ÌâÄ¿ÄÚÈÝ

700¡æÊ±£¬ÏòÈÝ»ýΪ2LµÄÃܱÕÈÝÆ÷ÖгäÈëÒ»¶¨Á¿µÄCOºÍH2O£¬·¢Éú·´Ó¦£ºCO£¨g£©+H2O£¨g£©CO2£¨g£© +H2£¨g£©£¬·´Ó¦¹ý³ÌÖвⶨµÄ²¿·ÖÊý¾Ý¼ûÏÂ±í£¨±íÖÐt1 <t2£©£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A£®·´Ó¦ÔÚt1 minÄ򵀮½¾ùËÙÂÊΪv£¨H2£©=0.40/t1moI/£¨L¡¤min£©

B£®±£³ÖÆäËûÌõ¼þ²»±ä£¬ÆðʼʱÏòÈÝÆ÷ÖгäÈë0.60 mol COºÍ1.20 mol H2£¬µ½´ïƽºâʱ£¬n£¨CO2£©=0.40 mol

C£®±£³ÖÆäËûÌõ¼þ²»±ä£¬ÏòƽºâÌåϵÖÐÔÙͨÈë0.20 mol CO£¬ÓëԭƽºâÏà±È£¬´ïµ½ÐÂÆ½ºâʱCOת»¯ÂʼõС£¬H2OµÄÌå»ý·ÖÊýÒ²¼õС

D£®Î¶ÈÉýÖÁ800¡æ£¬ÉÏÊö·´Ó¦Æ½ºâ³£ÊýΪ0.64£¬ÔòÕý·´Ó¦ÎªÎüÈÈ·´Ó¦

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

CO2µÄ»ØÊÕÀûÓöԼõÉÙÎÂÊÒÆøÌåÅÅ·Å¡¢¸ÄÉÆÈËÀàÉú´æ»·¾³¾ßÓÐÖØÒªÒâÒå¡£ÀûÓÃCO2ºÍCH4ÖØÕû¿ÉÖÆºÏ³ÉÆø£¨Ö÷Òª³É·ÖCO¡¢H2£©.ÖØÕû¹ý³ÌÖв¿·Ö·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ£º

¢ÙCH4£¨g£©=C£¨s£©+2H2£¨g£© ¡÷H=+75.0 kJ/mol

¢ÚCO2£¨g£©+H2£¨g£©=CO£¨g£©+H2O£¨g£© ¡÷H=+41.0 kJ/mol

¢ÛCO£¨g£©+H2£¨g£©=C£¨s£©+H2O£¨g£© ¡÷H=-131.0kJ/mol

£¨1£©·´Ó¦CO2£¨g£©+CH4£¨g£©=2CO£¨g£© + 2H2£¨g£©µÄ¡÷H=____________£»

£¨2£©¹Ì¶¨n£¨CO2£©= n£¨CH4£©£¬¸Ä±ä·´Ó¦Î¶ȣ¬CO2ºÍCH4µÄƽºâת»¯ÂʼûÏÂ×óͼ¡£

¢ÙͬζÈÏÂCO2µÄƽºâת»¯ÂÊ____________£¨Ìî¡°´óÓÚ¡±»ò¡°Ð¡ÓÚ¡±£©CH4µÄƽºâת»¯ÂÊ£¬ÆäÔ­ÒòÊÇ____________£»

¢Ú¸ßÎÂϽøÐи÷´Ó¦Ê±³£»áÒò·´Ó¦¢ÙÉú³É¡°»ý̼¡±£¨Ì¼µ¥ÖÊ£©£¬Ôì³É´ß»¯¼ÁÖж¾£¬¸ßÎÂÏ·´Ó¦¢ÙÄÜ×Ô·¢½øÐеÄÔ­ÒòÊÇ____________¡£

£¨3£©Ò»¶¨Ìõ¼þÏÂPd-Mg/SiO2´ß»¯¼Á¿ÉʹCO2¡°¼×Í黯¡±´Ó¶ø±ä·ÏΪ±¦£¬Æä·´Ó¦»úÀíÈçÉÏÓÒͼËùʾ£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ____________£¬·´Ó¦¹ý³ÌÖÐÌ¼ÔªËØµÄ»¯ºÏ¼ÛΪ-2¼ÛµÄÖмäÌåÊÇ____________¡£

£¨4£©Â±Ë®¿ÉÔÚÎüÊÕÑ̵ÀÆøÖÐCO2µÄͬʱ±»¾»»¯£¬ÊµÏÖÒÔ·ÏÖηϣ¬ÆäÖÐÉæ¼°µÄÒ»¸ö·´Ó¦ÊÇCaSO4 + Na2CO3=CaCO3 + Na2SO4£¬Ôò´ïµ½Æ½ºâºó£¬ÈÜÒºÖÐc£¨CO32-£©/c£¨SO42-£©=____________[ÓÃKsp£¨CaSO4£©ºÍKsp£¨CaCO3£©±íʾ]

Ìú¼°Æä»¯ºÏÎïÓÐÖØÒªÓÃ;£¬Èç¾ÛºÏÁòËáÌú[Fe2(OH)n(SO4)3n/2]mÊÇÒ»ÖÖÐÂÐ͸ßЧµÄË®´¦Àí»ìÄý¼Á£¬¶ø¸ßÌúËá¼Ø(ÆäÖÐÌúµÄ»¯ºÏ¼ÛΪ+6)ÊÇÒ»ÖÖÖØÒªµÄɱ¾úÏû¶¾¼Á£¬Ä³¿ÎÌâС×éÉè¼ÆÈçÏ·½°¸ÖƱ¸ÉÏÊöÁ½ÖÖ²úÆ·£º

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÈôAΪH2O(g)£¬Ð´³ö·´Ó¦·½³Ìʽ£º_______________________________£»

£¨2£©ÈôBΪNaClO3ÓëÏ¡ÁòËᣬд³öÆäÑõ»¯Fe2+µÄÀë×Ó·½³Ìʽ(»¹Ô­²úÎïΪCl-)___________________£»

£¨3£©ÈôCΪKNO3ºÍKOHµÄ»ìºÏÎд³öÆäÓëFe2O3¼ÓÈȹ²ÈÚÖÆµÃ¸ßÌúËá¼Ø(K2FeO4)µÄ»¯Ñ§·½³Ìʽ£¬²¢Å䯽£º

¡õFe2O3+¡õKNO3+¡õKOH¡õ_________+¡õKNO2+¡õ__________

£¨4£©Îª²â¶¨ÈÜÒº¢ñÖÐÌúÔªËØµÄ×ܺ¬Á¿£¬ÊµÑé²Ù×÷ÈçÏ£º×¼È·Á¿È¡20.00mLÈÜÒº¢ñÓÚ´øÈû×¶ÐÎÆ¿ÖУ¬¼ÓÈë×ãÁ¿H2O2£¬µ÷½ÚpH£¼3£¬¼ÓÈȳýÈ¥¹ýÁ¿H2O2£»¼ÓÈë¹ýÁ¿KI³ä·Ö·´Ó¦ºó£¬ÔÙÓÃ0.1000mol?L-1Na2S2O3±ê×¼ÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄ±ê×¼ÈÜÒº20.00mL£®

ÒÑÖª£º

2Fe3++2I-¨T2Fe2++I2

I2+2S2O32-¨T2I-+S4O42-

¢ÙµÎ¶¨Ñ¡ÓõÄָʾ¼Á¼°µÎ¶¨ÖÕµã¹Û²ìµ½µÄÏÖÏó_______________________________£»

¢ÚÈÜÒº¢ñÖÐÌúÔªËØµÄ×ܺ¬Á¿Îª____________g?L-1£®ÈôµÎ¶¨Ç°ÈÜÒºÖÐH2O2ûÓгý¾¡£¬Ëù²â¶¨µÄÌúÔªËØµÄº¬Á¿½«_________(Ìî¡°Æ«¸ß¡±¡°Æ«µÍ¡±»ò¡°²»±ä¡±)¡£

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

¾«Ó¢¼Ò½ÌÍø