ÌâÄ¿ÄÚÈÝ

(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______             ______________¡£

(14·Ö)(1)H©sH  £¨2·Ö£©      (2)3NO2+H2O=2H++2NO3-+NO  £¨3·Ö£©
(3)4NH3+6NO5N2+6H2O       £¨3·Ö£©
(4)¢Ù¢Ü   £¨2·Ö£©   (5)¼î     £¨2·Ö£©   c(NH4+)> c(Cl-)>c(OH-)>c(H+)£¨2·Ö£©

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

(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)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»£©¡£

 

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

¾«Ó¢¼Ò½ÌÍø