ÌâÄ¿ÄÚÈÝ

17£®¶ÌÖÜÆÚÔªËØA£¬B£¬C£¬DÔÚÖÜÆÚ±íÖеÄλÖÃÈçͼËùʾ£¬E2+ÓëDµÄ¼òµ¥ÒõÀë×ÓÓÐÏàͬµÄµç×Ó²ã½á¹¹£¬DÔªËØÔ­×ӵĴÎÍâ²ãµç×ÓÊýµÈÓÚÆäËû²ãµç×ÓÊýÖ®ºÍ£¬ÔòÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
 A B  C  
  D  
A£®CµÄ·Ç½ðÊôÐÔ×îÇ¿£¬¿ÉÓëEÔªËØÐγɹ²¼Û»¯ºÏÎï
B£®ÔªËØD¿ÉÐγÉÈýÖÖ¼Û̬µÄËᣬÇÒËá¸ùÀë×Ó¾ùÄÜ´Ù½øË®µÄµçÀë
C£®Ô­×Ó°ë¾¶ÓÉ´óµ½Ð¡µÄÅÅÁÐ˳ÐòΪ£ºE£¾D£¾B£¾A£¾C
D£®C¡¢DÔªËØÐγɵÄ×î¼òµ¥µÄÇ⻯ÎïµÄÈÈÎȶ¨ÐÔ£ºD£¼C

·ÖÎö ¶ÌÖÜÆÚÔªËØA¡¢B¡¢C¡¢DËÄÖÖÔªËØ£¬DÔªËØÔ­×ӵĴÎÍâ²ãµç×ÓÊýµÈÓÚÆäËû²ãµç×ÓÊýÖ®ºÍ£¬ÓÉλÖÿÉÖªDλÓÚµÚÈýÖÜÆÚ£¬Ôò´ÎÍâ²ãµç×ÓÊýΪ8£¬M²ãµç×ÓÊýΪ8-2=6£¬ÔòDΪS£¬BΪO£¬½áºÏλÖùØÏµAΪN£¬CΪF£¬E2+ÓëDµÄ¼òµ¥ÒõÀë×ÓÓÐÏàͬµÄµç×Ó²ã½á¹¹£¬ÔòEΪCa£¬È»ºó½áºÏÔªËØ»¯ºÏÎïÐÔÖÊÀ´½â´ð£®

½â´ð ½â£º¶ÌÖÜÆÚÔªËØA¡¢B¡¢C¡¢DËÄÖÖÔªËØ£¬DÔªËØÔ­×ӵĴÎÍâ²ãµç×ÓÊýµÈÓÚÆäËû²ãµç×ÓÊýÖ®ºÍ£¬ÓÉλÖÿÉÖªDλÓÚµÚÈýÖÜÆÚ£¬Ôò´ÎÍâ²ãµç×ÓÊýΪ8£¬M²ãµç×ÓÊýΪ8-2=6£¬ÔòDΪS£¬BΪO£¬½áºÏλÖùØÏµAΪN£¬CΪF£¬E2+ÓëDµÄ¼òµ¥ÒõÀë×ÓÓÐÏàͬµÄµç×Ó²ã½á¹¹£¬ÔòEΪCa£¬
A£®FµÄ·Ç½ðÊôÐÔ×îÇ¿£¬¿ÉÓëCaÔªËØÐγÉÀë×Ó»¯ºÏÎ¹ÊA´íÎó£»
B£®S¶ÔÓ¦µÄËáÓÐÑÇÁòËá¡¢ÁòËá¡¢ÇâÁòËᣬÁòËá¸ùÀë×Ó²»ÄÜ´Ù½øË®µÄµçÀ룬¹ÊB´íÎó£»
C£®µç×Ó²ãÔ½¶à£¬°ë¾¶Ô½´ó£¬Í¬ÖÜÆÚ´Ó×óÏòÓÒÔ­×Ó°ë¾¶¼õС£¬ÔòÔ­×Ó°ë¾¶ÓÉ´óµ½Ð¡µÄÅÅÁÐ˳ÐòΪ£ºE£¾D£¾A£¾B£¾C£¬¹ÊC´íÎó£»
D£®·Ç½ðÊôÐÔF£¾S£¬C¡¢DÔªËØÐγɵÄ×î¼òµ¥µÄÇ⻯ÎïµÄÈÈÎȶ¨ÐÔ£ºD£¼C£¬¹ÊDÕýÈ·£»
¹ÊÑ¡D£®

µãÆÀ ±¾Ì⿼²éλÖᢽṹÓëÐÔÖʹØÏµµÄÓ¦Óã¬Îª¸ßƵ¿¼µã£¬°ÑÎÕÔªËØÖÜÆÚ±í½á¹¹¡¢ÔªËØÖÜÆÚÂÉÄÚÈÝΪ½â´ðµÄ¹Ø¼ü£¬ÊÔÌâ²àÖØ¿¼²éѧÉúµÄ·ÖÎö¡¢Àí½âÄÜÁ¦¼°Áé»îÓ¦ÓÃÄÜÁ¦£¬ÌâÄ¿ÄѶȲ»´ó£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
7£®Ä³ÊµÑéС×éͬѧΪÁË̽¾¿Í­ÓëŨÁòËáµÄ·´Ó¦£¬½øÐÐÁËÈçÏÂʵÑ飬ʵÑé×°ÖÃÈçͼËùʾ£®
ʵÑé²½Ö裺
¢ÙÏÈÁ¬½ÓÈçͼËùʾµÄ×°Ö㬼ì²éºÃÆøÃÜÐÔ£¬ÔÙ¼ÓÈëÊÔ¼Á£»
¢Ú¼ÓÈÈAÊԹܣ¬´ýBÊÔ¹ÜÖÐÆ·ºìÈÜÒºÍÊÉ«ºó£¬Ï¨Ãð¾Æ¾«µÆ£»
¢Û½«CuË¿ÏòÉϳ鶯Àë¿ªÒºÃæ£®
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©AÊÔ¹ÜÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪCu+2H2SO4£¨Å¨£©$\frac{\underline{\;\;¡÷\;\;}}{\;}$CuSO4+SO2¡ü+2H2O£®
£¨2£©Äܹ»Ö¤Ã÷Í­ÓëŨÁòËá·´Ó¦Éú³ÉÆøÌåµÄʵÑéÏÖÏóÊÇBÊÔ¹ÜÖÐÆ·ºìÈÜÒºÍÊÉ«£®
£¨3£©ÔÚÊ¢ÓÐBaCl2ÈÜÒºµÄCÊÔ¹ÜÖУ¬³ýÁ˵¼¹Ü¿ÚÓÐÆøÅÝÍ⣬ÎÞÆäËûÃ÷ÏÔÏÖÏó£¬Èô½«ÆäÖеÄÈÜÒº·Ö³ÉÁ½·Ý£¬·Ö±ðµÎ¼ÓÏÂÁÐÈÜÒº£¬½«²úÉú³ÁµíµÄ»¯Ñ§Ê½ÌîÈë±íÖжÔÓ¦µÄλÖã®
µÎ¼ÓµÄÈÜÒºÂÈË®°±Ë®
³ÁµíµÄ»¯Ñ§Ê½
д³öÆäÖÐSO2±íÏÖ»¹Ô­ÐÔµÄÀë×Ó·´Ó¦·½³Ìʽ£ºSO2+Cl2+2H2O=4H++SO42-+2Cl-£¨»òBa2++SO2+Cl2+2H2O=BaSO4¡ý+4H++2Cl-£©£®
£¨4£©ÊµÑéÍê±Ïºó£¬ÏÈϨÃð¾Æ¾«µÆ£¬ÓÉÓÚµ¼¹ÜEµÄ´æÔÚ£¬ÊÔ¹ÜBÖеÄÒºÌå²»»áµ¹ÎüÈëÊÔ¹ÜAÖУ¬ÆäÔ­ÒòÊǵ±AÊÔ¹ÜÄÚÆøÌåѹǿ¼õСʱ£¬¿ÕÆø´ÓEµ¼¹Ü½øÈëAÊÔ¹ÜÖУ¬Î¬³ÖAÊÔ¹ÜÖÐѹǿƽºâ£®
£¨5£©ÊµÑéÍê±Ïºó£¬×°ÖÃÖвÐÁôµÄÆøÌåÓж¾£¬²»ÄÜ´ò¿ªµ¼¹ÜÉϵĽºÈû£®ÎªÁË·ÀÖ¹¸ÃÆøÌåÅÅÈë¿ÕÆøÖÐÎÛȾ»·¾³£¬²ð³ý×°ÖÃǰ£¬Ó¦µ±²ÉÈ¡µÄ²Ù×÷ÊÇ´ÓEµ¼¹Ü¿ÚÏòAÊÔ¹ÜÖлºÂýµØ¹ÄÈë×ãÁ¿µÄ¿ÕÆø£¬½«²ÐÁôµÄSO2ÆøÌå¸ÏÈëNaOHÈÜÒºÖУ¬Ê¹Ö®±»ÍêÈ«ÎüÊÕ£®
£¨6£©½«SO2ÆøÌåͨÈ뺬ÓÐn mol Na2SµÄÈÜÒºÖУ¬³ä·Ö·´Ó¦ºó£¬ÈÜÒºÖгöÏÖ»ÆÉ«»ë×Ç£¬ÊÔ·ÖÎö¸ÃÈÜÒº×î¶àÄÜÎüÊÕSO2ÆøÌå2.5n mol£¨²»¿¼ÂÇÈܽâµÄSO2£©£®
2£®2015Äê2ÔÂ16ÈÕÀî¿ËÇ¿×ÜÀíµ½¶«±±µ÷Ñо­¼ÃÇé¿ö£¬Öصã×ß·ÃÁ˸ÖÌú³§£¬¹ÄÀø¸ÖÌú³§Ìá¸ß¸ÖÌúÖÊÁ¿ºÍ²úÁ¿£¬Ìú¼°Æä»¯ºÏÎïÔÚÈÕ³£Éú»îÖÐÓ¦Óù㷺£®
£¨1£©ÀûÓÃFe2+¡¢Fe3+µÄ´ß»¯×÷Ó㬳£ÎÂÏ¿ɽ«SO2ת»¯ÎªSO42-£¬´Ó¶øÊµÏÖ¶ÔSO2µÄÖÎÀí£®ÒÑÖªº¬SO2µÄ·ÏÆøÍ¨È뺬Fe2+¡¢Fe3+µÄÈÜҺʱ£¬ÆäÖÐÒ»¸ö·´Ó¦µÄÀë×Ó·½³ÌʽΪ4Fe2++O2+4H+=4Fe3++2H2O£¬ÔòÁíÒ»·´Ó¦µÄÀë×Ó·½³ÌʽΪ2Fe3++SO2+2H2O=2Fe2++SO42-+4H+£»
£¨2£©Ñõ»¯ÌúÊÇÖØÒª¹¤ÒµÑÕÁÏ£¬ÏÂÃæÊÇÖÆ±¸Ñõ»¯ÌúµÄÒ»ÖÖ·½·¨£¬ÆäÁ÷³ÌÈçÏ£º

¢Ù²Ù×÷¢ñµÄÃû³ÆÊǹýÂË£»²Ù×÷¢òΪϴµÓ£¬Ï´µÓ²Ù×÷µÄ¾ßÌå·½·¨ÎªÑز£Á§°ôÍù©¶·ÖмÓÈëÊÊÁ¿ÕôÁóË®ÖÁ½þû³Áµí£¬ÈÃÕôÁóË®×ÔÈ»Á÷Ï£¬Öظ´2-3´Î£»
¢ÚÂËÒºAÖмÓÈëÉÔ¹ýÁ¿µÄNH4HCO3ÈÜÒºÉú³É³ÁµíͬʱÓÐÒ»ÖÖÆøÌå²úÉú£¬Ð´³öÆä»¯Ñ§·½³Ìʽ£ºFeSO4+2NH4 HCO3=FeCO3¡ý+£¨NH4£©2SO4+CO2¡ü+H2O£»
£¨3£©Èç¹ûìÑÉÕ²»³ä·Ö£¬²úÆ·Öн«ÓÐFeO´æÔÚ£¬³ÆÈ¡3.0gÑõ»¯Ìú²úÆ·£¬Èܽ⣬ÔÚ250mLÈÝÁ¿Æ¿Öж¨ÈÝ£»Á¿È¡25.00mL´ý²âÈÜÒºÓÚ×¶ÐÎÆ¿ÖУ¬ÓÃËữµÄ0.01000mol/L KMnO4ÈÜÒºµÎ¶¨ÖÁÖÕµã£¬ÖØ¸´µÎ¶¨2-3´Î£¬ÏûºÄKMnO4ÈÜÒºÌå»ýµÄƽ¾ùֵΪ20.00mL£¬
¢Ù¸ÃʵÑéÖеÄKMnO4ÈÜÒºÐèÒªËữ£¬ÓÃÓÚËữµÄËáÊÇc£¨Ìî×ÖĸÐòºÅ£©£®
a£®Ï¡ÏõËá    b£®Ï¡ÑÎËá    c£®Ï¡ÁòËá    d£®Å¨ÏõËá
¢Ú¼ÆËãÉÏÊö²úÆ·ÖÐFe2O3µÄÖÊÁ¿·ÖÊýΪ76%£®

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

¾«Ó¢¼Ò½ÌÍø