£¨1£©ÒÑÖª2H2£¨g£©+O2£¨g£©=2H2O£¨l£©¦¤H£½£­571.6 kJ/mol£¬
CO£¨g£©£«1/2O2£¨g£©£½CO2£¨g£©¦¤H£½£­283.0 kJ/mol¡£Ä³H2ºÍCOµÄ»ìºÏÆøÌåÍêȫȼÉÕʱ·Å³ö113.74 kJÈÈÁ¿£¬Í¬Ê±Éú³É3.6 gҺ̬ˮ£¬ÔòÔ­»ìºÏÆøÌåÖÐH2ºÍCOµÄÎïÖʵÄÁ¿Ö®±ÈΪ___________¡£
£¨2£©ÒÔ¼×´¼¡¢¿ÕÆø£¬ÇâÑõ»¯¼ØÈÜҺΪԭÁÏ£¬Ê¯Ä«Îªµç¼«£¬¿É¹¹³ÉȼÁϵç³Ø£»ÒÑÖª¸ÃȼÁϵç³ØµÄ×Ü·´Ó¦Ê½ÊÇ£º2CH3OH +3O2+4OH-=2CO32-+6H2O£¬¸ÃȼÁϵç³Ø·¢Éú·´Ó¦Ê±£¬Õý¼«ÇøÈÜÒºµÄPH__________ (Ìî¡°Ôö´ó¡±£¬ ¡°¼õС¡± »ò¡°²»±ä¡±)¸Ãµç³ØµÄ¸º¼«·´Ó¦Ê½Îª_________________¡£
£¨3£© ÓÃÉÏÊöȼÁϵç³Ø½øÐдÖÍ­µÄ¾«Á¶£¬´ÖÍ­Ó¦Á¬½ÓµçÔ´µÄ________¼«£¬¸Ã´ÖÍ­¾«Á¶µç½â³ØµÄÒõ¼«·´Ó¦Ê½Îª_________________¡£
£¨1£©1£º1......2·Ö
£¨2£©Ôö´ó¡­.1·Ö  CH3OH -6e£­£«8OH-£¨1£©£½CO32-£«6H2O ....2·Ö
£¨3£©Õý¡­..1·Ö¡£  Cu2+ + 2e- =Cu¡­.2·Ö

ÊÔÌâ·ÖÎö£º½â£ºË®µÄÎïÖʵÄÁ¿Îª=0.2mol£¬ÓÉ2H2+O2¨T2H2O¿ÉÖª£¬n£¨H2£©=n£¨H2O£©=0.2mol£¬ÓÉ2H2£¨g£©+O2£¨g£©¨T2H2O£¨l£©¡÷H=-571.6kJ?mol-1¿ÉÖª£¬0.2molH2ȼÉշųöµÄÈÈÁ¿Îª57.16KJ£¬ÔòCOȼÉշųöµÄÈÈÁ¿Îª113.74KJ-57.16KJ=56.58KJ£¬Éè»ìºÏÆøÌåÖÐCOµÄÎïÖʵÄÁ¿Îªx£¬Ôò
CO£¨g£©+O2£¨g£©=CO2£¨g£©¡÷H=-283kJ?mol-1
1                         283KJ
X                         56.58KJ

½âµÃx=0.2mol£¬¼´n£¨CO£©=0.20mol£¬ÔòÔ­»ìºÏÆøÌåÖÐH2ºÍCOµÄÎïÖʵÄÁ¿Ö®±ÈΪ1:1.£¨2£©ÒÔ¼×´¼¡¢¿ÕÆø¡¢ÇâÑõ»¯¼ØÈÜҺΪԭÁÏ£¬Ê¯Ä«Îªµç¼«¿É¹¹³ÉȼÁϵç³Ø£¬Õý¼«·´Ó¦Îª£º3O2+12H2O+12e-=12OH-£¬×Ü·´Ó¦Ê½Îª£º2CH3OH+3O2+4OH-=2CO32-+6H2O£¬Á½Ê½Ïà¼õ£¬¸º¼«·´Ó¦Îª£º2CH3OH-12e-+16OH-=2CO32-+12H2O£¬£¨3£©ºÍµçÔ´µÄÕý¼«ÏàÁ¬µÄµç¼«ÊÇÑô¼«£¬µç½â·½·¨¾«Á¶´ÖÍ­£¬µç½â³ØµÄÑô¼«²ÄÁÏÊÇ´ÖÍ­£¬µç¼«·´Ó¦Îª£ºCu-2e-=Cu2+
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

ÇâÑõÁ½ÖÖÔªËØÐγɵij£¼ûÎïÖÊÓÐH2OÓëH2O2£¬ÔÚÒ»¶¨Ìõ¼þϾù¿É·Ö½â¡£
£¨1£©ÒÑÖª£º
»¯Ñ§¼ü
¶Ï¿ª1mol»¯Ñ§¼üËùÐèµÄÄÜÁ¿£¨kJ£©
H¡ªH
436
O¡ªH
463
O=O
498
¢ÙH2OµÄµç×ÓʽÊÇ       ¡£
¢ÚH2O£¨g£©·Ö½âµÄÈÈ»¯Ñ§·½³ÌʽÊÇ       ¡£
¢Û11.2 L£¨±ê×¼×´¿ö£©µÄH2ÍêȫȼÉÕ£¬Éú³ÉÆø̬ˮ£¬·Å³ö       kJµÄÈÈÁ¿¡£
£¨2£©Ä³Í¬Ñ§ÒÔH2O2·Ö½âΪÀý£¬Ì½¾¿Å¨¶ÈÓëÈÜÒºËá¼îÐÔ¶Ô·´Ó¦ËÙÂʵÄÓ°Ïì¡£³£ÎÂÏ£¬°´ÕÕÈç±íËùʾµÄ·½°¸Íê³ÉʵÑé¡£
ʵÑé±àºÅ
·´Ó¦Îï
´ß»¯¼Á
a
50 mL 5% H2O2ÈÜÒº
 
1 mL 0.1 mol¡¤L-1 FeCl3ÈÜÒº
b
50 mL 5% H2O2ÈÜÒº
ÉÙÁ¿Å¨ÑÎËá
1 mL 0.1 mol¡¤L-1 FeCl3ÈÜÒº
c
50 mL 5% H2O2ÈÜÒº
ÉÙÁ¿Å¨NaOHÈÜÒº
1 mL 0.1 mol¡¤L-1 FeCl3ÈÜÒº
d
50 mL 5% H2O2ÈÜÒº
 
MnO2
¢Ù²âµÃʵÑéa¡¢b¡¢cÖÐÉú³ÉÑõÆøµÄÌå»ýËæʱ¼ä±ä»¯µÄ¹ØϵÈçͼ1Ëùʾ¡£
 
ͼ1                              Í¼2
ÓɸÃͼÄܹ»µÃ³öµÄʵÑé½áÂÛÊÇ_________¡£
¢Ú²âµÃʵÑédÔÚ±ê×¼×´¿öÏ·ųöÑõÆøµÄÌå»ýËæʱ¼ä±ä»¯µÄ¹ØϵÈçͼ2Ëùʾ¡£½âÊÍ·´Ó¦ËÙÂʱ仯µÄÔ­Òò£º         ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

ΪÁ˼õÉÙCO¶Ô´óÆøµÄÎÛȾ£¬Ä³Ñо¿ÐÔѧϰС×éÄâÑо¿COºÍH2O·´Ó¦×ª»¯ÎªÂÌÉ«ÄÜÔ´H2¡£ÒÑÖª£º
2CO(g)+O2(g)£½2CO2(g)   ¡÷H£½£­566kJ¡¤moL£­1
2H2(g)+O2(g)£½2H2O(g)   ¡÷H£½£­483.6KJ¡¤moL£­1
H2O (g)£½H2O(l)         ¡÷H£½£­44.0KJ¡¤moL£­1
£¨1£©ÇâÆøµÄ±ê׼ȼÉÕÈÈ¡÷H£½           kJ¡¤moL£­1
£¨2£©Ð´³öCOºÍH2O(g)×÷ÓÃÉú³ÉCO2ºÍH2µÄÈÈ»¯Ñ§·½³Ìʽ                 
£¨3£©Íù 1LÌå»ý²»±äµÄÈÝÆ÷ÖмÓÈë1.00mol COºÍ1.00mol H2O(g)£¬ÔÚt¡æʱ·´Ó¦²¢´ïµ½Æ½ºâ£¬Èô¸Ã·´Ó¦µÄ»¯Ñ§Æ½ºâ³£ÊýK£½1£¬Ôòt¡æʱCOµÄת»¯ÂÊΪ       £»·´Ó¦´ïµ½Æ½ºâºó£¬Éý¸ßζȣ¬´Ëʱƽºâ³£Êý½«      £¨Ìî¡°±ä´ó¡±¡¢¡°²»±ä¡±»ò¡°±äС¡±£©£¬Æ½ºâ½«Ïò        £¨Ìî¡°Õý¡±»ò¡°Ä桱£©·½ÏòÒƶ¯¡£
£¨4£©ÔÚCOºÍH2O·´Ó¦×ª»¯ÎªÂÌÉ«ÄÜÔ´H2ÖУ¬ÎªÁËÌá¸ßCOµÄת»¯ÂÊ£¬¿É²ÉÈ¡µÄ´ëÊ©ÊÇ        ¡£
A£®Ôö´óµÄCOŨ¶ÈB£®Ôö´óµÄH2O(g)Ũ¶ÈC£®Ê¹Óô߻¯¼ÁD£®½µµÍζÈ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨14·Ö£©ÇâÆøÊÇÒ»ÖÖÇå½àÄÜÔ´£¬ÇâÆøµÄÖÆÈ¡Óë´¢´æÊÇÇâÄÜÔ´ÀûÓÃÁìÓòµÄÑо¿Èȵ㡣
£¨1£©ÒÔ¼×ÍéΪԭÁÏÖÆÈ¡ÇâÆøÊǹ¤ÒµÉϳ£ÓõÄÖÆÇâ·½·¨¡£ÒÑÖª£º
CH4(g)£«H2O(g) ===CO(g)£«3H2(g)        ¦¤H£½+206.2 kJ/mol
CH4(g)£«CO2(g) ===2CO(g)£«2H2(g)       ¦¤H£½+247.4 kJ/mol
CH4(g)ÓëH2O(g)·´Ó¦Éú³ÉCO2(g)ºÍH2(g)µÄÈÈ»¯Ñ§·½³ÌʽΪ               ¡£
£¨2£©ÁòÌú¿ó(FeS2)ȼÉÕ²úÉúµÄSO2ͨ¹ýÏÂÁеâÑ­»·¹¤ÒÕ¹ý³Ì¼ÈÄÜÖÆH2SO4£¬ÓÖÄÜÖÆH2¡£

ÒÑÖª1g FeS2ÍêȫȼÉշųö7.1 kJÈÈÁ¿£¬FeS2ȼÉÕ·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ      ¡£
¸ÃÑ­»·¹¤ÒÕ¹ý³ÌµÄ×Ü·´Ó¦·½³ÌʽΪ      ¡£
£¨3£©µç½âÄòËØ[CO(NH2)2]µÄ¼îÐÔÈÜÒºÖÆÇâµÄ×°ÖÃʾÒâͼ¼ûͼ£¨µç½â³ØÖиôĤ½ö×èÖ¹ÆøÌåͨ¹ý£¬Òõ¡¢Ñô¼«¾ùΪ¶èÐԵ缫£©¡£µç½âʱ£¬Ñô¼«µÄµç¼«·´Ó¦Ê½Îª               ¡£

£¨4£©ÓÃÎüÊÕH2ºóµÄÏ¡ÍÁ´¢ÇâºÏ½ð×÷Ϊµç³Ø¸º¼«²ÄÁÏ(ÓÃMH±íʾ)£¬NiO(OH)×÷Ϊµç³ØÕý¼«²ÄÁÏ£¬KOHÈÜÒº×÷Ϊµç½âÖÊÈÜÒº£¬¿ÉÖƵøßÈÝÁ¿£¬³¤ÊÙÃüµÄÄøÇâµç³Ø¡£µç³Ø³ä·ÅµçʱµÄ×Ü·´Ó¦Îª£º
NiO(OH)£«MHNi(OH)2£«M
¢Ùµç³Ø·Åµçʱ£¬Õý¼«µÄµç¼«·´Ó¦Ê½Îª      ¡£
¢Ú³äµçÍê³Éʱ£¬Ni(OH)2È«²¿×ª»¯ÎªNiO(OH)¡£Èô¼ÌÐø³äµç½«ÔÚÒ»¸öµç¼«²úÉúO2£¬O2À©É¢µ½ÁíÒ»¸öµç¼«·¢Éúµç¼«·´Ó¦±»ÏûºÄ£¬´Ó¶ø±ÜÃâ²úÉúµÄÆøÌåÒýÆðµç³Ø±¬Õ¨£¬´Ëʱ£¬Òõ¼«µÄµç¼«·´Ó¦Ê½Îª      ¡£
£¨5£©Mg2CuÊÇÒ»ÖÖ´¢ÇâºÏ½ð¡£350¡æʱ£¬Mg2CuÓëH2·´Ó¦£¬Éú³ÉMgCu2ºÍ½öº¬Ò»ÖÖ½ðÊôÔªËصÄÇ⻯ÎÆäÖÐÇâµÄÖÊÁ¿·ÖÊýΪ0.077£©¡£Mg2CuÓëH2·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ           ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

ÌúÔªËØÊÇÖØÒªµÄ½ðÊôÔªËØ£¬µ¥ÖÊÌúÔÚ¹¤ÒµºÍÉú»îÖÐʹÓõÃ×îΪ¹ã·º¡£Ìú»¹ÓкܶàÖØÒªµÄ»¯ºÏÎï¼°Æ仯ѧ·´Ó¦¡£ÈçÌúÓëË®·´Ó¦£º3Fe(s)£«4H2O(g)£½Fe3O4(s)£«4H2(g)   ¡÷H
(1)ÉÏÊö·´Ó¦µÄƽºâ³£Êý±í´ïʽK£½_______¡£
(2) ÒÑÖª£º¢Ù3Fe(s)£«2O2(g)£½Fe3O4(s)  ¡÷H1£½£­1118.4kJ/mol
¢Ú2H2(g)£«O2(g)£½2H2O(g)  ¡÷H2£½£­483.8kJ/mol
¢Û2H2(g)£«O2(g)£½2H2O(l)  ¡÷H3£½£­571.8kJ/mol
Ôò¡÷H£½_______¡£
(3)ÔÚt0Cʱ,¸Ã·´Ó¦µÄƽºâ³£ÊýK£½16£¬ÔÚ2LºãκãÈÝÃܱÕÈÝÆ÷¼×ºÍÒÒÖÐ,·Ö±ð°´Ï±íËùʾ¼ÓÈëÎïÖÊ£¬·´Ó¦¾­¹ýÒ»¶Îʱ¼äºó´ïµ½Æ½ºâ¡£
 
Fe
H2O£¨g£©
Fe3O4
H2
¼×/mol
1.0
1.0
1.0
1.0
ÒÒ/mol
1.0
1.5
1.0
1.0
 
¢Ù¼×ÈÝÆ÷ÖÐH2OµÄƽºâת»¯ÂÊΪ_______ (½á¹û±£ÁôһλСÊý)¡£
¢ÚÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ_______ (Ìî±àºÅ)
A.ÈôÈÝÆ÷ѹǿºã¶¨£¬Ôò·´Ó¦´ïµ½Æ½ºâ״̬
B.ÈôÈÝÆ÷ÄÚÆøÌåÃܶȺ㶨£¬Ôò·´Ó¦´ïµ½Æ½ºâ״̬
C.¼×ÈÝÆ÷ÖÐH2OµÄƽºâת»¯ÂÊ´óÓÚÒÒÈÝÆ÷ÖÐH2OµÄƽºâת»¯ÂÊ
D.Ôö¼ÓFe3O4¾ÍÄÜÌá¸ßH2OµÄת»¯ÂÊ
(4)Èô½«(3)ÖÐ×°ÖøÄΪºãÈݾøÈÈ(²»ÓëÍâ½ç½»»»ÄÜÁ¿)×°Ö㬰´Ï±í³äÈëÆðʼÎïÖÊ£¬ÆðʼʱÓëƽºâºóµÄ¸÷ÎïÖʵÄÁ¿¼û±í£º
 
Fe
H2O£¨g£©
Fe3O4
H2
Æðʼ/mol
3.0
4.0
0
0
ƽºâ/mol
m
n
p
q
 
ÈôÔÚ´ïƽºâºóµÄ×°ÖÃÖмÌÐø¼ÓÈëA¡¢B¡¢CÈýÖÖ×´¿öϵĸ÷ÎïÖÊ£¬¼û±í£º
 
Fe
H2O£¨g£©
Fe3O4
H2
A/mol
3.0
4.0
0
0
B/mol
0
0
1
4
C/mol
m
n
p
q
 
µ±ÉÏÊö¿ÉÄæ·´Ó¦ÔÙÒ»´Î´ïµ½Æ½ºâ״̬ºó,ÉÏÊö¸÷×°ÖÃÖÐH2µÄ°Ù·Öº¬Á¿°´ÓÉ´óµ½Ð¡µÄ˳ÐòÅÅÁеĹØϵÊÇ
________(ÓÃA¡¢B¡¢C±íʾ£©¡£
(5)ÒÑÖªFe(OH)3µÄKsp£½2.79¡Á10£­39£¬¶øFeCl3ÈÜÒº×ÜÊÇÏÔʾ½ÏÇ¿µÄËáÐÔ£¬ÈôijFeCl3ÈÜÒºµÄpHΪ3,Ôò¸ÃÈÜÒºÖÐc(Fe3+)£½________mol ? L-1 (½á¹û±£Áô3λÓÐЧÊý×Ö)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

Ä¿Ç°£¬Ïû³ýµªÑõ»¯ÎïÎÛȾÓжàÖÖ·½·¨¡£
£¨1£©ÓÃCH4´ß»¯»¹Ô­µªÑõ»¯Îï¿ÉÒÔÏû³ýµªÑõ»¯ÎïµÄÎÛȾ¡£ÒÑÖª£º
¢ÙCH4(g)£«4NO2(g)£½4NO(g)£«CO2(g)£«2H2O(g) ¦¤H£½£­574 kJ¡¤mol£­1
¢ÚCH4(g)£«4NO(g)£½2N2(g)£«CO2(g)£«2H2O(g) ¦¤H£½£­1160 kJ¡¤mol£­1
¢ÛH2O(g)£½H2O(l) ¦¤H£½£­44£®0 kJ¡¤mol£­1
д³öCH4 (g)ÓëNO2 (g)·´Ó¦Éú³ÉN2 (g) ,CO2(g)ºÍH2O(l)µÄÈÈ»¯Ñ§·½³Ì ʽ_____________________
£¨2£©ÓûîÐÔÌ¿»¹Ô­·¨´¦ÀíµªÑõ»¯Îï¡£Óйط´Ó¦Îª£ºC(s)£«2NO(g)N2(g)£«CO2(g)ijÑо¿Ð¡×éÏòºãÈÝÃܱÕÈÝÆ÷¼ÓÈëÒ»¶¨Á¿µÄ»îÐÔÌ¿ºÍNO£¬ºãΣ¨T¡£C)Ìõ¼þÏ·´Ó¦£¬·´Ó¦½øÐе½²»Í¬Ê±¼ä²âµÃ¸÷ÎïÖʵÄŨ¶ÈÈçÏÂ

¢Ù²»ÄÜ×÷ΪÅжϷ´Ó¦´ïµ½»¯Ñ§Æ½ºâ״̬µÄÒÀ¾Ý   ÊÇ_______
A£®ÈÝÆ÷ÄÚCO2µÄŨ¶È±£³Ö²»±ä
B£®vÕý£¨N2£©="2" vÕý£¨NO£©
C£®ÈÝÆ÷ÄÚѹǿ±£³Ö²»±ä  
D£®»ìºÏÆøÌåµÄÃܶȱ£³Ö²»±ä
E£®»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿±£³Ö²»±ä
¢ÚÔÚT¡£Cʱ£®¸Ã·´Ó¦µÄƽºâ³£ÊýΪ_______(±£ÁôÁ½Î»Ð¡Êý)£»
¢ÛÔÚ30  min,¸Ä±äijһÌõ¼þ,·´Ó¦ÖØдﵽƽºâ,Ôò¸Ä±äµÄÌõ¼þÊÇ_______
£¨3£©¿Æѧ¼ÒÕýÔÚÑо¿ÀûÓô߻¯¼¼Êõ½«³¬ÒôËÙ·É»úβÆøÖеÄNOºÍCOת±ä³ÉCO2ºÍN2,Æ䷴ӦΪ£º
2CO£«2NON2£«2CO¦¤H<0Ñо¿±íÃ÷£ºÔÚʹÓõÈÖÊÁ¿´ß»¯¼Áʱ£¬Ôö´ó´ß»¯¼ÁµÄ±È±íÃæ»ý¿ÉÌá¸ß»¯Ñ§·´Ó¦ËÙÂÊ£®ÎªÁË·Ö±ðÑé֤ζȡ¢´ß»¯¼ÁµÄ±È±íÃæ»ý¶Ô»¯Ñ§·´ Ó¦ËÙÂʵÄÓ°Ïì¹æÂÉ¡¢Ä³Í¬Ñ§Éè¼ÆÁËÈý×éʵÑ飬²¿·ÖʵÑéÌõ¼þÒѾ­ÌîÔÚϱíÖС£
 
¢ÙÉϱíÖÐ:a=_______,b=________,e=________
¢ÚÇëÔÚ¸ø³öµÄ×ø±êͼÖУ¬»­³öÉϱíÖÐʵÑéIIºÍʵÑéIIIÌõ¼þÏ»ìºÏÆøÌåÖÐNOŨ¶ÈËæʱ¼ä±ä»¯µÄÇ÷ÊÆÇúÏßͼ,²¢±êÃ÷ÏàÓ¦µÄʵÑé±àºÅ
 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

µçÀëƽºâ³£Êý£¨ÓÃKa±íʾ)µÄ´óС¿ÉÒÔÅжϵç½âÖʵÄÏà¶ÔÇ¿Èõ¡£25¡æʱ£¬ÓйØÎïÖʵĵçÀëƽºâ³£ÊýÈçϱíËùʾ£º
»¯Ñ§Ê½
HF
H2CO3
HClO
µçÀëƽºâ³£Êý
£¨Ka£©
7.2¡Á10-4
K1=4.4¡Á10-7
K2=4.7¡Á10-11
3.0¡Á10-8
 
£¨1£©ÒÑÖª25¡æʱ£¬¢ÙHF(aq)+OH¡ª(aq)£½F¡ª(aq)+H2O(l)  ¦¤H£½£­67.7kJ/mol£¬
¢ÚH+(aq)+OH¡ª(aq)£½H2O(l)        ¦¤H£½£­57.3kJ/mol£¬
Çâ·úËáµÄµçÀë·½³Ìʽ¼°ÈÈЧӦ¿É±íʾΪ________________________¡£
£¨2£©½«Å¨¶ÈΪ0.1 mol/LHFÈÜÒº¼ÓˮϡÊÍÒ»±¶£¨¼ÙÉèζȲ»±ä£©£¬ÏÂÁи÷Á¿Ôö´óµÄÊÇ____¡£
A£®c(H+)       B£®c(H+)¡¤c(OH¡ª)     C£®   D£®
£¨3£©25¡æʱ£¬ÔÚ20mL0.1mol/LÇâ·úËáÖмÓÈëVmL0.1mol/LNaOHÈÜÒº£¬²âµÃ»ìºÏÈÜÒºµÄpH±ä»¯ÇúÏßÈçͼËùʾ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ_____¡£

A£®pH£½3µÄHFÈÜÒººÍpH£½11µÄNaFÈÜÒºÖУ¬ ÓÉË®µçÀë³öµÄc(H+)ÏàµÈ
B£®¢ÙµãʱpH£½6£¬´ËʱÈÜÒºÖУ¬c(F¡ª)£­c(Na+)£½9.9¡Á10-7mol/L
C£®¢Úµãʱ£¬ÈÜÒºÖеÄc(F¡ª)£½c(Na+)
D£®¢ÛµãʱV£½20mL£¬´ËʱÈÜÒºÖÐc(F¡ª)< c(Na+)£½0.1mol/L
£¨4£©ÎïÖʵÄÁ¿Å¨¶È¾ùΪ0.1mol/LµÄÏÂÁÐËÄÖÖÈÜÒº£º ¢Ù Na2CO3ÈÜÒº ¢Ú NaHCO3ÈÜÒº¢Û NaFÈÜÒº ¢ÜNaClOÈÜÒº¡£ÒÀ¾ÝÊý¾ÝÅжÏpHÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ______________¡£
£¨5£©Na2CO3ÈÜÒºÏÔ¼îÐÔÊÇÒòΪCO32¡ªË®½âµÄÔµ¹Ê£¬ÇëÉè¼Æ¼òµ¥µÄʵÑéÊÂʵ֤Ã÷Ö®
___________________________________________________________¡£
£¨6£©³¤ÆÚÒÔÀ´£¬Ò»Ö±ÈÏΪ·úµÄº¬ÑõËá²»´æÔÚ¡£1971ÄêÃÀ¹ú¿Æѧ¼ÒÓ÷úÆøͨ¹ýϸ±ùĩʱ»ñµÃHFO£¬Æä½á¹¹Ê½ÎªH¡ªO¡ªF¡£HFOÓëË®·´Ó¦µÃµ½HFºÍ»¯ºÏÎïA£¬Ã¿Éú³É1molHFתÒÆ     molµç×Ó¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÍƶÏÌâ

¶ÌÖÜÆÚÔªËØA¡¢B¡¢C¡¢D¡¢EÔ­×ÓÐòÊýÒÀ´ÎÔö´ó¡£AÊÇÖÜÆÚ±íÖÐÔ­×Ӱ뾶×îСµÄÔªËØ£¬BÔ­×ӵļ۵ç×ÓÊýµÈÓÚ¸ÃÔªËØ×îµÍ»¯ºÏ¼ÛµÄ¾ø¶ÔÖµ£¬CÓëDÄÜÐγÉD2CºÍD2C2Á½ÖÖ»¯ºÏÎ¶øDÊÇͬÖÜÆÚÖнðÊôÐÔ×îÇ¿µÄÔªËØ£¬EµÄ¸ºÒ»¼ÛÀë×ÓÓëCºÍAÐγɵÄijÖÖ»¯ºÏÎï·Ö×Óº¬ÓÐÏàͬµÄµç×ÓÊý¡£
£¨1£©A¡¢C¡¢DÐγɵĻ¯ºÏÎïÖк¬ÓеĻ¯Ñ§¼üÀàÐÍΪ              ¡£
£¨2£©ÒÑÖª£º¢ÙE£­E¡ú2E¡¤£»¡÷H£½£«a kJ¡¤mol-1 
¢Ú 2A¡¤¡úA£­A£»¡÷H=£­b kJ¡¤mol-1 
¢ÛE¡¤£«A¡¤¡úA£­E£»¡÷H=£­c kJ¡¤mol-1£¨¡°¡¤¡±±íʾÐγɹ²¼Û¼üËùÌṩµÄµç×Ó£© 
д³ö298Kʱ£¬A2ÓëE2·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ                                   ¡£
£¨3£©ÔÚijζÈÏ¡¢ÈÝ»ý¾ùΪ2LµÄÈý¸öÃܱÕÈÝÆ÷ÖУ¬°´²»Í¬·½Ê½Í¶Èë·´Ó¦Î±£³ÖºãκãÈÝ£¬Ê¹Ö®·¢Éú·´Ó¦£º2A2£¨g£©£«BC£¨g£©X£¨g£©£»¡÷H£½£­dJ¡¤mol£­1£¨d£¾0£¬XΪA¡¢B¡¢CÈýÖÖÔªËØ×é³ÉµÄÒ»ÖÖ»¯ºÏÎ¡£³õʼͶÁÏÓë¸÷ÈÝÆ÷´ïµ½Æ½ºâʱµÄÓйØÊý¾ÝÈçÏ£º
ʵÑé
¼×
ÒÒ
±û
³õʼͶÁÏ
2 molA2¡¢1 molBC
1 molX
4 molA2¡¢2 molBC
ƽºâʱn£¨X£©
0.5mol
n2
n3
·´Ó¦µÄÄÜÁ¿±ä»¯
·Å³öQ1kJ
ÎüÊÕQ2kJ
·Å³öQ3kJ
ÌåϵµÄѹǿ
P1
P2
P3
·´Ó¦ÎïµÄת»¯ÂÊ
¦Á1
¦Á2
¦Á3
 
¢ÙÔÚ¸ÃζÈÏ£¬¼ÙÉè¼×ÈÝÆ÷´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâËùÐèʱ¼äΪ4 min£¬Ôò¸Ãʱ¼ä¶ÎÄÚA2µÄƽ¾ù·´Ó¦ËÙÂÊv(A2)       ¡£
¢Ú¸ÃζÈÏ´˷´Ó¦µÄƽºâ³£ÊýKµÄֵΪ          ¡£
¢ÛÈý¸öÈÝÆ÷Öеķ´Ó¦·Ö±ð´ïƽºâʱ¸÷×éÊý¾Ý¹ØϵÕýÈ·µÄÊÇ      £¨ÌîÐòºÅ£©¡£
A£®¦Á1£«¦Á2£½1               B£®Q1£«Q2£½d            C£®¦Á3£¼¦Á1            
D£®P3£¼2P1£½2P2       E£®n2£¼n3£¼1.0mol           F£®Q3£½2Q1
¢ÜÔÚÆäËûÌõ¼þ²»±äµÄÇé¿öÏ£¬½«¼×ÈÝÆ÷µÄÌåϵÌå»ýѹËõµ½1L£¬ÈôÔÚµÚ8min´ïµ½ÐµÄƽºâʱA2µÄ×Üת»¯ÂÊΪ65.5%£¬ÇëÔÚÏÂͼÖл­³öµÚ5min µ½ÐÂƽºâʱXµÄÎïÖʵÄÁ¿Å¨¶ÈµÄ±ä»¯ÇúÏß¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÒÑÖª·´Ó¦A2(g)£«B2(g)=2AB(g)£¬¶Ï¿ª1molA2ÖеĻ¯Ñ§¼üÏûºÄµÄÄÜÁ¿ÎªQ1 kJ£¬¶Ï¿ª1molB2ÖеĻ¯Ñ§¼üÏûºÄµÄÄÜÁ¿ÎªQ2 kJ£¬Éú³É1molABÖеĻ¯Ñ§¼üÊͷŵÄÄÜÁ¿ÎªQ3kJ£¨Q1¡¢Q2¡¢Q3¾ù´óÓÚÁ㣩£¬ÔòÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ                                         £¨   £©
A£®ÈôA2ºÍB2µÄ×ÜÄÜÁ¿Ö®ºÍ´óÓÚÉú³ÉµÄ2ABµÄ×ÜÄÜÁ¿£¬Ôò·´Ó¦·ÅÈÈ
B£®ÈôA2ºÍB2µÄ×ÜÄÜÁ¿Ö®ºÍСÓÚÉú³ÉµÄ2ABµÄ×ÜÄÜÁ¿£¬Ôò·´Ó¦·ÅÈÈ
C£®Èô¸Ã·´Ó¦Îª·ÅÈÈ·´Ó¦£¬ÔòQ1£«Q2 < Q3
D£®Èô¸Ã·´Ó¦ÎªÎüÈÈ·´Ó¦£¬ÔòQ1£«Q2 < Q3

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸