Calculate the heat energy released when 17.7 g of liquid mercury at 25.00 °c is converted to solid mercury at its melting point.

Answer;

Q = 359.2-J  

Explanation;

Given that;

Constants for mercury at 1 atm  

Heat capacity of Hg(l) is 28.0 J/(mol*K)  

melting point is 234.32 K  

Enthalpy of fusion is 2.29 kJ/mol

17.7-g Hg / 200.6g/mol = 0.0882 mol Hg;

°C + 273 = 298 K;

2.29-kJ/mol = 2290-J/mol  

Q = (m x ΔT x Cp) + (m x Hf)  

Q = 0.0882-mol x (298 - 234.32) x 28.0-J/mol*K) + (0.0882-mol x 2290-J/mol)  

Q = 157.26-J + 201.978-J  

Q = 359.2-J  

Q=359-J (3 sig fig allowed due to 17.7-g given in problem)

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when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.