ÌâÄ¿ÄÚÈÝ

ÄÜÔ´Óë»·¾³³ÉΪÈËÃÇÈÕÒæ¹Ø×¢µÄÎÊÌ⣬ËüÓëÈËÀàµÄÉú»îºÍÉç»á·¢Õ¹ÃÜÇÐÏà¹Ø¡£

£¨1£©Ä¿Ç°£¬ÈËÀàÒÔ»¯Ê¯È¼ÁÏΪÖ÷ÒªÄÜÔ´£¬³£¼ûµÄ»¯Ê¯È¼ÁϰüÀ¨Ãº¡¢Ê¯ÓͺÍ_____£»

£¨2£©ÎªÌá¸ßúµÄÀûÓÃÂÊ£¬ÔÚÒ»¶¨Ìõ¼þÏ¿ɽ«Æäת»¯Îª¿ÉȼÐÔÆøÌ壬´Ë¹ý³ÌµÄ΢¹ÛʾÒâͼÈçͼËùʾ£º¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_____£»

£¨3£©½üÄêÀ´£¬ÎÒÊÐÅ©´å´óÁ¦ÍƹãʹÓÃÕÓÆø£¬ÕÓÆøµÄÖ÷Òª³É·ÖÊǼ×Í飬д³ö¼×ÍéȼÉյĻ¯Ñ§·½³Ìʽ_____¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

̽¾¿Ò»£º¹¤ÒµÌúºìµÄÖ÷Òª³É·ÖÊÇFe2O3£¬»¹º¬ÓÐÉÙÁ¿µÄFeO¡¢Fe3O4£®Ä³ÐËȤС×éÓòÝËá¾§ÌåÖÆÈ¡CO£¬²¢½øÐÐÁËÈçͼËùʾʵÑé¡£

ÊԻشð£º

£¨²éÔÄ×ÊÁÏ£©²ÝËá¾§Ìå(H2C2O4•2H2O)ÔÚŨH2SO4×÷ÓÃÏÂÊÜÈÈ·Ö½âÉú³ÉCO2ºÍCO¡£

(1)ʵÑéǰӦÏÈ_____¡£

(2)½øÈëDÖÐµÄÆøÌåÊÇ´¿¾»¡¢¸ÉÔïµÄCO£¬ÔòAÖÐÊÔ¼ÁÊÇ_____¡¢CÖеÄÊÔ¼ÁÊÇ_____(Ìî×Öĸ)¡£

a ŨÁòËá b ³ÎÇåµÄʯ»ÒË® c ÇâÑõ»¯ÄÆÈÜÒº

(3)F×°ÖõÄ×÷ÓÃÊÇ_____£¬ÈôȱÉÙ¸Ã×°Öã¬ÔòËù²âÑõÔªËØµÄÖÊÁ¿·ÖÊý½«_____(Ìî¡°Æ«´ó¡±¡°Æ«Ð¡¡±»ò¡°²»±ä¡±)

(4)ijͬѧÈÏΪͼʾװÖÃÓв»×ãÖ®´¦£¬¸Ä½øµÄ·½·¨ÊÇ_____¡£

(5)д³ö×°ÖÃDÖÐÑõ»¯ÌúÓëCO·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º_____¡£

̽¾¿¶þ£ºÇóij²ÝËáÑùÆ·ÖвÝËá¾§Ìå(H2C2O4•2H2O)µÄÖÊÁ¿·ÖÊý¡£

³ÆÈ¡8.75g²ÝËá¾§ÌåÑùÆ·ÅäÖÆ50.00gÈÜÒº£¬È¡10.00gÈÜÒº¼ÓÊÊÁ¿µÄÏ¡ÁòËᣬȻºóµÎ¼Ó25.00g3.16%KMnO4ÈÜÒº£¬Ç¡ºÃ·´Ó¦ÍêÈ«¡£(ÒÑÖª£º2KMnO4+5H2C2O4+3H2SO4£½K2SO4+2MnSO4+10CO2¡ü+8H2O)

ÔòKMnO4ÈÜÒºÏÔ_____É«£¬25.00g3.16%KMnO4ÈÜÒºÖÐKMnO4µÄÖÊÁ¿_____g£®Çë¼ÆËãÑùÆ·ÖвÝËá¾§Ìå(H2C2O4•2H2O)µÄÖÊÁ¿·ÖÊý_____¡£[д³ö¼ÆËã¹ý³Ì£¬M2(H2C2O4)£½90£¬M2(H2C2O4•2H2O)£½126£¬M2(KMnO4)£½158]¡£

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

¾«Ó¢¼Ò½ÌÍø