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£©

½âÎöÊÔÌâ·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¡£
¿¼µã£º¿¼²é³£¼ûÑõ»¯¼Á¡¢Ñõ»¯»¹Ô­·´Ó¦¡¢»¯Ñ§¼ÆËãºÍ¶ÔÎïÖÊÖƱ¸µÄÀí½â¡¢·ÖÎöÄÜÁ¦£¬¶Ô»¯Ñ§·½³ÌʽµÄ¼ÇÒä

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

(14·Ö) Mg(ClO3)2³£ÓÃ×÷´ßÊì¼Á¡¢³ý²Ý¼ÁµÈ£¬ÏÂͼΪÖƱ¸ÉÙÁ¿Mg(ClO3)2¡¤6H2OµÄ·½·¨£º

ÒÑÖª£º1¡¢Â±¿éÖ÷Òª³É·ÖΪMgCl2¡¤6H2O£¬º¬ÓÐMgSO4¡¢FeCl2µÈÔÓÖÊ¡£
2¡¢¼¸ÖÖ»¯ºÏÎïµÄÈܽâ¶È(S)ËæζÈ(T)±ä»¯ÇúÏßÈçÉÏͼ¡£
£¨1£©¼ÓÈëBaCl2µÄÄ¿µÄÊdzý_________£¬ÈçºÎÅжϸÃÀë×ÓÒѳý¾¡_____________________¡£
£¨2£©¼ÓÈëNaClO3±¥ºÍÈÜÒº»á·¢Éú·´Ó¦£ºMgCl2£«2NaClO3==Mg(ClO3)2£«2NaCl¡ý£¬ÇëÀûÓø÷´Ó¦£¬½áºÏÈܽâ¶Èͼ£¬ÖÆÈ¡Mg(ClO3)2¡¤6H2OµÄʵÑé²½ÖèÒÀ´ÎΪ£º
¢ÙÈ¡Ñù£¬¼ÓÈëNaClO3±¥ºÍÈÜÒº³ä·Ö·´Ó¦£»¢ÚÕô·¢Å¨Ëõ£»¢Û        £»¢ÜÀäÈ´½á¾§£»
¢Ý¹ýÂË¡¢Ï´µÓ£¬»ñµÃMg(ClO3)2¡¤6H2O¾§Ìå¡£
²úÆ·ÖÐMg(ClO3)2¡¤6H2Oº¬Á¿µÄ²â¶¨£º
²½Öè1£º×¼È·³ÆÁ¿3.50g²úÆ·ÈܽⶨÈݳÉ100mLÈÜÒº¡£
²½Öè2£ºÈ¡10mLÅäºÃµÄÈÜÒºÓÚ׶ÐÎÆ¿ÖУ¬¼ÓÈë10mLÏ¡ÁòËáºÍ20mL1.000mol/LµÄFeSO4ÈÜÒº£¬Î¢ÈÈ¡£
²½Öè3£ºÀäÈ´ÖÁÊÒΣ¬ÓÃ0.100mol/L K2Cr2O7ÈÜÒºµÎ¶¨ÖÁÖյ㣬´Ë¹ý³ÌÖз´Ó¦µÄÀë×Ó·½³ÌʽΪ£º
Cr2O72£­£«6Fe2£«£«14H£«==2Cr3£«£«6Fe3£«£«7H2O¡£
²½Öè4£º½«²½Öè2¡¢3Öظ´Á½´Î£¬Æ½¾ùÏûºÄK2Cr2O7ÈÜÒº15.00mL¡£
£¨3£©Ð´³ö²½Öè2ÖÐËù·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ                                      ¡£
£¨4£©²½Öè3ÖÐÈôµÎ¶¨Ç°Óñê×¼ÒºÈóÏ´µÎ¶¨¹Ü£¬»áµ¼ÖÂ×îÖÕ½á¹û     £¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°²»±ä¡±£©¡£
£¨5£©²úÆ·ÖÐMg(ClO3) 2¡¤6H2OµÄÖÊÁ¿·ÖÊýΪ        (¼ÆËã½á¹û±£ÁôÁ½Î»Ð¡Êý) ¡£

¹¤ÒµÉϳ£ÓøõÌú¿ó£¨ÓÐЧ³É·ÝΪFeO¡¤Cr2O3£¬Ö÷ÒªÔÓÖÊΪSiO2¡¢Al2O3£©ÎªÔ­ÁÏÉú²úÖظõËá¼Ø£¨K2Cr2O7£©£¬ÊµÑéÊÒÄ£Ä⹤ҵ·¨ÓøõÌú¿óÖÆÖظõËá¼ØµÄÖ÷Òª¹¤ÒÕÁ÷³ÌÈçÏÂͼ£¬Éæ¼°µÄÖ÷Òª·´Ó¦ÊÇ£º6FeO¡¤Cr2O3£«24NaOH£«7KClO3=12Na2CrO4£«3Fe2O3£«7KCl£«12H2O£¬ÊԻشðÏÂÁÐÎÊÌ⣺

£¨1£©¢ÝÖÐÈÜÒº½ðÊôÑôÀë×ӵļìÑé·½·¨ÊÇ                                ¡£
£¨2£©²½Öè¢Û±»³ÁµíµÄÀë×ÓΪ£¨ÌîÀë×Ó·ûºÅ£©                             ¡£
£¨3£©ÔÚ·´Ó¦Æ÷¢ÙÖУ¬¶þÑõ»¯¹èÓë´¿¼î·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º                  ¡£
£¨4£©Ñ̵ÀÆøÖеÄCO2¿ÉÓëH2ºÏ³É¼×´¼¡£CH3OH¡¢H2µÄȼÉÕÈÈ·Ö±ðΪ£º¦¤H=£­725.5 kJ/mol¡¢¦¤H=£­285.8 kJ/mol£¬Ð´³ö¹¤ÒµÉÏÒÔCO2¡¢H2ºÏ³ÉCH3OHµÄÈÈ»¯Ñ§·½³Ìʽ£º            ¡£
£¨5£©2011ÄêÔÆÄÏÇú¾¸µÄ¸õÎÛȾʼþ£¬ËµÃ÷º¬¸õ·ÏÔü£¨·ÏË®£©µÄËæÒâÅŷŶÔÈËÀàÉú´æ»·¾³Óм«´óµÄΣº¦¡£µç½â·¨ÊÇ´¦Àí¸õÎÛȾµÄÒ»ÖÖ·½·¨£¬½ðÊôÌú×÷Ñô¼«¡¢Ê¯Ä«×÷Òõ¼«µç½âº¬Cr2O72-µÄËáÐÔ·ÏË®£¬Ò»¶Îʱ¼äºó²úÉúFe(OH)3ºÍCr(OH)3³Áµí¡£
¢Ùд³öµç½â·¨´¦Àí·ÏË®µÄ×Ü·´Ó¦µÄÀë×Ó·½³Ìʽ                               ¡£
¢ÚÒÑÖªCr(OH)3µÄKsp=6.3¡Á10¨C31£¬ÈôµØ±íË®¸õº¬Á¿×î¸ßÏÞÖµÊÇ0.1 mg/L£¬ÒªÊ¹ÈÜÒºÖÐc(Cr3+)½µµ½·ûºÏµØ±íË®ÏÞÖµ£¬Ðëµ÷½ÚÈÜÒºµÄc(OH-)¡Ý     mol/L£¨Ö»Ð´¼ÆËã±í´ïʽ£©¡£

¶þÑõ»¯ÃÌ¿ÉÓÃ×÷¸Éµç³ØÈ¥¼«¼Á£¬ºÏ³É¹¤ÒµµÄ´ß»¯¼ÁºÍÑõ»¯¼Á£¬²£Á§¹¤ÒµºÍÌ´ɹ¤ÒµµÄ×ÅÉ«¼Á¡¢ÏûÉ«¼Á¡¢ÍÑÌú¼ÁµÈ¡£
£¨1£©¶þÑõ»¯ÃÌÔÚËáÐÔ½éÖÊÖÐÊÇÒ»ÖÖÇ¿Ñõ»¯¼Á£¬ÇëÓû¯Ñ§·½³Ìʽ֤Ã÷£º______________________¡£
£¨2£©Ð¿¡ªÃ̼îÐÔµç³Ø¾ßÓÐÈÝÁ¿´ó¡¢·ÅµçµçÁ÷´óµÄÌص㣬Òò¶øµÃµ½¹ã·ºÓ¦Óᣵç³ØµÄ×Ü·´Ó¦Ê½ÎªZn£¨s£©£«2MnO2£¨s£©£«H2O£¨l£©=Zn£¨OH£©2£¨s£©£«Mn2O3£¨s£©¡£
¢Ùµç³Ø¹¤×÷ʱ£¬MnO2·¢Éú________·´Ó¦¡£
¢Úµç³ØµÄÕý¼«·´Ó¦Ê½Îª________¡£
£¨3£©¹¤ÒµÉÏÒÔÈíÃÌ¿óΪԭÁÏ£¬ÀûÓÃÁòËáÑÇÌúÖƱ¸¸ß´¿¶þÑõ»¯Ã̵ÄÁ÷³ÌÈçÏ£º

ÒÑÖª£ºÈíÃÌ¿óµÄÖ÷Òª³É·ÖΪMnO2£¬»¹º¬Si£¨16.27%£©¡¢Fe£¨5.86%£©¡¢Al£¨3.42%£©¡¢Zn£¨2.68%£©ºÍCu£¨0.86%£©µÈÔªËصĻ¯ºÏÎ²¿·ÖÑôÀë×ÓÒÔÇâÑõ»¯Îï»òÁò»¯ÎïµÄÐÎʽÍêÈ«³ÁµíʱÈÜÒºµÄpH¼ûÏÂ±í¡£

³ÁµíÎï
Al£¨OH£©3
Fe£¨OH£©3
Fe£¨OH£©2
Mn£¨OH£©2
Cu£¨OH£©2
pH
5.2
3.2
9.7
10.4
6.7
³ÁµíÎï
Zn£¨OH£©2
CuS
ZnS
MnS
FeS
pH
8.0
£­0.42
2.5
7
7
 
»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÙÁòËáÑÇÌúÔÚËáÐÔÌõ¼þϽ«MnO2»¹Ô­ÎªMnSO4£¬Ëá½þʱ·¢ÉúµÄÖ÷Òª·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ________________________¡£
¢ÚÊÔ¼ÁXΪ________¡£
¢ÛÂËÔüAµÄÖ÷Òª³É·ÖΪ________¡£
¢Ü¼ÓÈëMnSµÄÄ¿µÄÖ÷ÒªÊdzýÈ¥ÈÜÒºÖеÄ________¡£

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

¾«Ó¢¼Ò½ÌÍø