£¨2013?±±¾©£©Ë®ÊÇ×îÆÕͨ¡¢×î³£¼ûµÄÎïÖÊÖ®Ò»£®
£¨1£©¡°Ë®¡±ÓкܶàÖÖ£®ÏÂÁС°Ë®¡±ÊôÓÚ´¿¾»ÎïµÄÊÇ
D
D
£¨Ìî×ÖĸÐòºÅ£©£®
A¡¢ºÓË®     B¡¢×ÔÀ´Ë®     C¡¢¿óȪˮ     D¡¢ÕôÁóË®£®
£¨2£©µç½âË®¿ÉÖ¤Ã÷Ë®ÓÉÇâ¡¢ÑõÁ½ÖÖÔªËØ×é³É£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
2H2O
 Í¨µç 
.
 
2H2¡ü+O2¡ü
2H2O
 Í¨µç 
.
 
2H2¡ü+O2¡ü
£®
£¨3£©Ë®ÊÇÖØÒªµÄÈܼÁºÍ»¯¹¤Ô­ÁÏ£®ÂȼҵÒÔ±¥ºÍʳÑÎˮΪԭÁÏ»ñµÃÉÕ¼îµÈ»¯¹¤²úÆ·£¬·´Ó¦Ô­ÀíΪ£º2NaCl+2H2O
 Í¨µç 
.
 
2NaOH+H2¡ü+Cl2¡ü£®
¢Ù20¡æʱ£¬NaClµÄÈܽâ¶ÈÊÇ36g£®¸ÃζÈÏ£¬±¥ºÍʳÑÎË®ÖÐÈÜÖÊÓëÈܼÁµÄÖÊÁ¿±ÈΪ
9£º25
9£º25
£®
¢ÚÉÕ¼î¿ÉÓÃÓÚ´¦ÀíÁòËáй©£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
2NaOH+H2SO4¨TNa2SO4+2H2O
2NaOH+H2SO4¨TNa2SO4+2H2O
£®
£¨4£©Ë®ÔÚ»¯Ñ§ÊµÑéÖоßÓÐÖØÒªµÄ×÷Ó㮽«ÌúË¿·ÅÔÚ³±ÊªµÄ¿ÕÆøÖУ¨ÈçͼËùʾ£©£¬Ò»¶Îʱ¼äºó£¬¹Û²ìµ½µ¼¹ÜÄÚÒºÃæÉÏÉý£»´ò¿ªK£¬µÎ¼ÓÏ¡ÑÎËᣬ¹Û²ìµ½µ¼¹ÜÄÚÒºÃæϽµ£¬µ¼¹Ü¿ÚÓÐÆøÅÝð³ö£¬¹Ø±ÕK£®Çë½âÊ͵¼¹ÜÄÚÒºÃæÉÏÉýºÍϽµµÄÔ­Òò£º
ÌúÔÚ³±ÊªµÄ¿ÕÆøÒ×ÉúÐ⣬ÏûºÄ¿ÕÆøÖеÄÑõÆø£¬Ê¹Æ¿ÄÚѹǿ¼õС£¬ËùÒԹ۲쵽µ¼¹ÜÄÚÒºÃæÉÏÉý£»ÑÎËáºÍÌú·´Ó¦Éú³ÉÇâÆø£¬ËùÒÔʹƿÄÚѹǿ±ä´ó£¬¹Û²ìµ½µ¼¹ÜÄÚÒºÃæϽµ£¬µ¼¹Ü¿ÚÓÐÆøÅÝð³ö
ÌúÔÚ³±ÊªµÄ¿ÕÆøÒ×ÉúÐ⣬ÏûºÄ¿ÕÆøÖеÄÑõÆø£¬Ê¹Æ¿ÄÚѹǿ¼õС£¬ËùÒԹ۲쵽µ¼¹ÜÄÚÒºÃæÉÏÉý£»ÑÎËáºÍÌú·´Ó¦Éú³ÉÇâÆø£¬ËùÒÔʹƿÄÚѹǿ±ä´ó£¬¹Û²ìµ½µ¼¹ÜÄÚÒºÃæϽµ£¬µ¼¹Ü¿ÚÓÐÆøÅÝð³ö
£®
·ÖÎö£º£¨1£©¸ù¾ÝºÓË®¡¢×ÔÀ´Ë®¡¢¿óȪˮº¬ÓÐÔÓÖʶøÕôÁóË®Öв»º¬ÓÐÔÓÖʽøÐнâ´ð£»
£¨2£©¸ù¾ÝˮͨµçÄÜ·Ö½âÉú³ÉÇâÆøºÍÑõÆø½øÐнâ´ð£»
£¨3£©¸ù¾ÝÈܽâ¶ÈµÄº¬ÒåÒÔ¼°ÉÕ¼îºÍÁòËá·´Ó¦Éú³ÉÁòËáÄƺÍË®½øÐнâ´ð£»
£¨4£©¸ù¾ÝÌúÔÚ³±ÊªµÄ¿ÕÆøÒ×ÉúÐâÒÔ¼°ÑÎËáºÍÌú·´Ó¦Éú³ÉÇâÆø½øÐнâ´ð£®
½â´ð£º½â£º£¨1£©ºÓË®¡¢×ÔÀ´Ë®¡¢¿óȪˮº¬ÓÐÔÓÖÊ£¬ÊôÓÚ»ìºÏÎ¶øÕôÁóË®Öв»º¬ÓÐÔÓÖÊ£¬ÊôÓÚ´¿¾»Î¹ÊÌD£»
£¨2£©Ë®Í¨µçÄÜ·Ö½âÉú³ÉÇâÆøºÍÑõÆø£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º2H2O
 Í¨µç 
.
 
2H2¡ü+O2¡ü£»¹ÊÌ2H2O
 Í¨µç 
.
 
2H2¡ü+O2¡ü£»
£¨3£©¢Ù20¡æʱ£¬NaClµÄÈܽâ¶ÈÊÇ36g£®ÊÇÖ¸¸ÃζÈÏÂ100gË®ÖÐ×î¶àÈܽâ36gÂÈ»¯ÄÆ£¬ËùÒÔ±¥ºÍʳÑÎË®ÖÐÈÜÖÊÓëÈܼÁµÄÖÊÁ¿±È=36£º100=9£º25£»¹ÊÌ9£º25£»
¢ÚÉÕ¼î¿ÉÓÃÓÚ´¦ÀíÁòËáй©£¬ÊÇÉÕ¼îºÍÁòËá·´Ó¦Éú³ÉÁòËáÄƺÍË®£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ2NaOH+H2SO4¨TNa2SO4+2H2O£»¹ÊÌ2NaOH+H2SO4¨TNa2SO4+2H2O£»
£¨4£©ÌúÔÚ³±ÊªµÄ¿ÕÆøÒ×ÉúÐ⣬ÏûºÄ¿ÕÆøÖеÄÑõÆø£¬Ê¹Æ¿ÄÚѹǿ¼õС£¬ËùÒԹ۲쵽µ¼¹ÜÄÚÒºÃæÉÏÉý£»ÑÎËáºÍÌú·´Ó¦Éú³ÉÇâÆø£¬ËùÒÔʹƿÄÚѹǿ±ä´ó£¬¹Û²ìµ½µ¼¹ÜÄÚÒºÃæϽµ£¬µ¼¹Ü¿ÚÓÐÆøÅÝð³ö£®¹ÊÌÌúÔÚ³±ÊªµÄ¿ÕÆøÒ×ÉúÐ⣬ÏûºÄ¿ÕÆøÖеÄÑõÆø£¬Ê¹Æ¿ÄÚѹǿ¼õС£¬ËùÒԹ۲쵽µ¼¹ÜÄÚÒºÃæÉÏÉý£»ÑÎËáºÍÌú·´Ó¦Éú³ÉÇâÆø£¬ËùÒÔʹƿÄÚѹǿ±ä´ó£¬¹Û²ìµ½µ¼¹ÜÄÚÒºÃæϽµ£¬µ¼¹Ü¿ÚÓÐÆøÅÝð³ö£®
µãÆÀ£º±¾Ì⿼²éÀûÓÿα¾ÖªÊ¶½â¾öÐÂÎÊÌâÄÜÁ¦£¬ÄܼÓÉîѧÉú¶Ô¿Î±¾ÖªÊ¶µÄÀí½â£¬ÑµÁ·Ñ§ÉúµÄ˼άÄÜÁ¦£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2013?±±¾©£©a¡¢b¡¢cÈýÖÖÎïÖʵÄÈܽâ¶ÈÇúÏßÈçͼËùʾ£®È¡µÈÖÊÁ¿t2¡æʱµÄa¡¢b¡¢cÈýÖÖÎïÖʵı¥ºÍÈÜÒº£¬·Ö±ðÕô·¢µÈÁ¿Ë®ºó»Ö¸´ÖÁt2¡æ£®ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2013?±±¾©Ò»Ä££©ÒÑÖªÏõËáÑÎÒ×ÈÜÓÚË®£®¶ÔÓÚ·´Ó¦£ºX+Ca£¨OH£©2=Y+Cu£¨OH£©2¡ý£¬ÏÂÁзÖÎöÖÐÕýÈ·µÄÊÇ£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2013?±±¾©¶þÄ££©°±»ùÄÆ£¨NaNH2£©ÊǺϳÉάÉúËØAµÄÔ­ÁÏ£®¹¤ÒµÉϽ«ÄÆÓÚ97¡«100¡æÈÛÈÚ£¬Ïò·´Ó¦ÈÝÆ÷ÖлºÂýͨÈëÎÞË®µÄÒº°±£¨NH3£©£¬ÔÙ¼ÓÈÈÖÁ350¡«360¡æÉú³É°±»ùÄƺÍÇâÆø£®ÏÂÁÐ˵·¨ÖУ¬ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
¢ÙÄÆ¿ÉÄÜÓëË®·´Ó¦
¢ÚÄƵÄÈÛµã±ÈÌúµÄÈÛµãµÍ
¢ÛάÉúËØAÖÐÒ»¶¨º¬ÓÐÄÆÔªËØ
¢Ü·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º2Na+2NH3
  ¡÷  
.
 
2NaNH2+H2¡ü

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2013?±±¾©Ä£Ä⣩ÏÂÁйØÓÚÑõµÄ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸