ÌâÄ¿ÄÚÈÝ

¸ßÃÌËá¼ØÊÇÖÐѧ³£ÓõÄÊÔ¼Á¡£¹¤ÒµÉÏÓÃÈíÃÌ¿óÖƱ¸¸ßÃÌËá¼ØÁ÷³ÌÈçÏ¡£

£¨1£©Ð´³öʵÑéÊÒÀûÓÃKMnO4·Ö½âÖÆÈ¡O2µÄ»¯Ñ§·½³Ìʽ                      
£¨2£©KMnO4Ï¡ÈÜÒºÊÇÒ»ÖÖ³£ÓõÄÏû¶¾¼Á¡£ÆäÏû¶¾Ô­ÀíÓëÏÂÁÐÎïÖÊÏàͬµÄÊÇ     
A£®84Ïû¶¾Òº(NaClOÈÜÒº)
B£®Ë«ÑõË®
C£®±½·Ó
D£®75%¾Æ¾«
£¨3£©Ôڵζ¨ÊµÑéÖУ¬³£Óà          £¨Ìî¡°Ëáʽ¡±»ò¡°¼îʽ¡±£©µÎ¶¨¹ÜÁ¿È¡KMnO4ÈÜÒº¡£
£¨4£©Ð´³ö·´Ó¦¢ÙµÄ»¯Ñ§·½³Ìʽ                                   
£¨5£©²Ù×÷¢ñµÄÃû³ÆÊÇ       £»²Ù×÷¢ò¸ù¾ÝKMnO4ºÍK2CO3Á½ÎïÖÊÔÚ     £¨Ìî
ÐÔÖÊ£©ÉϵIJîÒ죬²ÉÓà       £¨Ìî²Ù×÷²½Ö裩¡¢³ÃÈȹýÂ˵õ½KMnO4´Ö¾§Ìå¡£
£¨6£©ÉÏÊöÁ÷³ÌÖпÉÒÔÑ­»·Ê¹ÓõÄÎïÖÊÓР      ¡¢       £¨Ð´»¯Ñ§Ê½£©,¼øÓÚ´ËÏÖÓÃ100¶ÖÈíÃ̿󣨺¬MnO287.0%£©£¬ÀíÂÛÉÏ¿ÉÉú²úKMnO4¾§Ìå         ¶Ö£¨²»¿¼ÂÇÖƱ¸¹ý³ÌÖÐÔ­ÁϵÄËðʧ£©¡£
£¨1£©2KMnO4K2MnO4£«MnO2£«O2¡ü  £¨2·Ö£©
£¨2£©AB  £¨¸÷1·Ö£¬¹²2·Ö£¬Ñ¡´í²»µÃ·Ö£©
£¨3£©Ëáʽ £¨2·Ö£©
£¨4£©3K2MnO4£«2CO2 = 2KMnO4£«2K2CO3£«MnO2£¨2·Ö£©
£¨5£©¹ýÂË£¨1·Ö£©¡¡Èܽâ¶È£¨1·Ö£©¡¡Å¨Ëõ½á¾§£¨1·Ö£©
£¨6£©MnO2£¨1·Ö£©¡¡KOH£¨1·Ö£©   158£¨2·Ö£©

ÊÔÌâ·ÖÎö£º£¨1£©ÊµÑéÊÒÀûÓÃKMnO4·Ö½âÖÆÈ¡O2µÄ»¯Ñ§·½³Ìʽ2KMnO4 K2MnO4£«MnO2£«O2¡ü
£¨2£©KMnO4Ï¡ÈÜÒºÊÇÒ»ÖÖ³£ÓõÄÏû¶¾¼Á£¬ÆäÏû¶¾Ô­ÀíÊÇÀûÓÃÁËËüµÄÇ¿Ñõ»¯ÐÔ¡£ËùÒÔÓë¸ßÃÌËá¼ØÏû¶¾Ô­ÀíÏàͬµÄÊÇA¡¢84Ïû¶¾Òº£¬ÀûÓÃÁË´ÎÂÈËáµÄÇ¿Ñõ»¯ÐÔ£¬ÕýÈ·£»B¡¢Ë«ÑõË®¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬ÕýÈ·£»C¡¢±½·ÓÊÇÀûÓÃÁËÉø͸×÷ÓÃʹϸ¾úµ°°×ÖʱäÐÔ£¬´íÎó£»D¡¢75%¾Æ¾«ÊÇÀûÓÃÁËÉø͸×÷ÓÃʹϸ¾úµ°°×ÖʱäÐÔ£¬´íÎ󣬴ð°¸Ñ¡AB¡£
£¨3£©ÒòΪ¸ßÃÌËá¼Ø¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬ËùÒÔÓÃËáʽµÎ¶¨¹ÜÁ¿È¡£»
£¨4£©¸ù¾ÝËù¸ø²úÎд³ö·´Ó¦¢ÙµÄ»¯Ñ§·½³Ìʽ3K2MnO4£«2CO2 = 2KMnO4£«2K2CO3£«MnO2
£¨5£©½«ÈÜÒºÓë¹ÌÌå·ÖÀ룬ÓùýÂ˵ķ½·¨£¬ËùÒÔ²Ù×÷1Ϊ¹ýÂË£»KMnO4ºÍK2CO3Á½ÎïÖÊÔÚÈܽâ¶ÈÉϲÙ×÷²î±ð£¬K2CO3µÄÈܽâ¶È¸ü´ó£¬ËùÒÔÀûÓÃÈܽâ¶ÈµÄ²»Í¬£¬²ÉÓÃŨËõ½á¾§£¬³ÃÈȹýÂ˵õ½¸ßÃÌËá¼Ø½ºÌ壻
£¨6£©´ÓÁ÷³ÌͼÖпɿ´³ö£¬MnO2ÓëKOH¿ªÊ¼ÏûºÄ£¬ºóÓÖÉú³É£¬ËùÒÔ¿ÉÑ­»·Ê¹Ó㻶þÕßÑ­»·Ê¹Ó㬿ÉÈÏΪ¶þÑõ»¯ÃÌÈ«²¿×ª»¯Îª¸ßÃÌËá¼Ø£¬¼´MnO2¡«KMnO4£¬ÈíÃÌ¿óÖÐMnO2µÄÖÊÁ¿ÊÇ87¶Ö£¬ËùÒÔ¿ÉÉú³ÉKMnO4µÄÖÊÁ¿ÊÇ158¶Ö¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ijС×éͬѧ½«Ò»¶¨Å¨¶ÈNaHCO3ÈÜÒº¼ÓÈëµ½CuSO4ÈÜÒºÖз¢ÏÖÉú³ÉÁ˳Áµí¡£¼×ͬѧÈÏΪ³ÁµíÊÇCuCO3£»ÒÒͬѧÈÏΪ³ÁµíÊÇCuCO3ºÍCu(OH)2µÄ»ìºÏÎËûÃÇÉè¼ÆʵÑé²â¶¨³ÁµíÖÐCuCO3µÄÖÊÁ¿·ÖÊý¡£
(1)°´ÕÕ¼×ͬѧµÄ¹Ûµã£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ           ¡£
(2)ÒÒͬѧÈÏΪÓÐCu(OH)2Éú³ÉµÄÀíÂÛÒÀ¾ÝÊÇ       (ÓÃÀë×Ó·½³Ìʽ±íʾ)¡£
(3)Á½Í¬Ñ§ÀûÓÃÏÂͼËùʾװÖýøÐвⶨ

¢ÙÔÚÑо¿³ÁµíÎï×é³ÉÇ°£¬Ð뽫³Áµí´ÓÈÜÒºÖзÖÀë²¢¾»»¯¡£¾ßÌå²Ù×÷ÒÀ´ÎΪ       ¡¢Ï´µÓ¡¢¸ÉÔï¡£
¢Ú×°ÖÃEÖмîʯ»ÒµÄ×÷ÓÃÊÇ       ¡£
¢ÛʵÑé¹ý³ÌÖÐÓÐÒÔϲÙ×÷²½Ö裺
a£®¹Ø±ÕK1¡¢K3£¬´ò¿ªK2¡¢K4£¬³ä·Ö·´Ó¦
b£®´ò¿ªK1¡¢K4£¬¹Ø±ÕK2¡¢K3£¬Í¨Èë¹ýÁ¿¿ÕÆø
c£®´ò¿ªK1¡¢K3£¬¹Ø±ÕK2¡¢K4£¬Í¨Èë¹ýÁ¿¿ÕÆø
ÕýÈ·µÄ˳ÐòÊÇ       (ÌîÑ¡ÏîÐòºÅ£¬ÏÂͬ)¡£
Èôδ½øÐв½Öè       £¬½«Ê¹²âÁ¿½á¹ûÆ«µÍ¡£
¢ÜÈô³ÁµíÑùÆ·µÄÖÊÁ¿Îªm g£¬×°ÖÃDµÄÖÊÁ¿Ôö¼ÓÁËn g£¬Ôò³ÁµíÖÐCuCO3µÄÖÊÁ¿·ÖÊýΪ       ¡£
(4)±ûͬѧÈÏΪ»¹¿ÉÒÔͨ¹ý²âÁ¿CO2µÄÌå»ý»ò²âÁ¿       À´²â¶¨³ÁµíÖÐCuCO3µÄÖÊÁ¿·ÖÊý¡£
±½¼×Ëá¹ã·ºÓ¦ÓÃÓÚÖÆÒ©ºÍ»¯¹¤ÐÐÒµ£¬Ä³Í¬Ñ§³¢ÊÔÓüױ½µÄÑõ»¯·´Ó¦ÖƱ¸±½¼×Ëᣬ·´Ó¦Ô­
Àí£º

ʵÑé·½·¨£ºÒ»¶¨Á¿µÄ¼×±½ºÍÊÊÁ¿µÄKMnO4ÈÜÒºÔÚ100¡æ·´Ó¦Ò»¶Îʱ¼äºóÍ£Ö¹·´Ó¦£¬°´ÈçÏÂÁ÷³Ì·ÖÀë³ö±½¼×ËáºÍ»ØÊÕδ·´Ó¦µÄ¼×±½¡£

ÒÑÖª£º±½¼×ËáÏà¶Ô·Ö×ÓÖÊÁ¿122 £¬ÈÛµã122.4¡æ£¬ÔÚ25¡æºÍ95¡æʱÈܽâ¶È·Ö±ðΪ0.3 gºÍ6.9 g£»´¿¾»¹ÌÌåÓлúÎïÒ»°ã¶¼Óй̶¨È۵㡣
£¨1£©²Ù×÷¢ñΪ         £¬ÐèÒªÓõ½µÄÖ÷Òª²£Á§ÒÇÆ÷Ϊ          £»²Ù×÷¢òΪ          ¡£
£¨2£©ÎÞÉ«ÒºÌåAÊÇ        £¬¶¨ÐÔ¼ìÑéAµÄÊÔ¼ÁÊÇ            £¬ÏÖÏóÊÇ              ¡£
£¨3£©²â¶¨°×É«¹ÌÌåBµÄÈ۵㣬·¢ÏÖÆäÔÚ115¡æ¿ªÊ¼ÈÛ»¯£¬´ïµ½130¡æʱÈÔÓÐÉÙÁ¿²»ÈÛ£¬¸ÃͬѧÍƲâ°×É«¹ÌÌåBÊDZ½¼×ËáÓëKClµÄ»ìºÏÎÉè¼ÆÁËÈçÏ·½°¸½øÐÐÌá´¿ºÍ¼ìÑ飬ʵÑé½á¹û±íÃ÷ÍƲâÕýÈ·£®ÇëÔÚ´ðÌ⿨ÉÏÍê³É±íÖÐÄÚÈÝ¡£
ÐòºÅ
ʵÑé·½°¸
ʵÑéÏÖÏó
½áÂÛ
¢Ù
½«°×É«¹ÌÌåB¼ÓÈëË®ÖУ¬
¼ÓÈÈÈܽ⣬            
µÃµ½°×É«¾§ÌåºÍÎÞÉ«ÈÜÒº
 
¢Ú
È¡ÉÙÁ¿ÂËÒºÓÚÊÔ¹ÜÖУ¬
                  
Éú³É°×É«³Áµí
ÂËÒºº¬Cl-
¢Û
¸ÉÔï°×É«¾§Ì壬     
                   
°×É«¾§ÌåÊDZ½¼×Ëá
 
£¨4£©´¿¶È²â¶¨£º³ÆÈ¡1.220 g²úÆ·ÈܽâÔÚ¼×´¼ÖÐÅä³É100 mlÈÜÒº£¬ÒÆÈ¡2 5.00 mlÈÜÒº£¬µÎ¶¨£¬ÏûºÄKOHµÄÎïÖʵÄÁ¿Îª2.40 ¡Á 10 -3 mol£¬²úÆ·Öб½¼×ËáÖÊÁ¿·ÖÊýµÄ¼ÆËã±í´ïʽΪ               £¬¼ÆËã½á¹ûΪ             ¡££¨±£ÁôÁ½Î»ÓÐЧÊý×Ö£©¡£
ÂÈ»¯ÌúÊdz£¼ûµÄË®´¦Àí¼Á£¬ÀûÓ÷ÏÌúм¿ÉÖƱ¸ÎÞË®ÂÈ»¯Ìú¡£ÊµÑéÊÒÖƱ¸×°Öú͹¤ÒµÖƱ¸Á÷³ÌͼÈçÏ£º

ÒÑÖª£º(1)ÎÞË®FeCl3µÄÈÛµãΪ555 K¡¢·ÐµãΪ588 K¡£
(2) ·ÏÌúмÖеÄÔÓÖʲ»ÓëÑÎËá·´Ó¦
(3)²»Í¬Î¶ÈÏÂÁùË®ºÏÂÈ»¯ÌúÔÚË®ÖеÄÈܽâ¶ÈÈçÏ£º
ζÈ/¡æ
0
20
80
100
Èܽâ¶È(g/100 g H2O)
74.4
91.8
525.8
535.7
 
ʵÑéÊÒÖƱ¸²Ù×÷²½ÖèÈçÏ£º
¢ñ.´ò¿ªµ¯»É¼ÐK1£¬¹Ø±Õµ¯»É¼ÐK2£¬²¢´ò¿ª»îÈûa£¬»ºÂýµÎ¼ÓÑÎËá¡£
¢ò.µ±       Ê±£¬¹Ø±Õµ¯»É¼ÐK1£¬´ò¿ªµ¯»É¼ÐK2£¬µ±AÖÐÈÜÒºÍêÈ«½øÈëÉÕ±­ºó¹Ø±Õ»îÈûa¡£
¢ó.½«ÉÕ±­ÖÐÈÜÒº¾­¹ýһϵÁвÙ×÷ºóµÃµ½FeCl3¡¤6H2O¾§Ìå¡£
Çë»Ø´ð£º
£¨1£©ÉÕ±­ÖÐ×ãÁ¿µÄH2O2ÈÜÒºµÄ×÷ÓÃÊÇ       ¡£
£¨2£©ÎªÁ˲ⶨ·ÏÌúмÖÐÌúµÄÖÊÁ¿·ÖÊý£¬²Ù×÷¢òÖС°¡­¡­¡±µÄÄÚÈÝÊÇ__________¡£
£¨3£©´ÓFeCl3ÈÜÒºÖƵÃFeCl3?6H2O¾§ÌåµÄ²Ù×÷²½ÖèÊÇ£º¼ÓÈë _   ºó¡¢  _  ¡¢¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔ
£¨4£©ÊÔд³öÎüÊÕËþÖз´Ó¦µÄÀë×Ó·½³Ìʽ£º             ¡£
£¨5£©²¶¼¯Æ÷ζȳ¬¹ý673 Kʱ£¬´æÔÚÏà¶Ô·Ö×ÓÖÊÁ¿Îª325µÄÌúµÄÂÈ»¯Î¸ÃÎïÖʵķÖ×ÓʽΪ            ¡£
£¨6£©FeCl3µÄÖÊÁ¿·ÖÊýͨ³£¿ÉÓõâÁ¿·¨²â¶¨£º³ÆÈ¡m gÎÞË®ÂÈ»¯ÌúÑùÆ·£¬ÈÜÓÚÏ¡ÑÎËᣬÅäÖƳÉ100mLÈÜÒº£»È¡³ö10.00mL£¬¼ÓÈëÉÔ¹ýÁ¿µÄKIÈÜÒº£¬³ä·Ö·´Ó¦ºó£¬µÎÈ뼸µÎµí·ÛÈÜÒº£¬²¢ÓÃc mol?L-1 Na2S2O3ÈÜÒºµÎ¶¨£¬ÏûºÄV mL£¨ÒÑÖª£ºI2+2S2O32-¨T2I-+S4O62-£©¡£
¢ÙµÎ¶¨ÖÕµãµÄÏÖÏóÊÇ£º           _             
¢ÚÑùÆ·ÖÐÂÈ»¯ÌúµÄÖÊÁ¿·ÖÊý              _           

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

¾«Ó¢¼Ò½ÌÍø