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Hi Laurena,
Chemistry Thermochemistry Question
Rajah 3.18 menunjukkan eksperimen untuk menentukan haba pemendakan argentum klorida.
Rajah 3.18 Penentuan haba pemendakan argentum kloridaGunakan maklumat daripada Rajah 3.23 untuk menghitung haba pemendakan argentum klorida. [Muatan haba tentu larutan =4.2 \mathrm{~J} \mathrm{~g}^{-1}{ }^{\circ} \mathrm{C}^{-1}; ketumpatan larutan \left.=1 \mathrm{~g} \mathrm{~cm}^{-3}\right]
Chemistry Thermochemistry Solution
The question asks for heat of precipitation (haba pemendakan), which I assume must have a unit of kJ/mol as the final answer.
I will use the following steps to determine the heat of precipitation of silver chloride (haba pemendakan argentum klorida):
- Write down the chemical equation involved to identify the number of moles of AgCl formed.
- Calculate total mass of the solution, average initial temperature and temperature change
- Using Q = mcpΔT, calculate the total heat change
- Divide the total heat change, Q with the number of moles of AgCl formed to obtain heat of precipitation of silver chloride, ΔH
- AgNO3 + KCl → AgCl + KNO3
0.5 mol/dm3 x 50 cm3 = 0.5 mol/dm3 x 0.05 dm3 = 0.025 mol
Therefore, n = 0.025 moles of AgCl is produced.
- Total volume of the solution is 100 cm3. With a density of 1 g/cm3, this corresponds to m = 100 g of total mass.
The initial temperature is the average temperature of both solutions before mixing, which is:
T_initial = (28.5+29.5)/2 = 29.0 °C
The temperature change, ΔT = 32 - 29 = 3 °C
- Now, we plug in all our available values into Q = mcpΔT, using cp = 4.2 J/g°C
Q = 100(4.2)(3) = 1260 J = 1.26 kJ
- Finally, we can calculate for the heat of precipitation by dividing Q with n, assuming no heat loss to the surroundings.
ΔH = 1.26 kJ / 0.025 mol = 50.4 kJ/mol
Answer: 50.4 kJ/mol
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