ÏÖÓÐ D¡¢E¡¢FÁùÖÖÎïÖÊ£¬ÆäÖÐA¡¢B¡¢CÊÇÓɶÌÖÜÆÚÔªËØ×é³ÉµÄ³£¼ûµ¥ÖÊ£¬D·Ö×Óº¬ÓÐ10¸öµç×Ó£¬FÊÇÒ»ÖÖ²»º¬½ðÊôÔªËصÄÑΣ®Ò»¶¨Ìõ¼þÏÂת»¯¹ØϵÈçÏ£ºÇë»Ø´ð£º
£¨1£©×é³ÉAµÄÔªËØÔÚÖÜÆÚ±íÖеÄλÖÃÊÇ
 
£®
£¨2£©ÏÂÁйØÓÚDµÄ˵·¨ÕýÈ·µÄÊÇ
 
£¨ÌîÐòºÅ£©£®
a£®Óл¹Ô­ÐÔ                
b£®ÆäË®ÈÜÒºÄÜʹ·Ó̪ÈÜÒº±äºìÉ«
c£®¼«Ò×ÈÜÓÚË®              
d£®ÆäË®ÈÜÒºÄÜʹƷºìÈÜÒºÍÊÉ«
£¨3£©FÓëÏûʯ»Ò¹²ÈÈ·´Ó¦£¬¿ÉÓÃÓÚÖÆÈ¡DÆøÌ壬·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
 
£®
£¨4£©EµÄŨÈÜÒººÍÉÏÊöÎïÖÊÖÐ
 
ÈÜÒºËùº¬ÒõÀë×ÓÏàͬ£¬¸ÃÀë×ӵļìÑé·½·¨ÊÇ
 
£®
¿¼µã£ºÎÞ»úÎïµÄÍƶÏ
רÌ⣺ÍƶÏÌâ
·ÖÎö£ºFÊÇÒ»ÖÖ²»º¬½ðÊôÔªËصÄÑΣ¬ÔòFΪï§ÑΣ¬A¡¢B¡¢CÊÇÓɶÌÖÜÆÚÔªËØ×é³ÉµÄ³£¼ûµ¥ÖÊ£¬D·Ö×Óº¬ÓÐ10¸öµç×Ó£¬½áºÏת»¯¹Øϵ¿ÉÖª£¬BΪH2£¬AΪN2£¬DΪNH3£¬CΪCl2£¬EΪHCl£¬ÔòFΪNH4Cl£¬ÒÔ´ËÀ´½â´ð£®
½â´ð£º ½â£ºFÊÇÒ»ÖÖ²»º¬½ðÊôÔªËصÄÑΣ¬ÔòFΪï§ÑΣ¬A¡¢B¡¢CÊÇÓɶÌÖÜÆÚÔªËØ×é³ÉµÄ³£¼ûµ¥ÖÊ£¬D·Ö×Óº¬ÓÐ10¸öµç×Ó£¬½áºÏת»¯¹Øϵ¿ÉÖª£¬BΪH2£¬AΪN2£¬DΪNH3£¬CΪCl2£¬EΪHCl£¬ÔòFΪNH4Cl£¬
£¨1£©×é³ÉAµÄÔªËØΪN£¬Î»ÓÚµÚ¶þÖÜÆÚµÚVA×壬¹Ê´ð°¸Îª£ºµÚ¶þÖÜÆÚµÚVA×壻
£¨2£©a£®°±ÆøÖÐNÔªËصĻ¯ºÏ¼ÛΪ-3¼Û£¬Îª×îµÍ¼Û£¬Óл¹Ô­ÐÔ£¬¹ÊaÕýÈ·£»
b£®ÆäË®ÈÜÒºÖÐһˮºÏ°±µçÀë³öÇâÑõ¸ùÀë×Ó£¬ÏÔ¼îÐÔ£¬ÄÜʹ·Ó̪ÈÜÒº±äºìÉ«£¬¹ÊbÕýÈ·£»
c£®°±ÆøÓëË®ÐγÉÇâ¼ü£¬¼«Ò×ÈÜÓÚË®£¬¹ÊcÕýÈ·£»
d£®ÆäË®ÈÜÒº²»¾ßÓÐƯ°×ÐÔ£¬Ôò²»ÄÜʹƷºìÈÜÒºÍÊÉ«£¬¹Êd´íÎó£»
¹Ê´ð°¸Îª£ºabc£»
£¨3£©FÓëÏûʯ»Ò¹²ÈÈ·´Ó¦£¬¿ÉÓÃÓÚÖÆÈ¡DÆøÌ壬·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ£º2NH4Cl+Ca£¨OH£©2
  ¡÷  
.
 
CaCl2+2H2O+2NH3¡ü£¬¹Ê´ð°¸Îª£º2NH4Cl+Ca£¨OH£©2
  ¡÷  
.
 
CaCl2+2H2O+2NH3¡ü£»
£¨4£©EΪHCl£¬ÆäÈÜÒººÍNH4ClËùº¬ÒõÀë×ÓÏàͬΪCl-£¬¼ìÑéCl-µÄ·½·¨Îª£ºÈ¡ÈÜÒºÉÙÐíÓÚÊÔ¹ÜÖУ¬¼ÓÈëÏ¡ÏõËáËᣬÔÙ¼ÓÈëÏõËáÒøÈÜÒº£¬Óа×É«³ÁµíÉú³É£¬ËµÃ÷º¬ÓÐCl-£¬
¹Ê´ð°¸Îª£ºNH4Cl£»È¡ÈÜÒºÉÙÐíÓÚÊÔ¹ÜÖУ¬¼ÓÈëÏ¡ÏõËáËᣬÔÙ¼ÓÈëÏõËáÒøÈÜÒº£¬Óа×É«³ÁµíÉú³É£¬ËµÃ÷º¬ÓÐCl-£®
µãÆÀ£º±¾Ì⿼²éÎÞ»úÎïµÄÍƶϣ¬¸ù¾ÝFΪï§Ñμ°DΪ10µç×Ó΢Á£µÈ½øÐÐÍƶϣ¬²àÖطǽðÊô¼°Æ仯ºÏÎïÐÔÖÊ¡¢Ï໥ת»¯µÄ¿¼²é£¬ÌâÄ¿ÄѶÈÖеȣ®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¶ÔȼúÑÌÆøºÍÆû³µÎ²Æø½øÐÐÍÑÏõ¡¢ÍÑ̼ºÍÍÑÁòµÈ´¦Àí£¬¿ÉʵÏÖÂÌÉ«»·±£¡¢½ÚÄܼõÅŵÈÄ¿µÄ£®Æû³µÎ²ÆøÍÑÏõÍÑ̼µÄÖ÷ÒªÔ­ÀíΪ£º2NO£¨g£©+2CO£¨g£©?´ß»¯¼ÁN2£¨g£©+2CO2£¨g£©+Q£¨Q£¾0£©£®Ò»¶¨Ìõ¼þÏ£¬ÔÚÒ»ÃܱÕÈÝÆ÷ÖУ¬Óô«¸ÐÆ÷²âµÃ¸Ã·´Ó¦ÔÚ²»Í¬Ê±¼äµÄNOºÍCOŨ¶ÈÈçÏÂ±í£º
ʱ¼ä/s012345
c£¨NO£©/mol?L-11.00¡Á10-34.50¡Á10-42.50¡Á10-41.50¡Á10-41.00¡Á10-41.00¡Á10-4
c£¨CO£©/mol?L-13.60¡Á10-33.05¡Á10-32.85¡Á10-32.75¡Á10-32.70¡Á10-32.70¡Á10-3
£¨1£©Ð´³ö¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽ
 
£®
£¨2£©Ç°2sÄÚµÄƽ¾ù·´Ó¦ËÙÂʦԣ¨N2£©=
 
mol/£¨L?s£©£»´ïµ½Æ½ºâʱ£¬COµÄת»¯ÂÊΪ
 
£®
£¨3£©ÏÂÁÐÃèÊöÖÐÄÜ˵Ã÷ÉÏÊö·´Ó¦ÒÑ´ïƽºâµÄÊÇ
 

a£®2¦ÔÕý£¨NO£©=¦ÔÄ棨N2£©
b£®ÈÝÆ÷ÖÐÆøÌåµÄƽ¾ù·Ö×ÓÁ¿²»Ëæʱ¼ä¶ø±ä»¯
c£®ÈÝÆ÷ÖÐÆøÌåµÄÃܶȲ»Ëæʱ¼ä¶ø±ä»¯
d£®ÈÝÆ÷ÖÐCOµÄת»¯Âʲ»ÔÙ·¢Éú±ä»¯£®
£¨4£©²ÉÓõÍγôÑõÑõ»¯ÍÑÁòÍÑÏõ¼¼Êõ£¬Í¬Ê±ÎüÊÕSO2ºÍNOx£¬»ñµÃ£¨NH4£©2SO4µÄÏ¡ÈÜÒº£®Íù£¨NH4£©2SO4ÈÜÒºÖÐÔÙ¼ÓÈëÉÙÁ¿ £¨NH4£©2SO4¹ÌÌ壬
c(NH4+)
c(SO42-)
掙櫮
 
£¨Ìî¡°±ä´ó¡±¡¢¡°²»±ä¡±»ò¡°±äС¡±£©
£¨5£©ÓÐÎïÖʵÄÁ¿Å¨¶ÈÏàµÈµÄÈýÖÖï§ÑÎÈÜÒº£º¢ÙNH4Cl    ¢ÚNH4HCO3    ¢ÛNH4HSO4£¬ÕâÈýÖÖÈÜÒºÖÐË®µÄµçÀë³Ì¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ
 
£¨Ìî±àºÅ£©£®
£¨6£©ÏòBaCl2ÈÜÒºÖÐͨÈë×ãÁ¿SO2ÆøÌ壬ûÓгÁµíÉú³É£¬¼ÌÐøµÎ¼ÓÒ»¶¨Á¿µÄ°±Ë®ºó£¬Éú³ÉBaSO3³Áµí£®ÓõçÀëƽºâÔ­Àí½âÊÍÉÏÊöÏÖÏó£®
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐÓлúÎïÖУ¬ÊôÓڸ߷Ö×Ó»¯ºÏÎïµÄÊÇ£¨¡¡¡¡£©
A¡¢ÓÍÖ¬B¡¢ÕáÌÇ
C¡¢ÆÏÌÑÌÇD¡¢µ°°×ÖÊ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÓÃÊʵ±ÈÜÒº°Ñ3.31gijÌú¿óʯÑùÆ·Èܽ⣬Ȼºó¼Ó¹ýÁ¿¼îÈÜÒº£¬Éú³É³Áµí£¬ÔÙ×ÆÉÕ³Áµí£¬µÃ2.40g Fe2O3£®ÒÑÖª¸ÃÌú¿óʯÖÐÌúµÄÑõ»¯ÎïµÄÖÊÁ¿·ÖÊýΪ70%£®ÊÔ¼ÆË㣺
£¨1£©¸ÃÌú¿óʯÖÐÌúµÄÖÊÁ¿·ÖÊý
 
£®
£¨2£©¸ÃÌú¿óʯÖÐÌúµÄÑõ»¯ÎïµÄ»¯Ñ§Ê½
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÅðÔªËØÔÚ»¯Ñ§ÖÐÓкÜÖØÒªµÄµØ룬Åð¼°Æ仯ºÏÎï¹ã·ºÓ¦ÓÃÓÚÓÀ´Å²ÄÁÏ¡¢³¬µ¼²ÄÁÏ¡¢¸»È¼ÁϲÄÁÏ¡¢¸´ºÏ²ÄÁϵȸßвÄÁÏÁìÓò£®
£¨1£©Èý·ú»¯ÅðÔÚ³£Î³£Ñ¹ÏÂΪ¾ßÓд̱Ƕñ³ôºÍÇ¿´Ì¼¤ÐÔµÄÎÞÉ«Óж¾¸¯Ê´ÐÔÆøÌ壬Æä·Ö×ÓµÄÁ¢Ìå¹¹ÐÍΪ
 
£¬BÔ­×ÓµÄÔÓ»¯ÀàÐÍΪ
 
£®
£¨2£©Á×»¯ÅðÊÇÒ»ÖÖÊܵ½¸ß¶È¹Ø×¢µÄÄÍĥͿÁÏ£¬Ëü¿ÉÓÃ×÷½ðÊôµÄ±íÃæ±£»¤²ã£®Í¼1ÊÇÁ×»¯Åð¾§ÌåµÄ¾§°ûʾÒâͼ£¬ÔòÁ×»¯ÅðµÄ»¯Ñ§Ê½Îª
 
£¬¸Ã¾§ÌåµÄ¾§ÌåÀàÐÍÊÇ
 
£®

£¨3£©ÅðËᣨH3B03£©ÊÇÒ»ÖÖƬ²ã×´½á¹¹µÄ°×É«¾§Ì壬²ãÄÚµÄH3B03·Ö×Ó¼äͨ¹ýÇâ¼üÏàÁ¬£¨Èçͼ2£©£®
¢ÙÅðËá·Ö×ÓÖÐB×îÍâ²ãÓÐ
 
¸öµç×Ó£¬1molH3B03µÄ¾§ÌåÖÐÓÐ
 
 molÇâ¼ü£®
¢ÚÅðËáÈÜÓÚË®Éú³ÉÈõµç½âÖÊһˮºÏÅðËáB£¨OH£©3?H2 O£¬ËüµçÀëÉú³ÉÉÙÁ¿[B£¨OH£©4]Ò»ºÍH+£¬ÔòÅðËáΪ
 
 ÔªËᣬ[B£¨OH£©4]Ò»º¬ÓеĻ¯Ñ§¼üÀàÐÍΪ
 
£®
£¨4£©H3P04µÄK1¡¢K2¡¢K3·Ö±ðΪ7.6¡Á10-3¡¢6.3¡Á10-8¡¢4.4¡Á10-13£¬ÏõËáÍêÈ«µçÀ룬¶øÑÇÏõËáK=5.1¡Á10-4£¬Çë¸ù¾Ý½á¹¹ÓëÐÔÖʵĹØϵ½âÊÍ£º
¢ÙH3P04µÄK1Ô¶´óÓÚK2µÄÔ­Òò
 
£» ¢ÚÏõËá±ÈÑÇÏõËáËáÐÔÇ¿µÄÔ­Òò
 
£®
£¨5£©NiO¾§Ìå½á¹¹ÓëNaCl¾§ÌåÀàËÆ£¬Æ侧°ûµÄÀⳤΪa cm£¬Ôò¸Ã¾§ÌåÖоàÀë×î½üµÄÁ½¸öÑôÀë×Ӻ˼äµÄ¾àÀëΪ
 
£¨Óú¬ÓÐaµÄ´úÊýʽ±íʾ£©£®ÔÚÒ»¶¨Î¶ÈÏ£¬NiO¾§Ìå¿ÉÒÔ×Ô·¢µØ·ÖÉ¢²¢Ðγɡ°µ¥·Ö×Ӳ㡱£¨Èçͼ3£©£¬¿ÉÒÔÈÏΪÑõÀë×Ó×÷ÃÜÖµ¥²ãÅÅÁУ¬ÄøÀë×ÓÌî³äÆäÖУ¬ÁÐʽ²¢¼ÆËãÿƽ·½Ã×Ãæ»ýÉÏ·ÖÉ¢µÄ¸Ã¾§ÌåµÄÖÊÁ¿Îª
 
g£¨ÑõÀë×ӵİ뾶Ϊ1.40¡Á10-10m£¬
3
¡Ö1.732£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

±½ÒÒËáͬ·ÖÒì¹¹ÌåÖÐÊôÓÚõ¥ÀàµÄº¬Óб½»·µÄÓм¸ÖÖ£¨¡¡¡¡£©
A¡¢4ÖÖB¡¢5ÖÖC¡¢6ÖÖD¡¢7ÖÖ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐÖеÄʵÑéÒ»¶¨²»»á³öÏÖ³ÁµíÏÖÏóµÄÊÇ£¨¡¡¡¡£©
A¡¢CO2ÆøÌåͨÈëNaAlO2ÈÜÒºÖÐ
B¡¢SO2ÆøÌåͨÈëBaCl2ÈÜÒºÖÐ
C¡¢CO2ÆøÌåͨÈë±¥ºÍNa2CO3ÈÜÒºÖÐ
D¡¢SO2ÆøÌåͨÈëBa£¨OH£©2ÈÜÒºÖÐ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐÓйصç½âÖÊÈÜÒºÖÐ΢Á£µÄÎïÖʵÄÁ¿Å¨¶È¹ØϵÕýÈ·µÄÊÇ£¨¡¡¡¡£©
¢ÙÔÚO£®1mol/lNaHC03ÈÜÒºÖУºc£¨Na+£©£¾c£¨HCO3-£©£¾c£¨CO32-£©£¾c£¨H2C03£©
¢ÚÔÚ0.1mol/lNa2C03ÈÜÒºÖУºc£¨OH-£©-c£¨H+£©=c£¨HCO3-£©+2c£¨H2C03£©
¢ÛÏò0.1mol/lNaHC03ÈÜÒºÖмÓÈëµÈÌå»ýO£®1mol/lNaOHÈÜÒº£ºc£¨CO32-£©£¾c£¨HCO3-£©£¾c£¨OH-£©£¾c£¨H+£©
¢Ü³£ÎÂÏ£¬CH3COONaºÍCH3COOH»ìºÏÈÜÒº[PH=7£¬c£¨Na+£©=O£®1mol/l]£ºc£¨Na+£©=c£¨CH3COO-£©£¾c£¨H+£©=c£¨OH-£©
A¡¢¢Ù¢ÜB¡¢¢Ú¢ÜC¡¢¢Ù¢ÛD¡¢¢Ú¢Û

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁо§ÌåÄ£ÐͶÔÓ¦µÄÎïÖÊÈÛ»¯Ê±ÆÆ»µ¹²¼Û¼üµÄÊÇ£¨¡¡¡¡£©
A¡¢½ð¸ÕʯģÐÍ
B¡¢
¸É±ùÄ£ÐÍ
C¡¢
̼60Ä£ÐÍ
D¡¢
ÂÈ»¯ÄÆÄ£ÐÍ

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸