ÌâÄ¿ÄÚÈÝ
¢ÙÁ¿È¡50mL 0£®25 mol/LÁòËáµ¹ÈëСÉÕ±ÖУ¬²âÁ¿Î¶ȣ»
¢ÚÁ¿È¡50mL 0£®55mol/L NaOHÈÜÒº£¬²âÁ¿Î¶ȣ»
¢Û½«NaOHÈÜÒºµ¹ÈëСÉÕ±ÖÐ,»ìºÏ¾ùÔȺó²âÁ¿»ìºÏҺζȡ£
Çë»Ø´ð£º
£¨1£©NaOHÈÜÒºÉÔ¹ýÁ¿µÄÔÒò ____________¡£
£¨2£©¼ÓÈëNaOHÈÜÒºµÄÕýÈ·²Ù×÷ÊÇ_____________ £¨Ìî×Öĸ£©¡£
A£®Ñز£Á§°ô»ºÂý¼ÓÈë
B£®Ò»´ÎѸËÙ¼ÓÈë
C£®·ÖÈý´Î¼ÓÈë
£¨3£©Ê¹ÁòËáÓëNaOHÈÜÒº»ìºÏ¾ùÔȵÄÕýÈ·²Ù×÷ÊÇ__________ ¡£
£¨4£©ÉèÈÜÒºµÄÃܶȾùΪ1g¡¤cm-3,ÖкͺóÈÜÒºµÄ±ÈÈÈÈÝc=4.18 J¡¤£¨g¡¤¡æ£©-1£¬
Çë¸ù¾ÝʵÑéÊý¾Ýд³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ_____________ ¡£
£¨2£©B
£¨3£©Óû·Ðβ£Á§°ôÇáÇá½Á¶¯
£¨4£©H2SO4£¨aq£©+2NaOH£¨aq£©== Na2SO4£¨aq£©+2H2O£¨l£©£»¦¤H=-113£®6kJ¡¤mol-1
£¨5£©´óÓÚ£» ŨÁòËáÈÜÓÚË®·Å³öÈÈÁ¿
(12·Ö)ÀûÓÃÏÂͼװÖòⶨÖкÍÈȵÄʵÑé²½ÖèÈçÏ£º
¢ÙÓÃÁ¿Í²Á¿È¡50 mL 0.25 mol/LÁòËáµ¹ÈëСÉÕ±ÖУ¬²â³öÁòËáζȣ»
¢ÚÓÃÁíÒ»Á¿Í²Á¿È¡50 mL 0.55 mol/L NaOHÈÜÒº£¬²¢ÓÃÁíһζȼƲâ³öÆäζȣ»
![]()
¢Û½«NaOHÈÜÒºµ¹ÈëСÉÕ±ÖУ¬É跨ʹ֮»ìºÏ¾ùÔÈ£¬²â³ö»ìºÏÒº×î¸ß
ζȡ£»Ø´ðÏÂÁÐÎÊÌ⣺
(1)д³öÏ¡ÁòËáºÍÏ¡ÇâÑõ»¯ÄÆÈÜÒº·´Ó¦±íʾÖкÍÈȵÄÈÈ»¯Ñ§·½³Ìʽ(ÖкÍÈÈÊýֵΪ
57.3 kJ/mol)£º_______________________________________________¡£
(2)µ¹ÈëNaOHÈÜÒºµÄÕýÈ·²Ù×÷ÊÇ£º________¡£ (´ÓÏÂÁÐÑ¡³ö)¡£
A£®Ñز£Á§°ô»ºÂýµ¹Èë¡¡ B£®·ÖÈý´ÎÉÙÁ¿µ¹Èë C£®Ò»´ÎѸËÙµ¹Èë
(3)ʹÁòËáÓëNaOHÈÜÒº»ìºÏ¾ùÔȵÄÕýÈ·²Ù×÷ÊÇ£º________¡£ (´ÓÏÂÁÐÑ¡³ö)¡£
A£®ÓÃζȼÆÐ¡ÐĽÁ°è B£®½Ò¿ªÓ²Ö½Æ¬Óò£Á§°ô½Á°è
C£®ÇáÇáµØÕñµ´ÉÕ± D£®ÓÃÌ×ÔÚζȼÆÉϵĻ·Ðβ£Á§°ôÇáÇáµØ½Á¶¯
(4)ʵÑéÊý¾ÝÈçÏÂ±í£º
¢ÙÇëÌîдϱíÖеĿհףº
|
ÎÂ¶È ÊµÑé´ÎÊý¡¡ |
ÆðʼζÈt1¡æ |
ÖÕֹζÈt2/¡æ |
ÎÂ¶È²îÆ½¾ùÖµ (t2£t1)/¡æ |
||
|
H2SO4 |
NaOH |
ƽ¾ùÖµ |
|||
|
1 |
26.2 |
26.0 |
26.1 |
29.5 |
|
|
2 |
27.0 |
27.4 |
27.2 |
32.3 |
|
|
3 |
25.9 |
25.9 |
25.9 |
29.2 |
|
|
4 |
26.4 |
26.2 |
26.3 |
29.8 |
¢Ú½üËÆÈÏΪ0.55 mol/L NaOHÈÜÒººÍ0.25 mol/LÁòËáÈÜÒºµÄÃܶȶ¼ÊÇ1 g/cm3£¬ÖкͺóÉú³ÉÈÜÒºµÄ±ÈÈÈÈÝc£½4.18 J/(g¡¤¡æ)¡£ÔòÖкÍÈȦ¤H£½__________ ( ȡСÊýµãºóһλ)¡£
¢ÛÉÏÊöʵÑéÊýÖµ½á¹ûÓë57.3 kJ/molÓÐÆ«²î£¬²úÉúÆ«²îµÄÔÒò¿ÉÄÜÊÇ(Ìî×Öĸ)__________¡£
a£®ÊµÑé×°Öñ£Î¡¢¸ôÈÈЧ¹û²î
b£®ÓÃζȼƲⶨNaOHÈÜÒºÆðʼζȺóÖ±½Ó²â¶¨H2SO4ÈÜÒºµÄζÈ
c£®·Ö¶à´Î°ÑNaOHÈÜÒºµ¹ÈëÊ¢ÓÐÁòËáµÄСÉÕ±ÖÐ