ÌâÄ¿ÄÚÈÝ

10£®£¨1£©ÏÂÁÐÈý×éÎïÖÊÖУ¬¾ùÓÐÒ»ÖÖÎïÖʵÄÀà±ðÓëÆäËûÈýÖÖ²»Í¬£®
A¡¢MgO¡¢Na2O¡¢CO2¡¢CuO    B¡¢HCl¡¢H2O¡¢H2SO4¡¢HNO3 C¡¢NaOH¡¢Na2CO3¡¢KOH¡¢Cu£¨OH£©2
ÈýÖÖÎïÖÊÒÀ´ÎÊÇ£¨Ìѧʽ£©£ºACO2£»BH2O£»CNa2CO3£»
£¨2£©ÓÐÏÂÁÐÎïÖÊ£º¢ÙNaOH¡¡¢ÚCO2¡¡¢ÛNH4Cl¡¡¢ÜNa2O¡¡¢ÝMgCl2¡¡¢ÞCu¡¡¢ßÁòËá¢àC2H5OH£¨¾Æ¾«£©£¨ÓÃÐòºÅ×÷´ð£©£»
¢ñ°´×é³É·ÖÀ࣬ÊôÓÚµ¥ÖʵÄÊÇ¢Þ£¬ÊôÓÚËáÐÔÑõ»¯ÎïµÄÊÇ¢Ú£¬ÊôÓÚ¼îÐÔÑõ»¯ÎïµÄÊǢܣ»ÊôÓÚËáµÄÊǢߣ¬ÊôÓÚ¼îÊÇ¢Ù£¬ÊôÓÚÑεÄÊǢۢݣ»
¢ò°´ÊÇ·ñÊǵç½âÖÊ·ÖÀ࣬ÊôÓÚµç½âÖʵÄÊǢ٢ۢܢݢߣ¬ÊôÓڷǵç½âÖʵÄÊǢڢ࣮

·ÖÎö £¨1£©A£®MgO¡¢Na2O¡¢CuOÊôÓÚ½ðÊôÑõ»¯Î
B£®HCl¡¢H2O¡¢H2SO4¡¢HNO3ÊôÓÚË᣻
C£®NaOH¡¢KOH¡¢Cu£¨OH£©2ÊôÓڼ
£¨2£©¢ñµ¥ÖÊÊÇÓÐͬÖÖÔªËØ×é³ÉµÄ´¿¾»Î
ËáÐÔÑõ»¯ÎïÊǺͼӦÉú³ÉÑκÍË®µÄÑõ»¯Î
¼îÐÔÑõ»¯ÎïÊǺÍËá·´Ó¦Éú³ÉÑκÍË®µÄÑõ»¯Î
ËáÊÇÖ¸µçÀë³öµÄÑôÀë×ÓÈ«²¿ÊÇÇâÀë×ӵĻ¯ºÏÎ
¼îÊÇÖ¸µçÀë³öµÄÒõÀë×ÓÈ«²¿ÊÇÇâÑõ¸ùÀë×ӵĻ¯ºÏÎ
ÑÎÊÇÖ¸ÄܵçÀë³ö½ðÊôÀë×Ó£¨»ò笠ùÀë×Ó£©ºÍËá¸ùÀë×ӵĻ¯ºÏÎ
¢òÔÚÈÜÓÚË®»òÈÛÈÚ״̬ÏÂÄܵ¼µçµÄ»¯ºÏÎïÊǵç½âÖÊ£¬ÔÚÈÜÓÚË®ºÍÈÛÈÚ״̬϶¼²»Äܵ¼µçµÄ»¯ºÏÎïÊǷǵç½âÖÊ£®

½â´ð ½â£º£¨1£©A£®MgO¡¢Na2O¡¢CuOÊôÓÚ½ðÊôÑõ»¯ÎCO2ÊôÓڷǽðÊôÑõ»¯Î
B£®HCl¡¢H2O¡¢H2SO4¡¢HNO3ÊôÓÚËᣬH2OÊôÓÚÑõ»¯Î
C£®NaOH¡¢KOH¡¢Cu£¨OH£©2ÊôÓڼNa2CO3ÊôÓÚÑΣ»
¹Ê´ð°¸Îª£ºCO2£»H2O£»Na2CO3£»
£¨2£©¢ñ¢ÞCuÊǵ¥ÖÊ£»
CO2ºÍ¼î·´Ó¦Éú³ÉÑκÍË®£¬ÊôÓÚËáÐÔÑõ»¯Î
Na2OºÍËá·´Ó¦Éú³ÉÑκÍË®ÊôÓÚ¼îÐÔÑõ»¯Î
ËáÊÇÖ¸µçÀë³öµÄÑôÀë×ÓÈ«²¿ÊÇÇâÀë×ӵĻ¯ºÏÎ¹Ê¢ßH2SO4·ûºÏ£»¼îÊÇÖ¸µçÀë³öµÄÒõÀë×ÓÈ«²¿ÊÇÇâÑõ¸ùÀë×ӵĻ¯ºÏÎ¹ÊNaOH·ûºÏ£»
ÑÎÊÇÖ¸ÄܵçÀë³ö½ðÊôÀë×Ó£¨»ò笠ùÀë×Ó£©ºÍËá¸ùÀë×ӵĻ¯ºÏÎ¹ÊNH4Cl¡¢MgCl2·ûºÏ£»
¹Ê´ð°¸Îª£º¢Þ£»¢Ú£»¢Ü£¬¢ß£»¢Ù£»¢Û¢Ý£»
¢òÔÚÈÜÓÚË®»òÈÛÈÚ״̬ÏÂÄܵ¼µçµÄ»¯ºÏÎïÊǵç½âÖÊ£º¢ÙNaOH¡¡¢ÛNH4Cl¡¡¢ÜNa2O¡¡¢ÝMgCl2¡¡¢ßÁòËá·ûºÏ£»ÔÚÈÜÓÚË®ºÍÈÛÈÚ״̬϶¼²»Äܵ¼µçµÄ»¯ºÏÎïÊǷǵç½âÖÊ£¬¢ÚCO2¡¡¢àC2H5OH£¨¾Æ¾«£©·ûºÏ£¬
¹Ê´ð°¸Îª£º¢Ù¢Û¢Ü¢Ý¢ß£»¢Ú¢à£®

µãÆÀ ±¾Ì⿼²éÁËÎïÖʵķÖÀàÎÊÌ⣬Ðè¸ù¾ÝËá¡¢¼î¡¢ÑΡ¢µç½âÖÊ¡¢·Çµç½âÖʵĶ¨Òå¼°·ÖÀàÒÀ¾Ý½øÐÐÅжϣ¬ÄѶȲ»´ó£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
20£®ÑÎËáºÍÇâÑõ»¯ÄÆÊǹ¤ÒµÉÏÖØÒªµÄ»¯¹¤Ô­ÁÏ£¬Ò²ÊÇʵÑéÊÒÀï³£¼ûµÄÊÔ¼Á£®
¢ñ£®²â¶¨ÖкÍÈÈ£®
£¨1£©Ð´³öÏ¡ÑÎËáºÍÏ¡ÇâÑõ»¯ÄÆÈÜÒº·´Ó¦±íʾÖкÍÈȵÄÈÈ»¯Ñ§·½³Ìʽ£¨ÖкÍÈÈÊýֵΪ57.3kJ/mol£©£ºHCl £¨aq£©+NaOH£¨aq£©=NaCl£¨aq£©+H2O£¨l£©¡÷H=-57.3 kJ/mol
£¨2£©È¡50mL 0.5mol/L HClÈÜÒºÓë50mL0.55mol/L NaOHÈÜÒº½øÐвⶨ£¬µ«ÊµÑéÊýֵСÓÚ57.3kJ/mol£¬Ô­Òò¿ÉÄÜÊÇabdÌîÐòºÅ£©£®
a£®ÊµÑé×°Öñ£Î¡¢¸ôÈÈЧ¹û²î
b£®·Ö¶à´Î°ÑNaOHÈÜÒºµ¹ÈëÊ¢ÓÐÑÎËáµÄСÉÕ±­ÖÐ
c£®Á¿È¡ÑÎËáµÄÌå»ýʱÑöÊÓ¶ÁÊý
d£®ÓÃζȼƲⶨNaOHÈÜÒºÆðʼζȺóÖ±½Ó²â¶¨ÑÎËáµÄζÈ
¢ò£®Ëá¼îÖк͵樣®
£¨1£©Ä³Ñ§ÉúÓÃÒÑÖªÎïÖʵÄÁ¿Å¨¶ÈµÄÑÎËá²â¶¨Î´ÖªÎïÖʵÄÁ¿Å¨¶ÈµÄÇâÑõ»¯ÄÆÈÜÒº£¬Ñ¡Ôñ·Ó̪×÷ָʾ¼Á£®ÊµÑéÖв»±ØÓõ½µÄÊÇa
a£®ÈÝÁ¿Æ¿        b£®Ìú¼Ų̈¡¡¡¡¡¡¡¡¡¡ c£®×¶ÐÎÆ¿¡¡¡¡¡¡¡¡¡¡¡¡d£®ËáʽµÎ¶¨¹Ü
£¨2£©Óñê×¼µÄÑÎËáµÎ¶¨´ý²âµÄÇâÑõ»¯ÄÆÈÜҺʱ£¬×óÊÖ°ÑÎÕËáʽµÎ¶¨¹ÜµÄ»îÈû£¬ÓÒÊÖÒ¡¶¯×¶ÐÎÆ¿£¬ÑÛ¾¦×¢ÊÓ׶ÐÎÆ¿ÖÐÈÜÒºÑÕÉ«µÄ±ä»¯£®Ö±µ½Òò¼ÓÈëÒ»µÎÑÎËᣬÈÜÒºµÄÑÕÉ«ÓɺìÉ«±äΪÎÞÉ«£¬ÇÒ°ë·ÖÖÓ²»ÍÊÉ«£¬¼´Í£Ö¹µÎ¶¨£®
£¨3£©ÈôµÎ¶¨¿ªÊ¼ºÍ½áÊøʱ£¬ËáʽµÎ¶¨¹ÜÖеÄÒºÃæÈçͼËùʾ£ºÔòÆðʼ¶ÁÊýΪ9.00mL£¬ÖÕµã¶ÁÊýΪ26.10mL£®ÒÑÖªÓÃc£¨HCl£©=1.0¡Á10-2mol/LµÄÑÎËá±ê¶¨25mLµÄÇâÑõ»¯ÄÆÈÜÒº£¬²âµÃc£¨NaOH£©0.0068mol/L
£¨4£©ÏÂÁвÙ×÷ÖпÉÄÜʹËù²âÇâÑõ»¯ÄÆÈÜÒºµÄŨ¶ÈÊýֵƫµÍµÄÊÇad£¨ÌîÐòºÅ£©£®
a£®¼îʽµÎ¶¨¹ÜδÓôý²âÒºÈóÏ´¾ÍÖ±½Ó×¢Èë´ý²âÇâÑõ»¯ÄÆÈÜÒº
b£®×¶ÐÎÆ¿ÓÃÕôÁóˮϴ¾»ºóûÓиÉÔï
c£®ËáʽµÎ¶¨¹ÜÔڵζ¨Ç°ÓÐÆøÅÝ£¬µÎ¶¨ºóÆøÅÝÏûʧ
d£®¶ÁÈ¡ÑÎËáÌå»ýʱ£¬¿ªÊ¼ÑöÊÓ¶ÁÊý£¬µÎ¶¨½áÊøʱ¸©ÊÓ¶ÁÊý£®

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

¾«Ó¢¼Ò½ÌÍø