How much heat energy is required to boil 66.7 g of ammonia, NH3? The molar heat of vaporization of ammonia is 23.4 kJ/mol.

Chemistry

1 answer:

pochemuha2 years ago

6 0

Answer:

91.7 kJ

Explanation:

Step 1: Given data

Mass of ammonia (m): 66.7 g

Molar heat of vaporization of ammonia (ΔH°vap): 23.4 kJ/mol

Step 2: Calculate the moles (n) corresponding to 66.7 g of ammonia

The molar mass of ammonia is 17.03 g/mol.

66.7 g × 1 mol/17.03 g = 3.92 mol

Step 3: Calculate the heat (Q) required to boil 3.92 moles of ammonia

We will use the following expression.

Q = ΔH°vap × n

Q = 23.4 kJ/mol × 3.92 mol = 91.7 kJ

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3. A 31.2-g piece of silver (s = 0.237 J/(g · °C)), initially at 277.2°C, is added to 185.8 g of a liquid, initially at 24.4°C,

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Answer:

$Cp_{liquid}=2.54\frac{J}{g\°C}$

Explanation:

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In this case, since silver is initially hot as it cools down, the heat it loses is gained by the liquid, which can be thermodynamically represented by:

$Q_{Ag}=-Q_{liquid}$

That in terms of the heat capacities, masses and temperature changes turns out:

$m_{Ag}Cp_{Ag}(T_2-T_{Ag})=-m_{liquid}Cp_{liquid}(T_2-T_{liquid})$

Since no phase change is happening. Thus, solving for the heat capacity of the liquid we obtain:

$Cp_{liquid}=\frac{m_{Ag}Cp_{Ag}(T_2-T_{Ag})}{-m_{liquid}(T_2-T_{liquid})} \\\\Cp_{liquid}=\frac{31.2g*0.237\frac{J}{g\°C}*(28.3-227.2)\°C}{185.8g*(28.3-24.4)\°C}\\ \\Cp_{liquid}=2.54\frac{J}{g\°C}$