ÌâÄ¿ÄÚÈÝ
·Ö±ðÈ¡40mLµÄ0£®50mol/LÑÎËáÓë40mL 0£®55mol/LÇâÑõ»¯ÄÆÈÜÒº½øÐÐÖкͷ´Ó¦¡£Í¨¹ý²â¶¨·´Ó¦¹ý³ÌÖÐËù·Å³öµÄÈÈÁ¿¿É¼ÆËãÖкÍÈÈ¡£Çë»Ø´ðÏÂÁÐÎÊÌâ¡£
£¨1£©ÀíÂÛÉÏÏ¡Ç¿Ëᡢϡǿ¼î·´Ó¦Éú³É1molˮʱ·Å³ö57£®3kJµÄÈÈÁ¿£¬Ð´³ö±íʾϡÁòËáºÍÏ¡ÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄÖкÍÈȵÄÈÈ»¯Ñ§·½³Ìʽ £»
£¨2£©ÈçÓÒͼËùʾ£¬ÒÇÆ÷AµÄÃû³ÆÊÇ_______________£»ÔÚʵÑé¹ý³ÌÖУ¬Èç¹û²»°ÑζȼÆÉϵÄËáÓÃË®³åÏ´¸É¾»Ö±½Ó²âÁ¿NaOHÈÜÒºµÄζȣ¬Ôò²âµÃµÄ¡÷H £¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족£©£»
£¨3£©¼ÙÉèÑÎËáºÍNaOHÈÜÒºµÄÃܶȶ¼ÊÇ1g/cm3£¬ÓÖÖªÖкͺóÉú³ÉÈÜÒºµÄ±ÈÈÈÈÝc£½4£®18¡Á10£3 kJ/£¨g¡¤¡æ£©¡£ÎªÁ˼ÆËãÖкÍÈÈ£¬Ä³Ñ§ÉúʵÑé¼Ç¼Êý¾ÝÈçÏ£º
ÒÀ¾Ý¸ÃѧÉúµÄʵÑéÊý¾Ý¼ÆË㣬¸ÃʵÑé²âµÃµÄÖкÍÈÈ¡÷H£½___________£»£¨½á¹û±£ÁôһλСÊý£©
£¨4£©ÈôÓÃ0£®50mol/L´×Ëá´úÌæÑÎËáºÍÇâÑõ»¯ÄÆÈÜÒº·´Ó¦Ôò²âµÃÖкÍÈÈ»á £¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족£©¡£
£¨1£©1/2H2SO4£¨aq£©+ NaOH £¨aq£©== 1/2Na2SO4 £¨aq£©+H2O£¨1£©£»¡÷H £½£57£®3kJ¡¤mol£1£¨2·Ö£©
£¨2£©»·Ðβ£Á§½Á°è°ô£¨1·Ö£©Æ«´ó£¨1·Ö£©
£¨3£©£51£®8 kJ£¯mol£¨2·Ö£©
£¨4£©Æ«Ð¡£¨1·Ö£©
½âÎö:
ÂÔ
£¨7·Ö£©·Ö±ðÈ¡40mLµÄ0.50 mol/LÑÎËáÓë40mLµÄ0.55 mol/LÇâÑõ»¯ÄÆÈÜÒº½øÐÐÖкͷ´Ó¦¡£Í¨¹ý²â¶¨·´Ó¦¹ý³ÌÖÐËù·Å³öµÄÈÈÁ¿¿É¼ÆËãÖкÍÈÈ¡£Çë»Ø´ðÏÂÁÐÎÊÌâ¡£
(1)ÀíÂÛÉÏÏ¡Ç¿Ëᡢϡǿ¼î·´Ó¦Éú³É1 mol ˮʱ·Å³ö57.3 kJµÄÈÈÁ¿£¬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»¯ÄÆÈÜÒºµÄÃܶȶ¼ÊÇ1 g/cm3£¬ÓÖÖªÖкͺóÉú³ÉÈÜÒºµÄ±ÈÈÈÈÝc£½4.18 J/(g¡¤¡æ)¡£ÎªÁ˼ÆËãÖкÍÈÈ£¬ÊµÑéʱ»¹Ðè²âÁ¿µÄÊý¾ÝÓÐ(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)ijѧÉúʵÑé¼Ç¼Êý¾ÝÈçÏ£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 |
ÒÀ¾Ý¸ÃѧÉúµÄʵÑéÊý¾Ý¼ÆË㣬¸ÃʵÑé²âµÃµÄÖкÍÈȦ¤H£½________£»