ÌâÄ¿ÄÚÈÝ
·Ö±ðÈ¡40mLµÄ0.50mol/LÑÎËáÓë40mLµÄ0.55mol/LÇâÑõ»¯ÄÆÈÜÒº½øÐÐÖкͷ´Ó¦£®Í¨¹ý²â¶¨·´Ó¦¹ý³ÌÖÐËù·Å³öµÄÈÈÁ¿¿É¼ÆËãÖкÍÈÈ£®Çë»Ø´ðÏÂÁÐÎÊÌ⣮
£¨1£©ÀíÂÛÉÏÏ¡Ç¿Ëᡢϡǿ¼î·´Ó¦Éú³É1molˮʱ·Å³ö57.3kJµÄÈÈÁ¿£¬Ð´³ö±íʾϡÁòËáºÍÏ¡ÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄÖкÍÈȵÄÈÈ»¯Ñ§·½³Ìʽ______£®
£¨2£©ÈçͼËùʾ£¬AΪÅÝÄËÜÁϰ壬ÉÏÃæÓÐÁ½¸öС¿×£¬·Ö±ð²åÈëζȼƺͻ·Ðβ£Á§½Á°è°ô£¬Á½¸öС¿×²»ÄÜ¿ªµÃ¹ý´ó£¬ÆäÔÒòÊÇ______£»
£¨3£©¼ÙÉèÑÎËáºÍÇâÑõ»¯ÄÆÈÜÒºµÄÃܶȶ¼ÊÇ1g/cm3£¬ÓÖÖªÖкͺóÉú³ÉÈÜÒºµÄ±ÈÈÈÈÝc=4.18J/£¨g?¡æ£©£®ÎªÁ˼ÆËãÖкÍÈÈ£¬ÊµÑéʱ»¹Ðè²âÁ¿µÄÊý¾ÝÓУ¨ÌîÐòºÅ£©______£®
A£®·´Ó¦Ç°ÑÎËáÈÜÒºµÄζÈ
B£®·´Ó¦Ç°ÑÎËáÈÜÒºµÄÖÊÁ¿
C£®·´Ó¦Ç°ÇâÑõ»¯ÄÆÈÜÒºµÄζÈ
D£®·´Ó¦Ç°ÇâÑõ»¯ÄÆÈÜÒºµÄÖÊÁ¿
E£®·´Ó¦ºó»ìºÏÈÜÒºµÄ×î¸ßζÈ
F£®·´Ó¦ºó»ìºÏÈÜÒºµÄÖÊÁ¿
£¨4£©Ä³Ñ§ÉúʵÑé¼Ç¼Êý¾ÝÈçÏ£º
ÒÀ¾Ý¸ÃѧÉúµÄʵÑéÊý¾Ý¼ÆË㣬¸ÃʵÑé²âµÃµÄÖкÍÈÈ¡÷H=______£®
£¨1£©ÀíÂÛÉÏÏ¡Ç¿Ëᡢϡǿ¼î·´Ó¦Éú³É1molˮʱ·Å³ö57.3kJµÄÈÈÁ¿£¬Ð´³ö±íʾϡÁòËáºÍÏ¡ÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄÖкÍÈȵÄÈÈ»¯Ñ§·½³Ìʽ______£®
£¨2£©ÈçͼËùʾ£¬AΪÅÝÄËÜÁϰ壬ÉÏÃæÓÐÁ½¸öС¿×£¬·Ö±ð²åÈëζȼƺͻ·Ðβ£Á§½Á°è°ô£¬Á½¸öС¿×²»ÄÜ¿ªµÃ¹ý´ó£¬ÆäÔÒòÊÇ______£»
£¨3£©¼ÙÉèÑÎËáºÍÇâÑõ»¯ÄÆÈÜÒºµÄÃܶȶ¼ÊÇ1g/cm3£¬ÓÖÖªÖкͺóÉú³ÉÈÜÒºµÄ±ÈÈÈÈÝc=4.18J/£¨g?¡æ£©£®ÎªÁ˼ÆËãÖкÍÈÈ£¬ÊµÑéʱ»¹Ðè²âÁ¿µÄÊý¾ÝÓУ¨ÌîÐòºÅ£©______£®
A£®·´Ó¦Ç°ÑÎËáÈÜÒºµÄζÈ
B£®·´Ó¦Ç°ÑÎËáÈÜÒºµÄÖÊÁ¿
C£®·´Ó¦Ç°ÇâÑõ»¯ÄÆÈÜÒºµÄζÈ
D£®·´Ó¦Ç°ÇâÑõ»¯ÄÆÈÜÒºµÄÖÊÁ¿
E£®·´Ó¦ºó»ìºÏÈÜÒºµÄ×î¸ßζÈ
F£®·´Ó¦ºó»ìºÏÈÜÒºµÄÖÊÁ¿
£¨4£©Ä³Ñ§ÉúʵÑé¼Ç¼Êý¾ÝÈçÏ£º
ʵÑé ÐòºÅ | ÆðʼζÈt1/¡æ | ÖÕֹζÈt2/¡æ | |
ÑÎËá | ÇâÑõ»¯ÄÆ | »ìºÏÈÜÒº | |
1 | 20.0 | 20.1 | 23.2 |
2 | 20.2 | 20.4 | 23.4 |
3 | 20.5 | 20.6 | 23.6 |
£¨1£©ÒÑ֪ϡǿËᡢϡǿ¼î·´Ó¦Éú³É1molҺ̬ˮʱ·Å³ö57.3kJµÄÈÈÁ¿£¬Ï¡ÁòËáºÍÇâÑõ»¯ÄÆÏ¡ÈÜÒº¶¼ÊÇÇ¿ËáºÍÇ¿¼îµÄÏ¡ÈÜÒº£¬Ôò·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ£ºNaOH£¨aq£©+
H2SO4£¨aq£©¨T
Na2SO4£¨aq£©+H2O £¨l£©¡÷H=-57.3 kJ/mol£¬
¹Ê´ð°¸Îª£ºNaOH£¨aq£©+
H2SO4£¨aq£©¨T
Na2SO4£¨aq£©+H2O£¨l£©¡÷H=-57.3 kJ/mol£»
£¨2£©ÈçͼËùʾ£¬AΪÅÝÄËÜÁϰ壬ÉÏÃæÓÐÁ½¸öС¿×£¬·Ö±ð²åÈëζȼƺͻ·Ðβ£Á§½Á°è°ô£¬ÈôÁ½¸öС¿×¿ªµÃ¹ý´ó£¬»áµ¼ÖÂɢʧ½Ï¶àµÄÈÈÁ¿£¬Ó°Ïì²â¶¨½á¹û£¬
¹Ê´ð°¸Îª£º¼õÉÙÈÈÁ¿É¢Ê§£»
£¨3£©ÓÉQ=cm¡÷T¿ÉÖª£¬²â¶¨ÖкÍÈÈÐèÒª²â¶¨µÄÊý¾ÝΪ£ºA£®·´Ó¦Ç°ÑÎËáÈÜÒºµÄζȡ¢B£®·´Ó¦Ç°ÑÎËáÈÜÒºµÄÖÊÁ¿ºÍE£®·´Ó¦ºó»ìºÏÈÜÒºµÄ×î¸ßζȣ¬
¹ÊÑ¡ACE£»
£¨4£©µÚ1´ÎʵÑéÑÎËáºÍNaOHÈÜÒºÆðʼƽ¾ùζÈΪ20.05¡æ£¬·´Ó¦ºóζÈΪ£º23.2¡æ£¬·´Ó¦Ç°ºóζȲîΪ£º3.15¡æ£»
µÚ2´ÎʵÑéÑÎËáºÍNaOHÈÜÒºÆðʼƽ¾ùζÈΪ20.3¡æ£¬·´Ó¦Ç°ºóζȲîΪ£º3.1¡æ£»
µÚ3´ÎʵÑéÑÎËáºÍNaOHÈÜÒºÆðʼƽ¾ùζÈΪ20.55¡æ£¬·´Ó¦Ç°ºóζȲîΪ£º3.05¡æ£»
40mLµÄ0.50mol/LÑÎËáÓë40mLµÄ0.55mol/LÇâÑõ»¯ÄÆÈÜÒºµÄÖÊÁ¿ºÍΪm=80mL¡Á1g/cm3=80g£¬c=4.18J/£¨g?¡æ£©£¬´úÈ빫ʽQ=cm¡÷TµÃÉú³É0.05molµÄË®·Å³öÈÈÁ¿Q=4.18J/£¨g?¡æ£©¡Á80g¡Á
=1.036kJ£¬¼´Éú³É0.02molµÄË®·Å³öÈÈÁ¿Îª£º1.036kJ£¬ËùÒÔÉú³É1molµÄË®·Å³öÈÈÁ¿Îª£º1.036kJ¡Á
=-51.8kJ/mol£¬¼´¸ÃʵÑé²âµÃµÄÖкÍÈÈ¡÷H=-51.8kJ/mol£¬
¹Ê´ð°¸Îª£º-51.8kJ/mol£®
1 |
2 |
1 |
2 |
¹Ê´ð°¸Îª£ºNaOH£¨aq£©+
1 |
2 |
1 |
2 |
£¨2£©ÈçͼËùʾ£¬AΪÅÝÄËÜÁϰ壬ÉÏÃæÓÐÁ½¸öС¿×£¬·Ö±ð²åÈëζȼƺͻ·Ðβ£Á§½Á°è°ô£¬ÈôÁ½¸öС¿×¿ªµÃ¹ý´ó£¬»áµ¼ÖÂɢʧ½Ï¶àµÄÈÈÁ¿£¬Ó°Ïì²â¶¨½á¹û£¬
¹Ê´ð°¸Îª£º¼õÉÙÈÈÁ¿É¢Ê§£»
£¨3£©ÓÉQ=cm¡÷T¿ÉÖª£¬²â¶¨ÖкÍÈÈÐèÒª²â¶¨µÄÊý¾ÝΪ£ºA£®·´Ó¦Ç°ÑÎËáÈÜÒºµÄζȡ¢B£®·´Ó¦Ç°ÑÎËáÈÜÒºµÄÖÊÁ¿ºÍE£®·´Ó¦ºó»ìºÏÈÜÒºµÄ×î¸ßζȣ¬
¹ÊÑ¡ACE£»
£¨4£©µÚ1´ÎʵÑéÑÎËáºÍNaOHÈÜÒºÆðʼƽ¾ùζÈΪ20.05¡æ£¬·´Ó¦ºóζÈΪ£º23.2¡æ£¬·´Ó¦Ç°ºóζȲîΪ£º3.15¡æ£»
µÚ2´ÎʵÑéÑÎËáºÍNaOHÈÜÒºÆðʼƽ¾ùζÈΪ20.3¡æ£¬·´Ó¦Ç°ºóζȲîΪ£º3.1¡æ£»
µÚ3´ÎʵÑéÑÎËáºÍNaOHÈÜÒºÆðʼƽ¾ùζÈΪ20.55¡æ£¬·´Ó¦Ç°ºóζȲîΪ£º3.05¡æ£»
40mLµÄ0.50mol/LÑÎËáÓë40mLµÄ0.55mol/LÇâÑõ»¯ÄÆÈÜÒºµÄÖÊÁ¿ºÍΪm=80mL¡Á1g/cm3=80g£¬c=4.18J/£¨g?¡æ£©£¬´úÈ빫ʽQ=cm¡÷TµÃÉú³É0.05molµÄË®·Å³öÈÈÁ¿Q=4.18J/£¨g?¡æ£©¡Á80g¡Á
3.15¡æ+3.1¡æ+3.05 |
3 |
1mol |
0.02mol |
¹Ê´ð°¸Îª£º-51.8kJ/mol£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿