(14·Ö)A¡¢B¡¢C¡¢DΪËÄÖÖµ¥ÖÊ£¬³£ÎÂʱ£¬A¡¢BÊÇÆøÌ壬C¡¢DÊǹÌÌå¡£E¡¢F¡¢G¡¢H¡¢IΪÎåÖÖ»¯ºÏÎF²»ÈÜË®£¬EΪÆøÌåÇÒ¼«Ò×ÈÜË®³ÉΪÎÞÉ«ÈÜÒº£¬GÈÜË®µÃ»Æ×ØÉ«ÈÜÒº¡£Õâ¾ÅÖÖÎïÖʼ䷴ӦµÄת»¯¹ØϵÈçͼËùʾ

£¨1£©Ð´³öËÄÖÖµ¥ÖʵĻ¯Ñ§Ê½
A____ _____    B__________    C__    ____    D___  _____
£¨2£©Ð´³öE£«F¡úH£«IµÄÀë×Ó·½³Ìʽ                                    
£¨3£©Ð´³öG+I¡úH+D+EµÄ»¯Ñ§·½³Ìʽ                                   
£¨4£©Ä³¹¤³§ÓÃBÖÆƯ°×·Û¡£
¢Ùд³öÖÆƯ°×·ÛµÄ»¯Ñ§·½³Ìʽ                                       ¡£
¢ÚΪ²â¶¨¸Ã¹¤³§ÖƵõÄƯ°×·ÛÖÐÓÐЧ³É·ÖµÄº¬Á¿£¬Ä³¸ÃС×é½øÐÐÁËÈçÏÂʵÑ飺³ÆȡƯ°×·Û2.0g£¬ÑÐÄ¥ºóÈܽ⣬ÅäÖóÉ250mLÈÜÒº£¬È¡³ö25.00mL¼ÓÈ뵽׶ÐÎÆ¿ÖУ¬ÔÙ¼ÓÈë¹ýÁ¿µÄKIÈÜÒººÍ¹ýÁ¿µÄÁòËᣨ´Ëʱ·¢ÉúµÄÀë×Ó·½³ÌʽΪ£º                     £©£¬¾²ÖᣴýÍêÈ«·´Ó¦ºó£¬ÓÃ0.1mol¡¤L-1µÄNa2S2O3ÈÜÒº×ö±ê×¼ÈÜÒºµÎ¶¨·´Ó¦Éú³ÉµÄµâ£¬ÒÑÖª·´Ó¦Ê½Îª£º2Na2S2O3+I2=Na2S4O6+2NaI£¬¹²ÓÃÈ¥Na2S2O3ÈÜÒº20.00mL¡£Ôò¸ÃƯ°×·ÛÖÐÓÐЧ³É·ÖµÄÖÊÁ¿·ÖÊýΪ             £¨±£Áôµ½Ð¡ÊýµãºóÁ½Î»£©¡£

£¨14·Ö£¬Ç°4ÿ¸ö1·Ö¡¢ÆäÓàÿ¿Õ2·Ö£©
£¨1£©A£®H2£¬B£®Cl2£¬C£®Fe£¬D£®S    £¨2£©FeS+2H+=Fe2++H2S¡ü   £¨3£©2FeCl3+H2S=2FeCl2+S¡ý+2HCl
£¨4£©¢Ù2Cl2+2Ca(OH)2=CaCl2+Ca(ClO)2+2H2O     ¢ÚClO£­+2I£­+2H+=I2+Cl£­+H2O  £¬ 35.75%

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011½ì¸ÊËàÊ¡À¼ÖÝÒ»ÖиßÈýµÚÈý´ÎÄ£Ä⿼ÊÔ£¨Àí×Û£©»¯Ñ§²¿·Ö ÌâÐÍ£ºÌî¿ÕÌâ

(14·ÖA¡¢B¡¢C¡¢D ¡¢E¡¢FÊdz£¼ûµÄÆøÌ壬ÆäÖÐA¡¢B¡¢C¡¢DΪµ¥ÖÊ£¬ÓйصÄת»¯¹ØϵÈçÏÂͼËùʾ(·´Ó¦Ìõ¼þ¾ùÒÑÂÔÈ¥£©¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)DµÄµç×ÓʽΪ                      ¡£
(2)·´Ó¦¢ÛµÄÀë×Ó·½³ÌʽΪ                                              ¡£
(3)YºÍEÔÚÒ»¶¨Ìõ¼þÏ¿ɷ´Ó¦Éú³ÉBºÍZ£¬¿ÉÏû³ýE¶Ô»·¾³µÄÎÛȾ£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                          ¡£
(4)³£ÎÂÏÂ0.1mol/LµÄYÈÜÒºÖÐc(H+)/c(OH-)=1¡Á10-8£¬ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ           ¡£
¢Ù¸ÃÈÜÒºµÄpH=11
¢Ú¸ÃÈÜÒºÖеÄÈÜÖʵçÀë³öµÄÑôÀë×ÓŨ¶ÈΪ0.1mol/L
¢Û½«pH=11µÄYÈÜÒº¼ÓˮϡÊÍ100±¶ºó£¬pHֵΪ9
¢Ü¸ÃÈÜÒºÖÐË®µçÀë³öµÄc(H+)Óëc(OH-)³Ë»ýΪ1¡Á10-22
¢Ý0.1mol/LµÄÑÎËáÈÜÒºV1 LÓë¸Ã0.1mol/LµÄYÈÜÒºV2 L»ìºÏ£¬Èô»ìºÏÈÜÒºpH=7£¬Ôò:V1>V2
(5)³£ÎÂÏÂpH=aµÄXÈÜÒººÍpH=bµÄYÈÜÒºµÈÌå»ý»ìºÏ£¬Èôa+b=14£¬Ôò»ìºÏºóµÄÈÜÒº³Ê________ÐÔ£¬»ìºÏÈÜÒºÖи÷Àë×ÓŨ¶È´óС¹ØϵΪ______             ______________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010-2011ѧÄê¸ÊËàÊ¡¸ßÈýµÚÈý´ÎÄ£Ä⿼ÊÔ£¨Àí×Û£©»¯Ñ§²¿·Ö ÌâÐÍ£ºÌî¿ÕÌâ

(14·ÖA¡¢B¡¢C¡¢D ¡¢E¡¢FÊdz£¼ûµÄÆøÌ壬ÆäÖÐA¡¢B¡¢C¡¢DΪµ¥ÖÊ£¬ÓйصÄת»¯¹ØϵÈçÏÂͼËùʾ(·´Ó¦Ìõ¼þ¾ùÒÑÂÔÈ¥£©¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)DµÄµç×ÓʽΪ                       ¡£

(2)·´Ó¦¢ÛµÄÀë×Ó·½³ÌʽΪ                                               ¡£

(3)YºÍEÔÚÒ»¶¨Ìõ¼þÏ¿ɷ´Ó¦Éú³ÉBºÍZ£¬¿ÉÏû³ýE¶Ô»·¾³µÄÎÛȾ£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                           ¡£

(4)³£ÎÂÏÂ0.1mol/LµÄYÈÜÒºÖÐc(H+)/c(OH-)=1¡Á10-8£¬ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ            ¡£

¢Ù¸ÃÈÜÒºµÄpH=11

¢Ú¸ÃÈÜÒºÖеÄÈÜÖʵçÀë³öµÄÑôÀë×ÓŨ¶ÈΪ0.1mol/L

¢Û½«pH=11µÄYÈÜÒº¼ÓˮϡÊÍ100±¶ºó£¬pHֵΪ9

¢Ü¸ÃÈÜÒºÖÐË®µçÀë³öµÄc(H+)Óëc(OH-)³Ë»ýΪ1¡Á10-22

¢Ý0.1mol/LµÄÑÎËáÈÜÒºV1 LÓë¸Ã0.1mol/LµÄYÈÜÒºV2 L»ìºÏ£¬Èô»ìºÏÈÜÒºpH=7£¬Ôò:V1>V2

(5)³£ÎÂÏÂpH=aµÄXÈÜÒººÍpH=bµÄYÈÜÒºµÈÌå»ý»ìºÏ£¬Èôa+b=14£¬Ôò»ìºÏºóµÄÈÜÒº³Ê________ÐÔ£¬»ìºÏÈÜÒºÖи÷Àë×ÓŨ¶È´óС¹ØϵΪ______              ______________¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010-2011ѧÄê¹ãÎ÷¹ðÁÖÖÐѧ¸ßÈý1ÔÂÔ¿¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ

(14·Ö) £¨15·Ö£©A¡¢B¡¢C¡¢DÊǶÌÖÜÆÚÔªËØÐγɵÄËÄÖÖÆøÌåµ¥ÖÊ£¬¼×¡¢ÒÒ¡¢±û¡¢¶¡ÊÇ»¯ºÏÎÆäÖл¯ºÏÎïÒÒÊÇÀë×Ó¾§Ì壬DÔªËصÄÔ­×Ó×îÍâ²ãµç×ÓÊýÊÇ´ÎÍâ²ãµÄ3±¶£¬CµÄÑõ»¯ÐÔ±ÈDÇ¿£¬ËüÃǵÄת»¯¹ØϵÈçͼËùʾ£¨ËüÃǵÄÉú³ÉÎï¾ù¸ø³ö£¬·´Ó¦Ìõ¼þÂÔÈ¥£©¡£

£¨1£©Ð´³ö·Ö×Óʽ£ºA           ¡¢B           ¡¢

C           ¡¢D           £»

£¨2£©Ð´³öÒÒµÄÒõÀë×ÓË®½âµÄÀë×Ó·½³Ìʽ£º                                          £»

£¨3£©Ð´³ö·´Ó¦¢Ù¡¢¢Ú»¯Ñ§·½³Ìʽ£º¢Ù                   ¡¢¢Ú                       £»

£¨4£©¾Ù³öʵÀý˵Ã÷µ¥ÖÊC±ÈDÑõ»¯ÐÔÇ¿£¨Óû¯Ñ§·½³Ìʽ±íʾ£©£º                      ¡£

£¨5£©A¡¢C\¡¢DÓëB·´Ó¦Éú³ÉµÄÇ⻯ÎïµÄ·ÐµãÓɸߵ½µÍµÄ˳ÐòÊÇ                      ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2009-2010ѧÄêÕý¶¨ÖÐѧ¸ß¶þÏÂѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ ÌâÐÍ£ºÌî¿ÕÌâ

(14·Ö)A¡¢B¡¢C¡¢DΪËÄÖÖµ¥ÖÊ£¬³£ÎÂʱ£¬A¡¢BÊÇÆøÌ壬C¡¢DÊǹÌÌå¡£E¡¢F¡¢G¡¢H¡¢IΪÎåÖÖ»¯ºÏÎF²»ÈÜË®£¬EΪÆøÌåÇÒ¼«Ò×ÈÜË®³ÉΪÎÞÉ«ÈÜÒº£¬GÈÜË®µÃ»Æ×ØÉ«ÈÜÒº¡£Õâ¾ÅÖÖÎïÖʼ䷴ӦµÄת»¯¹ØϵÈçͼËùʾ

£¨1£©Ð´³öËÄÖÖµ¥ÖʵĻ¯Ñ§Ê½

A____  _____    B__________    C__     ____    D___   _____

£¨2£©Ð´³öE£«F¡úH£«IµÄÀë×Ó·½³Ìʽ                                     

£¨3£©Ð´³öG+I¡úH+D+EµÄ»¯Ñ§·½³Ìʽ                                    

£¨4£©Ä³¹¤³§ÓÃBÖÆƯ°×·Û¡£

¢Ùд³öÖÆƯ°×·ÛµÄ»¯Ñ§·½³Ìʽ                                        ¡£

¢ÚΪ²â¶¨¸Ã¹¤³§ÖƵõÄƯ°×·ÛÖÐÓÐЧ³É·ÖµÄº¬Á¿£¬Ä³¸ÃС×é½øÐÐÁËÈçÏÂʵÑ飺³ÆȡƯ°×·Û2.0g£¬ÑÐÄ¥ºóÈܽ⣬ÅäÖóÉ250mLÈÜÒº£¬È¡³ö25.00mL¼ÓÈ뵽׶ÐÎÆ¿ÖУ¬ÔÙ¼ÓÈë¹ýÁ¿µÄKIÈÜÒººÍ¹ýÁ¿µÄÁòËᣨ´Ëʱ·¢ÉúµÄÀë×Ó·½³ÌʽΪ£º                      £©£¬¾²ÖᣴýÍêÈ«·´Ó¦ºó£¬ÓÃ0.1mol¡¤L-1µÄNa2S2O3ÈÜÒº×ö±ê×¼ÈÜÒºµÎ¶¨·´Ó¦Éú³ÉµÄµâ£¬ÒÑÖª·´Ó¦Ê½Îª£º2Na2S2O3+I2=Na2S4O6+2NaI£¬¹²ÓÃÈ¥Na2S2O3ÈÜÒº20.00mL¡£Ôò¸ÃƯ°×·ÛÖÐÓÐЧ³É·ÖµÄÖÊÁ¿·ÖÊýΪ              £¨±£Áôµ½Ð¡ÊýµãºóÁ½Î»£©¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸