¡¾ÌâÄ¿¡¿Á×ÊÇÈËÌ庬Á¿½Ï¶àµÄÔªËØÖ®Ò»£¬Á׵Ļ¯ºÏÎïÔÚÒ©ÎïÉú²úºÍÅ©Ò©ÖÆÔìµÈ·½ÃæÓÃ;·Ç³£¹ã·º¡£»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©»ù̬Á×Ô­×ӵĺËÍâµç×ÓÅŲ¼Ê½Îª____________________¡£

£¨2£©P4S3¿ÉÓÃÓÚÖÆÔì»ð²ñ,Æä·Ö×ӽṹÈçͼ¼×Ëùʾ¡£

¢ÙµÚÒ»µçÀëÄÜ£ºÁ×_____________Áò;µç¸ºÐÔ£ºÁ×_____________Áò(Ìî¡°>¡±»ò¡°<¡±)¡£

¢ÚP4S3·Ö×ÓÖÐÁòÔ­×ÓµÄÔÓ»¯¹ìµÀÀàÐÍΪ_____________¡£

¢Ûÿ¸öP4S3·Ö×ÓÖк¬¹Âµç×Ó¶ÔµÄÊýĿΪ______________¡£

£¨3£©N¡¢P¡¢As¡¢Sb¾ùÊǵÚVA×åµÄÔªËØ¡£

¢ÙÉÏÊöÔªËصÄÇ⻯ÎïµÄ·Ðµã¹ØϵÈçͼÒÒËùʾ£¬·Ðµã£ºPH3<NH3,ÆäÔ­ÒòÊÇ____________________;·Ðµã£ºPH3<AsH3<SbH3£¬ÆäÔ­ÒòÊÇ_____________________________________________________¡£

¢ÚijÖÖ´ÅÐÔµª»¯ÌúµÄ¾§°û½á¹¹Èçͼ±ûËùʾ,¸Ã»¯ºÏÎïµÄ»¯Ñ§Ê½Îª_________________¡£

£¨4£©Á×»¯ÂÁÈÛµãΪ2000¡æ£¬ËüÓ뾧Ìå¹è»¥ÎªµÈµç×ÓÌ壬Á×»¯ÂÁ¾§°û½á¹¹Èçͼ¶¡Ëùʾ¡£

¢ÙͼÖÐAµãºÍBµãµÄÔ­×Ó×ø±ê²ÎÊýÈçͼ¶¡Ëùʾ£¬ÔòCµãµÄÔ­×Ó×ø±ê²ÎÊýΪ_______________¡£

¢ÚÁ×»¯ÂÁ¾§ÌåµÄÃܶÈΪ¦Ñg¡¤cm-3ÓÃNA±íʾ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÊýÖµ£¬Ôò¸Ã¾§°ûÖоàÀë×î½üµÄÁ½¸öÂÁÔ­×ÓÖ®¼äµÄ¾àÀëΪ_____________________________cm¡£

¡¾´ð°¸¡¿ 1s22s22p63s23p3»ò[Ne]3s23p3 > < sp3 10 NH3·Ö×Ó¼ä´æÔÚ·Ö×Ó¼äÇâ¼ü Ïà¶Ô·Ö×ÓÖÊÁ¿²»¶ÏÔö´ó£¬·Ö×Ó¼ä×÷ÓÃÁ¦²»¶ÏÔöÇ¿ Fe3N (£¬£¬)

¡¾½âÎö¡¿ÊÔÌâ·ÖÎö£º±¾Ì⿼²éÎïÖʽṹÓëÐÔÖÊ£¬Éæ¼°Ô­×ÓºËÍâµç×ÓÅŲ¼Ê½µÄÊéд£¬µÚÒ»µçÀëÄܺ͵縺ÐԵıȽϣ¬Ô­×ÓÔÓ»¯·½Ê½µÄÅжϣ¬ÎïÖʷеã¸ßµÍµÄ±È½Ï£¬¾§Ì廯ѧʽµÄÈ·¶¨ÒÔ¼°¾§°ûµÄ¼ÆËã¡£

£¨1£©PµÄºËµçºÉÊýΪ15£¬PÔ­×ÓºËÍâµç×ÓÊýΪ15£¬¸ù¾Ý¹¹ÔìÔ­Àí£¬»ù̬PÔ­×ӵĺËÍâµç×ÓÅŲ¼Ê½Îª1s22s22p63s23p3»ò[Ne] 3s23p3¡£

£¨2£©¢ÙPÔ­×ӵļ۵ç×ÓÅŲ¼Ê½Îª3s23p3£¬SÔ­×ӵļ۵ç×ÓÅŲ¼Ê½Îª3s23p4£¬PÔ­×ÓµÄ3p´¦ÓÚ°ë³äÂú½ÏÎȶ¨£¬µÚÒ»µçÀëÄÜ£ºPS¡£·Ç½ðÊôÐÔ£ºPS£¬µç¸ºÐÔ£ºPS¡£

¢ÚÓɽṹ֪£¬P4S3ÖÐÿ¸öSÔ­×ÓÐγÉ2¸ö¦Ò¼ü£¬SÔ­×ÓÉÏ»¹ÓÐ2¶Ô¹Âµç×Ó¶Ô£¬ÔòSÔ­×ÓΪsp3ÔÓ»¯¡£

¢ÛÓɽṹ֪£¬Ã¿¸öPÔ­×ÓÐγÉ3¶Ô¹²Óõç×Ó¶Ô£¬Ã¿¸öPÔ­×ÓÉÏÓÐ1¶Ô¹Âµç×Ó¶Ô£»Ã¿¸öSÔ­×ÓÐγÉ2¶Ô¹²Óõç×Ó¶Ô£¬Ã¿¸öSÔ­×ÓÉÏÓÐ2¶Ô¹Âµç×Ó¶Ô£»Ã¿¸öP4S3Öк¬Óеŵç×Ó¶ÔÊýΪ41+32=10¡£

£¨3£©¢Ù·Ðµã£ºPH3NH3µÄÔ­ÒòÊÇ£ºNH3·Ö×Ó¼ä´æÔÚÇâ¼ü£¬PH3·Ö×Ӽ䲻´æÔÚÇâ¼ü¡£·Ðµã£ºPH3<AsH3<SbH3µÄÔ­ÒòÊÇ£ºPH3¡¢AsH3¡¢SbH3¶¼ÊôÓÚ·Ö×Ó¾§Ì壬Ïà¶Ô·Ö×ÓÖÊÁ¿ÒÀ´ÎÔö´ó£¬·Ö×Ó¼ä×÷ÓÃÁ¦ÒÀ´ÎÔöÇ¿£¬·ÐµãÉý¸ß¡£

¢ÚÓá°¾ù̯·¨¡±£¬Fe£º12+2+3=6£¬NÈ«ÔÚ¾§°ûÄÚ²¿£¬N£º2£¬N£¨Fe£©£ºN£¨N£©=6:2=3:1£¬»¯Ñ§Ê½ÎªFe3N¡£

£¨4£©¢Ù¶ÔÕÕ¾§°ûͼʾ£¬×ø±êϵÒÔ¼°A¡¢Bµã×ø±ê£¬Ñ¡AµãΪ²ÎÕյ㣬¹Û²ìCµãÔÚ¾§°ûÖÐλÖã¨Ìå¶Ô½ÇÏß´¦£©£¬ÓÉA¡¢Bµã×ø±ê¿ÉÒÔÍÆÖªCµã×ø±êΪ£¨£¬ £¬ £©¡£

¢ÚÓá°¾ù̯·¨¡±£¬1¸ö¾§°ûÖÐAl£º8+6=4£¬P£º4£¬¾§ÌåµÄ»¯Ñ§Ê½ÎªAlP£¬1mol¾§ÌåµÄÖÊÁ¿Îª58g£»É辧°ûµÄ±ß³¤Îªx£¬Ôò¾§°ûµÄÌå»ýΪx3£¬Ôò¦Ñg/cm3NA=58g£¬½âµÃx=cm£¬¾§°ûÖоàÀë×î½üµÄÁ½¸öÂÁÔ­×ÓÖ®¼äµÄ¾àÀëΪx=cm¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿³£ÎÂÏ£¬ÏÂÁÐÈÜÒºÖи÷×éÀë×ÓÒ»¶¨ÄÜ´óÁ¿¹²´æµÄÊÇ£¨ £©
A.¼ÓÈëÂÁ·ÛÓÐÇâÆøÉú³ÉµÄÈÜÒºÖУºMg2+ £¬ Cl- £¬ NO3- £¬ K+
B.³£ÎÂÏ£¬c(H+) =0.1 mol/LµÄÈÜÒºÖУºNa+ £¬ AlO2-¡¢S2-¡¢ SO32-
C.º¬ÓÐ0.1 mol/LHCO3-µÄÈÜÒº£ºNa+ £¬ Fe3+ £¬ NO3- £¬ SCN-
D. =0.1 mol/LµÄÈÜÒº£ºNa+ £¬ K+ £¬ CO32- £¬ NO3-

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ä³»¯Ñ§Ñо¿ÐÔѧϰС×éÕë¶ÔÔ­µç³ØÐγÉÌõ¼þ£¬Éè¼ÆÁËʵÑé·½°¸£¬½øÐÐÈçÏÂ̽¾¿¡£

(1)ÇëÌîдÓйØʵÑéÏÖÏ󲢵óöÏà¹Ø½áÂÛ¡£

񅧏

ʵÑé×°ÖÃ

ʵÑéÏÖÏó

1

п°ôÖð½¥Èܽ⣬±íÃæÓÐÆøÌåÉú³É£»Í­°ô±íÃæÎÞÏÖÏó

2

Á½Ð¿°ôÖð½¥Èܽ⣬±íÃæ¾ùÓÐÆøÌåÉú³É£»µçÁ÷¼ÆÖ¸Õ벻ƫת

3

Í­°ô±íÃæµÄÏÖÏóÊÇ______________________£¬µçÁ÷¼ÆÖ¸Õë___________________

¢Ùͨ¹ýʵÑé2ºÍ3£¬¿ÉµÃ³öÔ­µç³ØµÄÐγÉÌõ¼þÊÇ______________________________¡£

¢Úͨ¹ýʵÑé1ºÍ3£¬¿ÉµÃ³öÔ­µç³ØµÄÐγÉÌõ¼þÊÇ______________________________¡£

¢ÛÈô½«3×°ÖÃÖÐÁòËá»»³ÉÒÒ´¼£¬µçÁ÷¼ÆÖ¸Õ뽫²»·¢Éúƫת£¬´Ó¶ø¿ÉµÃ³öÔ­µç³ØÐγÉÌõ¼þÊÇ___________________¡£

(2)·Ö±ðд³öʵÑé3ÖÐZn°ôºÍCu°ôÉÏ·¢ÉúµÄµç¼«·´Ó¦Ê½£º

Zn°ô£º______________________________¡£

Cu°ô£º______________________________¡£

(3)ʵÑé3µÄµçÁ÷ÊÇ´Ó________°ôÁ÷³ö(Ìî¡°Zn¡±»ò¡°Cu¡±)£¬·´Ó¦¹ý³ÌÖÐÈôÓÐ0.4molµç×Ó·¢ÉúÁËתÒÆ£¬ÔòZnµç¼«ÖÊÁ¿¼õÇá___________g¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁз´Ó¦¼ÈÊôÓÚÑõ»¯»¹Ô­·´Ó¦£¬ÓÖÊÇÎüÈÈ·´Ó¦µÄÊÇ£¨ £©
A.Éúʯ»ÒÓëË®×÷ÓÃÖÆÊìʯ»Ò
B.×ÆÈȵÄľ̿ÓëCO2·´Ó¦
C.¼×ÍéÔÚÑõÆøÖеÄȼÉÕ·´Ó¦
D.Ba£¨OH£©28H2O¾§ÌåÓëNH4Cl¾§ÌåµÄ·´Ó¦

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁÐÊôÓÚ´¿¾»ÎïµÄÊÇ

A.µ¨·¯B.ÇâÑõ»¯Ìú½ºÌåC.ÑÎËáD.Ư°×·Û

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁÐÈÈ»¯Ñ§·½³ÌʽÖЦ¤H´ú±íȼÉÕÈȵÄÊÇ (¡¡¡¡)

A. CH4(g)£«O2(g)===2H2O(l)£«CO(g)¡¡¦¤H1

B. H2(g)£«O2(g)===H2O(g)¡¡¦¤H2

C. C2H5OH(l)£«3O2(g)===2CO2(g)£«3H2O(l)¡¡¦¤H3

D. 2CO(g)£«O2(g)===2CO2(g)¡¡¦¤H4

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÔÚ²»Í¬Î¶ÈÏ£¬Ë®ÈÜÒºÖÐc(H£«)Óëc(OH-)¹ØϵÈçͼËùʾ¡£ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ£¨ £©

A.ͼÖÐÎåµãKw¼äµÄ¹Øϵ£ºB£¾C£¾A=D=E
B.E µã¶ÔÓ¦µÄË®ÈÜÒºÖУ¬¿ÉÄÜÓÐNH4+¡¢Ba 2+¡¢Cl£­¡¢I£­´óÁ¿Í¬Ê±´æÔÚ
C.Èô0.1 mol/L µÄNaHA ÈÜҺˮÈÜÒºÖÐc(H£«)Óëc(OH£­)¹ØϵÈçͼD µãËùʾ£¬ÔòÈÜÒºÖÐÓУºc(HA£­)£¾c(OH£­)£¾c(A2£­ )£¾c(H2A)
D.ÏòNaHSO4ÈÜÒºÖеÎÈëBa(OH)2ÈÜÒº£¬µ±c(H£«)Óëc(OH£­)¹ØϵÈçͼA µãËùʾ£¬ÔòÈÜÒºÖз´Ó¦£º2H£«+SO42-+Ba2£«+2OH£­=BaSO4¡ý+2H2O

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÑÇÏõËáÄÆÊÇÒ»ÖÖÖØÒªµÄ»¯¹¤Ô­ÁÏ£¬Çë»Ø´ðÏÂÁÐÎÊÌâ:

I.¸ßÌúËá¼Ø(K2FeO4)ÈÜÒº³Ê×ϺìÉ«£¬ÏòÆäÖмÓÈë¹ýÁ¿ÑÇÏõËáÄƺó£¬ÈÜÒº×ϺìÉ«Öð½¥ÍÊÈ¥£¬²¢³öÏÖºìºÖÉ«³Áµí£¬Çëд³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ__________________¡£

I1.ʵÑéÊÒÄ£ÄâÏÂͼËùʾÁ÷³ÌÖƱ¸ÑÇÏõËáÄÆ:

ÒÑÖª: ¢ÙÑõ»¯¹ý³ÌÖУ¬¿ØÖÆ·´Ó¦ÒºÎ¶ÈÔÚ35~60¡æÌõ¼þÏÂÖ÷Òª·¢Éú·´Ó¦:

C6H12O6+ 12HNO3=3HOOC-COOH+9NO2¡ü+3NO¡ü+9H2O

¢ÚÇâÑõ»¯ÄÆÈÜÒºÎüÊÕNO ºÍNO2·¢Éú·´Ó¦:

NO+NO2+2NaOH=2NaNO2+H2O¡¢2NO2+2NaOH=NaNO3+NaNO2+H2O

£¨1£©Ä¾Ð¼µÄÖ÷Òª³É·ÖÊÇÏËάËØ£¬½áºÏÒÑÖªÐÅÏ¢¢Ù,ÄãÈÏΪÏòľмÖмÓÏ¡ÁòËáµÄ×÷ÓÃÊÇ_____ £¬Ñõ»¯¹ý³ÌÖз´Ó¦Î¶Ȳ»Ò˸ßÓÚ60¡æ£¬Ô­ÒòÊÇ______________¡£

£¨2£©²»×öÈκδ¦Àí£¬°´´Ë¹ý³Ì½øÐУ¬ÇâÑõ»¯ÄÆÈÜÒºÎüÊÕºóµÄÈÜÒºÖгýÁËOH-Í⻹ÓÐÁ½ÖÖÒõÀë×Ó£¬ÆäÖÐÒ»ÖÖÊÇNO2-£¬NO2-ÓëÁíÒ»ÖÖÒõÀë×ÓµÄÎïÖʵÄÁ¿Ö®±ÈΪ________¡£

£¨3£©×°ÖÃBÓÃÓÚÖƱ¸NaNO2£¬Ê¢ÑbµÄÊÔ¼Á³ýNaOH(aq)Í⣬»¹¿ÉÒÔÊÇ_____ (Ìî×Öĸ)¡£

A.NaCl(aq) B.Na2CO3(aq) C.NaNO3( aq )

¢ó.²â¶¨²úÆ·´¿¶È:

[ʵÑé²½Öè] ¢Ù׼ȷ³ÆÁ¿ag ²úÆ·Åä³É200mLÈÜÒº

¢Ú´Ó²½Öè¢ÙÅäÖƵÄÈÜÒºÖÐÒÆÈ¡20.00mL ¼ÓÈë׶ÐÎÆ¿ÖÐ

¢ÛÓÃc mol/LËáÐÔKMnO4ÈÜÒºµÎ¶¨ÖÁÖÕµã

¢ÜÖظ´ÒÔÉϲÙ×÷3 ´Î£¬ÏûºÄËáÐÔKMnO4ÈÜÒºµÄƽ¾ùÌå»ýΪVmL

£¨1£©×¶ÐÎÆ¿Öз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ____________¡£

£¨2£©²úÆ·ÖÐNaNO2 µÄ´¿¶ÈΪ______________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿´ß»¯»¹Ô­CO2Êǽâ¾öÎÂÊÒЧӦ¼°ÄÜÔ´ÎÊÌâµÄÖØÒªÊÖ¶ÎÖ®Ò»¡£Ñо¿±íÃ÷£¬ÔÚCu/ZnO´ß»¯¼Á´æÔÚÏ£¬ÔÚCO2ÖÐͨÈëH2£¬¶þÕß¿É·¢ÉúÒÔÏÂÁ½¸öƽÐз´Ó¦£º

·´Ó¦¢ñ CO2(g)+3H2(g)CH3OH(g)+H2O(g) ¦¤H1=-53.7 kJ¡¤mol-1¡¡

·´Ó¦¢ò CO2(g)+H2(g)CO(g)+H2O(g)¡¡ ¡¡¦¤H2=+41.2 kJ¡¤mol-1

ijʵÑéÊÒ¿ØÖÆÒ»¶¨µÄCO2ºÍH2³õʼͶÁϱȣ¬ÔÚÏàͬѹǿÏÂ,¾­¹ýÏàͬ·´Ó¦Ê±¼ä²âµÃÈçÏÂʵÑéÊý¾Ý£¨ÆäÖС°¼×´¼Ñ¡ÔñÐÔ¡±ÊÇָת»¯µÄCO2ÖÐÉú³É¼×´¼µÄ°Ù·Ö±È£©£º

·´Ó¦ÐòºÅ

T/K

´ß»¯¼Á

CO2ת»¯ÂÊ/%

¼×´¼Ñ¡ÔñÐÔ/%

¢Ù

543

Cu/ZnOÄÉÃ×°ô

12.3

42.3

¢Ú

543

Cu/ZnOÄÉÃ×Ƭ

10.9

72.7

¢Û

553

Cu/ZnOÄÉÃ×°ô

15.3

39.1

¢Ü

553

Cu/ZnOÄÉÃ×Ƭ

12.0

71.6

£¨1£©CO2µÄµç×ÓʽÊÇ_____________¡£

£¨2£©·´Ó¦¢ñµÄƽºâ³£Êý±í´ïʽÊÇK=______¡£

£¨3£©¶Ô±È¢ÙºÍ¢Û¿É·¢ÏÖ£ºÍ¬Ñù´ß»¯¼ÁÌõ¼þÏ£¬Î¶ÈÉý¸ß£¬CO2ת»¯ÂÊÉý¸ß£¬ ¶ø¼×´¼µÄÑ¡ÔñÐÔÈ´½µµÍ£¬Çë½âÊͼ״¼Ñ¡ÔñÐÔ½µµÍµÄ¿ÉÄÜÔ­Òò_______________£»

¶Ô±È¢Ù¡¢¢Ú¿É·¢ÏÖ£¬ÔÚͬÑùζÈÏ£¬²ÉÓÃCu/ZnOÄÉÃ×ƬʹCO2ת»¯ÂʽµµÍ£¬ ¶ø¼×´¼µÄÑ¡ÔñÐÔÈ´Ìá¸ß£¬Çë½âÊͼ״¼µÄÑ¡ÔñÐÔÌá¸ßµÄ¿ÉÄÜÔ­Òò____________¡£

£¨4£©ÓÐÀûÓÚÌá¸ßCO2ת»¯ÎªCH3OHƽºâת»¯ÂʵĴëÊ©ÓÐ____¡£

a£®Ê¹ÓÃCu/ZnOÄÉÃ×°ô×ö´ß»¯¼Á

b£®Ê¹ÓÃCu/ZnOÄÉÃ×Ƭ×ö´ß»¯¼Á

c£®µÍ·´Ó¦Î¶È

d£®Í¶ÁϱȲ»±ä,Ôö¼Ó·´Ó¦ÎïµÄŨ¶È

e£®Ôö´óCO2ºÍH2µÄ³õʼͶÁϱÈ

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸