(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
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿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 ÎÂ¶È |
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>>
°Ù¶ÈÖÂÐÅ - 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