ÌâÄ¿ÄÚÈÝ

£¨10·Ö£©¿ÕÆø´µ³ö·¨¹¤ÒÕ£¬ÊÇÄ¿Ç°¡°º£Ë®Ìáä塱µÄ×îÖ÷Òª·½·¨Ö®Ò»¡£Æ乤ÒÕÁ÷³ÌÈçÏ£º

£¨1£©äåÔÚÖÜÆÚ±íÖÐλÓÚ_________ÖÜÆÚ£¬_________×å¡£
£¨2£©²½Öè¢ÜµÄÀë×Ó·½³Ìʽ£º________________________________________        ¡£
£¨3£©²½Öè¢ÞµÄÕôÁó¹ý³ÌÖУ¬äå³ö¿ÚζÈΪºÎÒª¿ØÖÆÔÚ80¡ª90¡æ¡£Î¶ȹý¸ß»ò¹ýµÍ¶¼²»ÀûÓÚÉú²ú£¬Çë½âÊÍÔ­Òò:___________________________________________     ¡£
£¨4£©²½Öè¢àÖÐäåÕôÆøÀäÄýºóµÃµ½ÒºäåÓëäåË®µÄ»ìºÏÎ¿ÉÀûÓÃËüÃǵÄÏà¶ÔÃܶÈÏà²îºÜ´óµÄÌصã½øÐзÖÀë¡£ÈôÔÚʵÑéÊÒ·ÖÀëÉÏÊö»ìºÏÎïµÄ·ÖÀëÒÇÆ÷µÄÃû³ÆÊÇ___________£¬·ÖÀëʱҺäå´Ó·ÖÀëÆ÷µÄ_________£¨Ìî¡°ÉÏ¿Ú¡±»ò¡°Ï¿ڡ±£©Åųö¡£
£¨5£©ÎªÊ²Ã´²»Ö±½ÓÓú¬äåµÄº£Ë®½øÐÐÕôÁóµÃµ½Òºä壬¶øÒª¾­¹ý¡°¿ÕÆø´µ³ö¡¢SO2ÎüÊÕ¡¢ÂÈ»¯¡±£º                                                             ¡£

£¨1£©µÚËÄ ¢÷A£¨2·Ö£©
£¨2£©Br2+SO2+2H2O=4H++2Br¨D+ SO42¨D£¨2·Ö£©
£¨3£©Î¶ȹý¸ß£¬´óÁ¿Ë®ÕôÆøÅųö£¬äåÆøÖÐË®Ôö¼Ó£»
ζȹýµÍ£¬äå²»ÄÜÍêÈ«Õô³ö£¬ÎüÊÕÂʵ͡£            £¨2·Ö£©
£¨4£©·ÖҺ©¶·   Ï¿ڠ                               £¨2·Ö£©
£¨5£©ÂÈ»¯ºóµÄº£Ë®ËäÈ»º¬ÓÐäåµ¥ÖÊ£¬µ«Å¨¶ÈµÍ£¬Èç¹ûÖ±½ÓÕôÁóÔ­ÁÏ£¬²úÆ·³É±¾¸ß
¡°¿ÕÆø´µ³ö¡¢SO2ÎüÊÕ¡¢ÂÈ»¯¡±µÄ¹ý³Ìʵ¼ÊÉÏÊÇÒ»¸öBr2µÄŨËõ¹ý³Ì¡££¨2·Ö£©
ÂÔ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨14·Ö£©±¾ÊÀ¼Í£¬ÈËÀàÉç»á½«Öð½¥²½ÈëÇâ¾­¼Ãʱ´ú¡£Ä¿Ç°´ó¹æÄ£²úÇⷽʽÈÔÊÇ»¯Ñ§ÖÆÇâ¡£
I¡¢´ß»¯ÖØÕûÖÆÇâ
ÒÔ¼×´¼ÎªÀý£¬·´Ó¦ÖÆÇâÆøµÄÒ»°ã;¾¶ÓУº
CH3OH£¨l£©=2H2£¨g£©£«CO£¨g£©             ¡÷H1=£«128 kJ¡¤mol£­1
CH3OH£¨l£©£«H2O£¨l£©=3H2£¨g£©£«CO2£¨g£©     ¡÷H2=" a" kJ¡¤mol£­1
ÒÑÖª£ºH2£¨g£©£«1/2O2£¨g£©=H2O£¨l£©         ¡÷H=£­286 kJ¡¤mol£­1
ΪÇóµÃ¡÷H2£¬»¹ÒªÖªµÀ        µÄȼÉÕÈÈ£¬ÈôÆäȼÉÕÈÈΪ¡÷H=Ò»283 kJ¡¤mol£­1£¬Ôò¡÷H2=            ¡£
¢ò¡¢½ðÊôÖû»ÖÆÇâ
£¨1£©Ñо¿±íÃ÷£¬¸ÕÇиîµÄ½ðÊô±íÃæ¾ßÓкܸߵķ´Ó¦»îÐÔ¡£µ±ÂÁ»òÂÁºÏ½ðÔÚË®Öб»Çиî»òÄëËéµÄʱºò£¬¿ÉÒÔ³ÖÐøµØÊͷųöÇâÆø¡£Ê¹ÓÃÂÁÓëË®·´Ó¦ÖÆÇâÆø±ÈʹÓÃÆäËü»îÆýðÊôÓëË®·´Ó¦ÖÆÇâÆøµÄÓŵãÓУº¢Ù¼ÛÁ®£¬³É±¾½ÏµÍ£»¢Ú                      ¡£
£¨2£©ÀûÓÃÌìÈ»ÆøÖØÕûµÃµ½µÄCO¡¢H2»ìºÏÆø¶Ô½ðÊôÑõ»¯Îï½øÐл¹Ô­£¬È»ºó½«½ðÊôÓëË®·´Ó¦·Å³öÇâÆø£¬ÓÉ´Ë´ï³ÉÒ»¸öÁ¼ÐÔÑ­»·¡£¸ù¾Ý¸ßÖÐËùѧµÄ֪ʶºÏÀíÔ¤²â¸Ã½ðÊôµ¥ÖÊ£¬²¢Ð´³ö¸Ã½ðÊôÔÚ¼ÓÈȵÄÌõ¼þÏÂÓëË®ÕôÆû·´Ó¦µÄ»¯Ñ§·½³Ìʽ                  ¡£
III¡¢Ì«ÑôÄÜÖÆÇâ
ÀûÓùâÄÜ·Ö½âË®±ØÐëÒªÓд߻¯¼ÁµÄ²ÎÓë¡£ÏÂÁÐÓйش߻¯¼ÁµÄ˵·¨ÕýÈ·µÄÊÇ               
A£®Ê¹ÓøßЧ´ß»¯¼Á·Ö½âË®ÖƵÃH2µÄͬʱ»¹¿ÉÒÔ»ñµÃÄÜÁ¿
B£®Ê¹ÓøßЧ´ß»¯¼Áºó³£ÎÂÏÂË®¿ÉÒÔ×Ô·¢·Ö½â
C£®¹è½º¾ßÓжà¿×½á¹¹£¬ÓнϴóµÄ±íÃæ»ý£¬³£ÓÃ×ö´ß»¯¼ÁµÄÔØÌå
D£®¶ÔÓÚ¿ÉÄæ·´Ó¦£¬´ß»¯¼ÁÔÚÔö´óÕý·´Ó¦ËÙÂʵÄͬʱҲÔö´óÄæ·´Ó¦ËÙÂÊ
¢ô¡¢ÅäλÇ⻯ÎïÖÆÇâ
ÔÚÅðÇ⻯ÄÆ£¨NaBH4£©Ë®ÈÜÒºÖмÓÈëÌض¨´ß»¯¼Áºó£¬¿ÉÒÔѸËٵط¢ÉúË®½â·´Ó¦Éú³ÉÆ«ÅðËáÄƺÍÇâÆø¡£Çëд³ö´ËË®½â·´Ó¦µÄ»¯Ñ§·½³Ìʽ                        ¡£
½«ÎÞˮƫÅðËáÄÆ¡¢Ç⻯þ£¨MgH2£©·ÅÈëÇòÄ¥É豸ÖУ¬Í¨Èëë²Æø»òÇâÆø²¢±£³Öѹǿ100~500 kPaÑÐÄ¥0£®5~4 h£¬¼´¿ÉµÃµ½ÅðÇ⻯ÄÆ¡£ÑÐÄ¥¹ý³ÌÖÐÐèҪͨÈëë²Æø»òÇâÆø²¢±£³Öѹǿ100~500 kPaµÄÄ¿µÄÊÇ                                        ¡£

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

¾«Ó¢¼Ò½ÌÍø