¡¾ÌâÄ¿¡¿¸ß¯Á¶Ìú¹ý³ÌÖз¢Éú·´Ó¦£º Fe2O3(s)£«CO(g)Fe(s)£«CO2(g)£¬¸Ã·´Ó¦ÔÚ²»Í¬Î¶ÈϵÄƽºâ³£Êý¼ûÓÒ±í¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨ £©

ζÈT/¡æ

1000

1150

1300

ƽºâ³£ÊýK

4.0

3.7

3.5

A. ÓɱíÖÐÊý¾Ý¿ÉÅжϸ÷´Ó¦£º·´Ó¦ÎïµÄ×ÜÄÜÁ¿£¼Éú³ÉÎïµÄ×ÜÄÜÁ¿

B. 1000¡æÏÂFe2O3ÓëCO·´Ó¦£¬t min´ïµ½Æ½ºâʱc(CO) =2¡Á10-3 mol/L£¬ÔòÓÃCO2±íʾ¸Ã·´Ó¦µÄƽ¾ùËÙÂÊΪ2¡Á10£­3/t mol¡¤L£­1¡¤min£­1

C. ΪÁËʹ¸Ã·´Ó¦µÄKÔö´ó£¬¿ÉÒÔÔÚÆäËûÌõ¼þ²»±äʱ£¬Ôö´óc(CO)

D. ÆäËûÌõ¼þ²»±äʱ£¬Ôö¼ÓFe2O3µÄÓÃÁ¿£¬²»ÄÜÓÐЧ½µµÍÁ¶ÌúβÆøÖÐCOµÄº¬Á¿

¡¾´ð°¸¡¿D

¡¾½âÎö¡¿

A£®ÓɱíÖÐÊý¾Ý¿ÉÅжϣ¬Æ½ºâ³£ÊýËæζÈÉý¸ß¼õС£¬ËµÃ÷·´Ó¦Îª·ÅÈÈ·´Ó¦£¬¸Ã·´Ó¦£º·´Ó¦ÎïµÄ×ÜÄÜÁ¿£¾Éú³ÉÎïµÄ×ÜÄÜÁ¿£¬¹ÊA´íÎó£»B£®1000¡æÏÂFe2O3ÓëCO·´Ó¦£¬tmin´ïµ½Æ½ºâʱc£¨CO£©=2¡Á10-3mol/L£¬ÉèCOÆðʼŨ¶Èx£¬

Fe2O3£¨s£©+CO£¨g£©Fe£¨s£©+CO2£¨g£©£¬

ÆðʼÁ¿£¨mol/L£©x 0

±ä»¯Á¿£¨mol/L£©x-2¡Á10-3 x-2¡Á10-3

ƽºâÁ¿£¨mol/L£©2¡Á10-3 x-2¡Á10-3

K==4£¬x=10-2mol/L£»ÓÃCO±íʾ¸Ã·´Ó¦µÄƽ¾ùËÙÂÊΪ£¬¹ÊB´íÎó£»C£®Æ½ºâ³£ÊýÖ»ÊÜζÈÓ°Ï죬Ôö´óc£¨CO£©²»Äܸı仯ѧƽºâ³£Êý£¬¹ÊC´íÎó£»D£®Ôö´ó¹ÌÌåµÄÁ¿£¬²»¸Ä±äƽºâµÄÒƶ¯£¬ÔòÆäËûÌõ¼þ²»±äʱ£¬Ôö¼ÓFe2O3µÄÓÃÁ¿£¬²»ÄÜÓÐЧ½µµÍÁ¶ÌúβÆøÖÐCOµÄº¬Á¿£¬¹ÊDÕýÈ·£»´ð°¸ÎªD¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÂÁ¡¢ÌúÔÚÉú»î¡¢Éú²úÖÐÓÐ׏㷺µÄÓÃ;£¬Çë»Ø´ðÏÂÁÐÎÊÌâ¡£

(1) Fe2+µÄ×îÍâ²ãµç×ÓÅŲ¼Ê½____________¡£ÔªËØFeÓëMnµÄµÚÈýµçÀëÄÜ·Ö±ðΪI3(Fe)¡¢I3(Mn)£¬ÔòI3(Fe)______I3(Mn)(Ìî¡°>¡±¡¢¡°<")¡£

(2)Æø̬ÂÈ»¯ÂÁµÄ·Ö×Ó×é³ÉΪ(AlCl3)2£¬ÆäÖÐAl¡¢Cl¾ù´ï8e-Îȶ¨½á¹¹£¬AlÔ­×ÓµÄÔÓ»¯·½Ê½Îª__________¡£¸ù¾ÝµÈµç×ÓÔ­Àí£¬AlO2-µÄ¿Õ¼ä¹¹ÐÍΪ_____¡£

(3) Fe(CO)5µÄÈÛµãΪ-20 ¡æ£¬·ÐµãΪ103 ¡æ£¬Ò×ÈÜÓÚÒÒÃÑ£¬Æ侧ÌåÀàÐÍΪ______£¬

(4) ¿Æѧ¼ÒÃÇ·¢ÏÖijЩº¬ÌúµÄÎïÖÊ¿É´ß»¯ÄòËغϳÉëÂ(N2H4)£¬·Ðµã£ºN2H4>C2H6µÄÖ÷ÒªÔ­ÒòΪ______________________¡£

(5) FeO¾§ÌåµÄ¾§°ûÈçͼËùʾ£¬¼ºÖª£ºFeO¾§ÌåµÄÃܶÈΪ¦Ñ g/cm3£¬NA´ú±í°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ¡£Ôڸþ§°ûÖУ¬ÓëFe2+½ôÁÚÇҵȾàÀëµÄFe2+ÊýĿΪ_____£»Fe2+ÓëO2-×î¶ÌºË¼ä¾àΪ______pm(ÓæѺÍNA±íʾ)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿£¨1£©»ù̬CuÔ­×ӵĺËÍâµç×ÓÅŲ¼Ê½Îª____________________

£¨2£©´ÓÔ­×Ó¹ìµÀÖصþ·½Ê½¿¼ÂÇ£¬µª·Ö×ÓÖеĹ²¼Û¼üÀàÐÍÓÐ____________£»

£¨3£©Ë®ÈÜÒºÖÐ

¢ÙË®·Ö×ÓÖÐÑõÔ­×ÓµÄÔÓ»¯ÀàÐÍÊÇ_____£¬¼ü¼ü½Ç____Ìî¡°¡±¡°¡±»ò¡°¡±

¢Ú²»´æÔÚµÄ΢Á£¼ä×÷ÓÃÁ¦ÓÐ______ ¡£

A.Àë×Ó¼ü ¼«ÐÔ¼ü Åäλ¼ü Çâ¼ü ·¶µÂ»ªÁ¦

£¨4£©»ÆÍ­¿óÒ±Á¶Í­Ê±²úÉúµÄ¿É¾­¹ý;¾¶ÐγÉËáÓê¡£

¢ÙµÄ¿Õ¼ä¹¹ÐÍΪ________¡£´Ó½á¹¹½Ç¶È£¬½âÊ͵ÄËáÐÔÇ¿ÓÚµÄÔ­ÒòÊÇ_______

¢ÚÒÑÖª£º¶àÔ­×Ó·Ö×ÓÖУ¬ÈôÔ­×Ó¶¼ÔÚͬһƽÃæÉÏÇÒÕâЩԭ×ÓÓÐÏ໥ƽÐеÄp¹ìµÀ£¬Ôòpµç×Ó¿ÉÔÚ¶à¸öÔ­×Ó¼äÔ˶¯£¬Ðγɡ°ÀëÓò¼ü¡±»ò´ó¼ü¡£´ó¼ü¿ÉÓñíʾ£¬ÆäÖÐm¡¢n·Ö±ð´ú±í²ÎÓëÐγɴó¼üµÄÔ­×Ó¸öÊýºÍµç×ÓÊý£¬Èç±½·Ö×ÓÖдó¼ü±íʾΪ¡£ÏÂÁÐ΢Á£ÖдæÔÚ¡°ÀëÓò¼ü¡±µÄÊÇ_____£»

A. B. C.

¢ÛÍ­¾§ÌåÖÐCuÔ­×ӵĶѻý·½Ê½Èçͼ¢ÙËùʾ£¬Æä¶Ñ»ý·½Ê½Îª_____£¬ÅäλÊýΪ_______£®

¢Ü½ðÍ­ºÏ½ðµÄ¾§°ûÈçͼ¢ÚËùʾ¡£½ðÍ­ºÏ½ð¾ßÓд¢Ç⹦ÄÜ£¬´¢ÇâºóAuÔ­×ÓλÓÚ¶¥µã£¬CuÔ­×ÓλÓÚÃæÐÄ£¬HÔ­×ÓÌî³äÔÚÓÉ1¸öAuÔ­×Ӻ;àAuÔ­×Ó×î½üµÄ3¸öCuÔ­×Ó¹¹³ÉµÄËÄÃæÌå¿Õ϶ÖУ¬ÈôCuÔ­×ÓÓëAuÔ­×ÓµÄ×î¶Ì¾àÀëΪ£¬°¢·ü¼ÓµÂÂÞ³£ÊýµÄλΪ£¬Ôò¸Ã¾§Ìå´¢ÇâºóÃܶÈΪ______Áгö¼ÆËãʽ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Îª·ÀÖκ¬ÁòúȼÉÕ²úÉúµÄ´óÆøÎÛȾ£¬Ä³¹¤³§Éè¼ÆÁËеÄÖÎÎÛ·½·¨£¬¹¤ÒÕÁ÷³ÌÈçͼËùʾ¡£ÏÂÁÐÐðÊö´íÎóµÄÊÇ£¨ £©

A.¸ÃÁ÷³Ì¿ÉÒÔ³ýȥúȼÉÕʱ²úÉúβÆøÖеÄSO2£¬±ä·ÏΪ±¦

B.¸ÃÁ÷³ÌÖÐH2SO4ºÍFe2£¨SO4£©3ÈÜÒº¿ÉÒÔÑ­»·ÀûÓÃ

C.Á÷³ÌÖÐÿÎüÊÕ11.2LSO2£¨±ê×¼×´¿ö£©Í¬Ê±²úÉú1molFe2+

D.Á÷³ÌÖÐÉæ¼°µÄFe2+±»Ñõ»¯µÄ·´Ó¦µÄÀë×Ó·½³ÌʽΪ4Fe2++O2+2H2O¨T4Fe3++4OH-

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿½«ÖÊÁ¿ÎªmgµÄͭмÍêÈ«ÈÜÓÚÊÊÁ¿Å¨ÏõËáÖУ¬·´Ó¦ºóµÃµ½NO2¡¢NOµÄ»ìºÏÆøÌ壬½«ËùµÃÆøÌåͨÈë300mL2molL-1NaOHÈÜÒºÖУ¬Ç¡ºÃÍêÈ«·´Ó¦£¬Éú³Éº¬NaNO3ºÍNaNO2µÄÑÎÈÜÒº£¬ÆäÖÐNaNO3µÄÎïÖʵÄÁ¿Îª0.2mol£¬ÔòmµÄֵΪ£¨ £©

A.12.8B.19.2C.25.6D.51.2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁÐÎïÖʵÄת»¯ÖУ¬A¡¢XÊdz£¼ûµ¥ÖÊ£¬YÔÚ³£ÎÂÏÂΪÎÞÉ«ÒºÌ壬B¡¢C¡¢D¾ùΪÖÐѧ»¯Ñ§Öг£¼ûµÄ»¯ºÏÎËüÃÇÖ®¼äµÄת»¯¹ØϵÈçͼËùʾ£¨²¿·Ö²úÎïºÍ·´Ó¦Ìõ¼þÒÑÂÔÈ¥£©¡£

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÈôBΪÄܵ¼ÖÂËáÓêµÄÆøÌ壬ÔòDµÄ»¯Ñ§Ê½¿ÉÄÜÊÇ__¡£

£¨2£©ÈôCΪµ­»ÆÉ«¹ÌÌ壬DΪǿ¼îÈÜÒº£¬ÔòC¡úDµÄÀë×Ó·½³ÌʽΪ__¡£

£¨3£©ÈôAΪµ­»ÆÉ«¹ÌÌ壬DΪǿËᣬÔòDµÄŨÈÜÒºÓë̼·´Ó¦Éú³ÉÆøÌåBµÄ»¯Ñ§·½³ÌʽΪ__¡£

£¨4£©ÈôAΪ»ÆÂÌÉ«ÆøÌ壬XΪ³£¼û½ðÊô£¬ÔòXÓëBÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪ__¡£ÏÂÁÐÊÔ¼ÁÄܼø±ðBÈÜÒºÓëCÈÜÒºµÄÊÇ__¡££¨Ìî±êºÅ£©

A.AgNO3ÈÜÒº B.KSCNÈÜÒº C.ÑÎËá D.NaOHÈÜÒº

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Àë×ÓÒºÌåÊÇÒ»ÖÖÖ»ÓÉÀë×Ó×é³ÉµÄÒºÌ壬ÔÚµÍÎÂÏÂÒ²ÄÜÒÔҺ̬Îȶ¨´æÔÚ£¬ÊÇÒ»ÖÖºÜÓÐÑо¿¼ÛÖµµÄÈܼÁ¡£¶ÔÀë×ÓÒºÌåµÄÑо¿ÏÔʾ×î³£¼ûµÄÀë×ÓÒºÌåÖ÷ÒªÓÉÒÔϵÄÕýÀë×Ӻ͸ºÀë×Ó×é³É£º

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÔÚÖÜÆÚ±íÖеÄλÖÃÊÇ______£¬Æä¼Ûµç×ÓÅŲ¼Ê½Îª______ͼ1ÖиºÀë×ӵĿռ乹ÐÍΪ______¡£

£¨2£©ÂÈ»¯ÂÁµÄÈÛµãΪ£¬µª»¯ÂÁµÄÈÛµã¸ß´ï£¬ËüÃǶ¼ÊÇ»îÆýðÊôºÍ·Ç½ðÊôµÄ»¯ºÏÎÈÛµãÏà²îÕâô´óµÄÔ­ÒòÊÇ______¡£

£¨3£©Í¼ÖÐÕýÀë×ÓÓÐÁîÈ˾ªÆæµÄÎȶ¨ÐÔ£¬ËüµÄµç×ÓÔÚÆä»·×´½á¹¹Öи߶ÈÀëÓò¡£¸ÃÕýÀë×ÓÖÐNÔ­×ÓµÄÔÓ»¯·½Ê½Îª______£¬CÔ­×ÓµÄÔÓ»¯·½Ê½Îª______¡£

£¨4£©ÎªÁËʹÕýÀë×ÓÒÔµ¥ÌåÐÎʽ´æÔÚÒÔ»ñµÃÁ¼ºÃµÄÈܽâÐÔÄÜ£¬ÓëNÔ­×ÓÏàÁ¬µÄ²»Äܱ»HÔ­×ÓÌæ»»£¬Çë½âÊÍÔ­Òò£º______¡£

£¨5£©¡¢Mg¡¢AlÈýÖÖÔªË÷µÄµÚÒ»µçÀëÄÜÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ______¡£

£¨6£©ÒÑÖªµª»¯ÂÁµÄ¾§°û½á¹¹Èçͼ2Ëùʾ¡£¾§ÌåÖеªÔ­×Ӷѻý·½Ê½Èçͼ3Ëùʾ£¬ÕâÖֶѻý·½Ê½³ÆΪ______¡£ÁùÀâÖùµ×±ß±ß³¤Îªacm£¬¸ßΪccm£¬°¢·ü¼ÓµÂÂÞ³£ÊýµÄֵΪN£¬µª»¯ÂÁ¾§ÌåµÄÃܶÈΪ______Áгö¼ÆËãʽ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÔÚÃܱÕÈÝÆ÷Öз¢Éú·´Ó¦X(g)£«3Y(g)2Z(g)£¬ÈôX¡¢Y¡¢ZµÄÆðʼŨ¶È·Ö±ðΪ0.1 mol¡¤L£­1¡¢0.3mol¡¤L£­1ºÍ0.2mol¡¤L£­1£¬Ôòƽºâʱ¸÷ÎïÖʵÄŨ¶È²»¿ÉÄÜÊÇ£¨ £©

A. XΪ0.2 mol¡¤L£­1

B. YΪ0.1 mol¡¤L£­1

C. ZΪ0.3 mol¡¤L£­1

D. ZΪ0.1 mol¡¤L£­1ʱ£¬YΪ0.45 mol¡¤L£­1

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ìú¼°Æ仯ºÏÎïÔÚÉú²ú¡¢Éú»îÖдæÔڹ㷺ÓÃ;¡£

£¨1£©ÈçͼÊÇÌúÔªËØÔÚÔªËØÖÜÆÚ±íÖеÄÓйØÐÅÏ¢¡£

²¹È«ÌúÔ­×ӵĺËÍâµç×ÓÅŲ¼Ê½_____________3p63d64s2¡£¡°55.85¡±ÊÇ________________________¡£×ÔÈ»½çÖдæÔÚµÄ54FeºÍ56Fe£¬ËüÃÇ»¥³ÆΪ_____________¡£

£¨2£©½«ÌúƬ·ÅÈëÀäŨÁòËáÖУ¬Æ¬¿Ìºó½«ÌúƬÒÆÈëÁòËáÍ­ÈÜÒºÖУ¬·¢ÏÖÌúƬ±íÃæÎÞÃ÷ÏԱ仯£¬Ô­ÒòÊÇ_____________¡£Í¨³£Ö¤Ã÷ijÈÜÒºÖк¬Fe2+µÄ»¯Ñ§·½·¨ÊÇ_____________________¡£

£¨3£©SO2 ÓÐÇ¿»¹Ô­ÐÔ£¬Ð´³öÆäÓë FeCl3 ÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ______________________¡£

ÈôÓÐ 0.4mol FeCl3 ·¢Éú·´Ó¦£¬ÔòÐèÒª±ê×¼×´¿öÏ嵀 SO2______________Éý¡£

£¨4£©·¢»ÆµÄÉˮͨ³£º¬ Fe3+£¬ÈôÏòË®ÖмÓÈë Na2CO3 ÈÜÒº£¬Ôò²úÉúºìºÖÉ«³ÁµíºÍÎÞÉ«ÎÞζµÄÆøÌå¡£Çë´ÓƽºâÒƶ¯½Ç¶È½âÊÍÕâÒ»ÏÖÏó___________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸