ÌâÄ¿ÄÚÈÝ

ÑÎËá¡¢ÁòËáºÍÏõËáÊÇÖÐѧ½×¶Î³£¼ûµÄÈýÖÖËᣮÇë¾ÍÈýÖÖËáÓë½ðÊôÍ­·´Ó¦µÄÇé¿ö£¬»Ø´ðÏÂÁÐÎÊÌ⣺£¨1£©Ï¡ÑÎËáÓëÍ­²»·´Ó¦£¬µ«ÏòÏ¡ÑÎËáÖмÓÈëH2O2ºó£¬Ôò¿ÉʹͭÈܽ⣬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ
Cu+H2O2+2H+=Cu2++2H2O
Cu+H2O2+2H+=Cu2++2H2O
£®
ijͬѧδ¼ÓÈëÑõ»¯¼Á£¬¶øÊÇÉè¼ÆÁËÒ»¸öʵÑé×°Öã¬Ò²ÄÜʹͭºÜ¿ìÈÜÓÚÏ¡ÑÎËᣮÇëÔÚ·½¸ñÖл­³ö¸Ã×°ÖÃ

£¨2£©ÓÉ4¸ùͬÖÊÁ¿µÄµç¼«²ÄÁÏ£¨Ag Cu Pt£©×é³ÉÏÂÁÐ×°Öã®ÏÂÁÐ×°ÖÃAΪ
Ô­µç
Ô­µç
³Ø£¨Ìîд¡±Ô­µç¡±»ò¡±µç½â¡±£©£¬·¢ÉúÑõ»¯·´Ó¦µÄµç¼«·´Ó¦Ê½Îª£º
Cu-2e-=Cu2+
Cu-2e-=Cu2+
£¬B³ØÖÐÑô¼«µÄµç¼«·´Ó¦Ê½Îª£º
4OH--4e-=2H2O+O2
4OH--4e-=2H2O+O2
£®µ±¹Ø±Õµç¼ükÒ»¶Îʱ¼ä£¬·¢ÏÖAg°ôÓë Cu°ôÖÊÁ¿²îΪ28gʱ£¬Ôò×°ÖÃB²úÉúÆøÌåµÄ×ÜÖÊÁ¿Îª
1.8
1.8
g£¨µç½âÖÊÈÜÒº×ãÁ¿£©£®

£¨3£©½«µÈÖÊÁ¿µÄͭƬ·Ö±ðÓëµÈÌå»ý¡¢¹ýÁ¿µÄŨÏõËáºÍÏ¡ÏõËá·´Ó¦£¬ËùµÃµ½µÄÈÜҺǰÕß³ÊÂÌÉ«£¬ºóÕß³ÊÀ¶É«£¬Ä³Í¬Ñ§Ìá³ö¿ÉÄÜÊÇCu2+Ũ¶È²»Í¬ÒýÆðµÄ£¬ÄãͬÒâÕâÖÖ¿´·¨Âð£¿
²»Í¬Òâ
²»Í¬Òâ
£¨ÌͬÒ⡱»ò¡°²»Í¬Ò⡱£©£¬ÀíÓÉÊÇ
µÈÖÊÁ¿µÄͭƬÓëµÈÌå»ý¡¢¹ýÁ¿µÄŨÏõËᡢϡÏõËá·´Ó¦£¬ËùµÃÈÜÒºÖÐCu2+µÄŨ¶È»ù±¾ÏàµÈ
µÈÖÊÁ¿µÄͭƬÓëµÈÌå»ý¡¢¹ýÁ¿µÄŨÏõËᡢϡÏõËá·´Ó¦£¬ËùµÃÈÜÒºÖÐCu2+µÄŨ¶È»ù±¾ÏàµÈ
£®
·ÖÎö£º£¨1£©Ë«ÑõË®¾ßÓÐÑõ»¯ÐÔ£¬ÔÚËáÐÔ»·¾³ÏÂÄܽ«½ðÊôÍ­Ñõ»¯£»Óõç½âÔ­ÀíʵÏÖÍ­ºÍÑÎËáµÄ·´Ó¦£¬¸ù¾Ýµç½âÔ­ÀíÀ´Ñ¡ÔñºÏÊʵĵç½â×°Öã»
£¨2£©Ô­µç³ØµÄ¹¹³ÉÌõ¼þ£ºÓлîÆÃÐÔ²»Í¬µÄÁ½¸öµç¼«¡¢Óеç½âÖÊÈÜÒº¡¢±ÕºÏ»Ø·£»¸º¼«·¢ÉúÑõ»¯·´Ó¦£¬Ê§µç×ӵļ«Îª¸º¼«£¬ºÍÔ­µç³ØÕý¼«ÏàÁ¬µÄΪµç½â³ØµÄÑô¼«£¬Õû¸öµç³ØÖеÃʧµç×ÓÊýÄ¿ÏàµÈ£»
£¨3£©µÈÖÊÁ¿µÄͭƬ·Ö±ðÓëµÈÌå»ý¡¢¹ýÁ¿µÄŨÏõËáºÍ¹ýÁ¿µÄÏ¡ÏõËᷴӦʱͭȫ²¿×ª»¯ÎªÀë×Ó£®
½â´ð£º½â£º£¨1£©Ë«ÑõË®¾ßÓÐÑõ»¯ÐÔ£¬ÔÚËáÐÔ»·¾³ÏÂÄܽ«½ðÊôÍ­Ñõ»¯£¬Àë×Ó·½³ÌʽΪ£ºCu+H2O2+2HCl=CuCl2+2H2O£¬Óõç½âÔ­ÀíʵÏÖÍ­ºÍÑÎËáµÄ·´Ó¦£¬¸ù¾Ýµç½âÔ­ÀíÑô¼«±ØÐëÊǽðÊôÍ­£¬µç½âÖʱØÐëÊÇÏ¡ÑÎËᣬ¹Ê´ð°¸Îª£ºCu+H2O2+2HCl=CuCl2+2H2O£»
£¨2£©¸ù¾Ý×°ÖõÄÌص㣬ȷ¶¨×°ÖÃAΪԭµç³Ø£¬×°ÖÃBΪµç½â³Ø£¬Ô­µç³ØÖУ¬Ê§µç×ӵļ«Îª¸º¼«£¬¸º¼«·¢ÉúÑõ»¯·´Ó¦£¬µç¼«·´Ó¦Îª£ºCu-2e-=Cu2+£¬ºÍÔ­µç³ØÕý¼«ÏàÁ¬µÄΪµç½â³ØµÄÑô¼«£¬ÒõÀë×ӷŵ磬µç¼«·´Ó¦Îª£º4OH--4e-=2H2O+O2£¬µ±Ag°ôÓë Cu°ôÖÊÁ¿²îΪ28gʱ£¬É跴ӦתÒƵç×ÓÊýΪx£¬ÔòCu°ôÖÊÁ¿¼õÉÙ32x£¬Ag°ôÖÊÁ¿Ôö¼Ó108x£¬ËùÒÔ108x+32x=28£¬x=0.2mol£¬×°ÖÃBÖÐÏ൱ÓÚµç½âË®£¬Ñô¼«·ÅÑõÆøµÄÖÊÁ¿Îª£º0.05mol¡Á32g/mol=1.6g£¬Òõ¼«·ÅÇâÆøµÄÖÊÁ¿Îª0.2g£¬¹Ê×°ÖÃB²úÉúÆøÌåµÄ×ÜÖÊÁ¿Îª1.8g£¬¹Ê´ð°¸Îª£º1.8£»
£¨3£©½«µÈÖÊÁ¿µÄͭƬ·Ö±ðÓëµÈÌå»ý¡¢¹ýÁ¿µÄŨÏõËáºÍ¹ýÁ¿µÄÏ¡ÏõËá·´Ó¦ºó£¬ËùµÃÈÜÒºÖÐCu2+µÄŨ¶È»ù±¾ÏàµÈ£¬ÑÕÉ«»ù±¾Ïàͬ£¬¹Ê´ð°¸Îª£º²»Í¬Ò⣻µÈÖÊÁ¿µÄͭƬÓëµÈÌå»ý¡¢¹ýÁ¿µÄŨÏõËᡢϡÏõËá·´Ó¦£¬ËùµÃÈÜÒºÖÐCu2+µÄŨ¶È»ù±¾ÏàµÈ£®
µãÆÀ£º±¾Ìâ½áºÏµç»¯Ñ§ÖªÊ¶¿¼²éÁ˽ðÊôÍ­µÄ»¯Ñ§ÐÔÖÊ£¬ÒªÇóѧÉú¾ß±¸·ÖÎöºÍ½â¾öÎÊÌâµÄÄÜÁ¦£¬ÄѶȽϴó£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÑÎËá¡¢ÁòËáºÍÏõËáÊÇÖÐѧ½×¶Î³£¼ûµÄ¡°Èý´óËᡱ£¬Çë¾Í¡°Èý´óËᡱµÄÐÔÖÊ£¬»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ï¡ÑÎËáÓëÍ­²»·´Ó¦£¬µ«ÏòÏ¡ÑÎËáÖмÓÈëH2O2ºó£¬Ôò¿ÉʹͭÈܽ⣮¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
Cu+H2O2+2HCl¨TCuCl2+2H2O
Cu+H2O2+2HCl¨TCuCl2+2H2O
£®
£¨2£©Ä³¿ÎÍâ»î¶¯Ð¡×éÉè¼ÆÁËÒÔÏÂʵÑé·½°¸ÑéÖ¤CuÓëŨÏõËá·´Ó¦µÄ¹ý³ÌÖпÉÄܲúÉúNO£®ÆäʵÑéÁ÷³ÌͼÈçÏ£º

¢ÙÈôÒª²â¶¨NOµÄÌå»ý£¬´ÓͼËùʾµÄ×°ÖÃÖУ¬ÄãÈÏΪӦѡÓÃ
A
A
×°ÖýøÐÐCuÓëŨÏõËᷴӦʵÑ飬ѡÓõÄÀíÓÉÊÇ
ÒòΪA×°ÖÿÉÒÔͨN2½«×°ÖÃÖеĿÕÆøÅž¡£¬·ÀÖ¹NO±»¿ÕÆøÖÐO2Ñõ»¯
ÒòΪA×°ÖÿÉÒÔͨN2½«×°ÖÃÖеĿÕÆøÅž¡£¬·ÀÖ¹NO±»¿ÕÆøÖÐO2Ñõ»¯
£®
¢ÚÑ¡ÓÃÈçͼËùʾÒÇÆ÷×éºÏÒ»Ì׿ÉÓÃÀ´²â¶¨Éú³ÉNOÌå»ýµÄ×°Öã¬ÆäºÏÀíµÄÁ¬½Ó˳ÐòÊÇ
123547
123547
£¨Ìî¸÷µ¼¹Ü¿Ú±àºÅ£©£®
123547
123547

¢ÛÔڲⶨNOµÄÌå»ýʱ£¬ÈôÁ¿Í²ÖÐË®µÄÒºÃæ±È¼¯ÆøÆ¿µÄÒºÃæÒªµÍ£¬´ËʱӦ½«Á¿Í²µÄλÖÃ
Éý¸ß
Éý¸ß
£¨¡°Ï½µ¡±»ò¡°Éý¸ß¡±£©£¬ÒÔ±£Ö¤Á¿Í²ÖеÄÒºÃæÓ뼯ÆøÆ¿ÖеÄÒºÃæ³Öƽ£®
£¨3£©¹¤ÒµÖÆÁòËáʱ£¬ÁòÌú¿ó£¨FeS2£©¸ßÎÂÏ¿ÕÆøÑõ»¯²úÉú¶þÑõ»¯Áò£º4FeS2+11O2=8SO2+2Fe2O3£¬Éè¿ÕÆøÖÐN2¡¢O2µÄº¬Á¿·Ö±ðΪ0.800ºÍ0.200£¨Ìå»ý·ÖÊý£©£¬4.8t FeS2ÍêÈ«ÖƳÉÁòËᣬÐèÒª¿ÕÆøµÄÌå»ý£¨±ê×¼×´¿ö£©Îª
1.68¡Á107
1.68¡Á107
L£®
ÑÎËá¡¢ÁòËáºÍÏõËáÊÇÖÐѧ½×¶Î³£¼ûµÄÈýÖÖÇ¿Ëᣮ
¢ñ£®Çë¾ÍÈýÕßÓë½ðÊôÍ­µÄ·´Ó¦Çé¿ö£¬»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÔÚ100mL¡¢18mol?L-1 µÄŨÁòËáÖмÓÈë¹ýÁ¿µÄͭƬ£¬¼ÓÈÈʹ֮³ä·Ö·´Ó¦£¬²úÉúµÄÆøÌåÔÚ±ê×¼Çé¿öϵÄÌå»ý¿ÉÄÜÊÇ
D
D

A.40.32L              B.30.24L          C.20.16L          D.6.72L
£¨2£©Èô½«¹ýÁ¿Í­·ÛÓëÒ»¶¨Á¿Å¨ÏõËá·´Ó¦£¬µ±·´Ó¦Íêȫֹͣʱ£¬¹²ÊÕ¼¯µ½µÄÆøÌå1.12L£¨±ê×¼×´¿ö£©£¬Ôò¸ÃÆøÌåµÄ³É·ÖÊÇ
NO2¡¢NO
NO2¡¢NO
£¬·´Ó¦ÖÐËùÏûºÄµÄÏõËáµÄÎïÖʵÄÁ¿¿ÉÄÜΪ
B
B
£®
A.0.1mol              B.0.15mol              C.0.2mol              D.0.25mol
£¨3£©Ä³Í¬Ñ§Ïò½þÅÝͭƬµÄÏ¡ÑÎËáÖмÓÈëH2O2ºó£¬Í­Æ¬Èܽ⣬²¢ÇҸ÷´Ó¦µÄ²úÎïÖ»ÓÐÂÈ»¯Í­ºÍË®£®¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
Cu+H2O2+2HCl=CuCl2+2H2O
Cu+H2O2+2HCl=CuCl2+2H2O
£®
¢ò£®ÓÃÈçÏÂ×°ÖÿÉÒÔÍê³ÉHClµÄÖÆÈ¡¼°ºóÐøһϵÁеÄÐÔÖÊʵÑ飨ͼÖмгÖ×°ÖÃÒÑÂÔÈ¥£©£®
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©×°ÖÃAÑ¡ÓÃŨÁòËáºÍŨÑÎËá»ìºÏÖÆÈ¡HClÆøÌ壬pÖÐÊ¢×°ÊÔ¼ÁΪ
ŨÁòËá
ŨÁòËá
£®
£¨2£©×°ÖÃBÖеÄËÄ´¦ÃÞ»¨ÒÀ´Î×öÁËÈçÏ´¦Àí£º¢Ù°üÓÐij³±ÊªµÄ¹ÌÌåÎïÖÊ  ¢ÚÕºÓÐKIÈÜÒº¢ÛÕºÓÐʯÈïÈÜÒº  ¢ÜÕºÓÐŨNaOHÈÜÒº£®·´Ó¦¿ªÊ¼ºó£¬¹Û²ìµ½¢Ú´¦ÓÐ×Ø»ÆÉ«ÎïÖʲúÉú£¬Ð´³ö¢Ú´¦·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ
CI2+2I-=2Cl-+I2
CI2+2I-=2Cl-+I2
£®ÔÚ¢Ù´¦°üÓеĹÌÌåÎïÖÊ¿ÉÄÜÊÇ
a£®MnO2        b£®KmnO4       C£®KCl              d£®Cu
£¨3£©ÔÚÕû¸öʵÑé¹ý³ÌÖУ¬ÔÚ¢Û´¦Äܹ۲쵽µÄÏÖÏóÊÇ
ÏȱäºìºóÍÊÉ«
ÏȱäºìºóÍÊÉ«
£®

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

¾«Ó¢¼Ò½ÌÍø