ÌâÄ¿ÄÚÈÝ

ÒÑÖªa¡¢I¡¢eΪÈýÖÖÓɶÌÖÜÆÚÔªËØ¹¹³ÉµÄÁ£×Ó£¬ËüÃǶ¼ÓÐ10¸öµç×Ó£¬Æä½á¹¹ÌصãÈçÏ£º

Á£×Ó´úÂë

a

I

e

Ô­×ÓºËÊý

µ¥ºË

ËĺË

Ë«ºË

Á£×ӵĵçºÉÊý

Ò»¸öµ¥Î»ÕýµçºÉ

0

Ò»¸öµ¥Î»¸ºµçºÉ

ÎïÖÊAÓÉa¡¢e¹¹³É£¬B¡¢C¡¢D¡¢K¶¼Êǵ¥ÖÊ£¬·´Ó¦¢Ù¨D¢Ý¶¼ÊÇÓÃÓÚ¹¤ÒµÉú²úµÄ·´Ó¦£¬¸÷ÓйØÎïÖÊÖ®¼äµÄÏ໥·´Ó¦×ª»¯¹ØÏµÈçÏÂͼËùʾ£º

ÇëÌîдÏÂÁпհףº

£¨1£©Ð´³öÏÂÁÐÎïÖʵĻ¯Ñ§Ê½£ºB                £»J£º                ¡£  

£¨2£©Ð´³öÏÂÁз´Ó¦µÄÀë×Ó·½³Ìʽ£º

¢ÙH+E£¨ÈÜÒº£©¡úM                                       ¡£

¢ÚIÈÜÓÚG                                               ¡£

£¨3£©ÔÚͨ³£×´¿öÏ£¬Èô1gCÆøÌåÔÚBÆøÌåÖÐÍêȫȼÉÕÉú³ÉHÆøÌåʱ·Å³ö92.3kJÈÈÁ¿£¬Ôò2molHÆøÌåÍêÈ«·Ö½âÉú³ÉCÆøÌåºÍBÆøÌåµÄÈÈ»¯Ñ§·½³ÌʽΪ:

                                                                          ¡£

£¨1£©B. C12   J.  NO

£¨2£©¢ÙH++C1O£­=HC1O

¢ÚNH3+H2O NH3?H2ONH4++OH-

£¨3£©2HC1£¨g£©=H2(g)+C12(g);¡÷H=+184.6kJ?mol-1

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø