(10·Ö)ÁòËáÔÚÉú²úÖÐÓй㷺µÄÓ¦Óã¬Ä³³§ÒÔÁòÌú¿óΪÖ÷ÒªÔ­ÁÏÉú²úÁòËá¡£

ÒÑÖª£º¢Ù550¡æʱ£¬  

    ¢ÚÒ»¶¨Î¶ÈÏ£¬ÁòÌú¿óÔÚ¿ÕÆøÖÐìÑÉÕ¿ÉÄÜ·¢ÉúÏÂÁз´Ó¦(Éè¿ÕÆøÖÐÓëµÄÌå»ý

    ±ÈΪ4£º1)£º,

    (1) 550¡æʱ£¬6£®4 g Óë×ãÁ¿³ä·Ö·´Ó¦Éú³É£¬·Å³öÈÈÁ¿  

 9£®83 kJ

  (Ìî¡°´óÓÚ¡±¡¢¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±)o

    (2)Óûʹ·´Ó¦¢ÙµÄƽºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬ÏÂÁдëÊ©¿ÉÐеÄÊÇ     ¡£(Ìî×Öĸ)

    a£®Ïòƽºâ»ìºÏÎïÖгäÈë    b£®Ïòƽºâ»ìºÏÎïÖгäÈë

    c£®¸Ä±ä·´Ó¦µÄ´ß»¯¼Á         d£®½µµÍ·´Ó¦µÄζÈ

  (3)ΪʹÍêÈ«Éú³É£¬Éú²úʱҪʹÓùýÁ¿µÄ

    ¿ÕÆø£¬Ôòµ±¿ÕÆø¹ýÁ¿50£¥Ê±£¬ËùµÃ¯ÆøÖеÄ

    Ìå»ý·ÖÊýÊǶàÉÙ?

  (4)720 g´¿¾»µÄÔÚ¿ÕÆøÖÐÍêÈ«ìÑÉÕ£¬ËùµÃ¹ÌÌå

    Öк͵ÄÎïÖʵÄÁ¿Ö®±Èn()£ºn()=6: £¬

    ´ËʱÏûºÄ¿ÕÆøΪmol¡£

    ¢ÙÊÔд³öÒ×Óë¿ÚµÄ¹Øϵʽ£º    ¡£

    ¢ÚÇëÔÚÓÒͼÖл­³öÓëµÄ¹ØϵÇúÏß¡£

¹²10·Ö¡£Ã¿¸öÎÊÌâ2·Ö¡£

£¨1£©Ð¡ÓÚ      £¨2£©bd

£¨3£©ÉèFeS2Ϊ1 mol£¬ÍêÈ«ìÑÉÕÐèÒªµÄn(O2) = 2.75 mol£¬Éú³Én(SO2) = 2 mol£»

¿ÕÆø¹ýÁ¿50%£¬ËùÐè¿ÕÆøΪ£º2.75 ¡Â 0.2 ¡Á 1.5 = 20.625 mol£»

SO2 Ìå»ý·ÖÊýΪ£º2 ¡Â ( 20.625 + 2£­ 2.75) = 10.0%

£¨4£©¢Ù b = 2.5a + 60

¢Ú ¶ËµãΪ£¨8£¬80£©¡¢£¨9£¬82.5£©Ö®¼äµÄÒ»ÌõÏ߶Î

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013½ìºþ±±Ê¡²¿·ÖÖصãÖÐѧ£¨ÌìÃÅÖÐѧµÈ£©¸ßÈýÉÏѧÆÚÆÚÖÐÁª¿¼»¯Ñ§ÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£ºÌî¿ÕÌâ

(6·Ö) ÂÈÆøÔÚÉú²úÉú»îÖÐÓ¦Óù㷺¡£
(1)¹¤ÒµÉÏ¿ÉÓÃMnSO4ÈÜÒºÎüÊÕÂÈÆø£¬»ñµÃMn2O3£¬Mn2O3¹ã·ºÓ¦ÓÃÓÚµç×Ó¹¤Òµ¡¢Ó¡È¾¹¤ÒµµÈÁìÓò¡£Çëд³ö¸Ã»¯Ñ§·´Ó¦µÄ»¯Ñ§·½³Ìʽ               ¡£
(2)º£µ×Ô̲Ø×ŷḻµÄÃ̽áºË¿ó£¬ÆäÖ÷Òª³É·ÖÊÇMnO2 ¡£1991ÄêÓÉAllenµÈÈËÑо¿£¬ÓÃÁòËáÁÜÏ´ºóʹÓò»Í¬µÄ·½·¨¿ÉÖƱ¸´¿¾»µÄMnO2£¬ÆäÖƱ¸¹ý³ÌÈçÏÂͼËùʾ£º

¢Ù²½ÖèIÖУ¬ÊÔ¼Á¼×±ØÐë¾ßÓеÄÐÔÖÊÊÇ         (ÌîÐòºÅ)¡£
a. Ñõ»¯ÐÔ       b.»¹Ô­ÐÔ         c.ËáÐÔ   
¢Ú²½Öè¢óÖУ¬ÒÔNaClO3ΪÑõ»¯¼Á£¬µ±Éú³É0.050 mol MnO2ʱ£¬ÏûºÄ0.10 mol¡¤L£­1µÄNaClO3ÈÜÒº200 mL £¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012-2013ѧÄêºþ±±Ê¡£¨ÌìÃÅÖÐѧµÈ£©¸ßÈýÉÏѧÆÚÆÚÖÐÁª¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

(6·Ö) ÂÈÆøÔÚÉú²úÉú»îÖÐÓ¦Óù㷺¡£

(1)¹¤ÒµÉÏ¿ÉÓÃMnSO4ÈÜÒºÎüÊÕÂÈÆø£¬»ñµÃMn2O3£¬Mn2O3¹ã·ºÓ¦ÓÃÓÚµç×Ó¹¤Òµ¡¢Ó¡È¾¹¤ÒµµÈÁìÓò¡£Çëд³ö¸Ã»¯Ñ§·´Ó¦µÄ»¯Ñ§·½³Ìʽ                ¡£

(2)º£µ×Ô̲Ø×ŷḻµÄÃ̽áºË¿ó£¬ÆäÖ÷Òª³É·ÖÊÇMnO2 ¡£1991ÄêÓÉAllenµÈÈËÑо¿£¬ÓÃÁòËáÁÜÏ´ºóʹÓò»Í¬µÄ·½·¨¿ÉÖƱ¸´¿¾»µÄMnO2£¬ÆäÖƱ¸¹ý³ÌÈçÏÂͼËùʾ£º

¢Ù²½ÖèIÖУ¬ÊÔ¼Á¼×±ØÐë¾ßÓеÄÐÔÖÊÊÇ          (ÌîÐòºÅ)¡£

a. Ñõ»¯ÐÔ        b.»¹Ô­ÐÔ          c.ËáÐÔ   

¢Ú²½Öè¢óÖУ¬ÒÔNaClO3ΪÑõ»¯¼Á£¬µ±Éú³É0.050 mol MnO2ʱ£¬ÏûºÄ0.10 mol¡¤L£­1 µÄNaClO3ÈÜÒº200 mL £¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ________________¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄêɽÎ÷Ê¡¸ßÈýµÚÒ»´Î½×¶ÎÐÔÕï¶Ï¿¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ

(10·Ö)µ¨í¶(CuSO4•5H20)¹ã·ºÓÃÓÚµç¶Æ¹¤ÒÕ£¬ÔÚÒ½Ò©ÉÏÓÃ×öÊÕÁ²¼Á¡¢·À¸¯¼ÁºÍ´ßͼÁ¡£ÒÔÏÂÊÇÓÃÍ­·ÛÑõ»¯·¨Éú²úµ¨·¯µÄÁ÷³Ìͼ£º

(1 )д³öÈܽâ¹ý³ÌÖз¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ£º____________

(2)1 £º1µÄH2SO4ÈÜÒºÊÇÓÉlÌå»ý98%µÄH2SO4Óë1Ìå»ýË®»ìºÏ¶ø³É¡£ÅäÖƸÃÁòËáËùÐèµÄ¹èËáÑÎÖÊÒÇÆ÷³ý²£Á§°ôÍâ.»¹ÐèÒª______¡¢______¡£

(3)ÒÑÖªÑõ»¯ÑÇÍ­(Cu2O)ÓëÏ¡H2SO4·´Ó¦ÓÐCuSO4ºÍCuÉú³É¡£¼ÙÉ豺ÉÕºó¹ÌÌåÖ»º¬Í­µÄÑõ»¯Îï.Ϊ¼ìÑé¸Ã¹ÌÌåµÄ³É·Ö.ÏÂÁÐʵÑéÉè¼ÆºÏÀíµÄÊÇ__________________(Ìî×Öĸ£©¡£

A. ¼ÓÈËÏ¡H2SO4£¬ÈôÈÜÒº³ÊÀ¶É«£¬ËµÃ÷¹ÌÌåÖÐÒ»¶¨ÓÐCuO

B. ¼ÓÈËÏ¡H2SO4£¬ÈôÓкìÉ«³ÁµíÎïÉú³É£¬ËµÃ÷¹ÌÌåÖÐÒ»¶¨ÓÐCu2O

C. ¼ÓÈËÏ¡HNO3£¬ÈôÓÐÎÞÉ«ÆøÌå(Ëæ¼´±ä³Éºì×ØÉ«£º)²úÉú£¬ËµÃ÷¹ÌÌåÖÐÓÐCu2O

D. ¼ÓÈËÏ¡HNO3£¬Èô¹ÌÌåÈ«²¿Èܽ⣬˵Ã÷¹ÌÌåÖÐûÓÐCu2O

(4)È¡2.50gµ¨í¶ÑùÆ·£¬Öð½¥ÉýμÓÈȷֽ⣬·Ö½â¹ý³ÌµÄÈÈÖØÇúÏß(ÑùÆ·ÖÊÁ¿Ëæζȱ仯µÄÇúÏß)ÈçͼËùʾ¡£

¢Ùaµãʱ¹ÌÌåÎïÖʵĻ¯Ñ§Ê½Îª______£¬cµãʱ¹ÌÌåÎïÖʵĻ¯Ñ§Ê½Îª______¡£

¢Ú10000Cʱ·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_________________________

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013½ìÖØÇìÊи߶þÉÏѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ

(14·Ö)ÁòËṤҵÖÐ2SO2(g)£«O2(g)´ß»¯¼Á¡÷2SO3(g)£»¦¤H<0(·ÅÈÈ·´Ó¦)ÓйØʵÑéÊý¾ÝÈçÏ£º

ѹǿ

 

SO2µÄ

ת»¯ÂÊ

ζÈ

1¡Á105 Pa

5¡Á105 Pa

10¡Á105 Pa

50¡Á105 Pa

100¡Á105 Pa

450 ¡æ

97.5%

98.9%

99.2%

99.6%

99.7%

550 ¡æ

85.6%

92.9%

94.9%

97.7%

98.3%

(1)ÔÚÉú²úÖг£ÓùýÁ¿µÄ¿ÕÆøÊÇΪÁË________¡£

(2)¸ßζԸ÷´Ó¦ÓкÎÓ°Ï죿________£¬Êµ¼ÊÉú²úÖвÉÓÃ400¡«500 ¡æµÄζȳýÁË¿¼ÂÇËÙÂÊÒòËØÍ⣬»¹¿¼Âǵ½________¡£

(3)Ôö´óѹǿ¶ÔÉÏÊö·´Ó¦ÓкÎÓ°Ï죿____£¬µ«¹¤ÒµÉÏÓÖ³£²ÉÓó£Ñ¹½øÐз´Ó¦£¬ÆäÔ­ÒòÊÇ______________¡£

(4)³£ÓÃŨH2SO4¶ø²»ÓÃË®ÎüÊÕSO3ÊÇÓÉÓÚ___ ___£¬Î²ÆøÖÐSO2±ØÐë»ØÊÕ£¬Ö÷ÒªÊÇΪÁË________¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸