ÌâÄ¿ÄÚÈÝ
A¡¢B¡¢C¡¢D¡¢E¡¢F ÊÇÁùÖÖ¶ÌÖÜÆÚµÄÖ÷×åÔªËØ£¬ËüÃǵÄÔ×ÓÐòÊýÒÀ´ÎµÝÔö£»AÔªËØÓëÆäËüÔªËؾù²»ÔÚͬһÖÜÆÚ£¬CÔªËØÓëB¡¢DÔªËØͬÖÜÆÚÏàÁÚ£»CÔªËصÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïÓëCÔªËصÄÆø̬Ç⻯ÎïXÄÜÏ໥·´Ó¦Éú³ÉÑÎY£»D¡¢FÔªËØͬ×壬EÊǶÌÖÜÆÚÖ÷×åÖÐÔ×Ӱ뾶×î´óµÄÔªËØ£®ÇëÍê³ÉÏÂÁÐÎÊÌ⣺
£¨1£©DλÓÚÔªËØÖÜÆÚ±íµÚ ÖÜÆÚ ×壮
£¨2£©EÔ×ӵĽṹʾÒâͼΪ £®
£¨3£©DºÍFÁ½ÔªËØÏà±È½Ï£¬·Ç½ðÊôÐÔ½ÏÇ¿µÄÊÇ £¨ÌîÔªËØÃû³Æ£©£¬¿ÉÒÔÑéÖ¤¸Ã½áÂÛµÄÊÂʵÊÇ
£¨Ö»Ð´Ò»¸ö¼´¿É£©£®
£¨4£©A¡¢DºÍEÈýÖÖÔªËØ×é³ÉµÄ»¯ºÏÎÆäËùº¬µÄ»¯Ñ§¼üÀàÐÍÓÐ £®
£¨5£©¼ìÑéXµÄÊÔ¼Á»òÊÔÖ½ÊÇ £®
£¨6£©ÓÉA¡¢C¡¢DÈýÖÖÔªËØ×é³ÉµÄÁ½ÖÖÇ¿µç½âÖʼ׺ÍÒÒ£¬ËüÃǵÄË®ÈÜÒº¾ù³ÊËáÐÔ£®Èô¼×ÒÖÖÆË®µÄµçÀ룬Ôò¼×µÄ»¯Ñ§Ê½Îª £»ÈôÒÒ´Ù½øË®µÄµçÀ룬ÔòÒÒµÄË®ÈÜÒº³ÊËáÐÔµÄÔÒòÊÇ£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£© £®
£¨7£©Ð´³öD¡¢EÁ½ÔªËØÐγɵÄÔ×Ó¸öÊý±ÈΪ1£º1µÄ»¯ºÏÎïÓëË®·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
£®
£¨8£©AÓëD¼ä¿ÉÐγɸºÒ»¼ÛË«Ô×ÓÒõÀë×Ó£¬ÓÐ10¸öµç×Ó£¬Ð´³ö¸ÃÒõÀë×ÓÓëÑÎËá·´Ó¦µÄÀë×Ó·½³Ìʽ
£®
£¨9£©ÒÑÖª0.4mol Һ̬C2 A4ÓëҺ̬˫ÑõË®·´Ó¦£¬Éú³ÉC2ºÍҺ̬ˮ£¬·Å³ö327.2kJµÄÈÈÁ¿£¬¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ £®
£¨10£©D¡¢E¡¢F¼òµ¥Àë×Ӱ뾶ÓÉ´óµ½Ð¡µÄ˳ÐòΪ£º £¨ÓÃÀë×Ó·ûºÅ±íʾ£©£®
£¨11£©Ð´³öA¡¢DÁ½ÖÖÔªËØÐγɵĻ¯ºÏÎïµÄÒ»ÖÖÓÃ; £®
£¨1£©DλÓÚÔªËØÖÜÆÚ±íµÚ
£¨2£©EÔ×ӵĽṹʾÒâͼΪ
£¨3£©DºÍFÁ½ÔªËØÏà±È½Ï£¬·Ç½ðÊôÐÔ½ÏÇ¿µÄÊÇ
£¨4£©A¡¢DºÍEÈýÖÖÔªËØ×é³ÉµÄ»¯ºÏÎÆäËùº¬µÄ»¯Ñ§¼üÀàÐÍÓÐ
£¨5£©¼ìÑéXµÄÊÔ¼Á»òÊÔÖ½ÊÇ
£¨6£©ÓÉA¡¢C¡¢DÈýÖÖÔªËØ×é³ÉµÄÁ½ÖÖÇ¿µç½âÖʼ׺ÍÒÒ£¬ËüÃǵÄË®ÈÜÒº¾ù³ÊËáÐÔ£®Èô¼×ÒÖÖÆË®µÄµçÀ룬Ôò¼×µÄ»¯Ñ§Ê½Îª
£¨7£©Ð´³öD¡¢EÁ½ÔªËØÐγɵÄÔ×Ó¸öÊý±ÈΪ1£º1µÄ»¯ºÏÎïÓëË®·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
£¨8£©AÓëD¼ä¿ÉÐγɸºÒ»¼ÛË«Ô×ÓÒõÀë×Ó£¬ÓÐ10¸öµç×Ó£¬Ð´³ö¸ÃÒõÀë×ÓÓëÑÎËá·´Ó¦µÄÀë×Ó·½³Ìʽ
£¨9£©ÒÑÖª0.4mol Һ̬C2 A4ÓëҺ̬˫ÑõË®·´Ó¦£¬Éú³ÉC2ºÍҺ̬ˮ£¬·Å³ö327.2kJµÄÈÈÁ¿£¬¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ
£¨10£©D¡¢E¡¢F¼òµ¥Àë×Ӱ뾶ÓÉ´óµ½Ð¡µÄ˳ÐòΪ£º
£¨11£©Ð´³öA¡¢DÁ½ÖÖÔªËØÐγɵĻ¯ºÏÎïµÄÒ»ÖÖÓÃ;
·ÖÎö£ºA¡¢B¡¢C¡¢D¡¢E¡¢F ÊÇÁùÖÖ¶ÌÖÜÆÚµÄÖ÷×åÔªËØ£¬ËüÃǵÄÔ×ÓÐòÊýÒÀ´ÎµÝÔö£»CÔªËصÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïÓëCÔªËصÄÆø̬Ç⻯ÎïXÄÜÏ໥·´Ó¦Éú³ÉÑÎY£¬ÔòCΪNÔªËØ¡¢XΪNH3¡¢YΪNH4NO3£»CÔªËØÓëB¡¢DÔªËØͬÖÜÆÚÏàÁÚ£¬Ô×ÓÐòÊýB£¼C£¼D£¬ÔòBΪCÔªËØ¡¢DΪOÔªËØ£»D¡¢FÔªËØͬ×壬ÔòFΪSÔªËØ£»EÊǶÌÖÜÆÚÖ÷×åÖÐÔ×Ӱ뾶×î´óµÄÔªËØ£¬ÔòEΪNaÔªËØ£»AÔªËØÓëÆäËüÔªËؾù²»ÔÚͬһÖÜÆÚ£¬ÇÒAµÄÔ×ÓÐòÊýСÓÚ̼ԪËØ£¬ÔòAΪHÔªËØ£¬¾Ý´Ë½â´ð£®
½â´ð£º½â£ºA¡¢B¡¢C¡¢D¡¢E¡¢F ÊÇÁùÖÖ¶ÌÖÜÆÚµÄÖ÷×åÔªËØ£¬ËüÃǵÄÔ×ÓÐòÊýÒÀ´ÎµÝÔö£»CÔªËصÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïÓëCÔªËصÄÆø̬Ç⻯ÎïXÄÜÏ໥·´Ó¦Éú³ÉÑÎY£¬ÔòCΪNÔªËØ¡¢XΪNH3¡¢YΪNH4NO3£»CÔªËØÓëB¡¢DÔªËØͬÖÜÆÚÏàÁÚ£¬Ô×ÓÐòÊýB£¼C£¼D£¬ÔòBΪCÔªËØ¡¢DΪOÔªËØ£»D¡¢FÔªËØͬ×壬ÔòFΪSÔªËØ£»EÊǶÌÖÜÆÚÖ÷×åÖÐÔ×Ӱ뾶×î´óµÄÔªËØ£¬ÔòEΪNaÔªËØ£»AÔªËØÓëÆäËüÔªËؾù²»ÔÚͬһÖÜÆÚ£¬ÇÒAµÄÔ×ÓÐòÊýСÓÚ̼ԪËØ£¬ÔòAΪHÔªËØ£¬
£¨1£©DΪOÔªËØ£¬Î»ÓÚÔªËØÖÜÆÚ±íµÚ¶þÖÜÆÚ¢öA×壬¹Ê´ð°¸Îª£º¶þ£»¢öA£»
£¨2£©EΪNaÔªËØ£¬ÆäÔ×ӵĽṹʾÒâͼΪ£¬¹Ê´ð°¸Îª£º£»
£¨3£©Í¬Ö÷×å×ÔÉ϶øÏ·ǽðÊôÐÔ¼õÈõ£¬¹Ê·Ç½ðÊôÐÔO£¾S£¬Ç⻯ÎïÎȶ¨ÐÔH2O£¾H2S£¬¿ÉÒÔ˵Ã÷OÔªËطǽðÊôÐÔ¸üÇ¿£¬
¹Ê´ð°¸Îª£ºÑõ£»Ç⻯ÎïÎȶ¨ÐÔH2O£¾H2S£»
£¨4£©H¡¢OºÍNaÈýÖÖÔªËØ×é³ÉµÄ»¯ºÏÎïΪNaOH£¬ÆäËùº¬µÄ»¯Ñ§¼üÓУºÀë×Ó¼ü¡¢¹²¼Û¼ü£¬
¹Ê´ð°¸Îª£ºÀë×Ó¼ü¡¢¹²¼Û¼ü£»
£¨5£©XΪNH3£¬ÆäË®ÈÜÒº³Ê¼îÐÔ£¬¼ìÑéNH3µÄÊÔ¼Á»òÊÔÖ½ÊǺìɫʯÈïÊÔÖ½»òºìɫʯÈïÊÔÒº£¬
¹Ê´ð°¸Îª£ººìɫʯÈïÊÔÖ½»òºìɫʯÈïÊÔÒº£»
£¨6£©ÓÉH¡¢N¡¢OÈýÖÖÔªËØ×é³ÉµÄÁ½ÖÖÇ¿µç½âÖʼ׺ÍÒÒ£¬ËüÃǵÄË®ÈÜÒº¾ù³ÊËáÐÔ£®Èô¼×ÒÖÖÆË®µÄµçÀ룬Ôò¼×µÄ»¯Ñ§Ê½ÎªHNO3£»ÈôÒÒ´Ù½øË®µÄµçÀ룬ÔòÒÒΪNH4NO3£¬ÆäË®ÈÜÒºÖÐNH4+Ë®½âNH4++H2O?NH3?H2O+H+£¬ÆÆ»µË®µÄµçÀëƽºâ£¬ÈÜÒº³ÊËáÐÔ£¬
¹Ê´ð°¸Îª£ºHNO3£»NH4++H2O?NH3?H2O+H+£»
£¨7£©O¡¢NaÁ½ÔªËØÐγɵÄÔ×Ó¸öÊý±ÈΪ1£º1µÄ»¯ºÏÎïΪNa2O2£¬ÓëË®·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º2Na2O2+2H2O¨T4NaOH+O2¡ü£¬
¹Ê´ð°¸Îª£º2Na2O2+2H2O¨T4NaOH+O2¡ü£»
£¨8£©HÓëO¼ä¿ÉÐγɸºÒ»¼ÛË«Ô×ÓÒõÀë×Ó£¬ÓÐ10¸öµç×Ó£¬¸ÃÒõÀë×ÓΪOH-£¬ÓëÑÎËá·´Ó¦µÄÀë×Ó·½³ÌʽΪ£ºH++OH-=H2O£¬
¹Ê´ð°¸Îª£ºH++OH-=H2O£»
£¨9£©ÒÑÖª0.4mol Һ̬N2H4ÓëҺ̬˫ÑõË®·´Ó¦£¬Éú³ÉN2ºÍҺ̬ˮ£¬·Å³ö327.2kJµÄÈÈÁ¿£¬Ôò1molN2H4·´Ó¦·Å³öµÄÈÈÁ¿Îª327.2kJ¡Á
=818kJ£¬¹Ê¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ£ºN2H4£¨l£©+2H2O2£¨l£©¨TN2£¨g£©+4H2O£¨l£©¡÷H=-818kJ?mol-1£¬
¹Ê´ð°¸Îª£ºN2H4£¨l£©+2H2O2£¨l£©¨TN2£¨g£©+4H2O£¨l£©¡÷H=-818kJ?mol-1£»
£¨10£©D¡¢E¡¢F¼òµ¥Àë×Ó·Ö±ðΪO2-¡¢Na+¡¢S2-£¬µç×Ó²ãÔ½¶àÀë×Ӱ뾶Խ¶à¡¢µç×Ó²ã½á¹¹ÏàͬºËµçºÉÊýԽСÀë×Ӱ뾶Խ´ó£¬¹ÊÀë×Ӱ뾶ÓÉ´óµ½Ð¡µÄ˳ÐòΪ£ºS2-£¾O2-£¾Na+£¬
¹Ê´ð°¸Îª£ºS2-£¾O2-£¾Na+£»
£¨11£©H¡¢OÁ½ÖÖÔªËØÐγɵĻ¯ºÏÎïH2O»òH2O2£¬H2O³£ÓÃ×÷ÈܼÁ£¬H2O2³£ÓÃ×÷ÂÌÉ«Ñõ»¯¼Á£¬
¹Ê´ð°¸Îª£ºH2O³£ÓÃ×÷ÈܼÁ»òH2O2³£ÓÃ×÷ÂÌÉ«Ñõ»¯¼Á£®
£¨1£©DΪOÔªËØ£¬Î»ÓÚÔªËØÖÜÆÚ±íµÚ¶þÖÜÆÚ¢öA×壬¹Ê´ð°¸Îª£º¶þ£»¢öA£»
£¨2£©EΪNaÔªËØ£¬ÆäÔ×ӵĽṹʾÒâͼΪ£¬¹Ê´ð°¸Îª£º£»
£¨3£©Í¬Ö÷×å×ÔÉ϶øÏ·ǽðÊôÐÔ¼õÈõ£¬¹Ê·Ç½ðÊôÐÔO£¾S£¬Ç⻯ÎïÎȶ¨ÐÔH2O£¾H2S£¬¿ÉÒÔ˵Ã÷OÔªËطǽðÊôÐÔ¸üÇ¿£¬
¹Ê´ð°¸Îª£ºÑõ£»Ç⻯ÎïÎȶ¨ÐÔH2O£¾H2S£»
£¨4£©H¡¢OºÍNaÈýÖÖÔªËØ×é³ÉµÄ»¯ºÏÎïΪNaOH£¬ÆäËùº¬µÄ»¯Ñ§¼üÓУºÀë×Ó¼ü¡¢¹²¼Û¼ü£¬
¹Ê´ð°¸Îª£ºÀë×Ó¼ü¡¢¹²¼Û¼ü£»
£¨5£©XΪNH3£¬ÆäË®ÈÜÒº³Ê¼îÐÔ£¬¼ìÑéNH3µÄÊÔ¼Á»òÊÔÖ½ÊǺìɫʯÈïÊÔÖ½»òºìɫʯÈïÊÔÒº£¬
¹Ê´ð°¸Îª£ººìɫʯÈïÊÔÖ½»òºìɫʯÈïÊÔÒº£»
£¨6£©ÓÉH¡¢N¡¢OÈýÖÖÔªËØ×é³ÉµÄÁ½ÖÖÇ¿µç½âÖʼ׺ÍÒÒ£¬ËüÃǵÄË®ÈÜÒº¾ù³ÊËáÐÔ£®Èô¼×ÒÖÖÆË®µÄµçÀ룬Ôò¼×µÄ»¯Ñ§Ê½ÎªHNO3£»ÈôÒÒ´Ù½øË®µÄµçÀ룬ÔòÒÒΪNH4NO3£¬ÆäË®ÈÜÒºÖÐNH4+Ë®½âNH4++H2O?NH3?H2O+H+£¬ÆÆ»µË®µÄµçÀëƽºâ£¬ÈÜÒº³ÊËáÐÔ£¬
¹Ê´ð°¸Îª£ºHNO3£»NH4++H2O?NH3?H2O+H+£»
£¨7£©O¡¢NaÁ½ÔªËØÐγɵÄÔ×Ó¸öÊý±ÈΪ1£º1µÄ»¯ºÏÎïΪNa2O2£¬ÓëË®·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º2Na2O2+2H2O¨T4NaOH+O2¡ü£¬
¹Ê´ð°¸Îª£º2Na2O2+2H2O¨T4NaOH+O2¡ü£»
£¨8£©HÓëO¼ä¿ÉÐγɸºÒ»¼ÛË«Ô×ÓÒõÀë×Ó£¬ÓÐ10¸öµç×Ó£¬¸ÃÒõÀë×ÓΪOH-£¬ÓëÑÎËá·´Ó¦µÄÀë×Ó·½³ÌʽΪ£ºH++OH-=H2O£¬
¹Ê´ð°¸Îª£ºH++OH-=H2O£»
£¨9£©ÒÑÖª0.4mol Һ̬N2H4ÓëҺ̬˫ÑõË®·´Ó¦£¬Éú³ÉN2ºÍҺ̬ˮ£¬·Å³ö327.2kJµÄÈÈÁ¿£¬Ôò1molN2H4·´Ó¦·Å³öµÄÈÈÁ¿Îª327.2kJ¡Á
1mol |
0.4mol |
¹Ê´ð°¸Îª£ºN2H4£¨l£©+2H2O2£¨l£©¨TN2£¨g£©+4H2O£¨l£©¡÷H=-818kJ?mol-1£»
£¨10£©D¡¢E¡¢F¼òµ¥Àë×Ó·Ö±ðΪO2-¡¢Na+¡¢S2-£¬µç×Ó²ãÔ½¶àÀë×Ӱ뾶Խ¶à¡¢µç×Ó²ã½á¹¹ÏàͬºËµçºÉÊýԽСÀë×Ӱ뾶Խ´ó£¬¹ÊÀë×Ӱ뾶ÓÉ´óµ½Ð¡µÄ˳ÐòΪ£ºS2-£¾O2-£¾Na+£¬
¹Ê´ð°¸Îª£ºS2-£¾O2-£¾Na+£»
£¨11£©H¡¢OÁ½ÖÖÔªËØÐγɵĻ¯ºÏÎïH2O»òH2O2£¬H2O³£ÓÃ×÷ÈܼÁ£¬H2O2³£ÓÃ×÷ÂÌÉ«Ñõ»¯¼Á£¬
¹Ê´ð°¸Îª£ºH2O³£ÓÃ×÷ÈܼÁ»òH2O2³£ÓÃ×÷ÂÌÉ«Ñõ»¯¼Á£®
µãÆÀ£º±¾Ì⿼²é½á¹¹ÐÔÖÊλÖùØϵӦÓã¬ÄѶȲ»´ó£¬ÍƶÏÔªËØÊǽâÌâ¹Ø¼ü£¬²àÖضԻù´¡ÖªÊ¶µÄ¹®¹Ì£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿