ÌâÄ¿ÄÚÈÝ

£¨10·Ö£©½«Ò»¶¨Á¿µÄSO2ºÍº¬0.7molÑõÆøµÄ¿ÕÆø(ºöÂÔCO2)·ÅÈë0.5 LÃܱÕÈÝÆ÷ÄÚ£¬550¡æʱ£¬ÔÚ´ß»¯¼Á×÷ÓÃÏ·¢Éú·´Ó¦£º£¨Õý·´Ó¦·ÅÈÈ£©¡£²âµÃn£¨O2£©Ëæʱ¼äµÄ±ä»¯Èçϱí

·´Ó¦´ïµ½5sºó£¬½«ÈÝÆ÷ÖеĻìºÏÆøÌåͨ¹ý¹ýÁ¿NaOHÈÜÒº£¬ÆøÌåÌå»ý¼õÉÙÁË22. 4L£¨´ËÌå»ýΪ±ê×¼×´¿öϵÄÌå»ý£©£»ÔÙ½«Ê£ÓàÆøÌåͨ¹ý½¹ÐÔûʳ×ÓËáµÄ¼îÐÔÈÜÒºÎüÊÕO2£¬ÆøÌåµÄÌå»ýÓÖ¼õÉÙÁË5.6L£¨´ËÌå»ýΪ±ê×¼×´¿öϵÄÌå»ý£©¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÓÃO2±íʾ´Ó0-lsÄڸ÷´Ó¦µÄƽ¾ù·´Ó¦ËÙÂÊΪ__________________¡£
£¨2£©O2µÄƽºâŨ¶Èc (O2)=____________________________£»
£¨3£© 4sʱ£¬SO2µÄÉú³ÉËÙÂÊ____________£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©O2µÄÏûºÄËÙÂÊ¡£
£¨4£©Çó¸Ã·´Ó¦´ïµ½Æ½ºâʱSO2µÄת»¯ÂÊÊÇ________£¨ÓðٷÖÊý±íʾ£©¡£
£¨5£©Èô½«Æ½ºâ»į̀ÆøÌåÖÐSO3µÄ5%ͨÈë¹ýÁ¿µÄBaCl2ÈÜÒº£¬Éú³É³Áµí_______¿Ë£¨¼ÆËã½á¹û±£ÁôһλСÊý£©¡£

£¨1£©0.6mol/(L¡¤s)
£¨2£©0.5mol/L
£¨3£©´óÓÚ
£¨4£©90%
£¨5£©10.5

½âÎöÊÔÌâ·ÖÎö£º£¨1£©0-lsÄÚO2µÄÎïÖʵÄÁ¿¼õÉÙ0.7-0.4=0.3mol£¬Å¨¶È¼õÉÙ0.3mol/0.5L=0.6mol/L£¬ËùÒÔÓÃO2±íʾ´Ó0-lsÄڸ÷´Ó¦µÄƽ¾ù·´Ó¦ËÙÂÊΪ0.6mol/L/1s=0.6mol/(L¡¤s)
£¨2£©¸ù¾ÝÌâÒâ¿ÉÖª£¬ÓëÇâÑõ»¯ÄƵķ´Ó¦µÄÊǶþÑõ»¯Áò¡¢ÈýÑõ»¯ÁòµÄ»ìºÏÆøÌ壬¶þÕßµÄÎïÖʵÄÁ¿Ö®ºÍÓ뿪ʼ¼ÓÈëµÄ¶þÑõ»¯ÁòµÄÎïÖʵÄÁ¿Ïàͬ£¬ËùÒÔ»ìºÏÆøÌåͨ¹ýÇâÑõ»¯ÄÆÈÜÒºÌå»ý¼õÉÙ£¨±ê×¼×´¿ö£©22.4L£¬ËµÃ÷¿ªÊ¼Í¨ÈëµÄ¶þÑõ»¯ÁòµÄÎïÖʵÄÁ¿ÊÇ1mol£¬Ê£ÓàÆøÌåͨ¹ý½¹ÐÔûʳ×ÓËáµÄ¼îÐÔÈÜÒºÎüÊÕO2£¬ÆøÌåµÄÌå»ýÓÖ¼õÉÙÁË5.6L£¨±ê×¼×´¿ö£©£¬ËµÃ÷Ê£ÓàÑõÆø5.6L£¬ÎïÖʵÄÁ¿ÊÇ0.25mol£¬Å¨¶ÈÊÇ5.6L/22.4L/mol/0.5L=0.5mol/L£¬ËùÒÔѹǿµÄƽºâŨ¶ÈÊÇ0.5mol/L£»
£¨3£©4sʱ£¬·´Ó¦ÒÑ´ïƽºâ£¬vÄæ(SO2)=2vÕý£¨O2£©£¬ËùÒÔSO2µÄÉú³ÉËÙÂÊ´óÓÚO2µÄÏûºÄËÙÂÊ¡£
£¨4£©ÑõÆøµÄƽºâÎïÖʵÄÁ¿ÊÇ0.25mol£¬ÔòÏûºÄÑõÆø0.7mol-0.25mol=0.45mol£¬ËùÒÔÏûºÄ¶þÑõ»¯Áò0.45mol¡Á2=0.9mol£¬¸ù¾Ý£¨2£©µÄ·ÖÎö¿ÉÖª£¬¶þÑõ»¯ÁòµÄ³õʼÁ¿ÊÇ1mol£¬ËùÒԸ÷´Ó¦´ïµ½Æ½ºâʱSO2µÄת»¯ÂÊÊÇ0.9mol/1mol¡Á100%=90%£»
£¨5£©»ìºÏÆøÌåÖеÄÈýÑõ»¯ÁòÓëBaCl2ÈÜÒº·´Ó¦Éú³ÉÁòËá±µ³Áµí¡£Í¬Àí¿É¼ÆËã³öÉú³ÉÈýÑõ»¯ÁòµÄÎïÖʵÄÁ¿ÊÇ0.9mol£¬Æä5%ÓëBaCl2ÈÜÒº·´Ó¦£¬Éú³ÉÁòËá±µ0.045mol£¬ÖÊÁ¿ÊÇ10.485g£¬±£ÁôһλСÊýÊÇ10.5g¡£
¿¼µã£º¿¼²é¶þÑõ»¯ÁòµÄ´ß»¯Ñõ»¯·´Ó¦£¬·´Ó¦ËÙÂʵļÆËãÓëÅжϣ¬»¯Ñ§·´Ó¦µÄÓйؼÆËã

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ÓÃͼÖÐËùʾʵÑé×°ÖÃÖ¤Ã÷Ñõ»¯Í­ÄܼӿìÔ¼7%µÄË«ÑõË®µÄ·Ö½â²¢Óë¶þÑõ»¯Ã̵Ĵ߻¯Ð§¹û½øÐбȽÏ(¼´±È½Ï·´Ó¦ËÙÂÊ)¡£ÓÃͼʾװÖòâÁ¿²úÉúÆøÌåµÄÌå»ý£¬ÆäËû¿ÉÄÜÓ°ÏìʵÑéµÄÒòËؾùÒѺöÂÔ£¬Ïà¹ØÊý¾ÝÈçÏ£º

£¨1£©´ÓʵÑéÔ­ÀíÀ´¿´£¬ÊµÑéÖеġ°´ý²âÊý¾Ý¡±¿ÉÒÔÖ¸¡¢Ò²¿ÉÒÔÖ¸                             ¡£
£¨2£©ÊµÑéʱÆøÌåÊÕ¼¯ÔÚBÖУ¬BÒÇÆ÷Ãû³ÆÊÇ__________¡£ÈôÒª¼ìÑé²úÉúµÄÆøÌåÊÇO2£¬´ýÆøÌåÊÕ¼¯½áÊø       ºó£¬Óõ¯»É¼Ð¼ÐסB϶ËÈ齺¹Ü£¬´ò¿ªµ¥¿×ÏðƤÈû£¬                   ¡£

£¨3£©ÎªÌ½¾¿CuOÔÚʵÑé¢ÚÖÐÊÇ·ñÆð´ß»¯×÷Ó㬳ýÓë¢Ù±È½ÏÍ⣬»¹Ðè²¹×öÏÂÁÐʵÑé²»±Øд¾ßÌå²½Öè)£ºa£®Ö¤Ã÷CuOµÄ»¯Ñ§ÐÔÖÊûÓб仯£¬b£®                   ¡£
£¨4£©ÎªÖ¤Ã÷Ñõ»¯Í­µÄ»¯Ñ§ÐÔÖÊÔÚ¼ÓÈëË«ÑõˮǰºóûÓз¢Éú¸Ä±ä£¬ÄãÉè¼ÆÑéÖ¤µÄʵÑéÊÇ                   ¡£
£¨5£©ÊµÑ鿪ʼʱ,µ±ÍùÈÝÆ÷ÖмÓÈëÒ»¶¨Á¿µÄË«ÑõË®ºó,ÓÉÓÚ¶Ìʱ¼äÄÚ²úÉú´óÁ¿ÆøÌå,·ÖҺ©¶·ÄÚµÄÒºÌå²»ÄÜ˳ÀûÁ÷ÏÂ,ΪÁ˽â¾öÕâ¸öÎÊÌâ,Äã²ÉÈ¡µÄ´ëÊ©ÊÇ                             ;ÔÚ²âÁ¿Éú³ÉµÄÆøÌåÌå»ýʱ,³ýÁËҪעÒâÊÓÏßÓë°¼ÒºÃæÏàƽÒÔÍâ,»¹Ó¦×¢Òâ                                

£¨±¾Ìâ16·Ö£©
£¨1£©ÎªÁ˼õÇáÆû³µÎ²ÆøÔì³ÉµÄ´óÆøÎÛȾ£¬ÈËÃÇ¿ªÊ¼Ì½Ë÷ÀûÓÃNOºÍCOÔÚÒ»¶¨Ìõ¼þÏÂת»¯ÎªÁ½ÖÖÎÞ¶¾ÆøÌåEºÍFµÄ·½·¨(ÒÑÖª¸Ã·´Ó¦¡÷H<0). ÔÚ2 LÃܱÕÈÝÆ÷ÖмÓÈëÒ»¶¨Á¿NOºÍCO£¬µ±Î¶ȷֱðÔÚT1ºÍT2ʱ£¬²âµÃ¸÷ÎïÖÊƽºâʱÎïÖʵÄÁ¿ÈçÏÂ±í£º

          ÎïÖÊ
T/¡æ     n/mol
NO
CO
E
F
³õʼ
0.100
0.100
0
0
T1
0.020
0.020
0.080
0.040
T2
0.010
0.010
0.090
0.045
 
¢ÙÇë½áºÏÉϱíÊý¾Ý£¬Ð´³öNOÓëCO·´Ó¦µÄ»¯Ñ§·½³Ìʽ                       .
¢Ú¸ù¾Ý±íÖÐÊý¾ÝÅжϣ¬Î¶ÈT1ºÍT2µÄ¹ØϵÊÇ(ÌîÐòºÅ)__________¡£
A£®T1>T2         B£®T1<T2   C£®T1=T2      D£®ÎÞ·¨±È½Ï
£¨2£©ÒÑÖª£º4NH3(g) + 3O2(g) = 2N­­2(g) + 6H2O(g);   ¦¤H=  - 1266.8 kJ/mol
N2(g) + O2(g) =" 2NO(g)" ;  ¦¤H =" +" 180.5kJ/mol£¬
Ôò°±´ß»¯Ñõ»¯µÄÈÈ»¯Ñ§·½³ÌʽΪ________________________________________¡£
£¨3£©500¡æÏ£¬ÔÚA¡¢BÁ½¸öÈÝÆ÷Öоù·¢ÉúºÏ³É°±µÄ·´Ó¦¡£¸ô°å¢ñ¹Ì¶¨²»¶¯£¬»îÈû¢ò¿É×ÔÓÉÒƶ¯¡£

µ±ºÏ³É°±ÔÚÈÝÆ÷BÖдïƽºâʱ£¬²âµÃÆäÖк¬ÓÐ1.0molN2£¬0.4molH2£¬0.4molNH3£¬´ËʱÈÝ»ýΪ2.0L¡£Ôò´ËÌõ¼þϵÄƽºâ³£ÊýΪ___________£»±£³ÖζȺÍѹǿ²»±ä£¬Ïò´ËÈÝÆ÷ÖÐͨÈë0.36molN2£¬Æ½ºâ½«___________£¨Ìî¡°ÕýÏò¡±¡¢¡°ÄæÏò¡±»ò¡°²»¡±£©Òƶ¯¡£
£¨4£©ÔÚÏàͬµÄÃܱÕÈÝÆ÷ÖУ¬Ó÷½·¨¢òºÍ·½·¨¢óÖƵõÄÁ½ÖÖCu2O·Ö±ð½øÐд߻¯·Ö½âË®µÄʵÑ飺
 ¦¤H >0
Ë®ÕôÆøµÄŨ¶È£¨mol¡¤L£­1£©Ëæʱ¼ät (min)±ä»¯ÈçÏÂ±í£º
ÐòºÅ
ζÈ
0
10
20
30
40
50
¢Ù
T1
0.050
0.0492
0.0486
0.0482
0.0480
0.0480
¢Ú
T1
0.050
0.0488
0.0484
0.0480
0.0480
0.0480
¢Û
T2
0.10
0.094
0.090
0.090
0.090
0.090
¿ÉÒÔÅжϣºÊµÑé¢ÙµÄÇ°20 minµÄƽ¾ù·´Ó¦ËÙÂÊ ¦Í(O2)£½              £»´ß»¯¼ÁµÄ´ß»¯Ð§ÂÊ£ºÊµÑé¢Ù   ÊµÑé¢Ú£¨Ìî¡°>¡±¡¢¡°<¡±£©¡£
£¨5£©×îÐÂÑо¿·¢ÏÖ£¬ÓøôĤµç½â·¨¿ÉÒÔ´¦Àí¸ßŨ¶ÈÒÒÈ©·ÏË®¡£Ô­Àí£ºÊ¹ÓöèÐԵ缫ºÍÒÒÈ©-Na2SO4ÈÜҺΪµç½âÖÊÈÜÒº£¬ÒÒÈ©·Ö±ðÔÚÒõ¡¢Ñô¼«×ª»¯ÎªÒÒ´¼ºÍÒÒËá¡£
×Ü·´Ó¦Îª:2CH3CHO+H2OCH3CH2OH+CH3COOH¡£
¹ý³ÌÖУ¬Á½¼«³ý·Ö±ðÉú³ÉÒÒËáºÍÒÒ´¼Í⣬¾ù²úÉúÎÞÉ«ÆøÌ壬Ñô¼«µç¼«·´Ó¦·Ö±ðΪ£º
4OH£­-4e£­¨TO2¡ü+2H2O£»                                   ¡£

£¨15·Ö£©Ì«ÑôÄܵç³ØÊÇÀûÓùâµçЧӦʵÏÖÄÜÁ¿±ä»¯µÄÒ»ÖÖÐÂÐÍ×°Öã¬Ä¿Ç°¶à²ÉÓõ¥¾§¹èºÍ¶à¾§¹è×÷Ϊ»ù´¡²ÄÁÏ¡£¸ß´¿¶ÈµÄ¾§Ìå¹è¿Éͨ¹ýÒÔÏ·´Ó¦»ñµÃ£º
·´Ó¦¢Ù£¨ºÏ³É¯£©£º
·´Ó¦¢Ú£¨»¹Ô­Â¯£©£º
ÓйØÎïÖʵķеãÈçϱíËùʾ£º

ÎïÖÊ
 
BCl3
 
PCl3
 
SiCl4
 
AsCl3
 
AlCl3
 
SiHCl3
 
·Ðµã
 
12£®1
 
73£®5
 
57£®0
 
129£®4
 
180£¨Éý»ª£©
 
31£®2
 
Çë»Ø´ðÒÔÏÂÎÊÌ⣺
£¨1£©Ì«ÑôÄܵç³ØµÄÄÜÁ¿×ª»¯·½Ê½Îª         £»ÓɺϳɯÖеõ½µÄSiHCl3ÍùÍù»ìÓÐÅð¡¢Áס¢Éé¡¢ÂÁµÈÂÈ»¯ÎïÔÓÖÊ£¬·ÖÀë³öSiHCl3µÄ·½·¨ÊÇ       ¡£
£¨2£©¶ÔÓÚÆøÏà·´Ó¦£¬ÓÃij×é·Ö(B)µÄƽºâѹǿ(PB)´úÌæÎïÖʵÄÁ¿Å¨¶È(cB)Ò²¿É±íʾƽºâ³£Êý£¨¼Ç×÷KP£©£¬Ôò·´Ó¦¢ÙµÄKP£½             £»
£¨3£©¶ÔÓÚ·´Ó¦¢Ú£¬ÔÚ0£®1MpaÏ£¬²»Í¬Î¶ȺÍÇâÆøÅä±È£¨H2/SiHCl3£©¶ÔSiHCl3Ê£ÓàÁ¿µÄÓ°ÏìÈçϱíËùʾ£º

¢Ù¸Ã·´Ó¦µÄ¡÷H2       0£¨Ìî¡°>¡±¡¢¡°<¡±¡¢¡°=¡±£©
¢Ú°´ÇâÆøÅä±È5:1ͶÈ뻹ԭ¯ÖУ¬·´Ó¦ÖÁ4minʱ²âµÃHClµÄŨ¶ÈΪ0£®12mol¡¤L¡ª1£¬ÔòSiHCl3ÔÚÕâ¶Îʱ¼äÄڵķ´Ó¦ËÙÂÊΪ               ¡£
¢Û¶ÔÉϱíµÄÊý¾Ý½øÐзÖÎö£¬ÔÚζȡ¢Åä±È¶ÔÊ£ÓàÁ¿µÄÓ°ÏìÖУ¬»¹Ô­Â¯Öеķ´Ó¦Î¶ÈÑ¡ÔñÔÚ1100¡æ£¬¶ø²»Ñ¡Ôñ775¡æ£¬ÆäÖеÄÒ»¸öÔ­ÒòÊÇÔÚÏàͬÅä±ÈÏ£¬Î¶ȶÔSiHCl3 Ê£ÓàÁ¿µÄÓ°Ï죬Çë·ÖÎöÁíÒ»Ô­ÒòÊÇ              ¡£
£¨4£©¶ÔÓÚ·´Ó¦¢Ú£¬ÔÚ1100¡æÏ£¬²»Í¬Ñ¹Ç¿ºÍÇâÆøÅä±È£¨H2/SiHCl3£©¶ÔSiHCl3Ê£ÓàÁ¿µÄÓ°ÏìÈçͼ27¡ª1Ëùʾ£º

¢Ù ͼÖÐP1        P2£¨Ìî¡°>¡±¡¢¡°<¡±¡¢¡°=¡±£©
¢ÚÔÚͼ27¡ª2Öл­³öÇâÆøÅä±ÈÏàͬÇé¿öÏ£¬1200¡æºÍ1100¡æµÄζÈÏ£¬ÏµÍ³ÖÐSiHCl3Ê£ÓàÁ¿Ëæѹǿ±ä»¯µÄÁ½Ìõ±ä»¯Ç÷ÊÆʾÒâͼ¡£

£¨15·Ö£©¡°µÍ̼ѭ»·¡±ÒÑÒýÆð¸÷¹ú¼ÒµÄ¸ß¶ÈÖØÊÓ£¬¶øÈçºÎ½µµÍ´óÆøÖÐCO2µÄº¬Á¿ºÍÓÐЧµØ¿ª·¢ÀûÓÃCO2Õý³ÉΪ»¯Ñ§¼ÒÑо¿µÄÖ÷Òª¿ÎÌâ¡£
(l)Óõ绡·¨ºÏ³ÉµÄ´¢ÇâÄÉÃ×̼¹Ü³£°éÓдóÁ¿µÄ̼ÄÉÃ׿ÅÁ££¨ÔÓÖÊ£©£¬ÕâÖÖ¿ÅÁ£¿ÉÓÃÈçÏÂÑõ»¯·¨Ìá´¿£¬ÇëÍê³É¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º

£¨2£©½«²»Í¬Á¿µÄCO(g)ºÍH2O(g)·Ö±ðͨÈëµ½Ìå»ýΪ2LµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬½øÐз´Ó¦£¬µÃµ½ÈçÏÂÁ½×éÊý¾Ý£º

¢ÙʵÑé2Ìõ¼þÏÂƽºâ³£ÊýK= __________¡£
¢ÚʵÑé3£¬Èô900¡æʱ£¬ÔÚ´ËÈÝÆ÷ÖмÓÈëCO¡¢H2O¡¢CO2¡¢H2¾ùΪ1mol£¬Ôò´Ëʱ
____________£¨Ìî¡°<¡±£¬¡°>¡±£¬¡°=¡±£©¡£
¢ÛÓÉÁ½×éʵÑé½á¹û£¬¿ÉÅжϸ÷´Ó¦µÄÕý·´Ó¦¡÷H_____________0£¨Ìî¡°<¡±£¬¡®¡®>¡±£¬¡°=¡±£©¡£
£¨3£©¼ºÖªÔÚ³£Î³£Ñ¹Ï£º

д³ö¼×´¼²»ÍêȫȼÉÕÉú³ÉÒ»Ñõ»¯Ì¼ºÍҺ̬ˮµÄÈÈ»¯Ñ§·½³Ìʽ£º________________________

¢ÙÒÑÖª¸Ã·´Ó¦µÄ¡÷H>0£¬¼òÊö¸ÃÉèÏëÄÜ·ñʵÏÖµÄÒÀ¾Ý£º________¡£
¢ÚÄ¿Ç°£¬ÔÚÆû³µÎ²ÆøϵͳÖÐ×°Öô߻¯×ª»¯Æ÷¿É¼õÉÙCOºÍNOµÄÎÛȾ£¬Æ仯ѧ·´Ó¦·½³ÌʽΪ__________¡£
£¨5£©CO2ÔÚ×ÔÈ»½çÑ­»·Ê±¿ÉÓëCaCO3·´Ó¦£¬CaCO3ÊÇÒ»ÖÖÄÑÈÜÎïÖÊ£¬ÆäKsp=2.8¡Á 10¡£CaCl2ÈÜÒºÓëNa2CO3ÈÜÒº»ìºÏ¿ÉÐγÉCaCO3³Áµí£¬ÏÖ½«µÈÌå»ýµÄCaCl2ÈÜÒºÓëNa2CO3ÈÜÒº»ìºÏ£¬ÈôNa2CO3ÈÜÒºµÄŨ¶ÈΪ1¡Á10mol/L£¬ÔòÉú³É³ÁµíËùÐèCaCl2ÈÜÒºµÄ×îСŨ¶ÈΪ__________mol/L¡£

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø