO2¡¢H2¡¢ÓëCH3OH¡¢H2OµÄƽ¾ùÄÜÁ¿ÓëºÏ³É¼×´¼·´Ó¦µÄ»î»¯ÄܵÄÇúÏßͼÈçͼËùʾ£¬¾Ýͼ»Ø´ðÎÊÌ⣺

¢Ùд³öºÏ³É¼×´¼·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ
CO2£¨g£©+3H2£¨g£©CH3OH£¨g£©+H2O£¨g£©¡÷H=-£¨n-m£©kJ?mol-1
CO2£¨g£©+3H2£¨g£©CH3OH£¨g£©+H2O£¨g£©¡÷H=-£¨n-m£©kJ?mol-1
£»
¢ÚÔÚͼÖÐÇúÏß
b
b
£¨Ìî¡°a¡±»ò¡°b¡±£©±íʾ¼ÓÈë´ß»¯¼ÁµÄÄÜÁ¿±ä»¯ÇúÏߣ¬´ß»¯¼ÁÄܼӿ췴ӦËÙÂʵÄÔ­ÀíÊÇ
´ß»¯¼ÁÄܽµµÍ¸Ã·´Ó¦µÄ»î»¯ÄÜ£¬Ìá¸ß»î»¯·Ö×ӵİٷÖÊý£¬»¯Ñ§·´Ó¦ËÙÂʼӿì
´ß»¯¼ÁÄܽµµÍ¸Ã·´Ó¦µÄ»î»¯ÄÜ£¬Ìá¸ß»î»¯·Ö×ӵİٷÖÊý£¬»¯Ñ§·´Ó¦ËÙÂʼӿì
£»
¢ÛÔÚÌå»ýΪl LµÄÃܱÕÈÝÆ÷ÖУ¬³äÈëlmol CO2ºÍ3mol H2£¬²âµÃCO2ºÍCH3OH£¨g£©µÄŨ¶ÈËæʱ¼ä±ä»¯ÈçͼËùʾ£®

´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâ£¬¼×´¼µÄƽ¾ù·´Ó¦ËÙÂÊv£¨CH3OH£©=
0.075mol?L-1?min-1
0.075mol?L-1?min-1
£»
¸Ã·´Ó¦µÄƽºâ³£Êýk=
16
3
16
3
£®
£¨2£©ÒÒ´¼ÊÇÖØÒªµÄ»¯¹¤²úÆ·ºÍÒºÌåȼÁÏ£¬Í¬Ñù¿ÉÒÔÀûÓÃCO2·´Ó¦ÖÆÈ¡ÒÒ´¼£º
2CO2£¨g£©+6H2£¨g£© CH3CH2OH£¨g£©+3H2O£¨g£©   25¡æʱ£¬K=2.95¡Á1011
ÔÚÒ»¶¨Ñ¹Ç¿Ï£¬²âµÃ·´Ó¦µÄʵÑéÊý¾ÝÈçÏÂ±í£®·ÖÎö±íÖÐÊý¾Ý»Ø´ðÏÂÁÐÎÊÌ⣺

500 600 700 800
1.5 45 33 20 12
2.0 60 43 28 15
3.0 83 62 37 22
¢ÙζÈÉý¸ß£¬KÖµ
¼õС
¼õС
£¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±¡¢»ò¡°²»±ä¡±£©£®
¢ÚÌá¸ßÇâ̼±È[n£¨H2£©/n£¨CO2£©]£¬KÖµ
²»±ä
²»±ä
£¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±¡¢»ò¡°²»±ä¡±£©£¬¶ÔÉú³ÉÒÒ´¼
ÓÐÀû
ÓÐÀû
£¨Ìî¡°ÓÐÀû¡±»ò¡°²»Àû¡±£©£®
¢ÛÔÚÓÒͼµÄ×ø±êϵÖÐ×÷ͼ˵Ã÷ѹǿ±ä»¯¶Ô·´Ó¦¢ÙµÄ»¯Ñ§Æ½ºâµÄÓ°Ï죮²¢¶ÔͼÖкá×ø±ê¡¢×Ý×ø±êµÄº¬Òå×÷±ØÒªµÄ±ê×¢£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

°ÑN2ºÍH2ÒÔ1£º1µÄÎïÖʵÄÁ¿±È»ìºÏºó·Ö³ÉËĵȷݣ¬·Ö±ðͬʱ³äÈëA¡¢B¡¢C¡¢DËĸö×°Óд߻¯¼ÁµÄÕæ¿ÕÈÝÆ÷ÖУ¨ËĸöÈÝÆ÷µÄÈÝ»ý¹Ì¶¨£¬ÇÒ²»µÈ£©£¬ÔÚ±£³ÖÏàͬζȵÄÌõ¼þÏ£¬ËÄÈÝÆ÷Öеĺϳɰ±·´Ó¦Ïà¼Ì´ïµ½Æ½ºâ״̬£®·ÖÎö±íÖеÄʵÑéÊý¾Ýºó»Ø´ðÓйØÎÊÌ⣨£¨2£©£¨3£©Ð¡ÌâÓÃA¡¢B¡¢C¡¢DÌî¿Õ£©£®
ÈÝÆ÷´úÂë A B C D
ƽºâʱ
.
M
£¨»ì£©
16 17
ƽºâʱN2ת»¯ÂÊ 20% ¢Ù ¢Ú ¢Û
ƽºâʱH2ת»¯ÂÊ 30%
£¨1£©¢Ù¢Ú¢Û·Ö±ðΪ
10%
10%
¡¢
6.25%
6.25%
¡¢
11.8%
11.8%
£®
£¨2£©¶¼´ïµ½Æ½ºâʱ£¬
A
A
ÈÝÆ÷µÄNH3µÄÎïÖʵÄÁ¿ËùÕ¼µÄ±ÈÀý×î´ó£®
£¨3£©ËĸöÈÝÆ÷µÄÈÝ»ýÓÉСµ½´óµÄ˳ÐòÊÇ
A£¼D£¼B£¼C
A£¼D£¼B£¼C
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£ºËÄ´¨Ê¡ÅÊÖ¦»¨ÊÐ2012£­2013ѧÄê¸ßÒ»ÏÂѧÆÚÆÚÄ©µ÷Ñмà²â»¯Ñ§ÊÔÌâ ÌâÐÍ£º022

ÏÂÁл¯ºÏÎ

¢ÙHCl

¢ÚNaOH

¢ÛCH3COOH

¢ÜNH3¡¤H2O

¢ÝCH3COONa

¢ÞNH4Cl

(1)ÊôÓÚÈõµç½âÖʵÄÊÇ________(ÌîÐòºÅ)£¬ÈÜÒº³ÊËáÐÔµÄÓÐ________(ÌîÐòºÅ)£®

(2)25¡æʱ£¬0.10 mol/LCH3COONaÈÜÒºpH£½11£¬¸ÃÈÜÒºÖÐÀë×ÓŨ¶È°´ÓÉ´óµ½Ð¡µÄ˳ÐòΪ________£®

(3)½«µÈpHµÈÌå»ýµÄHClºÍCH3COOH·Ö±ðÏ¡ÊÍm±¶ºÍn±¶£¬ÈôÏ¡ÊͺóÁ½ÈÜÒºµÄPHÈÔÏàµÈ£¬Ôòm________n(Ìî¡°´óÓÚ¡¢µÈÓÚ¡¢Ð¡ÓÚ¡±)£®

(4)25¡æʱ£¬ÔÚ20 mL0.1 mol/L´×ËáÈÜÒº¼ÓÈëVmL0.1 mol/LNaOHÈÜÒº£¬²âµÃ»ìºÏÈÜÒºµÄpH±ä»¯ÇúÏßÈçͼËùʾ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ________£®

A£®pH£½3µÄ´×ËáÈÜÒººÍpH£½11µÄ´×ËáÄÆÈÜÒºÖУ¬Ë®µÄµçÀë³Ì¶ÈÏàͬ

B£®¢ÙµãʱpH£½6£¬´ËʱÈÜÒºÖУ¬c(CH3COO£­)£­c(Na+)£½9.9¡Á10£­7 mol/L

C£®¢Úµãʱ£¬ÈÜÒºÖеÄc(CH3COO£­)£½c(Na+)

D£®¢ÛµãʱV£½20 mL£¬´ËʱÈÜÒºÖÐc(CH3COO£­)£¾c(Na+)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

£¨Ò»£©£¨6·Ö£© ½«2£®5g̼ËáÄÆ¡¢Ì¼ËáÇâÄƺÍÇâÑõ»¯ÄƹÌÌå»ìºÏÎïÍêÈ«ÈܽâÓÚË®£¬ÖƳÉÏ¡ÈÜÒº£¬È»ºóÏò¸ÃÈÜÒºÖÐÖðµÎ¼ÓÈë1mol/LµÄÑÎËᣬËù¼ÓÈëÑÎËáµÄÌå»ýÓë²úÉúCO2µÄÌå»ý£¨±ê×¼×´¿ö£©¹ØϵÈçÏÂͼËùʾ£º

£¨1£©Ð´³öOA¶ÎËù·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ                         

£¨2£©µ±¼ÓÈë35mLÑÎËáʱ£¬²úÉú¶þÑõ»¯Ì¼µÄÌå»ýΪ       mL£¨±ê×¼×´¿ö£©

£¨3£©Ô­»ìºÏÎïÖÐNa2CO3µÄÖÊÁ¿·ÖÊýΪ­­      ¡£

£¨¶þ£©£®°±ÊÇÖØÒªµÄµª·Ê£¬ÊDzúÁ¿×î´óµÄ»¯¹¤²úÆ·Ö®Ò»¡£¿Î±¾Àï½éÉܵĺϳɰ±¼¼Êõ½Ð¹þ²®·¨£¬Êǵ¹ú»¯Ñ§¼Ò¹þ²®ÔÚ1905Äê·¢Ã÷µÄ£¬ÆäºÏ³ÉÔ­ÀíΪ£ºN2(g) + 3H2(g)2NH3(g)£»

¡÷H£½¨D92.4 kJ/mol£¬ËûÒò´Ë»ñµÃÁË1918Äê¶Èŵ±´¶û»¯Ñ§½±¡£ÊԻشðÏÂÁÐÎÊÌ⣺

¢Å ÏÂÁз½·¨²»ÊʺÏʵÑéÊÒÖÆÈ¡°±ÆøµÄÊÇ             £¨ÌîÐòºÅ£©¡£

A£®ÏòÉúʯ»ÒÖеÎÈëŨ°±Ë®        B£®¼ÓÈÈŨ°±Ë®

C£®Ö±½ÓÓÃÇâÆøºÍµªÆøºÏ³É        D£®Ïò±¥ºÍÂÈ»¯ï§ÈÜÒºÖеÎÈëŨÇâÑõ»¯ÄÆÈÜÒº

¢Æ ºÏ³É°±¹¤ÒµÖвÉÈ¡µÄÏÂÁдëÊ©¿ÉÓÃÀÕÏÄÌØÁÐÔ­Àí½âÊ͵ÄÊÇ       £¨ÌîÐòºÅ£©¡£

A£®²ÉÓýϸßѹǿ£¨20 M Pa¡«50 M Pa£©

B£®²ÉÓÃ500¡æµÄ¸ßÎÂ

C£®ÓÃÌú´¥Ã½×÷´ß»¯¼Á

D£®½«Éú³ÉµÄ°±Òº»¯²¢¼°Ê±´ÓÌåϵÖзÖÀë³öÀ´£¬Î´·´Ó¦µÄN2ºÍH2Ñ­»·µ½ºÏ³ÉËþÖÐ

£¨3£© ÓÃÊý×Ö»¯ÐÅϢϵͳDIS£¨ÈçÏÂͼ¢ñËùʾ£ºËüÓÉ´«¸ÐÆ÷¡¢Êý¾Ý²É¼¯Æ÷ºÍ¼ÆËã»ú×é³É£©¿ÉÒԲⶨÉÏÊö°±Ë®µÄŨ¶È¡£ÓÃËáʽµÎ¶¨¹Ü׼ȷÁ¿È¡0.5000 mol/L´×ËáÈÜÒº25.00 mLÓÚÉÕ±­ÖУ¬ÒÔ¸ÃÖÖ°±Ë®½øÐе樣¬¼ÆËã»úÆÁÄ»ÉÏÏÔʾ³öÈÜÒºµ¼µçÄÜÁ¦Ë氱ˮÌå»ý±ä»¯µÄÇúÏßÈçÏÂͼ¢òËùʾ¡£

        

ͼ¢ñ                                Í¼¢ò

¢Ù Óõζ¨¹ÜÊ¢°±Ë®Ç°£¬µÎ¶¨¹ÜÒªÓà                          ÈóÏ´2¡«3±é£¬

¢Ú ÊÔ¼ÆËã¸ÃÖÖ°±Ë®µÄŨ¶È£º                      ¡£

¢Û ÏÂÁÐÇé¿öÏ£¬»áµ¼ÖÂʵÑé½á¹ûc(NH3¡¤H2O)Æ«µÍµÄÊÇ               ¡£

A£®µÎ¶¨½áÊøʱÑöÊÓ¶ÁÊý

B£®Á¿È¡25.00 mL´×ËáÈÜҺʱ£¬Î´ÓÃËùÊ¢ÈÜÒºÈóÏ´µÎ¶¨¹Ü

C£®µÎ¶¨Ê±£¬Òò²»É÷½«°±Ë®µÎÔÚÉÕ±­Íâ

£¨4£© 1998ÄêÏ£À°ÑÇÀïÊ¿¶àµÂ´óѧµÄMarnellosºÍStoukides²ÉÓøßÖÊ×Óµ¼µçÐÔµÄSCYÌÕ´É£¨ÄÜ´«µÝH+£©£¬ÊµÏÖÁ˸ßγ£Ñ¹Ï¸ßת»¯Âʵĵ绯ѧºÏ³É°±¡£ÆäʵÑé×°ÖÃÈçÏÂͼ¡£

Õý¼«µÄµç¼«·´Ó¦Ê½Îª£º                                         ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012½ìÕã½­Ê¡¸ß¶þÏÂѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÑ¡ÔñÌâ

ÓÒÏÂͼװÖÿÉÓÃÓÚ¶àÏÁ¿ÊµÑ顣ͼÖмг̶ֹ¨×°ÖÃÒÑÂÔÈ¥£¬¼×Óп̶ȣ¬¹©Á¿ÆøÓá£

( 1 )×°ÖÃÖÐÓп̶ȵļ׹ܿÉÒÔÓÃ______________´úÌæ (ÌîÒÇÆ÷Ãû³Æ)£¬°´Í¼Á¬½ÓºÃ×°Öú󣬼ì²é×°ÖÃÆøÃÜÐԵķ½·¨ÊÇ______________________________

( 2 )ijʵÑéС×éÓÃþ·Û¡¢ÑÎËá¡¢´×ËáÉè¼ÆʵÑéÀ´Ö¤Ã÷£ºÔÚͬÎÂͬѹÏ£¬µ±ÉÏÊöÁ½ÖÖËáµÄÎïÖʵÄÁ¿Ïàͬʱ£¬Óëþ·Û·´Ó¦Éú³ÉÇâÆøµÄÌå»ýÏàͬ¶ø·´Ó¦ËÙÂʲ»Í¬¡£×°ÖÃÈçÓÒͼËùʾ£¬

ÓйØʵÑéÊý¾Ý¼Ç¼ÓÚÏÂ±í£º

ËáÈÜÒº

ËáÈÜÒº

ÆøÌåÌå»ý£¯mL

·´Ó¦Ê±¼ä

(ʵÑéA)

(ʵÑéB)

(25¡æ¡¢101 kPa)

ʵÑéA

ʵÑéB

CH3COOH

0.1 mol£¯L

40.00mL

HClÈÜÒº

0.1 mol£¯L

40.00mL

5

t(a1)=155 s

t(b1)=7 s

10

t(a2)=310 s

t(b2)=16 s

15

t(a3)=465 s

t(b3)=30 s

20

t(a4)=665 s

t(b4)=54 s

¡­¡­

¡­¡­

¡­¡­

    Çë»Ø´ðÏÂÁÐÎÊÌ⣺

¢Ùÿ´ÎʵÑéÖÁÉÙÐèÒªÓõç×ÓÌìƽ(ÄܳÆ×¼1 mg) ³Æȡþ·Û___________________g£»

¢ÚÀäÈ´µ½25¡æºó£¬ÔÚ¶ÁÈ¡ÆøÌåÌå»ýʱ£¬Ê×ÏÈÓ¦ÈçºÎ²Ù×÷£º__________________________£»

¢Û·ÖÎöʵÑéÊý¾Ý£¬t(a1)Ô¶Ô¶´óÓÚt(b1)µÄÔ­ÒòÊÇ__________________________¡£

£¨3£©ÓÃͼʾװÖÃ,ijͬѧÉè¼ÆÁ˲ⶨ¶ÆпÌúƤ¶Æ²ãºñ¶ÈµÄʵÑé·½°¸£¬½«µ¥²àÃæ»ýΪS cm2¡¢ÖÊÁ¿Îªm gµÄ¶ÆпÌúƤÓë6mol¡¤L¡ª1  NaOHÈÜÒº·´Ó¦¡£»Ø´ðÏÂÁÐÎÊÌ⣺£¨ÒÑ֪пµÄÃܶÈΪ ¦Ñ  g/cm3£©

¢Ù д³öZn¶Æ²ãÓëNaOHÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ________________________________________

¢ÚΪÌá¸ß²â¶¨µÄ׼ȷÐÔ£¬Ð轫׶ÐÎÆ¿Éϵĵ¥¿×Ïð½ºÈû»»ÎªË«¿×Ïð½ºÈû£¬ÁíÒ»¿×²åÈë______£¨ÌîÒÇÆ÷Ãû³Æ)

ʵÑéʱÏÈÏò׶ÐÎÆ¿ÖмÓÈë¶ÆпÌúƤÑùÆ·£¬ÈûÉÏË«¿×Èû£¬ÔÙ¼ÓÈëNaOHÈÜÒº£»

¢Û ÒÑ֪ʵÑéÇ°ºó¼×¹ÜÖÐÒºÃæ¶ÁÊý²îΪV mL£¨ÊµÑéÌõ¼þµÄÆøÌåĦ¶ûÌå»ýΪVm mol¡¤L¡ª1£©¡£Ôò¶ÆпÌúƤµÄп¶Æ²ãºñ¶ÈΪ_________________________cm¡££¨Ð´³öÊýѧ±í´ïʽ£©

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸