ÔÚÏÂÁÐÓÐFeCl3ÈÜÒº²Î¼ÓµÄ·´Ó¦ÖУ¬ÓëFe3+Ë®½âÓйصķ´Ó¦ÊÇ
¢ÙFeCl3ÈÜÒºÓëCuµÄ·´Ó¦ ¢Ú½«FeCl3ÈÜÒº¼ÓÈÈÕô¸É£¬²¢×ÆÉÕ×îÖյõ½Fe2O3 ¢ÛFeCl3ÈÜÒºÓëKIµÄ·´Ó¦ ¢Ü±¥ºÍFeCl3ÈÜÒºµÎÈë·ÐË®ÖÐÖƱ¸Fe(OH)3½ºÌå ¢ÝFeCl3ÈÜÒºÓëH2SµÄ·´Ó¦¢ÞFeCl3ÈÜÒºÓëNaHCO3ÈÜÒºµÄ·´Ó¦ ¢ßÅäÖÆFeCl3ÈÜÒºÐè¼ÓÈëÒ»¶¨Á¿µÄÑÎËá

A£®¢Ù¢Ü¢ÞB£®¢Ú¢Û¢Ý¢ÞC£®¢Ú¢Ü¢Þ¢ßD£®¢Ù¢Ú¢Û¢Ü¢Ý¢Þ¢ß

C

½âÎöÕýÈ·´ð°¸£ºC
Ë®½â·½³ÌΪ;Fe3£«£«3H2OFe(OH)3£«3H£«
¢ÙÓëFe3+Ë®½âÎ޹أ¬FeCl3×÷Ñõ»¯¼Á ¢ÚÓëFe3+Ë®½âÓйأ¬½«FeCl3ÈÜÒº¼ÓÈÈ£¬HCl»Ó·¢£¬Fe(OH)3×ÆÉÕºó·Ö½â£¬×îÖյõ½Fe2O3 ¢ÛÓëFe3+Ë®½âÎ޹أ¬FeCl3ÏÔÑõ»¯ÐÔ ¢ÜÓëFe3+Ë®½âÓйأ¬±¥ºÍFeCl3ÈÜÒºµÎÈë·ÐË®ÖÐË®½â£¬ÖƱ¸Fe(OH)3½ºÌå ¢ÝÓëFe3+Ë®½âÎ޹أ¬FeCl3ÈÜÒº×÷Ñõ»¯¼Á¢ÞÓëFe3+Ë®½âÓйأ¬FeCl3Ë®½â³ÊËáÐÔ£¬NaHCO3Ë®½â³Ê¼îÐÔ£¬Ï໥´Ù½ø ¢ßÓëFe3+Ë®½âÓйأ¬ÅäÖÆFeCl3ÈÜÒºÐè¼ÓÈëÒ»¶¨Á¿µÄÑÎËᣬÒÖÖÆË®½â£¬·ÀÖ¹²úÉú³Áµí¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2012?Ö麣һģ£©Ã¾É°£¨MgO£©ÊÇÒ»ÖÖ¸ßÎÂÄÍ»ð²ÄÁÏ£®ÏÂͼÊÇ°±·¨ÖÆÈ¡¸ß´¿Ã¾É°µÄ²¿·Ö¹¤ÒÕÁ÷³Ì£º

Çë»Ø´ð£º
£¨1£©MgCl2?6H2OÈÜÓÚË®£¬ÈÜÒº³Ê
ËáÐÔ
ËáÐÔ
£¨Ìî¡°ËáÐÔ¡±£®¡°ÖÐÐÔ¡±£®»ò¡°¼îÐÔ¡±£©£®
£¨2£©Ð´³ö°±·Ö×ӵĵç×Óʽ
£®
£¨3£©Õô°±Êǽ«Ê¯»ÒÈé¼ÓÈëÂÈ»¯ï§ÈÜÒºÖУ¬²¢¼ÓÈÈ£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
Ca£¨OH£©2+2NH4Cl
  ¡÷  
.
 
2NH3¡ü+CaCl2+2H2O
Ca£¨OH£©2+2NH4Cl
  ¡÷  
.
 
2NH3¡ü+CaCl2+2H2O
£®
£¨4£©ÇáÉÕºóµÄ²úÎïΪ
MgO
MgO
£®
£¨5£©¹¤Òµ»ñÈ¡µÄMgCl2?6H2O³£»ìÓÐFeCl3£¬³ýÈ¥MgCl2ÈÜÒºÖеÄFe3+£¬¿ÉÑ¡ÓÃÏÂÁÐÊÔ¼Á
AC
AC

A£® MgO   ¡¡¡¡B£®NaOH ¡¡¡¡¡¡ C£®Mg£¨OH£©2     D£®KSCN
£¨6£©°±»¯·´Ó¦¹ý³ÌÖз´Ó¦µÄÀë×Ó·½³Ìʽ
Mg2++2NH3+2H2O=Mg£¨OH£©2¡ý+2 NH4+
Mg2++2NH3+2H2O=Mg£¨OH£©2¡ý+2 NH4+
£®
£¨7£©¹ýÂËÏ´µÓÖ÷ÒªÊdzýÈ¥¸½×ÅÔÚ¹ÌÌå±íÃæµÄÂÈÀë×Ó£¬¼ìÑé³ÁµíÒÑÏ´µÓ¸É¾»µÄ·½·¨ÊÇ
Ïò×îºóÒ»´ÎÏ´µÓºóµÄÈÜÒºÖеμÓÏõËáËữµÄAgNO3ÈÜÒº£¬ÈôÎÞ°×É«³ÁµíÉú³É£¬Ö¤Ã÷ÒѾ­Ï´µÓ¸É¾»
Ïò×îºóÒ»´ÎÏ´µÓºóµÄÈÜÒºÖеμÓÏõËáËữµÄAgNO3ÈÜÒº£¬ÈôÎÞ°×É«³ÁµíÉú³É£¬Ö¤Ã÷ÒѾ­Ï´µÓ¸É¾»
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨1£©£¨4·Ö£©ÏÂÁÐÓйØʵÑé²Ù×÷»ò½á¹ûµÄ˵·¨ÖÐÕýÈ·µÄÊÇ                 

A£®½«pHÊÔÖ½ÓÃÕôÁóˮʪÈóºó£¬²âÁ¿Ä³ÈÜÒºµÄpH

B£®°ÑͭƬºÍÌúƬ½ô¿¿ÔÚÒ»Æð½þÈëÏ¡ÁòËáÖУ¬Í­Æ¬±íÃæ³öÏÖÆøÅÝ

C£®µÎ¶¨¹ÜÏ´¾»ºó¾­ÕôÁóË®ÈóÏ´ºó£¬¼´¿É×¢Èë±ê×¼Òº½øÐÐÖк͵ζ¨ÊµÑé

D£®µç½âÑÎËáÈÜÒººó£¨ÑÎËá¹ýÁ¿£©£¬ÔÙͨÈëÒ»¶¨Á¿HClÆøÌåʱ£¬µç½âÖÊÈÜÒº¿É»Ö¸´µ½ºÍÔ­À´Ò»Ñù

E£®Óù㷺pHÊÔÖ½²âÁ¿Na2SÈÜÒºµÄpHʱ£¬µÃpH=10.5

F£®ÊµÑéÊÒÔÚÅäÖÆFeCl3ÈÜҺʱ£¬ÏȽ«FeCl3ÈÜÓÚÒ»¶¨Á¿µÄŨÑÎËáÖУ¬ÔÙ¼ÓÕôÁóˮϡÊÍÖÁËùÐèŨ¶È

£¨2£©£¨6·Ö£©Ä³Í¬Ñ§ÔÚ°ïÖúÀÏʦÕûÀíʵÑéÊҵĻ¯Ñ§ÊÔ¼Áʱ£¬·¢ÏÖһʢÓÐÎÞÉ«ÈÜÒºµÄÊÔ¼ÁÆ¿£¬±êÇ©ÆÆËð(ÈçÓÒͼ)£¬ÇëÄã¸ù¾ÝÒÑÕÆÎÕµÄ֪ʶ£¬¶Ô¸ÃÊÔ¼Á¿ÉÄÜÊÇʲôÎïÖʵÄÈÜÒº×÷³ö²ÂÏ룬²¢Éè¼ÆʵÑé¼ÓÒÔÑéÖ¤¡£

¢Ù²ÂÏ룺ÕâÖÖÊÔ¼Á¿ÉÄÜÊÇ_____________________£»

¢Ú¼òÊöÑéÖ¤µÄʵÑé·½°¸£º                                             ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010Äê½­ËÕÊ¡¹àÄÏÏصڶþ¸ß¼¶ÖÐѧ¸ß¶þÏÂѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔÌâ ÌâÐÍ£ºÊµÑéÌâ

£¨1£©£¨4·Ö£©ÏÂÁÐÓйØʵÑé²Ù×÷»ò½á¹ûµÄ˵·¨ÖÐÕýÈ·µÄÊÇ                 

A£®½«pHÊÔÖ½ÓÃÕôÁóˮʪÈóºó£¬²âÁ¿Ä³ÈÜÒºµÄpH
B£®°ÑͭƬºÍÌúƬ½ô¿¿ÔÚÒ»Æð½þÈëÏ¡ÁòËáÖУ¬Í­Æ¬±íÃæ³öÏÖÆøÅÝ
C£®µÎ¶¨¹ÜÏ´¾»ºó¾­ÕôÁóË®ÈóÏ´ºó£¬¼´¿É×¢Èë±ê×¼Òº½øÐÐÖк͵ζ¨ÊµÑé
D£®µç½âÑÎËáÈÜÒººó£¨ÑÎËá¹ýÁ¿£©£¬ÔÙͨÈëÒ»¶¨Á¿HClÆøÌåʱ£¬µç½âÖÊÈÜÒº¿É»Ö¸´µ½ºÍÔ­À´Ò»Ñù
E£®Óù㷺pHÊÔÖ½²âÁ¿Na2SÈÜÒºµÄpHʱ£¬µÃpH=10.5
F£®ÊµÑéÊÒÔÚÅäÖÆFeCl3ÈÜҺʱ£¬ÏȽ«FeCl3ÈÜÓÚÒ»¶¨Á¿µÄŨÑÎËáÖУ¬ÔÙ¼ÓÕôÁóˮϡÊÍÖÁËùÐèŨ¶È
£¨2£©£¨6·Ö£©Ä³Í¬Ñ§ÔÚ°ïÖúÀÏʦÕûÀíʵÑéÊҵĻ¯Ñ§ÊÔ¼Áʱ£¬·¢ÏÖһʢÓÐÎÞÉ«ÈÜÒºµÄÊÔ¼ÁÆ¿£¬±êÇ©ÆÆËð(ÈçÓÒͼ)£¬ÇëÄã¸ù¾ÝÒÑÕÆÎÕµÄ֪ʶ£¬¶Ô¸ÃÊÔ¼Á¿ÉÄÜÊÇʲôÎïÖʵÄÈÜÒº×÷³ö²ÂÏ룬²¢Éè¼ÆʵÑé¼ÓÒÔÑéÖ¤¡£

¢Ù²ÂÏ룺ÕâÖÖÊÔ¼Á¿ÉÄÜÊÇ_____________________£»
¢Ú¼òÊöÑéÖ¤µÄʵÑé·½°¸£º                                             ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010Ä꺣ÄÏÊ¡¸ß¶þÏÂѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§£¨Àí£©ÊÔÌâ ÌâÐÍ£ºÊµÑéÌâ

£¨6·Ö£©ÏÂÁÐÓйØʵÑé²Ù×÷»ò½á¹ûµÄ˵·¨ÖÐÕýÈ·µÄÊÇ                  £¨ÔÚ¸ø³öµÄÑ¡ÏîÖУ¬ÖÁÉÙÓÐÁ½ÏîÊÇ·ûºÏÌâÄ¿ÒªÇóµÄ£©

A£®½«pHÊÔÖ½ÓÃÕôÁóˮʪÈóºó£¬²âÁ¿Ä³ÈÜÒºµÄpH

B£®°ÑͭƬºÍÌúƬ½ô¿¿ÔÚÒ»Æð½þÈëÏ¡ÁòËáÖУ¬Í­Æ¬±íÃæ³öÏÖÆøÅÝ

C£®µÎ¶¨¹ÜÏ´¾»ºó¾­ÕôÁóË®ÈóÏ´ºó£¬¼´¿É×¢Èë±ê×¼Òº½øÐÐÖк͵ζ¨ÊµÑé

D£®µç½âÑÎËáÈÜÒººó£¨ÑÎËá¹ýÁ¿£©£¬ÔÙͨÈëÒ»¶¨Á¿HClÆøÌåʱ£¬µç½âÖÊÈÜÒº¿É»Ö¸´µ½ºÍÔ­À´Ò»Ñù

E£®Óù㷺pHÊÔÖ½²âÁ¿Na2SÈÜÒºµÄpHʱ£¬µÃpH=10.5

F£®ÊµÑéÊÒÔÚÅäÖÆFeCl3ÈÜҺʱ£¬ÏȽ«FeCl3ÈÜÓÚÒ»¶¨Á¿µÄŨÑÎËáÖУ¬ÔÙ¼ÓÕôÁóˮϡÊÍÖÁËùÐèŨ¶È

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010Äê½­ËÕÊ¡¸ß¶þÏÂѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔÌâ ÌâÐÍ£ºÊµÑéÌâ

£¨1£©£¨4·Ö£©ÏÂÁÐÓйØʵÑé²Ù×÷»ò½á¹ûµÄ˵·¨ÖÐÕýÈ·µÄÊÇ                 

A£®½«pHÊÔÖ½ÓÃÕôÁóˮʪÈóºó£¬²âÁ¿Ä³ÈÜÒºµÄpH

B£®°ÑͭƬºÍÌúƬ½ô¿¿ÔÚÒ»Æð½þÈëÏ¡ÁòËáÖУ¬Í­Æ¬±íÃæ³öÏÖÆøÅÝ

C£®µÎ¶¨¹ÜÏ´¾»ºó¾­ÕôÁóË®ÈóÏ´ºó£¬¼´¿É×¢Èë±ê×¼Òº½øÐÐÖк͵ζ¨ÊµÑé

D£®µç½âÑÎËáÈÜÒººó£¨ÑÎËá¹ýÁ¿£©£¬ÔÙͨÈëÒ»¶¨Á¿HClÆøÌåʱ£¬µç½âÖÊÈÜÒº¿É»Ö¸´µ½ºÍÔ­À´Ò»Ñù

E£®Óù㷺pHÊÔÖ½²âÁ¿Na2SÈÜÒºµÄpHʱ£¬µÃpH=10.5

F£®ÊµÑéÊÒÔÚÅäÖÆFeCl3ÈÜҺʱ£¬ÏȽ«FeCl3ÈÜÓÚÒ»¶¨Á¿µÄŨÑÎËáÖУ¬ÔÙ¼ÓÕôÁóˮϡÊÍÖÁËùÐèŨ¶È

£¨2£©£¨6·Ö£©Ä³Í¬Ñ§ÔÚ°ïÖúÀÏʦÕûÀíʵÑéÊҵĻ¯Ñ§ÊÔ¼Áʱ£¬·¢ÏÖһʢÓÐÎÞÉ«ÈÜÒºµÄÊÔ¼ÁÆ¿£¬±êÇ©ÆÆËð(ÈçÓÒͼ)£¬ÇëÄã¸ù¾ÝÒÑÕÆÎÕµÄ֪ʶ£¬¶Ô¸ÃÊÔ¼Á¿ÉÄÜÊÇʲôÎïÖʵÄÈÜÒº×÷³ö²ÂÏ룬²¢Éè¼ÆʵÑé¼ÓÒÔÑéÖ¤¡£

¢Ù²ÂÏ룺ÕâÖÖÊÔ¼Á¿ÉÄÜÊÇ_____________________£»

¢Ú¼òÊöÑéÖ¤µÄʵÑé·½°¸£º                                              ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸