¡¾ÌâÄ¿¡¿¡¶ÎÒÔڹʹ¬ÐÞÎÄÎչʾÁËר¼Ò¾«Õ¿µÄ¼¼ÒպͶԴ«Í³ÎÄ»¯µÄÈȰ®Óë¼áÊØ£¬Ò²ÁîÈËÌå»áµ½»¯Ñ§·½·¨ÔÚÎÄÎï±£»¤Öеľ޴ó×÷Óá£Ä³²©Îï¹ÝÐÞ¸´³öÍÁÌúÆ÷µÄ¹ý³ÌÈçÏ£º
£¨1£©¼ì²âÐâÊ´²úÎï
Ö÷Òª³É·ÖµÄ»¯Ñ§Ê½ | |||
Fe3O4 | Fe2O3¡¤H2O | FeO(OH) | FeOCl |
ÌúÆ÷ÔÚ¾ßÓÐO2¡¢________µÈ»·¾³ÖÐÈÝÒ×±»¸¯Ê´¡£
£¨2£©·ÖÎö¸¯Ê´ÔÀí£ºÒ»°ãÈÏΪ£¬Ìú¾¹ýÁËÈçϸ¯Ê´Ñ»·¡£
¢ñ.Feת»¯ÎªFe2£«¡£
¢ò. FeO(OH)ºÍFe2£«·´Ó¦ÐγÉÖÂÃܵÄFe3O4±£»¤²ã£¬Fe2£«µÄ×÷ÓÃÊÇ____(Ìî×Öĸ)¡£
a£®Ñõ»¯¼Á b£®»¹Ô¼Á c£®¼È²»ÊÇÑõ»¯¼ÁÒ²²»ÊÇ»¹Ô¼Á
£¨3£©Ñо¿·¢ÏÖ£¬Cl£¶ÔÌúµÄ¸¯Ê´»áÔì³ÉÑÏÖØÓ°Ïì¡£»¯Ñ§ÐÞ¸´£ºÍÑÂÈ¡¢»¹Ô£¬ÐγÉFe3O4±£»¤²ã£¬·½·¨ÈçÏ£º
½«ÌúÆ÷½þûÔÚÊ¢ÓÐ0.5 mol¡¤L£1 Na2SO3¡¢0.5 mol¡¤L£1 NaOHÈÜÒºµÄÈÝÆ÷ÖУ¬»ºÂý¼ÓÈÈÖÁ60¡«90 ¡æ¡£Ò»¶Îʱ¼äºó£¬È¡³öÆ÷ÎÓÃNaOHÈÜҺϴµÓÖÁÎÞCl£¡£
¢Ù¼ì²âÏ´µÓÒºÖÐCl£µÄ·½·¨ÊÇ________________________________________¡£
¢ÚÍÑÂÈ·´Ó¦£ºFeOCl£«OH£===FeO(OH)£«Cl£¡£Àë×Ó·´Ó¦µÄ±¾ÖÊÊÇÀë×ÓŨ¶ÈµÄ¼õС£¬±È½ÏFeOClÓëFeO(OH)Èܽâ¶ÈµÄ´óС£º________________¡£
¢ÛNa2SO3»¹ÔFeO(OH)ÐγÉFe3O4µÄÀë×Ó·½³ÌʽÊÇ_____SO32££«______FeO(OH)===______SO42££«____Fe3O4£«______H2O(½«·´Ó¦²¹³äÍêÈ«)¡£________
¡¾´ð°¸¡¿H2O£¨³±Êª£© c È¡ÉÙÁ¿Ï´µÓÒºÓÚÊÔ¹ÜÖУ¬¼ÓÈëÏ¡ÏõËáºÍÏõËáÒø»ìºÏÒº£¬ÈôÎÞ°×É«³Áµí²úÉú˵Ã÷Cl- S(FeOCl)>S[FeO(OH)] 1 6 1 2 3
¡¾½âÎö¡¿
£¨1£©¸ù¾ÝÌúÆ÷ÐâÊ´ºóµÄ³É·Ö¿ÉÖª£¬ÌúÆ÷ÔÚ¾ßÓÐO2¡¢Ë®(³±Êª)µÄ»·¾³ÏÂÒ×ÉúÐ⣻
£¨2£©FeO(OH)ÖÐFeµÄ¼Û̬Ϊ£«3¼Û£¬Fe3O4ÖÐÌúÓÐ1/3µÄFe2£«ºÍ2/3µÄFe3£«£¬Fe2£«µÄ»¯ºÏ¼Û²»±ä£¬¼´Fe2£«¼È²»ÊÇ»¹Ô¼ÁÒ²²»ÊÇÑõ»¯¼Á£¬¹ÊÑ¡ÏîcÕýÈ·£»
£¨3£©¢Ù¼ìÑéCl£Ò»°ãÓÃAgNO3ÈÜÒº£¬²Ù×÷·½·¨ÊÇÈ¡ÉÙÁ¿Ï´µÓÒºÓÚÊÔ¹ÜÖУ¬¼ÓÈëÏ¡ÏõËáºÍÏõËáÒøµÄ»ìºÏÒº£¬ÈôÎÞÉ«³Áµí²úÉú˵Ã÷ÎÞCl££¬·´Ö®ÓÐCl££»
¢Ú¸ù¾ÝÀë×Ó·´Ó¦µÄ±¾ÖÊ£¬ÍƳöS(FeOCl)>S[FeO(OH)]£»
¢ÛSO32£¡úSO42££¬SµÄ»¯ºÏ¼ÛÓÉ£«4¼Û¡ú£«6¼Û£¬Éý¸ß2¼Û£¬FeO(OH)3ÖÐFeµÄ»¯ºÏ¼ÛΪ£«3¼Û£¬Fe3O4ÖÐÌúµÄ¼Û̬ÈÏΪ£«8/3£¬FeµÄ»¯ºÏ¼Û½µµÍ£¬¸ù¾Ý»¯ºÏ¼ÛµÄÉý½µ·¨½øÐзÖÎö£¬µÃ³öSO32££«6FeO(OH)=SO42££«2Fe3O4£«3H2O¡£
| Äê¼¶ | ¸ßÖÐ¿Î³Ì | Äê¼¶ | ³õÖÐ¿Î³Ì |
| ¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁйØÓÚÓлú»¯ºÏÎïµÄ˵·¨´íÎóµÄÊÇ
A. ÃÞ¡¢Âé¡¢ÑòëÍêȫȼÉÕ¶¼Ö»Éú³ÉCO2ºÍH2O
B. ÊÒÎÂÏ£¬ÔÚË®ÖеÄÈܽâ¶ÈÒÒ´¼´óÓÚäåÒÒÍé
C. ÒÒÏ©ºÍ±½ÒÒÏ©·Ö×ÓÖÐËùÓÐÔ×Ó¾ùÄÜÔÚÍ¬Ò»Æ½ÃæÉÏ£¬¶¼ÄÜʹäåË®ÍÊÉ«
D. ÒÒÏ©Óë·Ö×ÓʽΪC3H6 µÄÓлúÎï²»Ò»¶¨»¥ÎªÍ¬ÏµÎï
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿I.£¨1£©´ÖÑÎÖк¬ÓÐCa2+¡¢Mg2+¡¢SO42-µÈÔÓÖÊÀë×Ó£¬´ÖÑξ«Öƹý³ÌÖÐҪʹÓÃNa2CO3ÈÜÒº£¬Çëд³ö¼ÓÈëNa2CO3ÈÜÒººóÏà¹Ø»¯Ñ§·´Ó¦µÄÀë×Ó·½³Ìʽ£º_________________________¡£
£¨2£©½«×ãÁ¿CO2ÆøÌåͨÈëË®²£Á§(Na2SiO3ÈÜÒº)ÖУ¬È»ºó¼ÓÈÈÕô¸É£¬ÔÙÔÚ¸ßÎÂϳä·ÖׯÉÕ£¬×îºóµÃµ½µÄ¹ÌÌåÎïÖÊÊÇ_________________________¡£
II.£¨1£©120 mLº¬ÓÐ0.20 mol̼ËáÄÆµÄÈÜÒººÍ200 mLÑÎËá,²»¹Ü½«Ç°ÕߵμÓÈëºóÕß,»¹Êǽ«ºóÕߵμÓÈëǰÕß,¶¼ÓÐÆøÌå²úÉú,µ«×îÖÕÉú³ÉµÄÆøÌåÌå»ý²»Í¬,ÔòÑÎËáµÄŨ¶ÈºÏÀíµÄ·¶Î§ÊÇ_________________________________________
£¨2£©È¡Ìå»ýÏàͬµÄKI¡¢Na2SO3¡¢FeBr2ÈÜÒº£¬·Ö±ðͨÈë×ãÁ¿ÂÈÆø£¬µ±Ç¡ºÃÍêÈ«·´Ó¦Ê±£¬ÈýÖÖÈÜÒºÏûºÄÂÈÆøµÄÎïÖʵÄÁ¿Ïàͬ£¬ÔòKI¡¢Na2SO3¡¢FeBr2ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈÖ®±ÈΪ_______¡£
£¨3£©ÔÚ·´Ó¦ 3BrF3+5H2O=HBrO3+Br2+9HF+O2 ÖУ¬µ±ÓÐ5 molË®·´Ó¦Ê±£¬ÓÉH2O»¹ÔµÄBrF3Ϊ______________mol¡£
III.½«Ò»¶¨ÖÊÁ¿µÄÌú·Û¼ÓÈëµ½×°ÓÐ100 mLijŨ¶ÈµÄÏ¡ÏõËáÈÜÒºÖгä·Ö·´Ó¦¡£
£¨1£©ÈÝÆ÷ÖÐÊ£ÓÐm gµÄÌú·Û£¬ÊÕ¼¯µ½NOÆøÌå448 mL(±ê×¼×´¿öÏÂ)¡£
¢ÙËùµÃÈÜÒºÖеÄÈÜÖʵĻ¯Ñ§Ê½Îª________________¡£
¢ÚÔÏõËáÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ__________________________¡£
£¨2£©ÏòÉÏÊö¹ÌÒº»ìºÏÎïÖÐÖ𽥵μÓÏ¡ÁòËáÖÁ¸ÕºÃ²»ÔÙ²úÉúÓö¿ÕÆø±ä³Éºì×ØÉ«ÆøÌåΪֹ£¬´ËʱÈÝÆ÷ÖÐÓÐÌú·Ûn g¡£
¢Ù´ËʱÈÜÒºÖÐÈÜÖʵĻ¯Ñ§Ê½Îª__________________¡£
¢Ú£¨m£n£©µÄֵΪ________________(¾«È·µ½0.1 g)¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Cl2¿ÉÓëÇ¿¼îÈÜÒº·´Ó¦£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺ ____Cl2+____KOH=____KCl+____KClO3+____H2O
£¨1£©ÇëÅ䯽ÉÏÊö»¯Ñ§·½³Ìʽ¡£____________
£¨2£©Cl2ÔÚ·´Ó¦ÖбíÏÖ³öÀ´µÄÐÔÖÊÊÇ________(Ìî±àºÅ)
¢ÙÖ»Óл¹ÔÐÔ ¢ÚÖ»ÓÐÑõ»¯ÐÔ¢Û¼ÈÓÐÑõ»¯ÐÔÓÖÓл¹ÔÐÔ
£¨3£©¸Ã·´Ó¦¹ý³ÌÓÐ0.5molµç×Ó×ªÒÆ£¬Ôò²Î¼Ó·´Ó¦µÄCl2Ϊ________mol£¬ÆäÌå»ýÔÚ±ê×¼×´¿öÏÂΪ_______L¡£
£¨4£©Ñõ»¯²úÎïÊÇ_____________£¬»¹Ô²úÎïÊÇ___________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Ç°20ºÅÖ÷×åÔªËØW¡¢X¡¢Y¡¢ZµÄÔ×ÓÐòÊýÒÀ´ÎÔö¼Ó¡£WµÄ×îÍâ²ãµç×ÓÊýÊÇÄÚ²ãµç×ÓÊýµÄ3±¶£¬W¡¢X¡¢Z×îÍâ²ãµç×ÓÊýÖ®ºÍΪ10£»WÓëYͬ×å¡£ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ
A. WµÄÇ⻯Îï·Ðµã´óÓÚYµÄÇ⻯Îï·Ðµã
B. ¼òµ¥Àë×Ó°ë¾¶£ºX > W
C. ZµÄÇ⻯ÎïÖк¬Àë×Ó¼ü
D. ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄ¼îÐÔ£ºZ > X
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ËáÐÔпÃÌ¸Éµç³ØÊÇÒ»ÖÖÒ»´ÎÐÔµç³Ø£¬Íâ¿ÇΪ½ðÊôп£¬ÖмäÊÇ̼°ô£¬ÆäÖÜΧÊÇÓÉ̼·Û¡¢MnO2¡¢ZnCl2ºÍNH4ClµÈ×é³ÉµÄºý×´Ìî³äÎï¡£¸Ãµç³Ø·Åµç¹ý³Ì²úÉúMnOOH¡£»ØÊÕ´¦Àí¸Ã·Ïµç³Ø¿ÉÒԵõ½¶àÖÖ»¯¹¤ÔÁÏ¡£ÓйØÊý¾ÝÈçϱíËùʾ£º
Èܽâ¶È/(g/100 gË®)
¡¡¡¡¡¡Î¶È/ ¡æ »¯ºÏÎï¡¡¡¡¡¡ | 0 | 20 | 40 | 60 | 80 | 100 |
NH4Cl | 29.3 | 37.2 | 45.8 | 55.3 | 65.6 | 77.3 |
ZnCl2 | 343 | 395 | 452 | 488 | 541 | 614 |
»¯ºÏÎï | Zn(OH)2 | Fe(OH)2 | Fe(OH)3 |
Ksp½üËÆÖµ | 10£17 | 10£17 | 10£39 |
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¸Ãµç³ØµÄÕý¼«·´Ó¦Ê½Îª________________________________________£¬
£¨2£©·Ïµç³Øºý×´Ìî³äÎï¼ÓË®´¦Àíºó£¬¹ýÂË£¬ÂËÒºÖÐÖ÷ÒªÓÐZnCl2ºÍNH4Cl£¬¶þÕß¿Éͨ¹ý¼ÓÈÈŨËõ¡¢ÀäÈ´½á¾§·ÖÀë»ØÊÕ£»ÂËÔüµÄÖ÷Òª³É·ÖÊÇMnO2¡¢CºÍMnOOH£¬Óû´ÓÖеõ½½Ï´¿µÄMnO2£¬×î¼ò±ãµÄ·½·¨ÊÇ_______£¬
£¨3£©ÓÃ·Ïµç³ØµÄÐ¿Æ¤ÖÆ±¸ZnSO4¡¤7H2OµÄ¹ý³ÌÖУ¬Ðè³ýȥпƤÖеÄÉÙÁ¿ÔÓÖÊÌú£¬Æä·½·¨ÊÇ£º¼ÓÏ¡H2SO4ºÍH2O2Èܽ⣬¼Ó¼îµ÷½ÚÖÁpHΪ________ʱ£¬Ìú¸ÕºÃ³ÁµíÍêÈ«(Àë×ÓŨ¶ÈСÓÚ1¡Á10£5 mol¡¤L£1ʱ£¬¼´¿ÉÈÏΪ¸ÃÀë×Ó³ÁµíÍêÈ«£¬ÈôÉÏÊö¹ý³Ì²»¼ÓH2O2ºó¹ûÊÇ__________ÔÒòÊÇ_____________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÔÚÈÜÒºÖмÓÈë×ãÁ¿Na2O2ºóÈÔÄÜ´óÁ¿¹²´æµÄÀë×Ó×éÊÇ( )
A. Ca2+¡¢Fe2+¡¢NO3£¡¢HCO3£B. K+¡¢AlO2£¡¢Cl£¡¢SO42£
C. H+¡¢Ba2+¡¢Cl£¡¢NO3£D. Na+¡¢Cl£¡¢CO32£¡¢SO32£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿NA´ú±í°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
A. ³£Î³£Ñ¹Ï£¬1mol P4ÖÐËùº¬P¡ªP¼üÊýĿΪ4NA
B. 100 mL 1 mol¡¤L1FeCl3ÈÜÒºÖÐËùº¬Fe3+µÄÊýĿΪ0.1NA
C. ³£Î³£Ñ¹Ï£¬2.24LCOºÍCO2»ìºÏÆøÌåÖк¬ÓеÄ̼Ô×ÓÊýĿΪ0.1NA
D. ³£Î³£Ñ¹Ï£¬18g H2Oº¬ÓеÄÔ×Ó×ÜÊýΪ3NA
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿»¯Ñ§Ñо¿Ö÷ÒªÓõÄÊÇʵÑé·½·¨£¬ËùÒÔѧϰ»¯Ñ§Àë²»¿ªÊµÑé¡£Çë»Ø´ð£º
(1)ÔÚÊÔ¹ÜÀï×¢ÈëÉÙÁ¿ÐÂÖÆ±¸µÄFeSO4ÈÜÒº£¬ÓýºÍ·µÎ¹ÜÎüÈ¡NaOHÈÜÒº£¬½«µÎ¹Ü¼â¶Ë²åÈëÊÔ¹ÜÀïÈÜÒºµ×²¿£¬ÂýÂý¼·³öNaOHÈÜÒº£¬¿ÉÒԹ۲쵽ÊÔ¹ÜÖвúÉú°×É«Ðõ×´³Áµí£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ____________________¡£Éú³ÉµÄ³ÁµíѸËÙ±ä³É»ÒÂÌÉ«£¬×îºó±ä³ÉºìºÖÉ«£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_______________________¡£
(2)ʵÑéÊÒÔÚ±£´æº¬Fe2+µÄÈÜҺʱ£¬¾³£ÏòÆäÖмÓÈëÌú·Û£¬ÆäÄ¿µÄÊÇ___________¡£ÎªÁ˼ìÑé¸ÃÈÜÒºÊÇ·ñ±äÖÊ£¬Ó¦²ÉÓõÄʵÑé²Ù×÷¼°ÏÖÏóÊÇ__________________________¡£
(3)³ýÈ¥Ìú·ÛÖлìÓеÄÂÁ·ÛӦѡÓõÄÊÔ¼ÁΪ_____£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ_____________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¹ú¼ÊѧУÓÅÑ¡ - 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