ºìÁ×P(s)ºÍCl2(g)·¢Éú·´Ó¦Éú³ÉPCl3(g)ºÍPCl5(g)¡£·´Ó¦¹ý³ÌºÍÄÜÁ¿¹ØϵÈçͼËùʾ£¨Í¼Öеġ÷H±íʾÉú³É1mol²úÎïµÄÊý¾Ý£©¡£
¸ù¾ÝÉÏͼ»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©PºÍCl2·´Ó¦Éú³ÉPCl3µÄÈÈ»¯Ñ§·½³ÌʽÊÇ
¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£»
£¨2£©PCl5·Ö½â³ÉPCl3ºÍCl2µÄÈÈ»¯Ñ§·½³ÌʽÊÇ
¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£»
ÉÏÊö·Ö½â·´Ó¦ÊÇÒ»¸ö¿ÉÄæ·´Ó¦¡£Î¶ÈT1ʱ£¬ÔÚÃܱÕÈÝÆ÷ÖмÓÈë0.80mol PCl5,·´Ó¦´ïƽºâʱPCl5»¹Ê£0.60 mol£¬Æä·Ö½âÂʦÁ1µÈÓÚ¡¡¡¡¡¡¡¡£»Èô·´Ó¦Î¶ÈÓÉT1Éý¸ßµ½T2£¬Æ½ºâʱPCl5µÄ·Ö½âÂÊΪ¦Á2£¬¦Á2 ¦Á1(Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±)£»
£¨3£©¹¤ÒµÉÏÖƱ¸PCl5ͨ³£·ÖÁ½²½½øÐУ¬ÏȽ«PºÍCl2·´Ó¦Éú³ÉÖмä²úÎïPCl3£¬È»ºó½µÎ£¬ÔÙºÍCl2·´Ó¦Éú³ÉPCl5¡£ÔÒòÊÇ________________________¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡
£¨4£©PºÍCl2·ÖÁ½²½·´Ó¦Éú³É1mol PCl5µÄ¡÷H3=____________£¬PºÍCl2Ò»²½·´Ó¦Éú³É1molPCl5µÄ¡÷H4__________¡÷H3(Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±)¡£
£¨5£©PCl5Óë×ãÁ¿Ë®³ä·Ö·´Ó¦£¬×îÖÕÉú³ÉÁ½ÖÖËᣬÆ仯ѧ·½³ÌʽÊÇ__________________________________¡£
¡¾´ð°¸¡¿
£¨1£©Cl2 (g)£« P(s)PCl3 (g) ¦¤H£½£306 kJ/ lmol¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡
£¨2£© PCl5(g)PCl3(g)£«Cl2(g) ¡÷H£½93kJ/ mol¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡
25£¥¡¡¡¡¡¡¡¡´óÓÚ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡
£¨3£©Á½²½·´Ó¦¾ùΪ·ÅÈÈ·´Ó¦£¬½µÎÂÓÐÀûÓÚÌá¸ß²úÂÊ£¬·ÀÖ¹²úÎï·Ö½â¡¡¡¡¡¡¡¡¡¡¡¡
£¨4£©£399kJ/ mol¡¡µÈÓÚ ¡¡¡¡¡¡
£¨5£©PCl5£«4H2O£½H3PO4£«5HCl
¡¾½âÎö¡¿£¨1£©ÓÉͼÏóÖªPºÍCl2·´Ó¦Éú³ÉPCl3µÄÈÈ»¯Ñ§·½³ÌʽÊÇP£¨s)+Cl2(g)PCl3(g) ¦¤H=-306 kJ¡¤mol-1¡£
£¨2£©¦¤H=Éú³ÉÎï×ÜÄÜÁ¿-·´Ó¦Îï×ÜÄÜÁ¿£¬·´Ó¦PCl3(g)+Cl2(g)PCl5(g) ¦¤H=-93 kJ¡¤mol-1£¬¹Ê·´Ó¦PCl5(g)PCl3(g)+Cl2(g) ¦¤H=+93 kJ¡¤mol-1¡£
·Ö½âÂÊ=¡Á100%=25%¡£
ÓÉÓÚ·´Ó¦PCl5PCl3+Cl2ÊÇÎüÈÈ·´Ó¦£¬Éý¸ßζȣ¬Æ½ºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬·Ö½âÂÊÔö´ó£¬¼´¦Á2´óÓÚ¦Á1¡£
£¨3£©3Cl2+2P2PCl3£»PCl3+Cl2PCl5Á½¸ö·´Ó¦¾ùΪ·ÅÈÈ·´Ó¦£¬½µµÍζȣ¬ÓÐÀûÓÚƽºâÏòÕýÏòÒƶ¯£¬Ìá¸ßÁËPCl3µÄת»¯ÂÊ£¬Í¬Ê±·ÀÖ¹PCl5·Ö½â¡£
£¨4£©¸ù¾Ý¸Ç˹¶¨ÂÉ£¬PºÍCl2·ÖÁ½²½·´Ó¦ºÍÒ»²½·´Ó¦Éú³ÉPCl5µÄ¦¤HÓ¦¸ÃÊÇÏàµÈµÄ¡£
£¨5£©PCl5ÓëH2O·´Ó¦Éú³ÉµÄËá¿ÉÄÜÊÇH3PO4¡¢HClºÍHClO£¬µ«PCl5H3PO4£¬PÔªËØ»¯ºÏ¼Ûû±ä£¬ÔòClÔªËØ»¯ºÏ¼ÛÒ²²»±ä£¬¼´¸Ã·´Ó¦Éú³ÉµÄÁ½ÖÖËáÊÇH3PO4ºÍHCl¡£ËùÒԸ÷´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇPCl5+4H2OH3PO4+5HCl¡£
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ijÂÈ»¯Ã¾ÈÜÒºµÄÃܶÈΪ1.18 g.cm-3£¬ÆäÖÐþÀë×ÓµÄÖÊÁ¿·ÖÊýΪ5.1%£¬300mL¸ÃÈÜÒºÖÐ Cl£Àë×ÓµÄÎïÖʵÄÁ¿Ô¼µÈÓÚ
A 0.37 mol B 1.5 mol C 0.74 mol D 0.63 mol
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ijÓлúÎïX£¨C4H6O5£©¹ã·º´æÔÚÓÚÐí¶àË®¹ûÖУ¬ÓÈÒÔÆ»¹û¡¢ÆÏÌÑ¡¢Î÷¹Ï¡¢É½é«ÈâΪ¶à£¬¾²âÊÔ£¬¸Ã»¯ºÏÎï¾ßÓÐÈçÏÂÐÔÖÊ£º ¢Ù1mol XÓë×ãÁ¿µÄ½ðÊôÄÆ·´Ó¦²úÉú1.5mol H2 ¢ÚXÓë´¼£¨ROH£©»òôÈËᣨRCOOH£©ÔÚŨH2SO4ºÍ¼ÓÈÈÌõ¼þϾù¿É·´Ó¦Éú³ÉÓÐÏãζµÄ²úÎï ¢ÛXÔÚÒ»¶¨Ìõ¼þϵķÖ×ÓÄÚÍÑË®²úÎ²»ÊÇ»·×´»¯ºÏÎ¿ÉºÍäåË®·¢Éú¼Ó³É·´Ó¦¡£¸ù¾ÝÉÏÊöÐÅÏ¢£¬ÏÂÁжÔXµÄ½á¹¹µÄÅжÏÕýÈ·µÄÊÇ£º£¨ £©
A.X·Ö×ÓÖп϶¨ÓÐ̼̼˫¼ü B. X·Ö×ÓÖпÉÄÜÓÐÈý¸öôÇ»ùºÍÒ»¸ö£COOR¹ÙÄÜÍÅ
C.X·Ö×ÓÖпÉÄÜÓÐÈý¸öôÈ»ù D. X·Ö×ÓÖпÉÄÜÓÐÒ»¸öôÇ»ùºÍ¶þ¸öôÈ»ù
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
25¡æºÍ101kPaʱ£¬ÒÒÍé¡¢ÒÒȲºÍ±ûÏ©×é³ÉµÄ»ìºÏÌþ8mL,Óë¹ýÁ¿ÑõÆø»ìºÏ²¢ÍêȫȼÉÕ£¬³ýȥˮÕôÆø£¬»Ö¸´µ½ÔÀ´µÄζȺÍѹǿ£¬ÆøÌå×ÜÌå»ýËõСÁË18mL£¬Ô»ìºÏÌþÖÐÒÒȲµÄÌå»ý·ÖÊýΪ( )
A. 12.5% B. 25% C. 50% D. 75%
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ÒÑÖª£º£¨1£©Zn£¨s£©+1/2O2£¨g£©==ZnO(s)£¬¦¤H=-348.3kJ/mol
(2) 2Ag(s)+1/2 O2£¨g£©== Ag2O(s)£¬¦¤H=-31.0kJ/mol
ÔòZn£¨s£©+ Ag2O(s)== ZnO(s)+ 2Ag(s)µÄ¦¤HµÈÓÚ
A.-317.3kJ/mol B.-379.3kJ/mol
C.-332.8 kJ/mol D.317.3 kJ/mol
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¶þÑõ»¯ÁòºÍµªµÄÑõ»¯ÎïÊdz£ÓõĻ¯¹¤ÔÁÏ£¬µ«Ò²ÊÇ´óÆøµÄÖ÷ÒªÎÛȾÎï¡£×ÛºÏÖÎÀíÆäÎÛȾÊÇ»µ¾³»¯Ñ§µ±Ç°µÄÖØÒªÑо¿ÄÚÈÝÖ®Ò»¡£
(1)ÁòËáÉú²úÖУ¬SO2´ß»¯Ñõ»¯Éú³ÉSO3£º
2SO2(s)+O2£¨g£©2SO3£¨g£©.
ijζÈÏ£¬SO2µÄƽºâת»¯ÂÊ(¦Á)ÓëÌåϵ×Üѹǿ(P)µÄ¹ØϵÈçÏÂͼËùʾ¡£¸ù¾Ýͼʾ»Ø´ðÏÂÁÐÎÊÌ⣺
¢Ù½«2.0 molSO2ºÍ1.0molO2ÖÃÓÚ10 LÃܱÕÈÝÆ÷ÖУ¬·´Ó¦´ïƽºâºó£¬Ìåϵ×ÜѹǿΪ0.10MPa¡£¸Ã·´Ó¦µÄƽºâ³£ÊýµÈÓÚ__________¡£
¢Úƽºâ״̬ÓÉA±äµ½Bʱ£¬Æ½ºâ³£ÊýK£¨A£©_______K(B)(Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±)¡£
(2)ÓÃCH4´ß»¯»¹ÔNOx¿ÉÒÔÏû³ýµªÑõ»¯ÎïµÄÎÛȾ¡£ÀýÈ磺
CH4(g)+4NO2(g)=4NO(g)+CO2(g)+2H2O(g) ¡÷H=-574kJ¡¤mol-1
CH4(g)+4NO(g)=2N2(g)+CO2(g)+2H2O(g) ¡÷H=-1160kJ¡¤mol-1
ÈôÓñê×¼×´¿öÏÂ4.48 L CH4»¹ÔNO2ÖÁN2£¬Õû¸ö¹ý³ÌÖÐתÒƵĵç×Ó×ÜÊýΪ__________(°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµÓÃNA±íʾ)£¬·Å³öµÄÈÈÁ¿Îª___________kJ¡£
(3)ÐÂÐÍÄÉÃײÄÁÏÑõȱλÌúËáÑÎ(MFe2Ox£¬3£¼x£¼4£¬M=Mn¡¢Co¡¢Zn»òNi)ÓÉÌúËáÑÎ(MFe2O4)¾¸ßλ¹Ô¶øµÃ£¬³£ÎÂÏ£¬ËüÄÜʹ¹¤Òµ·ÏÆøÖеÄËáÐÔÑõ»¯Îï·Ö½â³ýÈ¥¡£×ª»¯Á÷³ÌÈçͼËùʾ£º
Çëд³öMFe2Ox·Ö½âSO2µÄ»¯Ñ§·½³Ìʽ________________(²»±ØÅäƽ)¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
25 ¡æ£¬101 kPaʱ£¬Ç¿ËáÓëÇ¿¼îµÄÏ¡ÈÜÒº·¢ÉúÖкͷ´Ó¦µÄÖкÍÈÈΪ57.3 kJ/mol£¬ÐÁÍéµÄȼÉÕÈÈΪ5 518 kJ/mol¡£ÏÂÁÐÈÈ»¯Ñ§·½³ÌʽÊéдÕýÈ·µÄÊÇ£¨ £©
A.2H+£¨aq£©+£¨aq£©+Ba2+£¨aq£©+2OH-£¨aq£©====BaSO4£¨s£©+2H2O£¨l£©£»¦¤H=-57.3 kJ/mol
B.KOH£¨aq£©+H2SO4£¨aq£©====K2SO4£¨aq£©+H2O£¨l£©£»¦¤H=-57.3 kJ/mol
C.C8H18£¨l£©+O2£¨g£©====8CO2£¨g£©+9H2O£¨g£©£»¦¤H=-5 518 kJ/mol
D.2C8H18£¨g£©+25O2£¨g£©====16CO2£¨g£©+18H2O£¨l£©£»¦¤H=-5 518 kJ/mol
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ÏÂÁÐʵÑé²Ù×÷ÓëÔ¤ÆÚʵÑéÄ¿µÄ»òËùµÃʵÑé½áÂÛÒ»ÖµÄÊÇ
Ñ¡Ïî | ʵÑé²Ù×÷ | ʵÑéÄ¿µÄ»ò½áÂÛ |
A | ij¼ØÑÎÈÜÓÚÑÎËᣬ²úÉúÄÜʹ³ÎÇåʯ»ÒË®±ä»ë×ǵÄÎÞÉ«ÎÞζÆøÌå | ˵Ã÷¸Ã¼ØÑÎÊÇ |
B | Ïòº¬ÓÐÉÙÁ¿µÄÈÜÒºÖмÓÈë×ãÁ¿·ÛÄ©£¬½Á°èÒ»¶Îʱ¼äºó¹ýÂË | ³ýÈ¥ÈÜÒºÖÐÉÙÁ¿ |
C | ³£ÎÂÏ£¬Ïò±¥ºÍÈÜÒºÖмÓÉÙÁ¿·ÛÄ©£¬¹ýÂË£¬ÏòÏ´¾»µÄ³ÁµíÖмÓÏ¡ÑÎËᣬÓÐÆøÅݲúÉú | ˵Ã÷³£ÎÂÏÂ
|
D | ÓëŨÁòËá170¡æ¹²ÈÈ£¬ÖƵõÄÆøÌåͨÈËËáÐÔÈÜÒº | ¼ìÑéÖƵÃÆøÌåÊÇ·ñΪÒÒÏ© |
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com