ÒÒϩͨÈëäåË®¼°äåÒÒÍéµÄË®½â¶¼ÊÇÖÐѧÓлú»¯Ñ§µÄÖØÒª·´Ó¦¡£

£¨1£©Çëд³öʵÑéÊÒÖÆÒÒÏ©µÄ»¯Ñ§·½³Ìʽ                                           ¡£

£¨2£©ÔÚʵÑéÊÒÀѧϰС×éµÄͬѧ°ÑÒÒÏ©ÆøÌå³ÖÐøÍ¨Èë100ºÁÉý±¥ºÍµÄäåË®£¨Ô¼0.2mol/L£©Ö±ÖÁäåË®ÍêÈ«ÍÊÉ«£¬×Ðϸ¹Û²ìµÃµ½µÄÎÞÉ«ÈÜÒº£¬·¢ÏÖÁËÉÙÁ¿µÄÓÍ×´ÒºµÎ¡£

¶ÔÓÚäåË®ÍÊÉ«µÄ½âÊÍ£¬Í¬Ñ§ÃÇÓÐÁ½Öֹ۵㣬һÖÖÈÏΪ·¢ÉúÁËÈ¡´ú·´Ó¦£¬ÁíÒ»ÖÖÈÏΪ·¢ÉúÁ˼ӳɷ´Ó¦¡£ÇëÄãһͬ²ÎÓëʵÑé²¢»Ø´ðÓйØÎÊÌâ¡£

¢ÙÓÍ×´ÒºµÎ´¦ÓÚ       ²ã£¨Ìî¡°ÉÏ¡±»ò¡°Ï¡±£©¡£

¢ÚÓÐÈ˵óö½áÂÛ£º²â¶¨ÎÞÉ«ÈÜÒºµÄpH¼´¿ÉÅжϸ÷´Ó¦ÊÇÈ¡´ú·´Ó¦»¹ÊǼӳɷ´Ó¦¡£ÆäÀíÓÉÊÇ£º

                                                                            

                                                        ¡£

¢Û²âµÃÎÞÉ«ÈÜÒºµÄpH=7£¬ÔòÒÒÏ©ÓëäåË®·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                       

                                               ¡£

£¨3£©Éè¼ÆÊµÑé¼ìÑéäåÒÒÍéË®½â²úÉúµÄäåÀë×Ó¡£

¢ÙʵÑéÊÔ¼Á£ºpHÊÔÖ½¡¢äåÒÒÍé¡¢NaOHÈÜÒº¡¢AgNO3Ë®ÈÜÒº¡¢                  £»

  ʵÑéÒÇÆ÷£º²£Á§°ô¡¢±íÃæÃó¡¢½ºÍ·µÎ¹Ü¡¢ÊԹܼ°ÊԹܼС¢                 ¡£

¢ÚäåÒÒÍéË®½âµÄ»¯Ñ§·½³ÌʽΪ                                                 £»

ÅжÏäåÒÒÍéÒѾ­Íêȫˮ½âµÄÏÖÏóÊÇ                                           ¡£

¢ÛÈ¡ÉÙÁ¿äåÒÒÍéÍêȫˮ½âºóËùµÃµÄÈÜÒºÓÚÊÔ¹ÜÖУ¬¾­¹ýһϵÁÐʵÑ飬×îºó²úÉúµ­»ÆÉ«³Áµí£¬ËµÃ÷äåÒÒÍéË®½â²úÉúÁËäåÀë×Ó¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


Ó¡Ë¢µç·°åÊÇÓÉËÜÁϺÍÍ­²­¸´ºÏ¶ø³É£¬£¬¿ÌÖÆÓ¡Ë¢µç·ʱҪÓÃÈÜÒº×÷Ϊ¡°¸¯Ê´Òº¡±

ÈܽâÍ­¡£

£¨1£©Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º________________________________________;

£¨2£©Ð´³öFeCl3Óë×ãÁ¿µÄZn·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º__________________________________;

£¨3£©¸¯Ê´ÒºÓþûáʧЧ£¬µ«¿ÉÒÔ»ØÊÕÀûÓãº

¢ÙÈôÒªµÃµ½µ¥ÖÊÍ­£¬ÏÂÁÐÊÔ¼ÁÄÜʵÏÖµÄÊÇ£¨ÌîÐòºÅ£©

A£®ÂÈÆø      B£®ÂÁ      C£®ÏõËá      D£®Ï¡ÁòËá      E£®Ìú

¢ÚÈôÒª½«×ª»¯Îª£¬ÏÂÁÐÊÔ¼ÁÄÜʵÏÖµÄÊÇ£¨ÌîÐòºÅ£©

A£®ÂÈÆø      B¡¢ÂÁ      C¡¢ÏõËá      D£®Ï¡ÁòËá      E£®Ìú

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁÐÏà¹Ø·´Ó¦µÄÀë×Ó·½³ÌʽÊéдÕýÈ·µÄÊÇ

A£®ÇâÑõ»¯ÌúÈÜÓÚÇâµâË᣺Fe(OH)3+3H+ = Fe3++3H2O

B£®ÁòËáÍ­ÈÜÒºÏÔËáÐÔ£ºCu2+ + 2H2O == Cu(OH)2¡ý+ 2H+

C£®Ïò̼ËáÇâï§ÈÜÒºÖмӹýÁ¿Ê¯»ÒË®²¢¼ÓÈÈ£ºNH4++OH¡ªNH3¡ü+H2O

D£®ÓÃËữµÄ¸ßÃÌËá¼ØÈÜÒºÑõ»¯Ë«ÑõË®£º2MnO4¡ª+6H++5H2O2 = 2Mn2++5O2¡ü+8H2O

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÈôNA±íʾ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ£¬ÏÂÁÐÐðÊöÖÐÕýÈ·µÄÊÇ

A£®±ê×¼×´¿öÏ£¬22.4LÒÑÍéÖй²¼Û¼üÊýĿΪ19NA

B£®26gÒÒȲÓë±½µÄ»ìºÏÎïÖÐËùº¬Á¿Ì¼Ô­×ÓÊýΪ0.2NA

C£®ÔÚ±ê×¼×´¿öÏ£¬2.24LÒÒȲÖеĦҼüÊýΪ0.3NA

D£®1 mol±½ÓëÇâÆø¼Ó³É×î¶àÐèÒªÇâ·Ö×ÓÊýΪ6NA

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÔÚÒ»¸öÃܱÕÈÝÆ÷ÖзÅÈëµÈÎïÖʵÄÁ¿µÄNa2O2¹ÌÌåºÍNaHCO3¹ÌÌ壬³ä·Ö¼ÓÈȺóÀäÈ´£¬Åųö²úÉúµÄÆøÌå¡£ÏÂÁжÔÈÝÆ÷ÖеÄÊ£Óà¹ÌÌåµÄ˵·¨ÕýÈ·µÄÊÇ

A£®Ê£Óà¹ÌÌåÖ»ÓÐNa2CO3

B£®È¡ÉÙÁ¿Ê£Óà¹ÌÌåÅä³ÉÈÜÒº£¬ÖðµÎ¼ÓÈëÏ¡ÑÎËᣬÁ¢¼´²úÉúÆøÅÝ

C£®Ê£Óà¹ÌÌå¿ÉÄÜÊÇNaHCO3ºÍNaOHµÄ»ìºÏÎï

D£®Ê£Óà¹ÌÌåÊǵÈÎïÖʵÄÁ¿µÄNa2CO3ºÍNaOHµÄ»ìºÏÎï

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


Äܹ»Ö¤Ã÷¼×Íé¹¹ÐÍÊÇËÄÃæÌåµÄÊÂʵÊÇ                                   £¨    £©

    A£®¼×ÍéµÄËĸö¼ü¼üÄÜÏàͬ             B£®¼×ÍéµÄËĸö¼ü¼ü³¤ÏàµÈ

    C£®¼×ÍéµÄËùÓÐC-H¼ü¼ü½ÇÏàµÈ         D£®¶þÂȼ×ÍéûÓÐͬ·ÖÒì¹¹Ìå

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ͬ·ÖÒì¹¹ÌåÏÖÏóÔÚÓлú»¯Ñ§ÖÐÊǷdz£ÆÕ±éµÄ£¬ÏÂÁÐÓлúÎﻥΪͬ·ÖÒì¹¹ÌåµÄÊÇ£¨    £©¢ÙCH2£½CHCH3   ¢Ú  ¢ÛCH3CH2CH3  ¢ÜHCCCH3  ¢Ý ¢ÞCH3CH£½CHCH3              

    A£®¢ÙºÍ¢Ú         B£®¢ÙºÍ¢Û         C£®¢ÙºÍ¢Ü         D£®¢ÝºÍ¢Þ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁл¯Ñ§ÓÃÓïÕýÈ·µÄÊÇ

A£®ÒÒÏ©·Ö×ӵĵç×Óʽ

B£®ÁòÔ­×ӵĽṹʾÒâͼ

C£®µí·ÛµÄ»¯Ñ§Ê½  (C6H10O5)n

D£®ÁòËáÌúµÄµçÀë·½³Ìʽ Fe2(SO4)3 = 2Fe2+ + 3SO42¡¥

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÉèNA±íʾ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ£¬ÏÂÁÐÐðÊöÖÐÕýÈ·µÄÊÇ

A£®³£ÎÂÏ£¬1 mol Cl2Óë×ãÁ¿NaOHÈÜÒºÍêÈ«·´Ó¦£¬×ªÒƵĵç×ÓÊý2NA

B£®³£Î³£Ñ¹Ï£¬11.2 L CO2Öк¬ÓеķÖ×ÓÊýÊÇ0.5NA

C£®±ê×¼×´¿öÏ£¬22.4 LË®Ëùº¬Ô­×ÓÊýΪ3NA

D£®³£Î³£Ñ¹Ï£¬48 g O3ºÍO2µÄ»ìºÏÆøÌåÖÐÑõÔ­×ÓÊýΪ3NA

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸