ˮúÆø²»½öÊǺϳɰ±µÄÔ­ÁÏÆø£¬Ò²ÊǺϳÉÌþ¼°Æ仯¹¤²úÆ·µÄÔ­ÁÏ£®
£¨1£©¡°Ö±½ÓˮúÆøȼÁϵç³Ø¡±ÖУ¬Í¨CO¡¢H2µÄ¼«Îªµç³ØµÄ
 
¼«£¨Ñ¡Ì¡°Õý¡±£¬¡°¸º¡±£©£®
£¨2£©Ë®ÃºÆø±ä»»·´Ó¦£ºCO£¨g£©+H2O£¨g£©?CO2£¨g£©+H2£¨g£©¡÷H£¼0£¬ÏÂÁдëÊ©ÄÜÌá¸ß·´Ó¦ËÙÂʵÄÓÐ
 
£¨²»¶¨ÏîÑ¡Ôñ£©£®
a£®Éý¸ßζȠ   b£®¼ÓÈë´ß»¯¼Á     c£®Ôö´óѹǿ     d£®½µµÍŨ¶È
£¨3£©H2ºÍN2ÔÚ´ß»¯¼Á¡¢¸ßθßѹÌõ¼þϺϳɰ±µÄ»¯Ñ§·½³ÌʽΪ
 
£®
£¨4£©°±ÆøµÄË®ÈÜÒº¿ÉÓÃÓÚÎüÊÕÑ̵ÀÆøÖеĶþÑõ»¯Áò£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
 
£®
£¨5£©½«±ê×¼×´¿öÏÂ582.4LºÏ³ÉÆø£¨ÒÑÖª£ºn£¨CO£©£ºn£¨H2£©=4£º9£©Í¨ÈëºÏ³ÉËþ£¬Ò»¶¨Ìõ¼þÏ¿ɷ¢Éú2CO£¨g£©+4H2£¨g£©¡úCH2=CH2£¨g£©+2H2O£¨g£©£»CO£¨g£©+3H2¡úCH4£¨g£©+H2O£¨g£©£¬³ä·Ö·´Ó¦ºó£¬¾­²â¶¨²úÆ·ÖÐÖ»Óм×Íé¡¢ÒÒÏ©ºÍË®ÕôÆø£¨¼Ù¶¨COºÍH2¾ùÎÞÊ£Óࣩ£¬ÊÔ¼ÆËãÒݳöµÄÆøÌåÖÐÒÒÏ©µÄÎïÖʵÄÁ¿£¨Áгö¼ÆËã¹ý³Ì£©£®
¿¼µã£º»¯Ñ§Æ½ºâµÄ¼ÆËã,»¯Ñ§µçÔ´ÐÂÐ͵ç³Ø,»¯Ñ§·´Ó¦ËÙÂʵÄÓ°ÏìÒòËØ
רÌ⣺»ù±¾¸ÅÄîÓë»ù±¾ÀíÂÛ
·ÖÎö£º£¨1£©ÒÀ¾ÝȼÁϵç³ØÖÐȼÁÏÔÚ¸º¼«Ê§µç×Ó·¢ÉúÑõ»¯·´Ó¦£»
£¨2£©ÒÀ¾ÝÓ°Ïì·´Ó¦ËÙÂʵÄÒòËØ·ÖÎöÅжϣ»
£¨3£©H2ºÍN2ÔÚ´ß»¯¼Á¡¢¸ßθßѹÌõ¼þϺϳɰ±Æø£¬½áºÏÔ­ÔòÊغãÅäƽÊéд£»
£¨4£©°±ÆøµÄË®ÈÜÒº¿ÉÓÃÓÚÎüÊÕÑ̵ÀÆøÖеĶþÑõ»¯ÁòÉú³ÉÑÇÁòËá炙òÑÇÁòËáÇâ泥»
£¨5£©ÒÀ¾ÝÔªËØÊغã½áºÏ»¯Ñ§·½³Ìʽ¶¨Á¿¹Øϵ¼ÆËã·ÖÎöÓ¦Óã®
½â´ð£º ½â£º£¨1£©È¼Áϵç³ØÖÐȼÁÏÔÚ¸º¼«Ê§µç×Ó·¢ÉúÑõ»¯·´Ó¦£¬£©¡°Ö±½ÓˮúÆøȼÁϵç³Ø¡±ÖУ¬Í¨CO¡¢H2µÄ¼«Îªµç³ØµÄ¸º¼«£¬¹Ê´ð°¸Îª£º¸º£»
£¨2£©CO£¨g£©+H2O£¨g£©?CO2£¨g£©+H2£¨g£©¡÷H£¼0£¬·´Ó¦ÊÇÆøÌåÌå»ý²»±äµÄ·ÅÈÈ·´Ó¦£¬
a£®Éý¸ßζÈƽºâÄæÏò½øÐУ¬·´Ó¦ËÙÂÊÔö´ó£¬¹ÊaÕýÈ·£»   
b£®¼ÓÈë´ß»¯¼ÁÄÜÔö´ó·´Ó¦ËÙÂÊ£¬¹ÊbÕýÈ·£»     
c£®Ôö´óѹǿ£¬·´Ó¦ËÙÂÊÔö´ó£¬¹ÊÕýÈ·£»    
d£®½µµÍŨ¶È£¬·´Ó¦ËÙÂʼõС£¬¹Êd´íÎó£»
¹Ê´ð°¸Îª£ºabc£» 
£¨3£©H2ºÍN2ÔÚ´ß»¯¼Á¡¢¸ßθßѹÌõ¼þϺϳɰ±µÄ»¯Ñ§·½³ÌʽΪ£ºN2+3H2
  ´ß»¯¼Á  
.
¸ßθßѹ
2NH3£¬¹Ê´ð°¸Îª£ºN2+3H2
  ´ß»¯¼Á  
.
¸ßθßѹ
2NH3£»
£¨4£©°±ÆøµÄË®ÈÜÒº¿ÉÓÃÓÚÎüÊÕÑ̵ÀÆøÖеĶþÑõ»¯ÁòÉú³ÉÑÇÁòËá炙òÑÇÁòËáÇâ泥¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºSO2+2NH3+H2O=£¨NH4£©2SO3»òSO2+NH3+H2O=NH4HSO3£»
¹Ê´ð°¸Îª£ºSO2+2NH3+H2O=£¨NH4£©2SO3»òSO2+NH3+H2O=NH4HSO3£»
£¨5£©ÒÀ¾ÝÌâÒ⣺n£¨CO£©+n£¨H2£©=
582.4L
22.4L/mol
=26mol£»
n£¨CO£©=26mol¡Á4/£¨4+9£©=8mol£¬n£¨H2£©=26mol-8mol=18mol
ÓÉ·½³Ìʽ£º2CO£¨g£©+4H2£¨g£©¡úC2H4£¨g£©+2H2O£¨g£©ºÍCO£¨g£©+3H2£¨g£©¡úCH4£¨g£©+H2O£¨g£©
n£¨CH4£©+2n£¨C2H4£©=8mol£¬3n£¨CH4£©+4n£¨C2H4£©=18mol£¬½âµÃn£¨C2H4£©=3mol£»
´ð£ºÒݳöµÄÆøÌåÖÐÒÒÏ©µÄÎïÖʵÄÁ¿Îª3mol£®
µãÆÀ£º±¾Ì⿼²éÁË»¯Ñ§·´Ó¦ËÙÂÊ£¬»¯Ñ§Æ½ºâµÄÓ°ÏìÒòËØ·ÖÎöÅжϣ¬»¯Ñ§·½³ÌʽÊéд£¬Ô­µç³ØÔ­Àí·ÖÎöÓ¦Óã¬ÕÆÎÕ»ù´¡Êǹؼü£¬ÌâÄ¿½Ï¼òµ¥£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐʵÑé²Ù×÷»áÒýÆð²â¶¨½á¹ûÆ«¸ßµÄÊÇ£¨¡¡¡¡£©
A¡¢ÓÃÒÑ֪Ũ¶ÈµÄÑÎËáµÎ¶¨Î´ÖªÅ¨¶ÈµÄÇâÑõ»¯ÄÆÈÜҺʱ£¬¶ÁÈ¡µÎ¶¨¹ÜÄ©¶ÁÊýʱ£¬ÑöÊӿ̶ÈÏß
B¡¢²â¶¨ÁòËáÍ­¾§ÌåÖнᾧˮº¬Á¿µÄʵÑéʱ£¬¼ÓÈÈʱ¼ä¹ý¶ÌδÍêÈ«±ä°×
C¡¢Öк͵ζ¨Ê±£¬¼Ó´ý²âҺǰ׶ÐÎÆ¿ÄÚÓÐÉÙÁ¿Ë®
D¡¢²â¶¨1molÇâÆøÌå»ýµÄ²Ù×÷ÖУ¬·´Ó¦½áÊøºóδ³éÆø

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Ñо¿µªÑõ»¯ÎïÓëÐü¸¡ÔÚ´óÆøÖк£ÑÎÁ£×ÓµÄÏ໥×÷ÓÃʱ£¬Éæ¼°ÈçÏ·´Ó¦£º
2NO2£¨g£©+NaCl£¨s£©?NaNO3£¨s£©+ClNO£¨g£©  K1¡÷H£¼0  £¨¢ñ£©
2NO£¨g£©+Cl2£¨g£©?2ClNO£¨g£©  K2¡÷H£¼0  £¨¢ò£©
£¨1£©4NO2£¨g£©+2NaCl£¨s£©?2NaNO3£¨s£©+2NO£¨g£©+Cl2£¨g£©µÄƽºâ³£ÊýK=
 
£¨ÓÃK1¡¢K2±íʾ£©£®
£¨2£©ÎªÑо¿²»Í¬Ìõ¼þ¶Ô·´Ó¦£¨¢ò£©µÄÓ°Ï죬ÔÚºãÎÂÌõ¼þÏ£¬Ïò2LºãÈÝÃܱÕÈÝÆ÷ÖмÓÈë0.2mol NOºÍ0.1mol Cl2£¬10minʱ·´Ó¦£¨¢ò£©´ïµ½Æ½ºâ£®²âµÃ10minÄÚv£¨ClNO£©=7.5¡Á10-3mol?L-1?min-1£¬Ôòƽºâºón£¨Cl2£©=
 
mol£¬NOµÄת»¯ÂʧÑ1=
 
£®ÆäËüÌõ¼þ±£³Ö²»±ä£¬·´Ó¦£¨¢ò£©ÔÚºãѹÌõ¼þϽøÐУ¬Æ½ºâʱNOµÄת»¯ÂʧÑ2
 
§Ñ1£¨Ìî¡°£¾¡±¡°£¼¡±»ò¡°=¡±£©£¬Æ½ºâ³£ÊýK2
 
£¨Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±£®ÈôҪʹK2¼õС£¬¿É²ÉÓõĴëÊ©ÊÇ
 
£®
£¨3£©ÒÑÖªt¡æʱ£¬·´Ó¦FeO£¨s£©+CO£¨g£©?Fe£¨s£©+CO2£¨g£©µÄƽºâ³£ÊýK=0.25£®
¢Ùt¡æʱ£¬·´Ó¦´ïµ½Æ½ºâʱn£¨CO£©£ºn£¨CO2£©=
 
£®
¢ÚÈôÔÚ1LÃܱÕÈÝÆ÷ÖмÓÈë0.02mol FeO£¨s£©£¬²¢Í¨ÈëxmolCO£¬t¡æʱ·´Ó¦´ïµ½Æ½ºâ£®´ËʱFeO£¨s£©×ª»¯ÂÊΪ50%£¬Ôòx=
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Ϊ³ýÈ¥´ÖÑÎÖеÄCa2+¡¢Mg2+¡¢Fe3+¡¢SO
 
2-
4
ÒÔ¼°ÄàɳµÈÔÓÖÊ£¬Ä³Í¬Ñ§Éè¼ÆÁËÒ»ÖÖÖƱ¸¾«ÑεÄʵÑé·½°¸£¬²½ÖèÈçÏ£¨ÓÃÓÚ³ÁµíµÄÊÔ¼ÁÉÔ¹ýÁ¿£©£º
³ÆÈ¡´ÖÑÎ
Èܽâ
¢Ù
BaCl2
¢Ú
NaOH
¢Û
Na2CO3
¢Ü
¹ýÂË
¢Ý
ÂËÒº
ÊÊÁ¿Ê³ÑÎ
¢Þ
Õô·¢¡¢½á¾§¡¢ºæ¸É
¢ß
¾«ÑÎ
£¨1£©ÅжÏBaCl2ÒѹýÁ¿µÄ·½·¨ÊÇ
 
£®
£¨2£©µÚ¢Ü²½ÖУ¬Ïà¹ØµÄ»¯Ñ§·½³ÌʽÊÇ
 
£®
£¨3£©ÅäÖÆNaClÈÜҺʱ£¬Èô³öÏÖÏÂÁвÙ×÷£¬Æä½á¹ûÆ«¸ß»¹ÊÇÆ«µÍ£¿
A£®³ÆÁ¿Ê±NaClÒѳ±½â
 
           B£®ÌìƽµÄíÀÂëÒÑÐâÊ´
 

C£®¶¨ÈÝÒ¡ÔÈʱ£¬ÒºÃæϽµÓÖ¼ÓË®
 
   D£®¶¨ÈÝʱ¸©Êӿ̶ÈÏß
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¹¤ÒµÉÏÉè¼Æ½«VOSO4ÖеÄK2SO4¡¢SiO2ÔÓÖʳýÈ¥²¢»ØÊյõ½V2O5µÄÁ÷³ÌÈçͼ£ºÒÑÖª£ºVO2+Äܱ»KClO3Ñõ»¯³ÉVO3-
Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©²½Öè¢ÙËùµÃ·ÏÔüµÄ³É·ÖÊÇ
 
£¨Ð´»¯Ñ§Ê½£©£¬²Ù×÷IµÄÃû³Æ
 
£®
£¨2£©²½Öè¢Ú¡¢¢ÛµÄ±ä»¯¹ý³Ì¿É¼ò»¯Îª£¨ÏÂʽR±íʾVO2+£¬HA±íʾÓлúÝÍÈ¡¼Á£©£º
R2£¨SO4£©n£¨Ë®²ã£©+2nHA£¨Óлú²ã£©?2RAn£¨Óлú²ã£©+nH2SO4 £¨Ë®²ã£©
¢ÚÖÐÝÍȡʱ±ØÐë¼ÓÈëÊÊÁ¿¼î£¬ÆäÔ­ÒòÊÇ
 
£®
¢ÛÖÐXÊÔ¼ÁΪ
 
£®
£¨3£©¢ÝµÄÀë×Ó·½³ÌʽΪ
 
£®
£¨4£©25¡æʱ£¬È¡Ñù½øÐÐÊÔÑé·ÖÎö£¬µÃµ½·°³ÁµíÂʺÍÈÜÒºpHÖ®¼ä¹ØϵÈçÏÂ±í£º
pH1.31.41.51.61.71.81.92.02.1
·°³ÁµíÂÊ%88.194.896.598.098.898.896.493.189.3
½áºÏÉÏ±í£¬ÔÚʵ¼ÊÉú²úÖУ¬¢ÝÖмÓÈ백ˮ£¬µ÷½ÚÈÜÒºµÄ×î¼ÑpHΪ
 
£»Èô·°³ÁµíÂÊΪ93.1%ʱ²»²úÉúFe£¨OH£©3³Áµí£¬ÔòÈÜÒºÖÐc£¨Fe3+£©£¼
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐÎïÖÊ£º¢ÙN2      ¢ÚNa2O      ¢ÛNH3     ¢ÜCO2       ¢ÝNa2O2 ¢ÞNaOH              ¢ßCaBr2     ¢àH2O2     ¢áNH4Cl     ¢âHBr£®
»Ø´ðÏÂÁÐÎÊÌ⣺£¨ÌîÐòºÅ£©
£¨1£©Ö»º¬Àë×Ó¼üµÄÊÇ
 
£»
£¨2£©Ö»º¬¼«ÐÔ¼üµÄÊÇ
 
£»
£¨3£©º¬Óм«ÐÔ¼üºÍ·Ç¼«ÐÔ¼üµÄÊÇ
 
£»
£¨4£©º¬ÓзǼ«ÐÔ¼üµÄÀë×Ó»¯ºÏÎïÊÇ
 
£»
£¨5£©º¬Óм«ÐÔ¼üµÄÀë×Ó»¯ºÏÎïÊÇ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ij»¯Ñ§ÐËȤС×éÀûÓÃÈçͼËùʾԭµç³Ø×°ÖýøÐÐʵÑ飬Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©µç¼«·´Ó¦
¸º¼«£º
 

Õý¼«£º
 

Ô­µç³Ø×Ü·´Ó¦£º
 

£¨2£©µç×ÓÁ÷Ïò£ºÍâµç·ÖУºµç×Ó´Ó
 
¼«¡ú
 
_¼«£»
£¨3£©µçÁ÷·½Ïò£ºÍâµç·ÖУºµçÁ÷´Ó
 
¼«¡ú
 
_¼«£»
£¨4£©
 
¼«·¢ÉúÑõ»¯·´Ó¦£¬
 
¼«·¢Éú»¹Ô­·´Ó¦
£¨5£©µç½âÖÊÈÜÒºÖÐÀë×Ó¶¨ÏòÒƶ¯·½Ïò£ºÑôÀë×Ó¡ú
 
£¬ÒõÀë×Ó¡ú
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨1£©»ð¼ý²Ðº¡Öг£ÏÖºì×ØÉ«ÆøÌ壬ԭÒòΪ£ºN2O4£¨g£©?2NO2£¨g£©µ±Î¶ÈÉý¸ßʱ£¬ÆøÌåÑÕÉ«±äÉÔò·´Ó¦Îª
 
£¨Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±£©·´Ó¦£®
£¨2£©Ò»¶¨Î¶ÈÏ£¬ÉÏÊö·´Ó¦µÄìʱäΪ¡÷H£®ÏÖ½«1mol N2O4³äÈëÒ»ºãѹÃܱÕÈÝÆ÷ÖУ¬ÏÂÁÐʾÒâͼÕýÈ·ÇÒÄÜ˵Ã÷·´Ó¦´ïµ½Æ½ºâ״̬µÄÊÇ
 
£®ÈôÔÚÏàͬζÈÏ£¬ÉÏÊö·´Ó¦¸ÄÔÚÌå»ýΪ1LµÄºãÈÝÃܱÕÈÝÆ÷ÖнøÐУ¬Æ½ºâ³£Êý
 
£¨Ìî¡°Ôö´ó¡±¡°²»±ä¡±»ò¡°¼õС¡±£©£¬·´Ó¦3sºóNO2µÄÎïÖʵÄÁ¿Îª0.6mol£¬Ôò0¡«3sÄÚµÄƽ¾ù·´Ó¦ËÙÂÊv£¨N2O4£©=
 
mol?L-1?s-1£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

II£®£¨ÎïÖʽṹÌ⣩ϱíΪ³¤Ê½ÖÜÆÚ±íµÄÒ»²¿·Ö£¬ÆäÖеÄ×Öĸ´ú±íÏàÓ¦µÄÔªËØ

ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©a·Ö±ðÓëg¡¢dÐγɵĻ¯ºÏÎïÖзеã½Ï¸ßµÄÊÇ
 
£¨Ìѧʽ£©£®
¸Ã·Ö×ÓÖÐdÔ­×ÓµÄÔÓ»¯·½Ê½Îª
 
£®
£¨2£©c¡¢d¡¢e¡¢fÔªËصĵÚÒ»µçÀëÄÜ£¨I1£©ÓÉСµ½´óµÄ˳ÐòΪ
 
£¨ÔªËØ·ûºÅ±íʾ£©£®
£¨3£©ÓɱíÖÐÔªËØÐγɵÄÒ»ÖÖÀë×ÓÓëµ¥ÖÊd3»¥ÎªµÈµç×ÓÌ壬Æ仯ѧʽΪ
 
£®
£¨4£©ÔªËØhµÄ¶þ¼ÛÑôÀë×ӵĻù̬µç×ÓÅŲ¼Ê½Îª
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸