ʱ¼ä:2019-07-04 00:59:02
1¡¢Ñ¡ÔñÌâ ÈçͼËùʾ£¬¡÷H1=-393.5kJ-mol-1£¬¡÷H2=-395.4kJ-mol-1£¬ÏÂÁÐ˵·¨»ò±íʾʽÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®C£¨s¡¢Ê¯Ä«£©¨TC£¨s¡¢½ð¸Õʯ£©£¬¸Ã·´Ó¦µÄìʱäΪ¸ºÖµ
B£®Ê¯Ä«µÄÎȶ¨ÐÔÈõÓÚ½ð¸Õʯ
C£®Ê¯Ä«ºÍ½ð¸ÕʯµÄת»¯ÊÇÎïÀí±ä»¯
D£®1?molʯīµÄ×ܼüÄܱÈ1?mol½ð¸ÕʯµÄ×ܼüÄÜ´ó1.9?kJ
²Î¿¼´ð°¸£ºÓÉͼµÃ£º¢ÙC£¨S£¬Ê¯Ä«£©+O2£¨g£©=CO2£¨g£©¡÷H=-393.5kJ?mol-1
? ¢ÚC£¨S£¬½ð¸Õʯ£©+O2£¨g£©=CO2£¨g£©¡÷H=-395.4kJ?mol-1£¬
A¡¢ÀûÓøÇ˹¶¨Âɽ«¢Ù-¢Ú¿ÉµÃ£ºC£¨S£¬Ê¯Ä«£©=C£¨S£¬½ð¸Õʯ£©¡÷H=+1.9kJ?mol-1£¬¹ÊA´íÎó£»
B¡¢ÓÉC£¨S£¬Ê¯Ä«£©=C£¨S£¬½ð¸Õʯ£©¡÷H=+1.9kJ?mol-1£¬½ð¸ÕʯÄÜÁ¿´óÓÚʯīµÄ×ÜÄÜÁ¿£¬ÎïÖʵÄÁ¿ÄÜÁ¿Ô½´óÔ½²»Îȶ¨£¬Ôòʯī±È½ð¸ÕʯÎȶ¨£¬¹ÊB´íÎó£»
C¡¢Ê¯Ä«ºÍ½ð¸ÕʯÊDz»Í¬µÄÁ½ÖÖÎïÖÊ£¬Ê¯Ä«×ª»¯Îª½ð¸ÕʯÊôÓÚ»¯Ñ§±ä»¯£¬¹ÊC´íÎó£»
D¡¢ÓÉ C£¨S£¬Ê¯Ä«£©=C£¨S£¬½ð¸Õʯ£©¡÷H=+1.9kJ?mol-1£¬1 molʯīµÄ×ܼüÄܱÈ1 mol½ð¸ÕʯµÄ×ܼüÄÜ´óÓÚ1.9 kJ£¬¹ÊDÕýÈ·£»
¹ÊÑ¡£ºD£®
±¾Ìâ½âÎö£º
±¾ÌâÄѶȣºÒ»°ã
2¡¢Ñ¡ÔñÌâ ÒÑÖªC(ʯī)¡úC(½ð¸Õʯ)£¬¦¤H=+119KJ£¯mol£¬Ôò¿ÉÅжÏ
[? ]
A£®½ð¸Õʯ±ÈʯīÎȶ¨
B£®Ò»ÑùÎȶ¨
C£®Ê¯Ä«±È½ð¸ÕʯÎȶ¨
D£®ÎÞ·¨ÅжÏ
²Î¿¼´ð°¸£ºC
±¾Ìâ½âÎö£º
±¾ÌâÄѶȣº¼òµ¥
3¡¢Ñ¡ÔñÌâ ÒÑÖªµ¨·¯ÈÜÓÚˮʱÈÜҺζȽµµÍ£¬µ¨·¯·Ö½âµÄÈÈ»¯Ñ§·½³ÌʽΪ£ºCuSO4¡¤5H2O(s)£½CuSO4(s)£«5H2O(l)¦¤H£½Q1kJ¡¤mol-1£»ÊÒÎÂÏ£¬Èô½«1molÎÞË®ÁòËáÍÈܽâΪÈÜҺʱ·ÅÈÈQ2kJ£¬Ôò£º
[? ]
A£®Q1>Q2¡¡¡¡¡¡¡¡
B£®Q1£½Q2¡¡¡¡¡¡¡¡
C£®Q 1<Q2
D£®ÎÞ·¨±È½Ï
²Î¿¼´ð°¸£ºA
±¾Ìâ½âÎö£º
±¾ÌâÄѶȣºÒ»°ã
4¡¢Ñ¡ÔñÌâ ÒÑÖªÈÈ»¯Ñ§·½³Ìʽ£º2H2(g)+O2(g)=2H2O(g) ¡÷H1=- 483.6 kJ¡¤mol-1£¬Ôò¶ÔÓÚÈÈ»¯Ñ§·½³Ìʽ£º
2H2O(l)=2H2(g)+O2(g) ¡÷H2=b£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
[? ]
²Î¿¼´ð°¸£ºD
±¾Ìâ½âÎö£º
±¾ÌâÄѶȣºÒ»°ã
5¡¢Ñ¡ÔñÌâ ÅжϷ´Ó¦¹ý³ÌµÄ×Ô·¢ÐÔµÄÄ¿µÄÊÇ£¨¡¡¡¡£©
A£®ÅжϷ´Ó¦µÄ·½Ïò
B£®È·¶¨·´Ó¦ÊÇ·ñÒ»¶¨»á·¢Éú
C£®ÅжϷ´Ó¦¹ý³Ì·¢ÉúµÄËÙÂÊ
D£®ÅжϷ´Ó¦¹ý³ÌµÄÈÈЧӦ
²Î¿¼´ð°¸£º»¯Ñ§·´Ó¦µÄ×Ô·¢ÐÔÖ»ÄÜÓÃÓÚÅжϷ´Ó¦µÄ·½Ïò£¬²»ÄÜÈ·¶¨·´Ó¦ÊÇ·ñÒ»¶¨»á·¢ÉúºÍ¹ý³Ì·¢ÉúµÄËÙÂÊ£¬Ò²ÎÞ·¨ÅжϷ´Ó¦¹ý³ÌµÄÈÈЧӦ£¬¹ÊÑ¡£ºA£»
±¾Ìâ½âÎö£º
±¾ÌâÄѶȣº¼òµ¥