ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿µç½â¾«Á¶Í­µÄÑô¼«ÄàÖÐÖ÷Òªº¬Ag¡¢AuµÈ¹óÖؽðÊô¡£ÒÔÏÂÊÇ´Ó¾«Á¶Í­µÄÑô¼«ÄàÖлØÊÕÒø¡¢½ðµÄÁ÷³Ìͼ£º

(1)ÂȽðËá(HAuCl4)ÖеÄAuµÄ»¯ºÏ¼ÛΪ________¡£

(2)Í­Ñô¼«ÄàÑõ»¯Ê±£¬²ÉÓá°µÍαºÉÕ¡±¶ø²»²ÉÓá°¸ßαºÉÕ¡±µÄÔ­ÒòÊÇ_____________________¡£

(3)¡°±ºÉÕÔü¡±ÔÚ¡°¢ÙËá½þ¡±Ê±·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ______________________________________¡£

(4)¡°¢Ú½þ½ð¡±·´Ó¦ÖУ¬H2SO4µÄ×÷ÓÃΪ___________________________________________£¬¸Ã²½ÖèµÄ·ÖÀë²Ù×÷ÖУ¬ÐèÒª¶ÔËùµÃµÄAgCl½øÐÐˮϴ¡£¼òÊöÈçºÎÅжÏAgClÒѾ­Ï´µÓ¸É¾»£¿ _____________________________________________________¡£

(5)ÂȽðËá(HAuCl4)ÔÚpHΪ2¡«3µÄÌõ¼þϱ»²ÝËỹԭΪAu£¬Í¬Ê±·Å³ö¶þÑõ»¯Ì¼ÆøÌ壬Ôò¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_____________________________________________________¡£

(6)¼×È©»¹Ô­·¨³Á»ýÒø£¬Í¨³£ÊÇÔÚ½Á°èÏÂÓÚÊÒμ°Èõ¼îÐÔÌõ¼þϽøÐУ¬¼×È©±»Ñõ»¯ÎªÌ¼ËáÇâ¸ùÀë×Ó£¬Ôò¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ___________________________________________¡£

¡¾´ð°¸¡¿£«3 ¸ßαºÉÕʱ£¬Éú³ÉµÄAg2OÓÖ·Ö½âΪAgºÍO2(»ò2Ag2O4Ag£«O2¡ü) Ag2O£«2H£«£«2Cl£­=2AgCl£«H2O ÌṩH£«£¬ÔöÇ¿NaClO3µÄÑõ»¯ÐÔ È¡×îºóÒ»´ÎÏ´µÓÒºÉÙÐíÓÚÊÔ¹ÜÖУ¬µÎÈëBa(NO3)2ÈÜÒº£¬ÈôûÓа×É«³Áµí²úÉú£¬ÔòÒѾ­Ï´µÓ¸É¾»£¬·´Ö®£¬ÔòÐèÒª¼ÌÐøÏ´µÓ 2HAuCl4£«3H2C2O4=2Au¡ý£«8HCl£«6CO2¡ü 4Ag(SO3)23-£«HCHO£«5OH£­=4Ag¡ý£«8SO32-£«3H2O£«HCO3-

¡¾½âÎö¡¿

Í­Ñô¼«ÄàµÍÎÂÑõ»¯±ºÉÕ£¬È»ºó¼ÓÈëÁòËáºÍÂÈ»¯ÄƵĻìºÏÎ¹ýÂË£¬ÂËÔüΪAgClºÍAu£¬ÔÚÂËÔüÖмÓÁòËá¡¢NaClºÍÂÈËáÄƵĻìºÏÎ¹ýÂËÂËÔüΪAgCl£¬ÂËҺΪÂȽðËᣬAgClÖмÓNa2SO3ÈÜҺת»¯ÎªAg£¨SO3£©23-£¬ÔÙÓÃHCHO»¹Ô­µÃµ½´ÖAg£¬HAuCl4ÈÜÒºÖмÓH2C2O4Éú³É´Ö½ð¡£

£¨1£©ÂȽðËᣨHAuCl4£©ÖÐClΪ-1¼Û£¬HΪ+1¼Û£¬ÔòAuµÄ»¯ºÏ¼ÛΪ+3£»¹Ê´ð°¸Îª£º+3£»

£¨2£©µÍαºÉÕʱ£¬AgÓëÑõÆøת»¯ÎªAg2O£¬¸ßÎÂʱ£¬Ñõ»¯Òø·Ö½âÓÖÉú³ÉAgºÍÑõÆø£»¹Ê´ð°¸Îª£º¸ßαºÉÕʱ£¬Éú³ÉµÄAg2OÓÖ·Ö½âΪAgºÍO2(»ò2Ag2O4Ag£«O2¡ü)£»

£¨3£©Ëá½þʱ£¬Ñõ»¯ÒøÓëÑÎËá·´Ó¦Éú³ÉÂÈ»¯Òø£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£ºAg2O+2H++2Cl-=2AgCl+H2O£»¹Ê´ð°¸Îª£ºAg2O+2H++2Cl-=2AgCl+H2O£»

£¨4£©¡°¢Ú½þ½ð¡±·´Ó¦ÖУ¬ËáÐÔÌõ¼þÏ£¬AuÓëÂÈËáÄÆ·´Ó¦£¬¼ÓÁòËáÄÜÌṩH+£¬ÔöÇ¿NaClO3µÄÑõ»¯ÐÔ£»ÈÜÒºÖк¬ÓÐÁòËá¸ùÀë×Ó£¬¼ìÑé×îºóÒ»´ÎÏ´ÒºÖÐÊÇ·ñº¬ÓÐÁòËá¸ùÀë×Ó£¬Æä²Ù×÷Ϊȡ×îºóÒ»´ÎÏ´µÓÒºÉÙÐíÓÚÊÔ¹ÜÖУ¬µÎÈëBa£¨NO3£©2ÈÜÒº£¬ÈôûÓа×É«³Áµí²úÉú£¬ÔòÒѾ­Ï´µÓ¸É¾»£¬·´Ö®£¬ÔòÐèÒª¼ÌÐøÏ´µÓ£»¹Ê´ð°¸Îª£ºÌṩH£«£¬ÔöÇ¿NaClO3µÄÑõ»¯ÐÔ £»È¡×îºóÒ»´ÎÏ´µÓÒºÉÙÐíÓÚÊÔ¹ÜÖУ¬µÎÈëBa(NO3)2ÈÜÒº£¬ÈôûÓа×É«³Áµí²úÉú£¬ÔòÒѾ­Ï´µÓ¸É¾»£¬·´Ö®£¬ÔòÐèÒª¼ÌÐøÏ´µÓ£»

£¨5£©ÂȽðËáÓë²ÝËá·´Ó¦Éú³ÉAu¡¢HClºÍ¶þÑõ»¯Ì¼£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º2HAuCl4+3H2C2O4=2Au+8HCl+6CO2¡ü£»¹Ê´ð°¸Îª£º2HAuCl4+3H2C2O4=2Au+8HCl+6CO2¡ü£»

£¨6£©Èõ¼îÐÔÌõ¼þÏ£¬Ag£¨SO3£©23-±»¼×È©»¹Ô­ÎªAg£¬Í¬Ê±Éú³É̼ËáÇâ¸ùÀë×Ó£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º4Ag£¨SO3£©23-+HCHO+5OH-=4Ag+8SO32-+3H2O+HCO3-¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿Ä³Í¬Ñ§Ì½¾¿CuÓëNOµÄ·´Ó¦£¬²éÔÄ×ÊÁÏ£º¢ÙCuÓëNO·´Ó¦¿ÉÉú³ÉCuOºÍN2£¬¢ÚËáÐÔÌõ¼þÏ£¬NO»òNO2¨C¶¼ÄÜÓëMnO4¨C·´Ó¦Éú³ÉNO3¨CºÍMn2+

£¨1£©ÊµÑéÊÒÀûÓÃCuºÍÏ¡HNO3ÖƱ¸NO£¬Ð´³ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ_____________¡£

£¨2£©Ñ¡ÓÃÈçͼËùʾװÖÃÍê³ÉCuÓëNOµÄʵÑé¡£(¼Ð³Ö×°ÖÃÂÔ) ʵÑ鿪ʼǰ£¬Ïò×°ÖÃÖÐͨÈëÒ»¶Îʱ¼äµÄN2¡£»Ø´ðÏÂÁÐÎÊÌ⣺

¢ÙʹÓÃÍ­Ë¿µÄÓŵãÊÇ_____________________×°ÖÃEµÄ×÷ÓÃΪ_______________¡£

¢Ú×°ÖÃCÖÐÊ¢·ÅµÄÒ©Æ·¿ÉÄÜÊÇ_________£»

¢Û×°ÖÃDÖеÄÏÖÏóÊÇ_______________£»×°ÖÃFÖз´Ó¦µÄÀë×Ó·½³ÌʽÊÇ_______________¡£

£¨3£©²â¶¨NaNO2ºÍNaNO3 »ìºÏÈÜÒºÖÐNaNO2µÄŨ¶È¡£ È¡25.00mL»ìºÏÈÜÒºÓÚ׶ÐÎÆ¿ÖУ¬ÓÃ0.1000mol¡¤L£­1ËáÐÔKMnO4ÈÜÒº½øÐе樣¬ÊµÑéËùµÃÊý¾ÝÈçϱíËùʾ£º

µÎ¶¨´ÎÊý

1

2

3

4

ÏûºÄKMnO4ÈÜÒºÌå»ý/mL

20.90

20.12

20.00

19.88

¢ÙµÚÒ»´ÎʵÑéÊý¾Ý³öÏÖÒì³££¬Ôì³ÉÕâÖÖÒì³£µÄÔ­Òò¿ÉÄÜÊÇ_________£¨Ìî×Öĸ´úºÅ£©¡£

a£®×¶ÐÎÆ¿Ï´¾»ºóδ¸ÉÔï

b£®ËáʽµÎ¶¨¹ÜÓÃÕôÁóˮϴ¾»ºóδÓñê×¼ÒºÈóÏ´

c£®µÎ¶¨ÖÕµãʱÑöÊÓ¶ÁÊý

d£®ËáÐÔKMnO4ÈÜÒºÖк¬ÓÐÆäËûÑõ»¯ÐÔÊÔ¼Á

e£®×¶ÐÎÆ¿Ï´¾»ºóÓôý²âÒºÈóÏ´

¢ÚËáÐÔKMnO4ÈÜÒºµÎ¶¨ÑÇÏõËáÄÆÈÜÒºµÄÀë×Ó·½³ÌʽΪ___________________¡£

¢ÛNaNO2 µÄÎïÖʵÄÁ¿Å¨¶ÈΪ__________

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

¾«Ó¢¼Ò½ÌÍø