ÌâÄ¿ÄÚÈÝ

X¡¢YºÍZ¾ùΪ¶ÌÖÜÆÚÔªËØ£¬Ô­×ÓÐòÊýÒÀ´ÎÔö´ó£¬XµÄµ¥ÖÊΪÃܶÈ×îСµÄÆøÌ壬YÔ­×Ó×îÍâ²ãµç×ÓÊýÊÇÆäÖÜÆÚÊýµÄÈý±¶£¬ZÓëXÔ­×Ó×î´¦²ãµç×ÓÊýÏàͬ¡£»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©X¡¢YºÍZµÄÔªËØ·ûºÅ·Ö±ðΪ        ¡¢            ¡¢       ¡£
£¨2£©ÓÉÉÏÊöÔªËØ×é³ÉµÄ»¯ºÏÎïÖУ¬¼Èº¬Óй²¼Û¼üÓÖº¬ÓÐÀë×Ó¼üµÄÓР         ¡¢       ¡£
£¨3£©XºÍY×é³ÉµÄ»¯ºÏÎïÖУ¬¼Èº¬Óм«ÐÔ¹²¼Û¼üÓÖº¬ÓзǼ«ÐÔ¹²¼Û¼üµÄÊÇ       ¡£
´Ë»¯ºÏÎïÔÚËáÐÔÌõ¼þÏÂÓë¸ßÃÌËá¼Ø·´Ó¦µÄÀë×Ó·½³ÌʽΪ                        £»´Ë»¯ºÏÎﻹ¿É½«¼îÐÔ¹¤Òµ·ÏË®ÖеÄCN-Ñõ»¯ÎªÌ¼ËáÑκͰ±£¬ÏàÓ¦µÄÀë×Ó·½³ÌʽΪ                          ¡£


£¨1£©H  O   Na 
£¨2£©NaOH   Na2O2   
£¨3£©H2O2    5 H2O2+2MnO4-+6H+=2Mn2++8 H2O+5O2¡ü      H2O2+CN-+OH-=CO32-+NH3

½âÎö£¨1£©¸ù¾ÝÌâÒ⣬ÓÉÃܶÈ×îСµÄÆøÌåÍÆ³ö£ºXΪH£¬YΪO£¬ZÓëHͬ×åÇÒÔ­×ÓÐòÊý´óÓÚOÍÆ³öΪNa¡£
£¨2£©¸ù¾ÝËùѧ֪ʶ£¬ºÜÈÝÒ׵óö³£¼ûÎïÖÊNaOH¡¢Na2O2ÖУ¬¼Èº¬Óй²¼Û¼üÓÖº¬ÓÐÀë×Ó¼ü£»
£¨3£©H2O2¼Èº¬Óм«ÐÔ¹²¼Û¼üÓÖº¬ÓзǼ«ÐÔ¹²¼Û¼ü£»ËáÐÔÌõ¼þϸßÃÌËá¼Ø½«Ë«ÑõË®Ñõ»¯³ÉÑõÆø£»ÒÑÖª·´Ó¦ÎïΪH2O2¡¢CN-¡¢OH-£¬²úÎïΪCO32-¡¢NH3£¬Å䯽¼´¿É¡£
¡¾¿¼µã¶¨Î»¡¿±¾Ì⿼²éÔªËØÖÜÆÚ±íºÍÔªËØÖÜÆÚÂÉ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

°´ÒªÇóÌî¿Õ

£¨Ò»£©¡¢X¡¢YºÍZ¾ùΪ¶ÌÖÜÆÚÔªËØ£¬Ô­×ÓÐòÊýÒÀ´ÎÔö´ó£¬XµÄµ¥ÖÊΪÃܶÈ×îСµÄÆøÌ壬YµÄÒ»ÖÖµ¥ÖʾßÓÐÌØÊâ³ô棬ZÓëXÔ­×Ó×î´¦²ãµç×ÓÊýÏàͬ¡£»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÓÉÉÏÊöÔªËØ×é³ÉµÄ»¯ºÏÎïÖУ¬¼Èº¬Óм«ÐÔ¹²¼Û¼üÓÖº¬ÓÐÀë×Ó¼üµÄ»¯ºÏÎïµÄµç×Óʽ                         £»

£¨2£©XºÍY×é³ÉµÄ»¯ºÏÎïÖУ¬ÓÐÒ»ÖּȺ¬Óм«ÐÔ¹²¼Û¼üÓÖº¬ÓзǼ«ÐÔ¹²¼Û¼ü¡£´Ë»¯ºÏÎï¿É½«¼îÐÔ¹¤Òµ·ÏË®ÖеÄCN¡¥Ñõ»¯ÎªÌ¼ËáÑκͰ±£¬ÏàÓ¦µÄÀë×Ó·½³ÌʽΪ           

£¨¶þ£©¡¢ÔÚÒ»¶¨Ìõ¼þÏ£¬RO3n¡¥ ºÍI¡¥ ·¢Éú·´Ó¦£¬Àë×Ó·½³ÌʽΪ£º RO3n¡¥+6I¡¥+6H+==R¡¥+3I2+3H2

RO3n¡¥-ÖÐRÔªËØµÄ»¯ºÏ¼ÛΪ        £¬RÔªËØµÄÔ­×Ó×îÍâ²ãµç×ÓÓР       ¸ö¡£

£¨Èý£©¡¢Na2SxÔÚ¼îÐÔÈÜÒºÖпɱ»NaClOÑõ»¯ÎªNa2SO4£¬¶øNaClO±»»¹Ô­ÎªNaCl£¬Èô·´Ó¦ÖÐNa2SxÓëNaClOµÄÎïÖʵÄÁ¿Ö®±ÈΪ1©U16£¬ÔòxÖµÊÇ           

£¨ËÄ£©¡¢ÒÑÖªM2On2¡¥¿ÉÓëR2¡¥×÷Óã¬R2¡¥±»Ñõ»¯ÎªRµÄµ¥ÖÊ£¬M2On2¡¥µÄ»¹Ô­²úÎïÖУ¬MΪ+3¼Û£¬ÓÖÖªc(M2On2¡¥)=0£®3mol/LµÄÈÜÒº100mL¿ÉÓëc(R2¡¥)=0£®6mol/LµÄÈÜÒº150mLÇ¡ºÃÍêÈ«·´Ó¦£¬ÔònֵΪ             ¡£

 

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

¾«Ó¢¼Ò½ÌÍø