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Elan Coil [88]
2 years ago
7

One mole of ice at 0°C is added to two moles of water at 50°C under a constant external pressure of 1 atm. This process is carri

ed out in a thermally insulated vessel, such that no heat can escape the vessel. When equilibrium is reached, all of the ice has melted. Given: Δ H f u s i o n ∘ ( H 2 O a t 0 ∘ C ) = 6.01 k J m o l C p ( H 2 O , l i q u i d ) = 75 J m o l K a) (7 Points) Calculate the final temperature of the mixture (in °C). Show all work to support your answer.
Chemistry
1 answer:
kotykmax [81]2 years ago
8 0

Answer:

T2 = 29.79°C

Explanation:

Equliibrium signifies that heat loss = heat gained

Heat gained by Ice;

H = ML

Mass, M = Number of moles * Molar mass = 1 * 18 = 18g

l = 6.01 k J m o l = 334 J/g

C = 4.186 J/g

H =  18(334)

H = 6012

Heat lost by water

H = MCΔT

H = 18 * 4.186 * (50 - T2)

H = 3767.4 - 75.348T2

Since H = H, we have;

6012 = 3767.4 - 75.348T2

- 75.348T2 = 3767 - 6012

T2 = 2245 / 75.348

T2 = 29.79°C

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Answer:

C. 1.35

Explanation:

                                                     2NH3 (g) <-->          N2 (g) +             3H2 (g)

Initial concentration                2.2 mol/0.95L       1.1 mol/0.95L           0

change in concentration        2x                             x                           3x

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Equilibrium                       1.4 mol/0.95L=1.47M        1.58 M                   1.26 M

concentration

Change in concentration(NH3) = (2.2-1.4)mol/0.95 L = 0.84M

Equilibrium concentration (N2) = 1.1/0.95 +0.42=1.58 M

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