ÌâÄ¿ÄÚÈÝ

ÓÃÏÂͼװÖÿÉÒÔ½øÐвⶨSO2ת»¯³ÉSO3µÄת»¯ÂÊ¡£ÒÑÖªSO3µÄÈÛµãÊÇ16.8¡æ£¬·ÐµãÊÇ44.8¡æ¡£ÆøÌå·¢Éú×°ÖÃÖÐËùÉæ¼°·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºNa2SO3(s) + H2SO4 ¡ú Na2SO4 + H2O + SO2¡ü¡£

¸ù¾ÝÌâÒâÍê³ÉµÚ35~40Ì⣺
35£®¸ù¾ÝʵÑéÐèÒª£¬Ó¦¸ÃÔÚ¢ñ¡¢¢ò¡¢¢ó´¦Á¬½ÓºÏÊʵÄ×°Öá£Çë´ÓÏÂͼA¡«E×°ÖÃÖÐÑ¡Ôñ×îÊʺÏ×°Öò¢½«ÆäÐòºÅÌîÈëÏÂÃæµÄ¿Õ¸ñÖС£
¢ñ¡¢¢ò¡¢¢ó´¦Á¬½ÓµÄ×°Ö÷ֱðÊÇ________¡¢_________¡¢_________¡£

36£®´ÓÒÒ´¦¾ùÔÈͨÈëO2£¬ÎªÊ¹SO2ÓнϸߵÄת»¯ÂÊ£¬ÊµÑéÖÐÔÚ£º¢Ù¼ÓÈÈ´ß»¯¼Á£»¢ÚµÎ¼ÓŨÁòËáµÄ˳ÐòÖУ¬Ó¦²ÉÈ¡µÄ²Ù×÷ÊÇÏÈ_______ºó_________(Ìî±àºÅ)¡£
37£®ÔÚÖÆÈ¡SO3µÄ¹ý³ÌÖУ¬Èôζȹý¸ß£¬SO2µÄת»¯Âʻᠠ        (Ìî¡°Éý¸ß¡±¡¢¡°²»±ä¡±»ò¡°½µµÍ¡±£©¡£
38£®ÊµÑé½áÊøºó£¬Èç¹ûÈ¡³öÊÕ¼¯SO3µÄÒÇÆ÷²¢³¨¿Ú¶ÖÃÓÚ¿ÕÆøÖУ¬Äܹ»¿´µ½µÄÏÖÏóÊÇ£º
              __________¡£
39£®¸ÃʵÑéÒ»°ãʹÓýÏŨµÄÁòËáÖÆSO2£¬²»¿ÉʹÓÃÏ¡ÁòËᣬ¼òÊöÆäÔ­Òò£º___________¡£
40£®ÓÃa mol Na2SO3·ÛÄ©Óë×ãÁ¿½ÏŨµÄÁòËá½øÐдËʵÑ飬µ±·´Ó¦½áÊøʱ£¬¼ÌÐøͨÈëO2Ò»¶Îʱ¼äºó£¬²âµÃ×°ÖâóÔöÖØÁËb g£¬ÔòʵÑéÖÐSO2µÄת»¯ÂÊΪ             £¨Óú¬a¡¢bµÄ´úÊýʽÌîд£©¡£

35£®B¡¢A ¡¢E £¨3·Ö£©
36£®¢Ù£»¢Ú¡££¨2·Ö£©
37£®½µµÍ¡££¨1·Ö£©
38£®ÊԹܿÚÓа×Îí¡££¨2·Ö£©
39£®Ï¡ÁòËẬˮ½Ï¶à£¬SO2Ò×ÈÜÓÚË®¡££¨2·Ö£©
40£®(1£­) ¡Á100% £¨2·Ö£©

35£®¢ñ×°ÖñØÐëÒª¶Ô¶þÑõ»¯Áò½øÐиÉÔ¿ÉÒÔÓÃŨÁòËáÀ´¸ÉÔï¶þÑõ»¯ÁòÆøÌ壻SO3µÄÈÛµãÊÇ16£¬8¡æ£¬¿ÉÒÔÓñùË®ÀäÈ´À´»ñµÃÈýÑõ»¯Áò£»Î´·´Ó¦µôµÄ¶þÑõ»¯Áò¶Ô¿ÕÆø»á²úÉúÎÛȾ£¬¿ÉÒÔÓüîʯ»Ò»òÕßÇâÑõ»¯ÄÆÈÜÒºÀ´½øÐÐβÆø´¦Àí£¬¹Ê´ð°¸Îª£ºB£» A£»E£¨»òC£©£»
36£®Îª±£Ö¤²úÉúµÄ¶þÑõ»¯Áò¾¡¿ÉÄܶàµÄת»¯ÎªÈýÑõ»¯Áò£¬Ó¦ÏȼÓÈÈ´ß»¯¼ÁÔÙµÎÈëŨÁòË᣻37£®SO2ת»¯³ÉSO3µÄ¹ý³ÌΪ·ÅÈÈ£¬Éý¸ßƽºâÏòÄæÏòÒƶ¯£¬×ª»¯ÂÊSO2µÄת»¯ÂʻήµÍ£¬
¹Ê´ð°¸Îª£º36£®¢Ù£»¢Ú£»37£®½µµÍ£»
38£®ÊµÑé½áÊøºó£¬Èç¹ûÈ¡³öÊÕ¼¯SO3µÄÒÇÆ÷²¢³¨¿Ú¶ÖÃÓÚ¿ÕÆøÖУ¬Äܹ»¿´µ½µÄÏÖÏóÊÇ£ºÊԹܿÚÓа×Îí¡£¹Ê´ð°¸Îª£ºÊԹܿÚÓа×Îí¡£
39£®¸ÃʵÑéÒ»°ãʹÓýÏŨµÄÁòËáÖÆSO2£¬²»¿ÉʹÓÃÏ¡ÁòËᣬ¼òÊöÆäÔ­Òò£ºÏ¡ÁòËẬˮ½Ï¶à£¬SO2Ò×ÈÜÓÚË®¡£¹Ê´ð°¸Îª£ºÏ¡ÁòËẬˮ½Ï¶à£¬SO2Ò×ÈÜÓÚË®¡£
40£®¸ù¾ÝÁòÔ­×ÓÊغ㣬Na2SO3¡«SO2¡«SO3£¬amolNa2SO3·ÛÄ©Óë×ãÁ¿Å¨ÁòËá½øÐдËʵÑ飬µ±·´Ó¦½áÊøʱ£¬Ó¦²úÉú¶þÑõ»¯ÁòµÄÖÊÁ¿Îª64ag£¬²âµÃ×°ÖâóÔöÖØÁËbg£¬¼´ÎªÊ£Óà¶þÑõ»¯ÁòµÄÖÊÁ¿£¬ËùÒԲμӷ´Ó¦µÄ¶þÑõ»¯ÁòµÄÎïÖʵÄÁ¿Îª£º64ag-bgת»¯ÂʦÁ= £¨64a¨Db£©/64¡Á100%=£¨1600a-25b£©/16a%£¬¹Ê´ð°¸Îª£º £¨1600a-25b£©/16a»ò(1£­) ¡Á100%£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
(6·Ö)ÓÒͼÊÇijѧУʵÑéÊÒ´Ó»¯Ñ§ÊÔ¼ÁÉ̵êÂò»ØµÄÁòËáÊÔ¼Á±êÇ©ÉϵIJ¿·ÖÄÚÈÝ¡£

¢ñ.ÁòËá¾ßÓÐA-DËùʾµÄÐÔÖÊ£¬ÒÔϹý³ÌÖ÷Òª±íÏÖÁËŨÁòËáµÄÄÇЩÐÔÖÊ£¿Ç뽫ѡÏî×ÖĸÌîÔÚÏÂÁи÷СÌâµÄÀ¨ºÅÄÚ£º
AÇ¿ËáÐÔ     B ÎüË®ÐÔ     C ÍÑË®ÐÔ    D Ç¿Ñõ»¯ÐÔ
£¨1£©Å¨ÁòËá¿ÉÒÔ¸ÉÔïÇâÆø£¨       £©
£¨2£©Å¨ÁòËáʹľÌõ±äºÚ £¨        £©
£¨3£©ÈȵÄŨÁòËáÓëͭƬ·´Ó¦£¨        £©
¢ò£®ÏÖÓøÃŨÁòËáÅäÖÆ100 mL¡¡1 mol/LµÄÏ¡ÁòËá¡£¿É¹©Ñ¡ÓõÄÒÇÆ÷ÓУº ¢Ù½ºÍ·µÎ¹Ü£» ¢ÚÉÕÆ¿£»¢ÛÉÕ±­£»¢Ü Ò©³×£»¢ÝÁ¿Í²£»¢ÞÍÐÅÌÌìƽ£»¢ß²£Á§°ô£» ¢à100mLÈÝÁ¿Æ¿¡£Çë»Ø´ð£º
£¨1£©ÅäÖÆÏ¡ÁòËáʱ£¬ÉÏÊöÒÇÆ÷Öв»ÐèҪʹÓõÄÓР        £¨Ñ¡ÌîÐòºÅ£©¡£
£¨2£©¾­¼ÆË㣬ÅäÖÆ100mL1mol/LµÄÏ¡ÁòËáÐèÒªÓÃÁ¿Í²Á¿È¡ÉÏÊöŨÁòËáµÄÌå»ýΪ  ¡¡ mL¡£
£¨3£©ÏÂÁвÙ×÷»áʹÅäÖƵÄÈÜҺŨ¶ÈÆ«¸ßµÄÊÇ£¨   £©
A£®Á¿È¡Å¨H2SO4ʱ£¬¸©Êӿ̶ÈÏßB£®¶¨ÈÝʱ£¬¸©ÊÓÈÝÁ¿Æ¿¿Ì¶ÈÏß
C£®ÅäÖÆÇ°£¬ÈÝÁ¿Æ¿ÖÐÓÐË®ÖéD£®¶¨ÈݺóÒ¡ÔÈ·¢ÏÖÒºÃæϽµ¶øδÏòÆäÖÐÔÙ¼ÓË®
ijÐËȤС×é̽¾¿SO2ÆøÌ廹ԭFe3+¡¢I2£¬ËûÃÇʹÓõÄÒ©Æ·ºÍ×°ÖÃÈçͼËùʾ£º

£¨1£©ÏÂÁÐʵÑé·½°¸ÊÊÓÃÓÚÔÚʵÑéÊÒÖÆÈ¡ËùÐèSO2µÄÊÇ                          £»
A£®Na2SO3ÈÜÒºÓëHNO3¡¡             
B£®Na2SO3¹ÌÌåÓëŨÁòËá
C£®¹ÌÌåÁòÔÚ´¿ÑõÖÐȼÉÕ¡¡             
D£®Ì¼ÓëÈÈŨH2SO4
£¨2£©Ð´³öSO2ÆøÌåͨÈëAÖеÄÀë×Ó·´Ó¦___________________________________________£»
£¨3£©ÔÚÉÏÊö×°ÖÃÖÐͨÈë¹ýÁ¿µÄSO2£¬ÎªÁËÑéÖ¤AÖÐSO2ÓëFe3+·¢ÉúÁËÑõ»¯»¹Ô­·´Ó¦£¬
È¡AÖеÄÈÜÒº£¬·Ö³ÉÁ½·Ý£¬²¢Éè¼ÆÁËÈçÏÂʵÑ飺
·½°¸¢Ù£ºÍùµÚÒ»·ÝÊÔÒºÖмÓÈëÉÙÁ¿ËáÐÔKMnO4ÈÜÒº£¬×ϺìÉ«ÍÊÈ¥
·½°¸¢Ú£ºÍùµÚ¶þ·ÝÊÔÒº¼ÓÈëKSCNÈÜÒº£¬²»±äºì£¬ÔÙ¼ÓÈëÐÂÖƵÄÂÈË®£¬ÈÜÒº±äºì
ÉÏÊö·½°¸²»ºÏÀíµÄÊÇ¡¡   ¡¡£¬Ô­ÒòÊÇ                                                         £»
£¨4£©ÄܱíÃ÷I-µÄ»¹Ô­ÐÔÈõÓÚSO2µÄÏÖÏóÊÇ                                                         ¡£
£¨5£©ÏÂͼΪSO2µÄÖƱ¸ºÍÊÕ¼¯×°ÖÃͼ£º£¨¼Ð³ÖÒÇÆ÷Ê¡ÂÔ£©

ÒÇÆ÷AÃû³Æ             £¬Í¼ÖеÄ×°ÖôíÎóµÄÊÇ       (Ìî¡°A¡¢B¡¢C¡¢D¡±µÄÒ»¸ö»ò¶à¸ö)
¸Ã×°ÖÃÆøÃÜÐÔ¼ì²éµÄ²Ù×÷ÊÇ                                                   
                                                                         ¡£

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

¾«Ó¢¼Ò½ÌÍø