ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÊµÑéÊÒÖƱ¸±½¼×´¼ºÍ±½¼×ËáµÄ»¯Ñ§Ô­ÀíÊÇ

ÒÑÖª±½¼×È©Ò×±»¿ÕÆøÑõ»¯£¬±½¼×´¼µÄ·ÐµãΪ205.3 ¡æ£¬±½¼×ËáµÄÈÛµãΪ121.7 ¡æ£¬·ÐµãΪ249 ¡æ£¬Èܽâ¶ÈΪ0.34 g£»ÒÒÃѵķеãΪ34.8 ¡æ£¬ÄÑÈÜÓÚË®¡£ÖƱ¸±½¼×´¼ºÍ±½¼×ËáµÄÖ÷Òª¹ý³ÌÈçÏÂËùʾ£º

ÊÔ¸ù¾ÝÉÏÊöÐÅÏ¢£¬ÅжÏÒÔÏÂ˵·¨´íÎóµÄÊÇ(¡¡¡¡)

A. ²Ù×÷¢ñÊÇÝÍÈ¡·ÖÒº

B. ÒÒÃÑÈÜÒºÖÐËùÈܽâµÄÖ÷Òª³É·ÖÊDZ½¼×´¼

C. ²Ù×÷¢òÕôÁóËùµÃ²úÆ·¼×ÊDZ½¼×´¼

D. ²Ù×÷¢ó¹ýÂ˵õ½²úÆ·ÒÒÊDZ½¼×Ëá¼Ø

¡¾´ð°¸¡¿D

¡¾½âÎö¡¿ÊÔÌâ·ÖÎö£ºA¡¢´Ó¹ý³ÌÉÏ¿´²Ù×÷¢ñµÃµ½ÒÒÃÑÈÜÒººÍË®ÈÜÒº£¬ÕâÁ½¸öÊÇ»¥²»ÏàÈܵÄÒºÌ壬²ÉÓÃÝÍÈ¡ºÍ·ÖÒºµÄ·½·¨£¬ÕýÈ·£»B¡¢±½¼×È©ÔÚÇâÑõ»¯¼ØÖÐÉú³É±½ÒÒ´¼ºÍ±½¼×Ëá¼Ø£¬¸ù¾ÝÖƱ¸±½¼×ËáºÍ±½ÒÒ´¼µÄÖƱ¸¹ý³Ì£¬ÒÒÃÑÈÜÒºÈܽâµÄÖ÷Òª³É·ÖÊDZ½¼×´¼£¬ÕýÈ·£»C¡¢²Ù×÷¢òÀûÓÃÁ˱½¼×´¼µÄ·ÐµãΪ205.3¡æ£¬ÒÒÃѵķеãΪ34.8¡æ£¬Á½Õß»¥ÈÜ£¬ÀûÓ÷е㲻ͬ£¬²ÉÓÃÕôÁóµÄ·½·¨µÃµ½ÒÒÃѺͱ½ÒÒ´¼£¬ÕýÈ·£»D¡¢²Ù×÷¢óË®ÈÜÒºÖмÓÈëÑÎËᣬ±½¼×Ëá¼Øת±ä³É±½¼×Ëᣬ±½¼×Ëá²»ÈÜÓÚË®£¬²ÉÓùýÂ˵ķ½·¨½øÐзÖÀ룬´íÎó¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿ÎªÓÐЧÌáÉý¿ÕÆøÖÊÁ¿£¬¹ú¼ÒÇ¿ÖƸ÷ȼúÆóÒµÒª¶ÔȼúÑÌÆø½øÐÐÍÑÁò¡¢ÍÑÏõ´¦ÀíºóÅÅ·Å¡£»Ø´ðÏÂÁÐÎÊÌâ:

(1)ȼúÑÌÆø¡°ÍÑÏõ¡±ÖÐÉæ¼°µ½µÄ²¿·Ö·´Ó¦ÈçÏÂ:

a.4NH3(g)+6NO(g) 5N2(g)+6H2O(g) ¡÷H 1

b.4NH3(g)+5O2(g) 4NO(g)+6 H2O(g)¡÷H2 =-925kJ¡¤mol-1

c.N2(g)+O2(g) 2NO(g)¡÷H 3=+175kJ¡¤mol-1

Ôò¡÷H 1=_____kJ/mol¡£

(2)ÏòijºãÈÝÃܱÕÈÝÆ÷ÖмÓÈë2molNH3¡¢3molNO£¬ÔÚÊʵ±Ìõ¼þÏ·¢Éú(1)Öз´Ó¦a£¬·´Ó¦¹ý³ÌÖÐNOµÄƽºâת»¯ÂÊËæζÈT¡¢Ñ¹Ç¿pµÄ±ä»¯ÇúÏßÈçͼËùʾ:

¢Ùp1________p2(Ìî¡°>¡±¡°<¡±»ò¡°=¡±)¡£

¢ÚÈôÔÚѹǿΪ¼Óp2¡¢Î¶ÈΪ600¡æʱ£¬´ïµ½Æ½ºâʱ·´Ó¦ÌåϵÄÚ»¯Ñ§ÄܽµµÍÁË300kJ£¬ÔòNOµÄת»¯ÂÊΪ_______________________¡£

¢Û600¡æʱ£¬ÏÂÁи÷ÏîÊý¾ÝÖÐÄܱíÃ÷¸Ã·´Ó¦ÒÑ´ïµ½»¯Ñ§Æ½ºâ״̬µÄÊÇ______¡£

a.Ë®ÓëNOµÄÉú³ÉËÙÂÊÏàµÈ b.»ìºÏÆøÌåµÄÃܶȱ£³Ö²»±ä

c. NH3¡¢NOÉú³ÉËÙÂʱÈΪ5:4 d.ÈÝÆ÷µÄ×Üѹǿ±£³Ö²»±ä

(3)½«2molN2¡¢3molH2O¡¢0.5molNOµÄ»ìºÏÆøÌåÖÃÓÚx¡¢y¡¢zÈý¸öÈÝ»ýÏàͬµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬¿ØÖÆÊʵ±µÄÌõ¼þ·¢Éú·´Ó¦£¬·´Ó¦¹ý³ÌÖÐc(NO)Ëæʱ¼äµÄ±ä»¯ÈçͼËùʾ¡£

¢ÙÔÚ½¨Á¢Æ½ºâµÄ¹ý³ÌÖУ¬Èý¸öÈÝÆ÷Öз´Ó¦ËÙÂʵÄÏà¶Ô´óСΪ______________¡£

¢ÚÓëyÈÝÆ÷Öз´Ó¦Ïà±È£¬zÈÝÆ÷Öз´Ó¦¸Ä±äµÄÌõ¼þ¼°ÅжÏÒÀ¾Ý·Ö±ðÊÇ_____________________¡£

¡¾ÌâÄ¿¡¿µØϺÚ×÷·»Óò¡ËÀÖíÈâëçÖƵÄÀ°ÈâÍùÍùº¬ÓдóÁ¿µÄϸ¾ú,¿ÉÀûÓá°Ó«¹âËØ¡ª¡ªÓ«¹âËØøÉúÎï·¢¹â·¨¡±¶ÔÊг¡ÖÐÀ°È⺬ϸ¾ú¶àÉÙ½øÐмì²â:¢Ù½«À°ÈâÑÐÄ¥ºóÀëÐÄ´¦Àí,È¡Ò»¶¨Á¿ÉÏÇåÒº·ÅÈë·Ö¹â¹â¶È¼Æ(²â¶¨·¢¹âÇ¿¶ÈµÄÒÇÆ÷)·´Ó¦ÊÒÄÚ,¼ÓÈëÊÊÁ¿µÄÓ«¹âËغÍÓ«¹âËØø,ÔÚÊÊÒËÌõ¼þϽøÐз´Ó¦;¢Ú¼Ç¼·¢¹âÇ¿¶È²¢¼ÆËãATPº¬Á¿;¢Û²âËã³öϸ¾úÊýÁ¿¡£·ÖÎö²¢»Ø´ðÏÂÁÐÎÊÌâ:

(1)Ó«¹âËؽÓÊÜ____ÌṩµÄÄÜÁ¿ºó¾Í±»¼¤»î,ÔÚÓ«¹âËØøµÄ×÷ÓÃÏÂÐγÉÑõ»¯Ó«¹âËز¢ÇÒ·¢³öÓ«¹â¡£¸ù¾Ý·¢¹âÇ¿¶È¿ÉÒÔ¼ÆËã³öÉúÎï×éÖ¯ÖÐATPµÄº¬Á¿,Ô­ÒòÊÇ·¢¹âÇ¿¶ÈÓëATPº¬Á¿³É____________(Õý±È/·´±È);¸ù¾ÝATPº¬Á¿½ø¶ø²âËã³öϸ¾úÊýÁ¿µÄÒÀ¾ÝÊÇ:ÿ¸öϸ¾úϸ°ûÖÐATPº¬Á¿__________¡£

(2)¡°Ó«¹âËØ¡ª¡ªÓ«¹âËØøÉúÎï·¢¹â·¨¡±ÖÐÉæ¼°µÄÄÜÁ¿×ª»»ÊÇ_________;ÉúÎïϸ°ûÖÐATPµÄË®½âÒ»°ãÓë________(ÎüÄÜ·´Ó¦»ò·ÅÄÜ·´Ó¦)ÏàÁªÏµ¡£

(3)Ñо¿ÈËÔ±Óò»Í¬Ìõ¼þ´¦ÀíÓ«¹âËØøºó,²â¶¨Ã¸Å¨¶ÈÓë·¢¹âÇ¿¶ÈÈçͼËùʾ¡£

ÆäÖиßŨ¶ÈÑÎÈÜÒº¾­Ï¡Êͺóø»îÐÔ¿ÉÒÔ»Ö¸´,¸ßκÍHg2+´¦Àíºóø»îÐÔ²»¿É»Ö¸´¡£ÈôÒª½ÚÊ¡Ó«¹âËØøµÄÓÃÁ¿,¿ÉÒÔʹÓÃ____´¦Àí;Hg2+´¦Àíºóø»îÐÔ½µµÍ¿ÉÄÜÊÇÒòΪ______________________________¡£

¡¾ÌâÄ¿¡¿ÊµÑéÊÒ³£ÀûÓü×È©·¨²â¶¨(NH4)2SO4ÑùÆ·ÖеªµÄÖÊÁ¿·ÖÊý£¬Æä·´Ó¦Ô­ÀíΪ£º4NH4£«+6HCHO=3H£«+6H2O+(CH2)6N4H£« £¬È»ºóÓÃNaOH±ê×¼ÈÜÒºµÎ¶¨·´Ó¦Éú³ÉµÄËᣬijÐËȤС×éÓü×È©·¨½øÐÐÁËÈçÏÂʵÑ飺

²½ÖèI ³ÆÈ¡ÑùÆ·1.500g¡£

²½ÖèII ½«ÑùÆ·Èܽâºó£¬ÍêȫתÒƵ½250 mLÈÝÁ¿Æ¿ÖУ¬¶¨ÈÝ£¬³ä·ÖÒ¡ÔÈ¡£

²½Öè¢ó ÒÆÈ¡25.00mLÑùÆ·ÈÜÒºÓÚ250mL׶ÐÎÆ¿ÖУ¬¼ÓÈë10mL20£¥µÄÖÐÐÔ¼×È©ÈÜÒº£¬Ò¡ÔÈ¡¢¾²ÖÃ5 minºó,¼ÓÈë1~2µÎ·Ó̪ÊÔÒº£¬ÓÃNaOH±ê×¼ÈÜÒºµÎ¶¨ÖÁÖյ㡣°´ÉÏÊö²Ù×÷·½·¨ÔÙÖظ´2´Î¡£

(1)¸ù¾Ý²½Öè¢óÌî¿Õ£º

¢Ùµ½´ïµÎ¶¨ÖÕµãʱ£¬¸©ÊÓ¼îʽµÎ¶¨¹Ü¶ÁÊý£¬Ôò²âµÃÑùÆ·ÖеªµÄÖÊÁ¿·ÖÊý________(Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족)¡£

¢Ú´ïµ½µÎ¶¨ÖÕµãʱ£¬·¢ÏÖ¼îʽµÎ¶¨¹ÜµÄ¼â¶ËÓÐÆøÅÝ£¬ÔòµÎ¶¨Ê±ÓÃÈ¥NaOH±ê×¼ÈÜÒºµÄÌå»ý__________ (Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족)

¢ÛµÎ¶¨Ê±±ßµÎ±ßÒ¡¶¯×¶ÐÎÆ¿£¬ÑÛ¾¦Ó¦¹Û²ì___________

¢ÜµÎ¶¨´ïµ½ÖÕµãÅжϣº__________________________________________

(2)µÎ¶¨½á¹ûÈçϱíËùʾ£º

ÈôNaOH±ê×¼ÈÜÒºµÄŨ¶ÈΪ0.1010mol¡¤L£­1Ôò¸ÃÑùÆ·ÖеªµÄÖÊÁ¿·ÖÊýΪ______________

¡¾ÌâÄ¿¡¿ÖÎÀíSO2¡¢CO¡¢NOxÎÛȾÊÇ»¯Ñ§¹¤×÷ÕßÑо¿µÄÖØÒª¿ÎÌâ¡£

¢ñ.ÁòË᳧´óÁ¿Åŷź¬SO2µÄβÆø»á¶Ô»·¾³Ôì³ÉÑÏÖØΣº¦¡£

£¨1£©¹¤ÒµÉÏ¿ÉÀûÓ÷ϼîÒº£¨Ö÷Òª³É·ÖΪNa2CO3£©´¦ÀíÁòË᳧βÆøÖеÄSO2£¬µÃµ½Na2SO3ÈÜÒº£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ__________________________________¡£

¢ò.Á¤Çà»ìÄýÍÁ¿É×÷Ϊ·´Ó¦£»2CO(g)+O2(g)2CO2(g)µÄ´ß»¯¼Á¡£Í¼¼×±íʾÔÚÏàͬµÄºãÈÝÃܱÕÈÝÆ÷¡¢ÏàͬÆðʼŨ¶È¡¢Ïàͬ·´Ó¦Ê±¼ä¶ÎÏÂ,ʹÓÃͬÖÊÁ¿µÄ²»Í¬Á¤Çà»ìÄýÍÁ£¨¦ÁÐÍ¡¢¦ÂÐÍ£©´ß»¯Ê±£¬COµÄת»¯ÂÊÓëζȵĹØϵ¡£

£¨2£©a¡¢b¡¢c¡¢d ËĵãÖУ¬´ïµ½Æ½ºâ״̬µÄÊÇ__________________________________¡£

£¨3£©ÒÑÖªcµãʱÈÝÆ÷ÖÐO2Ũ¶ÈΪ0.02 mol/L£¬Ôò50¡æʱ£¬ÔÚ¦ÁÐÍÁ¤Çà»ìÄýÍÁÖÐCOת»¯·´Ó¦µÄƽºâ³£ÊýK=____________£¨Óú¬xµÄ´úÊýʽ±íʾ£©¡£

£¨4£©ÏÂÁйØÓÚͼ¼×µÄ˵·¨ÕýÈ·µÄÊÇ_____________¡£

A.COת»¯·´Ó¦µÄƽºâ³£ÊýK(a)

B.ÔÚ¾ùδ´ïµ½Æ½ºâ״̬ʱ£¬Í¬ÎÂϦÂÐÍÁ¤Çà»ìÄýÍÁÖÐCOת»¯ËÙÂʱȦÁÐÍÒª´ó

C.bµãʱCOÓëO2·Ö×ÓÖ®¼ä·¢ÉúÓÐЧÅöײµÄ¼¸ÂÊÔÚÕû¸öʵÑé¹ý³ÌÖÐ×î¸ß

D.eµãת»¯ÂʳöÏÖÍ»±äµÄÔ­Òò¿ÉÄÜÊÇζÈÉý¸ßºó´ß»¯¼Áʧȥ»îÐÔ

¢ó.ijº¬îÜ´ß»¯¼Á¿ÉÒÔ´ß»¯Ïû³ý²ñÓͳµÎ²ÆøÖеÄ̼ÑÌ(C)ºÍNOx¡£²»Í¬Î¶ÈÏ£¬½«Ä£ÄâβÆø£¨³É·ÖÈçϱíËùʾ£©ÒÔÏàͬµÄÁ÷ËÙͨ¹ý¸Ã´ß»¯¼Á£¬²âµÃËùÓвúÎCO2¡¢N2¡¢N2O£©ÓëNOµÄÏà¹ØÊý¾Ý½á¹ûÈçͼÒÒËùʾ¡£

Ä£ÄâβÆø

ÆøÌå(10mol)

̼ÑÌ

NO

O2

He

ÎïÖʵÄÁ¿(mol)

0.025

0.5

9.475

n

£¨5£©375¡æʱ£¬²âµÃÅųöµÄÆøÌåÖк¬0.45 molO2ºÍ0.0525 mol CO2£¬ÔòYµÄ»¯Ñ§Ê½Îª______________¡£

£¨6£©ÊµÑé¹ý³ÌÖвÉÓÃNOÄ£ÄâNOx£¬¶ø²»²ÉÓÃNO2µÄÔ­ÒòÊÇ____________________________¡£

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

¾«Ó¢¼Ò½ÌÍø