ÌâÄ¿ÄÚÈÝ

Na2CO3ºÍH2O2½áºÏ³É°ô×´¾§Ì壬Na2CO3¡¤xH2O2£¨¹ýÑõ»¯ÇâºÏÏ൱ÓÚË®ºÏ¡£×¢Ò⣺ʹÓøßŨ¶ÈH2O2ʱһ¶¨ÒªÐ¡ÐÄ£¬·ÀÖ¹±¬Õ¨Éú³ÉË®ºÍÑõÆø£©£¬¿ÉÏû¶¾¡¢Æ¯°×»ò×÷O2Ô´¡­¡£ÏÖ³ÆÈ¡Ò»¶¨ÖÊÁ¿µÄNa2CO3¡¤xH2O2¾§Ìå¼ÓÈÈ¡£ÊµÑé½á¹ûÒÔζȺÍÏà¶ÔÖÊÁ¿·ÖÊýʾÓÚÏÂͼ¡£
£¨1£©¸Ã°ô×´¾§ÌåµÄ»¯Ñ§Ê½Na2CO3¡¤xH2O2ÖÐx=_____________¡£
£¨2£©¼ÓÈȹý³ÌÖУ¬ÔÚ141¡æʱ±íÏÖΪ·ÅÈÈ£¬ÆäÔ­Òò¿ÉÄÜÊÇ£º_______________ 
a. Na2CO3¡¤xH2O2·Ö½â·ÅÈÈ                 
b. ²úÉúµÄH2O2·Ö½â·ÅÈÈ  
c. Na2CO3¡¤xH2O2·Ö½âÎüÊÕÈÈÁ¿Ð¡ÓÚ²úÉúµÄH2O2·Ö½â·Å³öµÄÈÈÁ¿
£¨3£©ÊÂʵÉÏ£¬Na2CO3¡¤H2O2£¨x£½1ʱ£©ÊÇNa2CO4¡¤H2O£¨Na2CO4½Ð×ö¹ýÑõ̼ËáÄÆ£©¡£Ï´Ò·ÛÖмÓÈëÊÊÁ¿µÄNa2CO4¿ÉÒÔÌá¸ßÏ´µÓÖÊÁ¿£¬ÆäÄ¿µÄÊǶÔÒÂÎï½øÐÐƯ°×¡¢Ïû¶¾¡£ÊÔÓû¯Ñ§·½³Ìʽ±íʾÉÏÊöÏ´µÓÔ­Àí______________________¡£
£¨4£©Ð´³öNa2CO4ÈÜÒºÓëÏ¡ÁòËá·´Ó¦µÄÀë×Ó·½³Ìʽ__________________¡£
£¨5£©ÏÂÁÐÎïÖʲ»»áʹ¹ý̼ËáÄÆʧЧµÄÊÇ_____________      
A¡¢ MnO2          B¡¢H2S          C¡¢CH3COOH         D¡¢NaHCO3
£¨6£©Na2O2¡¢K2O2¡¢CaO2ÒÔ¼°BaO2¶¼¿ÉÓëËá×÷ÓÃÉú³É¹ýÑõ»¯Ç⣬ĿǰʵÑéÊÒÖÆÈ¡¹ýÑõ»¯ÇâµÄË®ÈÜÒº¿Éͨ¹ýÉÏÊöijÖÖ¹ýÑõ»¯ÎïÓëÊÊÁ¿Ï¡ÁòËá×÷Óã¬×îÊʺϵĹýÑõ»¯ÎïÊÇ___________£¬Ô­ÒòÊÇ_______________¡£ÒªÊ¹»ñµÃµÄ¹ýÑõ»¯Çâ´ÓÆäË®ÈÜÒºÖзÖÀë³öÀ´£¬²ÉÈ¡µÄ´ëÊ©ÊÇ___________________ ¡£
£¨7£©Ð´³ö¹ýÑõËá¸ùµÄµÈµç×ÓÌå____________________£¨Ò»µ½¶þÖÖ£©
£¨1£©x£½1.5 
£¨2£©c
£¨3£©2Na2CO4£½2Na2CO3£«O2¡ü
£¨4£©2CO42£­£«4H£«£½2CO2¡ü£«O2¡ü£«H2O
£¨5£©D
£¨6£©BaO2£»Éú³É¹ýÑõ»¯ÇâºÍÁòËá±µ³Áµí£¬ÂËÈ¥³Áµí¼´µÃ½ÏΪ´¿¾»µÄ¹ýÑõ»¯ÇâË®ÈÜÒº£»¼õѹÕôÁó
£¨7£©F2¡¢Br2¡¢I2¡¢ClO-¡¢BrO-¡¢IO- £¨ÈÎÒâÒ»µ½Á½ÖÖ¼´¿É£©
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ººÃܶû¶Ù£®Ê·ÃÜ˹³öÉúÓÚ1931Ä꣬²»Ðޱ߷ùµÄËû²¢²»ÈÏΪ×Ô¼ºÊÇÒ»ÃûÊÀ½ç¼¶µÄ¿Æѧ¼Ò£¬×Ô³ÆÊÇ¡°³ø·¿»¯Ñ§Ê¦¡±£®¡°³ø·¿»¯Ñ§¡±¿ÉÒÔÀí½âΪÀûÓüÒÍ¥Éú»îÓÃÆ·À´×ö»¯Ñ§ÊµÑ飬¶Ô»¯Ñ§½øÐÐѧϰºÍÑо¿µÄ»î¶¯£®Ä³³ø·¿ÄÚÓÐÒ»º¬NaCl¡¢Na2CO3?10H2OºÍNaHCO3µÄ»ìºÏÎijͬѧÉè¼ÆÈçÏÂʵÑ飬ͨ¹ý²âÁ¿·´Ó¦Ç°ºóC¡¢D×°ÖÃÖÊÁ¿µÄ±ä»¯£¬²â¶¨¸Ã»ìºÏÎïÖи÷×é·ÖµÄÖÊÁ¿·ÖÊý£®

£¨1£©ÊµÑéʱ£¬BÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
Na2CO3?10H2O
  ¡÷  
.
 
Na2CO3+10 H2O¡¢2NaHCO3=Na2CO3+CO2¡ü+H2O
Na2CO3?10H2O
  ¡÷  
.
 
Na2CO3+10 H2O¡¢2NaHCO3=Na2CO3+CO2¡ü+H2O
£®
£¨2£©×°ÖÃC¡¢DÖÐÊ¢·ÅµÄÊÔ¼Á·Ö±ðΪ£ºC
ÎÞË®CaCl2
ÎÞË®CaCl2
£¬D
¼îʯ»Ò
¼îʯ»Ò
£®£¨¹©Ñ¡ÊÔ¼ÁΪ£ºÅ¨ÁòËá¡¢ÎÞË®CaCl2¡¢¼îʯ»Ò£©
£¨3£©E×°ÖÃÖеÄÒÇÆ÷Ãû³ÆÊÇ
¸ÉÔï¹Ü
¸ÉÔï¹Ü
£¬ËüÔÚ¸ÃʵÑéÖеÄÖ÷Òª×÷ÓÃÊÇ
·ÀÖ¹¿ÕÆøÖеÄH2OºÍCO2½øÈëDÖÐ
·ÀÖ¹¿ÕÆøÖеÄH2OºÍCO2½øÈëDÖÐ
£®
£¨4£©Èô½«A×°Öû»³ÉÊ¢·ÅNaOHÈÜÒºµÄÏ´ÆøÆ¿£¬Ôò²âµÃµÄNaClº¬Á¿½«
Æ«µÍ
Æ«µÍ
£¨Ìî¡°Æ«¸ß¡±¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족£©£®
£¨5£©·´Ó¦Ç°£¬ÔÚB×°ÖÃÖÐͨÈë¿ÕÆø¿É¼õÉÙʵÑéÎó²î£¬²Ù×÷·½·¨ÊÇ
¹Ø±Õµ¯»É¼Ðb£¬´ò¿ªµ¯»É¼Ða£¬Ïò×°ÖÃÖÐͨÈë¿ÕÆø
¹Ø±Õµ¯»É¼Ðb£¬´ò¿ªµ¯»É¼Ða£¬Ïò×°ÖÃÖÐͨÈë¿ÕÆø
£®
£¨2012?ÃÅÍ·¹µÇøһģ£©¡°¸»Ãº¡¢Æ¶ÓÍ¡¢ÉÙÆø¡±ÊÇÎÒ¹úÄÜÔ´·¢Õ¹ÃæÁÙµÄÏÖ×´£®Ëæ×ÅÄÜÔ´µÄÈÕÒæ½ôÕÅ£¬·¢Õ¹¡°Ãº»¯¹¤¡±¶ÔÎÒ¹úÄÜÔ´½á¹¹µÄµ÷Õû¾ßÓÐÖØÒªÒâÒ壮ÏÂͼÊÇú»¯¹¤²úÒµÁ´Ö®Ò»£®

¡°½à¾»Ãº¼¼Êõ¡±Ñо¿ÔÚÊÀ½çÉÏÏ൱Æձ飬¿ÆÑÐÈËԱͨ¹ýÏòµØÏÂú²ãÆø»¯Â¯Öн»Ìæ¹ÄÈë¿ÕÆøºÍË®ÕôÆøµÄ·½·¨£¬Á¬Ðø²ú³öÈÈÖµºÜ¸ßµÄú̿ºÏ³ÉÆø£¬ÆäÖ÷Òª³É·ÖÊÇCOºÍH2£®COºÍH2¿É×÷ΪÄÜÔ´ºÍ»¯¹¤Ô­ÁÏ£¬Ó¦ÓÃÊ®·Ö¹ã·º£®
£¨1£©ÒÑÖª£º
C£¨s£©+O2£¨g£©=CO2£¨g£©¡÷H1=-393.5kJ/mol¢Ù
C£¨s£©+H2O£¨g£©=CO£¨g£©+H2£¨g£©¡÷H2=+131.3kJ/mol¢Ú
Ôò·´Ó¦CO£¨g£©+H2£¨g£©+O2£¨g£©=H2O£¨g£©+CO2£¨g£©£¬¡÷H=
-524.8
-524.8
kJ/mol£®ÔÚ±ê×¼×´¿öÏ£¬33.6LµÄú̿ºÏ³ÉÆøÓëÑõÆøÍêÈ«·´Ó¦Éú³ÉCO2ºÍH2O£¬·´Ó¦¹ý³ÌÖÐתÒÆ
3
3
mol e-£®
£¨2£©ÔÚÒ»ºãÈݵÄÃܱÕÈÝÆ÷ÖУ¬ÓÉCOºÍH2ºÏ³É¼×´¼£ºCO£¨g£©+2H2£¨g£©?CH3OH£¨g£©
¢ÙÏÂÁÐÇéÐÎÄÜ˵Ã÷ÉÏÊö·´Ó¦ÒѴﵽƽºâ״̬µÄÊÇ
ad
ad

a£®Ìåϵѹǿ±£³Ö²»±ä
b£®ÃܱÕÈÝÆ÷ÖÐCO¡¢H2¡¢CH3OH£¨g£©3ÖÖÆøÌå¹²´æ
c£®CH3OHÓëH2ÎïÖʵÄÁ¿Ö®±ÈΪ1£º2
d£®Ã¿ÏûºÄ1mol COµÄͬʱÉú³É2molH2
¢ÚCOµÄƽºâת»¯ÂÊ£¨¦Á£©Óëζȡ¢Ñ¹Ç¿µÄ¹ØϵÈçÏÂͼËùʾ£®

A¡¢BÁ½µãµÄƽºâ³£Êý
Ò»Ñù
Ò»Ñù
£¨Ìî¡°Ç°Õß¡±¡¢¡°ºóÕß¡±»ò¡°Ò»Ñù¡±£©´ó£»´ïµ½A¡¢CÁ½µãµÄƽºâ״̬ËùÐèµÄʱ¼ätA
´óÓÚ
´óÓÚ
tC£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©£®ÔÚ²»¸Ä±ä·´Ó¦ÎïÓÃÁ¿µÄÇé¿öÏ£¬ÎªÌá¸ßCOµÄת»¯ÂʿɲÉÈ¡µÄ´ëÊ©ÊÇ
½µÎ¡¢¼Óѹ¡¢½«¼×´¼´Ó»ìºÏÌåϵÖзÖÀë³öÀ´
½µÎ¡¢¼Óѹ¡¢½«¼×´¼´Ó»ìºÏÌåϵÖзÖÀë³öÀ´
£¨´ð³öÁ½µã¼´¿É£©£®
£¨3£©¹¤×÷ζÈ650¡æµÄÈÛÈÚÑÎȼÁϵç³Ø£¬ÊÇÓÃú̿Æø£¨CO¡¢H2£©×÷¸º¼«È¼Æø£¬¿ÕÆøÓëCO2µÄ»ìºÏÆøÌåΪÕý¼«È¼Æø£¬ÓÃÒ»¶¨±ÈÀýµÄLi2CO3ºÍNa2CO3µÍÈÛµã»ìºÏÎï×öµç½âÖÊ£¬ÒÔ½ðÊôÄø£¨È¼Áϼ«£©Îª´ß»¯¼ÁÖƳɵģ®¸º¼«µÄµç¼«·´Ó¦Ê½Îª£ºCO+H2-4e-+2CO32-=3CO2+H2O£»Ôò¸Ãµç³ØµÄÕý¼«·´Ó¦Ê½Îª
O2+4e-+2CO2=2CO32-
O2+4e-+2CO2=2CO32-
£®

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø