ÌâÄ¿ÄÚÈÝ

£¨12·Ö£© ijͬѧÔÚʵÑéÊÒÓûÅäÖÆÎïÖʵÄÁ¿Å¨¶È¾ùΪ1£®0 mol/LµÄNaOHÈÜÒººÍÏ¡H2SO4¸÷450mL¡£ÌṩµÄÊÔ¼ÁÊÇ£ºNaOH¹ÌÌåºÍ98%µÄŨH2SO4£¨ÃܶÈΪ1£®84 g/cm3£©¼°ÕôÁóË®¡£
£¨1£©ÅäÖÆÁ½ÖÖÈÜҺʱ¶¼ÐèÒªµÄ²£Á§ÒÇÆ÷ÊÇ____________________________________ ¡£
£¨2£©Ó¦ÓÃÍÐÅÌÌìÆ½³ÆÁ¿NaOH ___________g£¬Ó¦ÓÃÁ¿Í²Á¿È¡Å¨H2SO4________mL¡£
£¨3£©ÅäÖÆÊ±£¬ÏÈÒª¼ì²éÈÝÁ¿Æ¿ÊÇ·ñ©ˮ£¬Æä·½·¨ÊÇ                            
                                                                      ¡£
£¨4£©Å¨ÁòËáÈÜÓÚË®µÄÕýÈ·²Ù×÷·½·¨ÊÇ_____                     ___________¡£
£¨5£©ÔÚÅäÖÆÉÏÊöÈÜҺʵÑéÖУ¬ÏÂÁвÙ×÷ÒýÆð½á¹ûÆ«µÍµÄÓÐ_________________
A£®¸ÃѧÉúÔÚÁ¿È¡Å¨ÁòËáʱ£¬¸©Êӿ̶ÈÏß
B£®³ÆÁ¿¹ÌÌåNaOHʱ£¬½«íÀÂëºÍÎïÆ·µÄλÖõߵ¹£¨Ã»ÓÐʹÓÃÓÎÂ룩
C£®ÈܽâH2SO4²Ù×÷ʱûÓÐÀäÈ´ÖÁÊÒξÍÁ¢¼´Íê³ÉºóÃæµÄÅäÖÆ²Ù×÷¡£
D£®ÔÚÉÕ±­ÖÐÈܽâ½Á°èʱ£¬½¦³öÉÙÁ¿ÈÜÒº
E£®Ã»ÓÐÓÃÕôÁóˮϴÉÕ±­2¡ª3´Î£¬²¢½«Ï´ÒºÒÆÈëÈÝÁ¿Æ¿ÖÐ
F£®½«Á¿Í²Ï´µÓ2¡ª3´Î£¬²¢È«²¿×ªÒÆÖÁÈÝÁ¿Æ¿ÖÐ
G£®ÈÝÁ¿Æ¿ÖÐÔ­À´´æÓÐÉÙÁ¿ÕôÁóË®
H£®½ºÍ·µÎ¹Ü¼ÓË®¶¨ÈÝʱ¸©Êӿ̶È
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨16·Ö£©Ä³»¯Ñ§Ñо¿ÐÔС×éͬѧÌá³ö»ØÊÕº¬Í­µçÀ·ÏÁÏÖÐÌáȡͭµÄÁ½ÖÖ·½°¸£¬²¢ÏòÀÏʦ×Éѯ£¬ÀÏʦ¾ÍÒÔϼ¸¸ö·½ÃæÓëͬѧÃÇÕ¹¿ªÌÖÂÛ£º
·½°¸¼×£º

£¨1£©Á½¸ö·½°¸Äĸö·ûºÏµ±Ç°Éú²úµÄÂÌÉ«ÀíÄΪʲô£¿
                                                                  £®
£¨2£©Ð´³ö·½°¸¼×ÖеĢڢ۲½ÖèÓëÍ­ÓйصÄÀë×Ó·´Ó¦µÄÀë×Ó·½³Ìʽ¡£
                             __________________ _       __________
                                _________________       _________¡£
£¨3£©ÀÏʦ½¨Ò飬ÎÞÂÛÊǼ׻¹ÊÇÒҵķ½°¸£¬ÔÚ¡°¼ÓÌúм¡±ÕâÒ»²½Ê±£¬Ó¦¸Ã¼ÓÈëÂÔ¹ýÁ¿µÄÌúм£¬Ä¿µÄÊÇ£º                                   ¡£
£¨4£©ÀÏʦÇëͬѧÃÇÓû¯Ñ§·½·¨¼ìÑéºìÉ«·ÛÄ©ÖÐÊÇ·ñº¬ÓйýÁ¿µÄÌúм¡£ÇëÄãÌîдÏÂÁбí¸ñд³öʵÑé²Ù×÷¡¢Ô¤ÆÚʵÑéÏÖÏó¡¢½áÂÛ¼°²Ù×÷£¨¢ÚÖÐÏà¹ØµÄÀë×Ó·½³Ìʽ£©¡£
ÐòºÅ
ʵÑé²Ù×÷
ʵÑéÏÖÏó
½áÂÛ¼°Àë×Ó·½³Ìʽ
¢Ù
È¡ÊÊÁ¿µÄÑùÆ·ÓÚÊÔ¹ÜÄÚ
¡ª¡ª
¡ª¡ª
¢Ú
ÓõιܵÎÈë×ãÁ¿ÁòËáÈÜÒº£¬²¢³ä·ÖÕñµ´ÊÔ¹Ü
 
 
¢Û
 
ÈÜÒºÏÈ»ÆÉ«£¬µÎÈëKSCNÈÜÒººóÏÔºìÉ«
 
 
 
 
 
£¨5£©ÎªÁËÌá¸ßÔ­ÁϵÄ,ÀÏʦ½¨Òé°Ñ×îºóÒ»²½ËùµÃdzÂÌÉ«ÂËҺͨ¹ýÕô·¢£®ÀäÈ´½á¾§£®¹ýÂË£®Ï´µÓ£®×ÔÈ»¸ÉÔïµÃµ½Ò»ÖÖ´ø½á¾§Ë®µÄÁòËáÑÇÌú¾§Ìå¡£
Ñо¿Ð¡×é°´ÀÏʦµÄ½¨ÒéÍê³ÉÉÏÃæ²Ù×÷²½Ö裬»ñµÃ¾§Ìåºó¶ÔÆä½øÐмì²â
¢ÙÏÈÈ¡a gµÄ¾§Ìå½øÐÐÍÑˮʵÑ飬»ñµÃÎÞË®¹ÌÌåΪ£¨a¡ª1£®26£©g
¢Ú½«ÎÞË®¹ÌÌåÈÜÓÚ×ãÁ¿µÄË®Åä³ÉÈÜÒººóµÎ¼Ó1£®00mol/LµÄÂÈ»¯±µÈÜÒº£¬µ±µÎ¼Ó10£®00mLÈÜҺʱ£¬³ÁµíÇ¡ºÃÍêÈ«¡£Ñо¿Ð¡×éͨ¹ý¼ÆËã²âÖª¸Ã¾§ÌåµÄ»¯Ñ§Ê½ÊÇ              ¡£
£¨17·Ö£©¾ßÓл¹Ô­ÐÔµÄÎÞË®²ÝËáÊÇÎÞÉ«ÎÞ³ôµÄ͸Ã÷½á¾§»ò°×É«·ÛÄ©¡£²ÝËáÔÚŨÁòËá²¢¼ÓÈÈÌõ¼þÏÂÈÝÒ×ÍÑȥˮ·Ö£¬·Ö½âΪ¶þÑõ»¯Ì¼ºÍÒ»Ñõ»¯Ì¼¡£
£¨1£©²ÝËᣨH2C2O4£©·Ö½âµÄ»¯Ñ§·½³Ìʽ Îª£º                        £¬
ÏÂÁÐ×°ÖÃÖУ¬¿ÉÓÃÓÚ²ÝËá·Ö½âÖÆÈ¡ÆøÌåµÄÊÇ       ¡££¨Ìî×Öĸ£©

£¨2£©Ä³Ì½¾¿Ð¡×éÀûÓòÝËá·Ö½â²úÉúµÄ»ìºÏÆøÌåºÍÌúÐâ·´Ó¦À´²â¶¨ÌúÐâÑùÆ·×é³É£¨¼Ù¶¨ÌúÐâÖÐÖ»ÓÐFe2O3¡¤nH2OºÍFeÁ½Öֳɷݣ©£¬ÊµÑé×°ÖÃÈçÏÂͼËùʾ£¬Çë»Ø´ð£º

¢Ù    ÎªµÃµ½¸ÉÔï¡¢´¿¾»µÄCOÆøÌå£¬Ï´ÆøÆ¿A¡¢BÖÐÊ¢·ÅµÄÊÔ¼Á·Ö±ðÊÇ       ¡¢      ¡£
¢Ú ÔÚµãȼ¾Æ¾«µÆÖ®Ç°Ó¦½øÐеIJÙ×÷ÊÇ£º£¨a£©     £»£¨b£©Í¨Èë»ìºÏÆøÌåÒ»¶Îʱ¼ä¡£
¢Û     ×¼È·³ÆÁ¿ÑùÆ·µÄÖÊÁ¿10.00 gÖÃÓÚÓ²Öʲ£Á§¹ÜÖУ¬³ä·Ö·´Ó¦ºóÀäÈ´¡¢³ÆÁ¿£¬Ó²Öʲ£Á§                        ¹ÜÖÐÊ£Óà¹ÌÌåÖÊÁ¿Îª8.32 g£¬DÖÐŨÁòËáÔöÖØ0.72 g£¬Ôòn =     £¨¼Ù¶¨FeºÍH2O                      ²»·¢Éú·´Ó¦£¬ÊµÑé¹ý³ÌÖÐÿ²½¾ùÍêÈ«ÎüÊÕ»ò·´Ó¦£©¡£
¢Ü     ÔÚ±¾ÊµÑéÖУ¬ÏÂÁÐÇé¿ö»áʹ²â¶¨½á¹ûnÆ«´óµÄÊÇ       £¨Ìî×Öĸ£©¡£
a£®È±ÉÙÏ´ÆøÆ¿B                     b£®È±ÉÙ×°ÖÃE
c£®·´Ó¦ºó¹ÌÌåÊÇÌúºÍÉÙÁ¿Fe2O3        d£®·´Ó¦ºó¹ÌÌåÊÇÌúºÍÉÙÁ¿Fe2O3¡¤nH2O
£¨3£©¸Ã̽¾¿Ð¡×黹ÀûÓÃKMnO4ËáÐÔÈÜÒºÓëH2C2O4ÈÜÒº·´Ó¦¹ý³ÌÖÐÈÜÒº×ÏÉ«ÏûʧµÄ·½·¨£¬Ñо¿Ó°Ïì·´Ó¦ËÙÂʵÄÒòËØ¡£
¢Ù    ÇëÍê³ÉÒÔÏÂʵÑéÉè¼Æ±í£¨±íÖв»ÒªÁô¿Õ¸ñ£©£º
£¨Ã¿´ÎʵÑéKMnO4ËáÐÔÈÜÒºµÄÓÃÁ¿¾ùΪ4mL¡¢H2C2O4ÈÜÒºµÄÓÃÁ¿¾ùΪ2mL£¬´ß»¯¼Á                   µÄÓÃÁ¿¿ÉÑ¡Ôñ0.5g¡¢0g£©
ʵÑé
񅧏
ʵÑéÄ¿µÄ
T/K
´ß»¯¼ÁÓÃÁ¿/g
C/mol¡¤l-1:]
KMnO4
H2C2O4
¢Ù
ΪÒÔÏÂʵÑé×÷²Î¿¼
298
0.5
0.01
0.1
¢Ú
̽¾¿KMnO4ËáÐÔÈÜÒºµÄŨ¶È¶Ô¸Ã·´Ó¦ËÙÂʵÄÓ°Ïì
298
0.5
0.001
0.1
¢Û
 
323
0.5
0.01
0.1
¢Ü
̽¾¿´ß»¯¼Á¶Ô·´Ó¦ËÙÂʵÄÓ°Ïì
 
 
 
0.1
¢Ú ÈôҪ׼ȷ¼ÆËã·´Ó¦ËÙÂÊ£¬¸ÃʵÑéÖл¹Ðè²â¶¨ÈÜÒº×ÏÉ«ÏûʧËùÐèÒªµÄʱ¼ä¡£ÇëÄãÉè¼Æ³ö
ͨ¹ý²â¶¨ÍÊɫʱ¼ä³¤¶ÌÀ´ÅжÏŨ¶È´óСÓë·´Ó¦ËÙÂʹØÏµµÄʵÑé·½°¸           ¡£
£¨8·Ö£©ÊµÑéÀ´Ô´ÓÚÉú»îÇÒ·þÎñÓÚÉú»î£¬Çë»Ø´ðЩÁÐÎÊÌ⣺
£¨1£©ÕýÈ·µÄʵÑé²Ù×÷ÊÇʵÑé³É¹¦µÄÖØÒªÒòËØ£¬ÏÂÁÐʵÑé²Ù×÷´íÎóµÄÊÇ£¨   £©

£¨2£©ÓÉͼµÄ×°ÖÃÖУ¬¸ÉÔïÉÕÆ¿ÄÚÊ¢ÓÐijÖÖÆøÌ壬ÉÕ±­ºÍµÎ¶¨¹ÜÄÚÊ¢·ÅijÖÖÒºÌå¡£

¼·Ñ¹µÎ¹ÜµÄ½ºÍ·£¬ÏÂÁÐÓëʵÑéÊÂʵ²»Ïà·ûµÄÊÇ£¨   £©
A£®CO2(NaHCO3ÈÜÒº)/ÎÞÉ«ÅçȪ             B£®NH3(H2Oº¬·Ó̪)/ºìÉ«ÅçȪ
C£®H2S(CuSO4ÈÜÒº)/ºÚÉ«ÅçȪ              D£®HCl(AgNO3ÈÜÒº)/°×É«ÅçȪ
£¨3£©Éè¼ÆÑ§ÉúʵÑéҪעÒⰲȫ¡¢ÎÞÎÛȾ¡¢ÏÖÏóÃ÷ÏÔ¡£¸ù¾ÝÆôÆÕ·¢ÉúÆ÷Ô­Àí£¬¿ÉÓõײ¿ÓÐС¿×µÄÊÔ¹ÜÖÆ¼òÒ׵įøÌå·¢ÉúÆ÷(¼ûÓÒͼ)¡£

Èô¹Ø±ÕK£¬²»ÄÜʹ·´Ó¦Í£Ö¹£¬¿É½«ÊԹܴÓÉÕ±­ÖÐÈ¡³ö(»áÓв¿·ÖÆøÌåÒÝÉ¢)¡£ÏÂÁÐÆøÌåµÄÖÆÈ¡ÒËʹÓøÃ×°ÖõÄÊÇ(   )
A£®ÓöþÑõ»¯ÃÌ(·ÛÄ©)ÓëË«ÑõË®ÖÆÑõÆø
B£®ÓÃпÁ£ÓëÏ¡ÁòËáÖÆÇâÆø
C£®ÓÃ̼ËáÄÆ¹ÌÌåÓëÑÎËáÖÆ¶þÑõ»¯Ì¼
D£®ÓÃ̼Ëá¸Æ(¿é×´)ÓëÏ¡ÁòËáÖÆ¶þÑõ»¯Ì¼
£¨4£©ÎªÌá´¿ÏÂÁÐÎïÖÊ£¨À¨ºÅÄÚΪÔÓÖÊ£©£¬Ñ¡ÓõÄÊÔ¼ÁºÍ·ÖÀë·½·¨¶¼ÕýÈ·µÄÊÇ£¨  £©
 
ÎïÖÊ
ÊÔ¼Á
·ÖÀë·½·¨
¢Ù
ÏõËá¼Ø£¨ÂÈ»¯ÄÆ£©
ÕôÁóË®
ÖØ½á¾§
¢Ú
¶þÑõ»¯Ì¼£¨ÂÈ»¯Ç⣩
±¥ºÍ̼ËáÄÆÈÜÒº
Ï´Æø
¢Û
ÒÒ´¼£¨Ë®£©
Éúʯ»Ò
ÕôÁó
¢Ü
±½£¨±½·Ó£©
ŨäåË®
·ÖÒº
    A£®¢Ù¢Ú         B£®¢Ù¢Û        C£®Ö»ÓТ۠      D£®¢Û¢Ü

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

¾«Ó¢¼Ò½ÌÍø