L ( 31.5g Libr 350 H₂0 g₁ x T + I molli Be 879 LiBr 87 g 17665600J = +20 31.5g LiBr = 381.5 g = 5 من ما 5 ماما 17- 387.5 g (4.184 89(OC)) -11067,3 = TF S 381.5g (41184 25°c) (TF - 20) (TF-20) -11647.3 = TE 1362 mol Libr 420 Solution 362 m LiBr-48.8kJ) 2 -17665645/ 10008 (1K5) 9 = -176656005 m² 38.5 S= 4.184 (9) વ. ૪C) TIF 20.0°C TF= °C =717668.6 KT -17665600 J
L ( 31.5g Libr 350 H₂0 g₁ x T + I molli Be 879 LiBr 87 g 17665600J = +20 31.5g LiBr = 381.5 g = 5 من ما 5 ماما 17- 387.5 g (4.184 89(OC)) -11067,3 = TF S 381.5g (41184 25°c) (TF - 20) (TF-20) -11647.3 = TE 1362 mol Libr 420 Solution 362 m LiBr-48.8kJ) 2 -17665645/ 10008 (1K5) 9 = -176656005 m² 38.5 S= 4.184 (9) વ. ૪C) TIF 20.0°C TF= °C =717668.6 KT -17665600 J
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Transcribed Image Text:L
(
31.5g Libr
350 H₂0
g₁
x
T
+
I molli Be
879 LiBr
87 g
17665600J =
+20
31.5g LiBr = 381.5
g
= 5 من ما 5 ماما 17-
387.5 g (4.184 89(OC))
-11067,3 =
TF
S
381.5g (41184 25°c) (TF - 20)
(TF-20)
-11647.3 = TE
1362 mol Libr
420
Solution
362 m LiBr-48.8kJ)
2
-17665645/ 10008
(1K5)
9 = -176656005
m² 38.5
S= 4.184 (9)
વ. ૪C)
TIF 20.0°C
TF=
°C
=717668.6
KT
-17665600 J
Expert Solution

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First value of energy is calculated and using that final temperature can be calculated.
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