ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿Ä³Fe 2(SO4) 3ÑùÆ·º¬ÓÐÉÙÁ¿FeSO4ÔÓÖÊ£¬ÎªÁ˲ⶨ¸ÃÑùÆ·ÖÐÌúÔªËصĺ¬Á¿£¬Éè¼ÆÈçÏÂʵÑ飺

¢ñ£®³ÆÈ¡ÑùÆ·m g£¬¼ÓÈëÏ¡H2SO4£¬ºó¼ÓË®ÅäÖƳÉ250.00 mLÈÜÒº£»

¢ò£®È¡25.00 mLÈÜÒº£¬ÏȼÓÈëH2O2£¬È»ºóÔÙ¼Ó¹ýÁ¿µÄ°±Ë®£¬¹ýÂË£»

¢ó£®½«³ÁµíÓÃÕôÁóˮϴµÓÊý´Îºó£¬ºæ¸É£»

¢ô£®×ÆÉÕÖÁÖÊÁ¿²»ÔÙ¼õÉÙΪֹ£¬µÃµ½ºì×ØÉ«¹ÌÌ壬ÀäÈ´ºó³ÆÁ¿£¬ÖÊÁ¿Îªn g¡£

Çë¸ù¾ÝÉÏÃæ²Ù×÷Á÷³Ì£¬»Ø´ðÒÔÏÂÎÊÌ⣺

¢Ù²½Öè¢ñÖгÆÁ¿Ê¹ÓõÄÒÇÆ÷ÊÇ________________£¬ÅäÖÆÓõ½µÄÒÇÆ÷ÓÐÉÕ±­¡¢²£Á§°ô¡¢½ºÍ·µÎ¹Ü£¬»¹Òª²¹³äµÄ²£Á§ÒÇÆ÷ÊÇ____________________¡£

¢Ú²½Öè¢òÖйýÂËÓõÄÒÇÆ÷ÓÐÂËÖ½¡¢Ìú¼Ų̈¡¢ÌúȦºÍÉÕ±­£¬»¹Òª²¹³äµÄ²£Á§ÒÇÆ÷ÊÇ________£»¼ÓÈëH2O2µÄÖ÷ҪĿµÄÊÇ_______________________________¡£

¢ÛÈôÒª¼ìÑéÈÜÒºÖеÄFe3+£¬ÔòÓ¦¸Ã¼ÓÈë________________ÊÔ¼Á¡£

¢Ü²½Öè¢óÖмìÑé¸Ã³ÁµíÒѾ­Ï´µÓ¸É¾»µÄ²Ù×÷ÊÇ________________¡£

¢Ý¼ÆËã¸ÃÑùÆ·ÖÐÌúÔªËصÄÖÊÁ¿·ÖÊýµÄ±í´ïʽÊÇ________________¡£

¡¾´ð°¸¡¿ÍÐÅÌÌìƽ 250mLÈÝÁ¿Æ¿Â©¶·ºÍ²£Á§°ô½«Fe2+Ñõ»¯ÎªFe3+KSCNÈ¡×îºóÒ»´ÎÏ´µÓÒºÉÙÐíÓÚÊÔ¹ÜÖУ¬µÎ¼ÓÑÎËáËữµÄBaCl2ÈÜÒº(»òBaCl2ÈÜÒº)£¬ÈôÎÞ°×É«³Áµí²úÉú£¬Ö¤Ã÷ÒÑÏ´µÓ¸É¾»7n/m¡Á100%»ò700n/m%

¡¾½âÎö¡¿

¢Ù²½Öè¢ñÖгÆÁ¿Fe 2(SO4) 3ÑùÆ·£¬¿ÉÒÔʹÓÃÍÐÅÌÌìƽ£¬ÅäÖÆ250.00 mLÈÜÒº£¬Óõ½µÄÒÇÆ÷ÓÐÉÕ±­¡¢²£Á§°ô¡¢½ºÍ·µÎ¹Ü¡¢250mLÈÝÁ¿Æ¿£¬¹Ê´ð°¸Îª£ºÍÐÅÌÌìƽ£»250mLÈÝÁ¿Æ¿£»

¢Ú¹ýÂËÓõÄÒÇÆ÷ÓÐÂËÖ½¡¢Ìú¼Ų̈¡¢ÌúȦºÍÉÕ±­£¬»¹Òª²¹³äµÄ²£Á§ÒÇÆ÷ÓЩ¶·ºÍ²£Á§°ô£»¼ÓÈëH2O2¿ÉÒÔ½«Fe2+Ñõ»¯ÎªFe3+£¬¹Ê´ð°¸Îª£ºÂ©¶·ºÍ²£Á§°ô£»½«Fe2+Ñõ»¯ÎªFe3+£»

¢ÛÒª¼ìÑéÈÜÒºÖеÄFe3+£¬¿ÉÒÔÑ¡ÓÃKSCN£¬¼ÓÈëKSCN£¬ÈÜÒºÏÔѪºìÉ«£¬¹Ê´ð°¸Îª£ºKSCN£»

¢Ü²½Öè¢óÖеijÁµíΪÇâÑõ»¯Ìú£¬ÈÜÒºÖеÄÈÜÖÊΪ(NH4)2SO4£¬¼ìÑé¸Ã³ÁµíÒѾ­Ï´µÓ¸É¾»¿ÉÒÔͨ¹ý¼ìÑéÊÇ·ñº¬ÓÐÁòËá¸ùÀë×ÓÀ´Íê³É£¬·½·¨Îª£ºÈ¡×îºóÒ»´ÎÏ´µÓÒºÉÙÐíÓÚÊÔ¹ÜÖУ¬µÎ¼ÓÑÎËáËữµÄBaCl2ÈÜÒº(»òBaCl2ÈÜÒº)£¬ÈôÎÞ°×É«³Áµí²úÉú£¬Ö¤Ã÷ÒÑÏ´µÓ¸É¾»£¬¹Ê´ð°¸Îª£ºÈ¡×îºóÒ»´ÎÏ´µÓÒºÉÙÐíÓÚÊÔ¹ÜÖУ¬µÎ¼ÓÑÎËáËữµÄBaCl2ÈÜÒº(»òBaCl2ÈÜÒº)£¬ÈôÎÞ°×É«³Áµí²úÉú£¬Ö¤Ã÷ÒÑÏ´µÓ¸É¾»£»

¢Ý×îºóµÄºì×ØÉ«¹ÌÌåΪÑõ»¯Ìú£¬ÖÊÁ¿Îªng£¬º¬ÓеÄÌúÔªËصÄÎïÖʵÄÁ¿Îª¡Á2=mol£¬ÔòÑùÆ·ÖÐÌúÔªËصÄÖÊÁ¿Îªmol¡Á¡Á56g/mol=7ng£¬¸ÃÑùÆ·ÖÐÌúÔªËصÄÖÊÁ¿·ÖÊý=¡Á100%=¡Á100%£¬¹Ê´ð°¸Îª£º¡Á100%¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿Ä³°àͬѧÓÃÈçÏÂʵÑé̽¾¿Fe2+¡¢Fe3+µÄÐÔÖÊ¡£»Ø´ðÏÂÁÐÎÊÌâ:

£¨1£©·Ö±ðÈ¡Ò»¶¨Á¿ÂÈ»¯Ìú¡¢ÂÈ»¯ÑÇÌú¹ÌÌå,¾ùÅäÖƳÉ0.1 mol¡¤L-1µÄÈÜÒº¡£ÔÚFeCl2ÈÜÒºÖÐÐè¼ÓÈëÉÙÁ¿Ìúм,ÆäÄ¿µÄÊÇ___________¡£

£¨2£©¼××éͬѧȡ2 mL FeCl2ÈÜÒº,¼ÓÈ뼸µÎÂÈË®,ÔÙ¼ÓÈë1µÎKSCNÈÜÒº,ÈÜÒº±äºì,˵Ã÷Cl2¿É½«Fe2+Ñõ»¯¡£FeCl2ÈÜÒºÓëÂÈË®·´Ó¦µÄÀë×Ó·½³ÌʽΪ____________¡£

£¨3£©ÒÒ×éͬѧÈÏΪ¼××éµÄʵÑé²»¹»ÑϽ÷,¸Ã×éͬѧÔÚ2 mL FeCl2ÈÜÒºÖÐÏȼÓÈë0.5 mLúÓÍ,ÔÙÓÚÒºÃæÏÂÒÀ´Î¼ÓÈ뼸µÎÂÈË®ºÍ1µÎKSCNÈÜÒº,ÈÜÒº±äºì,úÓ͵Ä×÷ÓÃÊÇ_____¡£

£¨4£©±û×éͬѧȡ10 mL 0.1 mol¡¤L-1 KIÈÜÒº,¼ÓÈë6 mL 0.1 mol¡¤L-1 FeCl3ÈÜÒº»ìºÏ¡£·Ö±ðÈ¡2 mL ´ËÈÜÒºÓÚ3Ö§ÊÔ¹ÜÖнøÐÐÈçÏÂʵÑé:

¢ÙµÚÒ»Ö§ÊÔ¹ÜÖмÓÈë1 mL CCl4³ä·ÖÕñµ´¡¢¾²ÖÃ,CCl4²ãÏÔ×ÏÉ«;

¢ÚµÚ¶þÖ§ÊÔ¹ÜÖмÓÈë1µÎK3[Fe(CN)6]ÈÜÒº,Éú³ÉÀ¶É«³Áµí;

¢ÛµÚÈýÖ§ÊÔ¹ÜÖмÓÈë1µÎKSCNÈÜÒº,ÈÜÒº±äºì¡£

ʵÑé¢Ú¼ìÑéµÄÀë×ÓÊÇ____(ÌîÀë×Ó·ûºÅ);ʵÑé¢ÙºÍ¢Û˵Ã÷:ÔÚI-¹ýÁ¿µÄÇé¿öÏÂ,

ÈÜÒºÖÐÈÔº¬ÓÐ___(ÌîÀë×Ó·ûºÅ),ÓÉ´Ë¿ÉÒÔÖ¤Ã÷¸ÃÑõ»¯»¹Ô­·´Ó¦Îª________¡£

£¨5£©¶¡×éͬѧÏòÊ¢ÓÐH2O2ÈÜÒºµÄÊÔ¹ÜÖмÓÈ뼸µÎËữµÄFeCl2ÈÜÒº,ÈÜÒº±ä³É×Ø»ÆÉ«,·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ___________________;Ò»¶Îʱ¼äºó,ÈÜÒºÖÐÓÐÆøÅݳöÏÖ,²¢·ÅÈÈ,ËæºóÓкìºÖÉ«³ÁµíÉú³É¡£²úÉúÆøÅݵÄÔ­ÒòÊÇ_______________;Éú³É³ÁµíµÄÔ­ÒòÊÇ__________________(ÓÃƽºâÒƶ¯Ô­Àí½âÊÍ)¡£

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

¾«Ó¢¼Ò½ÌÍø