ÌâÄ¿ÄÚÈÝ

ÌúÊÇÈËÌå²»¿ÉȱÉÙµÄ΢Á¿ÔªËØ£¬ÉãÈ뺬ÌúµÄ»¯ºÏÎï¿É²¹³äÌú¡£¡°ËÙÁ¦·Æ¡±ÊÇÊг¡ÉÏÒ»ÖÖ³£¼ûµÄ²¹ÌúÒ©Æ·£¬Ï±íÊÇ˵Ã÷ÊéµÄ²¿·ÖÄÚÈÝ¡£
[¹æ¸ñ]ÿƬº¬çúçêËáÑÇÌú
[ÊÊÓ¦Ö¢]ÓÃÓÚȱÌúÐÔƶѪ֢£¬Ô¤·À¼°ÖÎÁÆÓá£
[ÓÃÁ¿Ó÷¨]³ÉÈËÔ¤·ÀÁ¿/ÈÕ£¬³ÉÈËÖÎÁÆÁ¿¡ª/ÈÕ¡£
С¶ùÓÃÁ¿Ô¤·ÀÁ¿¡ª /ÈÕ£¬ÖÎÁÆÁ¿¡ª/ÈÕ
[Öü²Ø]±Ü¹â¡¢ÃÜ·â¡¢ÔÚ¸ÉÔï´¦±£´æ¡£
 
£¨1£©¸ÃÒ©Æ·ÖÐFe2+»á»ºÂýÑõ»¯¡£¹ú¼Ò¹æ¶¨¸ÃÒ©ÎïÖÐFe2+µÄÑõ»¯ÂÊ£¨ÒѾ­±»Ñõ»¯Fe2+µÄÖÊÁ¿ÓëFe2+×ÜÖÊÁ¿µÄ±ÈÖµ£©³¬¹ý10.00% ¼´²»ÄÜÔÙ·þÓá£
¢ÙʵÑéÊҿɲÉÓÃH2SO4ËữµÄKMnO4ÈÜÒº£¬¶Ô¡°ËÙÁ¦·Æ¡±ÖеÄFe2+½øÐеζ¨(¼ÙÉèÒ©Æ·ÖÐÆäËû³É·Ý²»ÓëKMnO4·´Ó¦)¡£Çëд³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ£º                                  ¡£
¢ÚʵÑéÇ°£¬Ê×ÏÈÒª¾«È·ÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄKMnO4ÈÜÒº250 mL£¬ÅäÖÆʱÐèÒªµÄ²£Á§ÒÇÆ÷³ý²£Á§°ô¡¢ÉÕ±­¡¢½ºÍ·µÎ¹ÜÍ⣬»¹Ðè          ¡£
¢ÛijͬѧÉè¼ÆÁËÏÂÁеζ¨·½Ê½£¨¼Ð³Ö²¿·ÖÂÔÈ¥£©£¬×îºÏÀíµÄÊÇ       ¡££¨Ìî×ÖĸÐòºÅ£©

£¨2£©³ÆÁ¿ÉÏÊöº¬ÌúÔªËØÖÊÁ¿·ÖÊýΪ20.00%µÄ¡°ËÙÁ¦·Æ¡±10.00 g£¬½«ÆäÈ«²¿ÈÜÓÚÏ¡H2SO4ÖУ¬ÅäÖƳÉ1000 mlÈÜÒº£¬È¡³ö20.00 ml£¬ÓÃ0.01000 mol?L-1µÄKMnO4ÈÜÒºµÎ¶¨£¬ÓÃÈ¥KMnO4ÈÜÒº12.00 ml £¬¸ÃÒ©Æ·ÖÐFe2+µÄÑõ»¯ÂÊΪ                 ¡£
£¨3£©ÒÑÖªçúçêËáΪ¶þÔªÓлúôÈËᣬº¬23.6 gçúçêËáµÄÈÜÒºÓë4.0 mol?L-1 100.0 mlµÄÇâÑõ»¯ÄÆÈÜҺǡºÃÍêÈ«Öк͡£ºË´Å¹²ÕñÇâÆ×·ÖÎöÏÔʾ£¬çúçêËá·Ö×ÓÆ×ͼÉÏÖ»ÓÐÁ½×éÎüÊշ塣д³öçúçêËáÈÜÒºÓëÇâÑõ»¯ÄÆÈÜÒºÍêÈ«Öк͵Ļ¯Ñ§·½³Ìʽ(ÓлúÎïд½á¹¹¼òʽ)                                        ¡£
£¨1£©¢Ù5Fe2++MnO-4+8H+£½5Fe3++Mn2++4H2O(2·Ö) ¢Ú250mlÈÝÁ¿Æ¿(2·Ö)¢Ûb(2·Ö)
£¨2£©16.00% (3·Ö)
£¨3£©HOOC£­CH2CH2£­COOH£«2NaOH¡úNaOOC£­CH2CH2£­COONa£«2H2O (3·Ö)

ÊÔÌâ·ÖÎö£º£¨1£©¢ÙFe2+¾ßÓл¹Ô­ÐÔ£¬±¶Ñõ»¯»áÉú³É³ÉFe3+£¬¸ßÃÌËá¼ØÈÜÒº¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬ÄÜÑõ»¯ÑÇÌúÀë×Ó£¬ÆäÖÐMnO4-ÖÐMnµÄ»¯ºÏ¼ÛÓÉ+7¼Û½µÎª+2¼Û£¬±ä»¯5£¬Fe2+ÖÐFeÓÉ+2¼ÛÉýΪ+3¼Û£¬±ä»¯1£¬¸ù¾Ý»¯ºÏ¼ÛÉý½µ×ÜÊýÏàµÈºÍÖÊÁ¿ÊغãµÃ·´Ó¦µÄÀë×Ó·½³ÌʽΪ5Fe2++MnO-4+8H+£½5Fe3++Mn2++4H2O¡£
¢ÚÒª¾«È·ÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄKMnO4ÈÜÒº250 mL£¬ÅäÖÆʱÐèÒªµÄ²£Á§ÒÇÆ÷³ý²£Á§°ô¡¢ÉÕ±­¡¢½ºÍ·µÎ¹ÜÍ⣬»¹Ðè250mlÈÝÁ¿Æ¿¡£
¢Ûa¡¢¸ßÃÌËá¼ØÈÜÒº¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬Äܸ¯Ê´¼îʽµÎ¶¨¹ÜµÄÏðƤ¹Ü£¬¹Êa´íÎó£»b¡¢¸ßÃÌËá¼ØÈÜÒº¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬Äܸ¯Ê´¼îʽµÎ¶¨¹ÜµÄÏðƤ¹Ü£¬Ó¦ÓÃËáʽµÎ¶¨¹ÜÊ¢×°£¬µÎ¶¨Ê±Îª±ãÓÚ¹Û²ìÑÕÉ«±ä»¯£¬µÎ¶¨ÖÕµãÑÕÉ«ÓÉdz±äÉîÒ×Óڹ۲죬Ӧ½«¸ßÃÌËá¼ØµÎµ½´ý²âÒºÖУ¬¹ÊbÕýÈ·£»c¡¢ÁòËáÑÇÌúÈÜÒºÏÔËáÐÔ£¬Ó¦¸Ã·ÅÔÚËáʽµÎ¶¨¹ÜÖУ¬Í¬Ê±µÎ¶¨ÖÕµãÑÕÉ«ÓÉÉî±ädzÉî²»Ò×Óڹ۲죬¹Êc´íÎ󣬴ð°¸Ñ¡b¡£
£¨2£©ÓÉMnO4-¡«5Fe2+£¬¿ÉµÃ1000.00mLÈÜÒºº¬ÓеÄFe2+µÄÎïÖʵÄÁ¿n(Fe2+)£½0.01 mol/L¡Á12.00¡Á10-3L¡Á5¡Á£½0.03 mol£¬m(Fe2+)£½0.03 mol¡Á56 g/mol£½1.68 g£¬ËùÒÔFe2+µÄÑõ»¯ÂÊ£½£½16.00%¡£
£¨3£©ÏûºÄÇâÑõ»¯ÄƵÄÎïÖʵÄÁ¿ÊÇ0.4mol£¬µ±çúçêËáΪһԪËáʱ£¬¸ù¾ÝÓë¼î·´Ó¦µÄ¹Øϵ£¬ÆäÏà¶Ô·Ö×ÓÖÊÁ¿Îª23.6¡Â0.4£½59£¬²»·ûºÏÓлúÎï×é³ÉÔ­Ôò£»µ±çúçêËáΪ¶þÔªËáʱ£¬ÆäÏà¶Ô·Ö×ÓÖÊÁ¿Îª23.6¡Â0.2£½118£¬ÓÖçúçêËá·Ö×ÓÖÐÓÐÁ½ÖÖλÖò»Í¬µÄÇâÔ­×Ó£¬ËùÒÔçúçêËáµÄ½á¹¹¼òʽΪHOOC-CH2-CH2-COOH£¬Òò´ËçúçêËáÈÜÒºÓëÇâÑõ»¯ÄÆÈÜÒºÍêÈ«Öк͵ÄÀë×Ó·½³ÌʽΪHOOC£­CH2CH2£­COOH£«2NaOH¡úNaOOC£­CH2CH2£­COONa£«2H2O¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨1£©ÏÂÁÐʵÑé²Ù×÷»ò¶ÔʵÑéÊÂʵµÄÃèÊö²»ÕýÈ·µÄÓР         £¨ÌîÐòºÅ£©¡£
A£®ÓÃÍÐÅÌÌìƽ³ÆÁ¿17.55gÂÈ»¯Äƾ§Ìå
B£®Ì¼ËáÄÆÈÜÒº±£´æÔÚ´ø²£Á§ÈûµÄÊÔ¼ÁÆ¿
C£®ÓøÉÔïµÄpHÊÔÖ½²â¶¨ÐÂÖÆÂÈË®µÄpH
D£®Ê¹ÓÃÈÝÁ¿Æ¿ÅäÖÆÈÜҺʱ£¬¸©Êӿ̶ÈÏ߶¨ÈݺóŨ¶ÈÆ«´ó
E£®Ïò±¥ºÍFeCl3ÈÜÒºÖеμÓÉÙÁ¿NaOHÈÜÒº£¬¿ÉÖÆÈ¡Fe(OH)3½ºÌå
F£®³ýÈ¥CO2ÆøÌåÖлìÓеÄÉÙÁ¿HCl£¬¿ÉÒÔ½«ÆøÌåͨÈë±¥ºÍ̼ËáÇâÄÆÈÜÒº
£¨2£©ÏÂͼΪÖÐѧ»¯Ñ§ÊµÑéÖг£¼ûµÄʵÑé×°ÖÃ
             
A             B               C
ʵÑéÊÒ³£ÓÃ×°ÖÃAÖƱ¸Ï±íÖÐÆøÌ壬Ç뽫·ÖҺ©¶·ºÍÔ²µ×ÉÕÆ¿ÖÐӦװµÄ»¯Ñ§ÊÔ¼ÁÌîдÍêÕû¡£
ÆøÌå
O2
Cl2
NH3
·ÖҺ©¶·ÖÐÊÔ¼Á
 
 
Ũ°±Ë®
Ô²µ×ÉÕÆ¿ÖÐÊÔ¼Á
 
KMnO4
 
 
¿ÉÓÃB×°ÖÃÅÅÒºÊÕ¼¯ÆøÌ壬ÆøÌåÓ¦´Ó¸Ã×°ÖÃ________(Ìî¡°×󡱡°ÓÒ¡±)¹Ü¿Úµ¼½ø£¬ÌÈÈôÀûÓøÃ×°ÖÃÊÕ¼¯Cl2£¬ÊÔ¼ÁÆ¿ÖÐÊ¢·ÅµÄÊÔ¼ÁΪ                 ¡£
C×°ÖÃÓÃÓÚ´¦Àí¶àÓàÆøÌå¶Ô»·¾³µÄÎÛȾ£¬ÈôÀûÓøÃ×°ÖÃÎüÊÕCl2£¬´ËʱÉÕ±­Öз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ           ¡£ÌÈÈô¸Ã×°ÖÃÖнö½öÊ¢·ÅÏ¡ÁòËᣬͨ³£²»ÊʺÏÎüÊÕ°±ÆøµÄÔ­ÒòÊÇ                  £¬ÈôÏòÉÕ±­ÖжîÍâÔÙ¼ÓÈëÒ»ÖÖҺ̬ÓлúÎïÔò¿É°²È«ÎüÊÕ°±Æø£¬ÕâÖÖÓлúÎïΪ            ¡£
CuSO4¡¤5H2OÊÇÍ­µÄÖØÒª»¯ºÏÎÓÐ׏㷺µÄÓ¦Óá£ÒÔÏÂÊÇCuSO4¡¤5H2OµÄʵÑéÊÒÖƱ¸Á÷³Ìͼ¡£

¸ù¾ÝÌâÒâÍê³ÉÏÂÁÐÌî¿Õ£º
£¨1£©¡°¼î½þ¡±µÄÄ¿µÄÊÇ                                                      £¬Ð´³öÓйصÄÀë×Ó·½³Ìʽ                                                  ¡£
£¨2£©ÏòÂËÔüÖÐÏȼÓÈë×ãÁ¿Ï¡ÁòËᣬȻºóÔٵμÓÉÙÁ¿Å¨ÏõËᣬÔÚ·ÏÔüÈܽâʱ¿ÉÒԹ۲쵽µÄʵÑéÏÖÏóÓР                                                                           ¡£
£¨3£©²Ù×÷aµÄÃû³ÆΪ           £¬ÖƵõÄCuSO4¡¤5H2OÖпÉÄÜ´æÔÚÏõËáÍ­ÔÓÖÊ£¬³ýÈ¥ÕâÖÖÔÓÖʵÄʵÑé²Ù×÷Ãû³ÆΪ                 ¡£
£¨4£©ÒÑÖª£ºCuSO4+2NaOH=Cu£¨OH£©2¡ý+ Na2SO4¡£³ÆÈ¡0£®26 gÌá´¿ºóµÄCuSO4¡¤5H2OÊÔÑùÓÚ׶ÐÎÆ¿ÖУ¬¼ÓÈë0£®1000 mol/LÇâÑõ»¯ÄÆÈÜÒº28£®00 mL£¬·´Ó¦ÍêÈ«ºó£¬¹ýÁ¿µÄÇâÑõ»¯ÄÆÓÃ0£®1000 mol/LÑÎËáµÎ¶¨ÖÁÖյ㣬ºÄÓÃÑÎËá8£®00 mL£¬Ôò¸ÃÊÔÑùÖÐCuSO4¡¤5H2OµÄÖÊÁ¿·ÖÊýΪ       £»ÉÏÊöµÎ¶¨ÖУ¬µÎ¶¨¹ÜÔÚ×¢ÈëÑÎËá֮ǰ£¬ÏÈÓÃÕôÁóˮϴ¾»£¬ÔÙÓà                      ¡£
£¨5£©ÔÚ¡°Ëá½þ¡±µÄ²½ÖèÖУ¬¢ÙÈôÖ»¼ÓÈëŨÁòËᣬд³ö¼ÓÈÈʱµÄ»¯Ñ§·½³Ìʽ                 ¡£
¢ÚÈô½«Å¨ÏõËá»»³É¹ýÑõ»¯Ç⣬³£ÎÂʱ¼´¿ÉÉú³ÉÁòËáÍ­£¬Ö¸³ö´ËÖÖ·½·¨µÄÓŵ㠠             ¡£

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

¾«Ó¢¼Ò½ÌÍø