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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
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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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
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
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