As the ball descends down the hill, its potential energy diminishes while its kinetic energy rises.
The ball's potential energy will decrease as it moves down the slope, and its kinetic energy will experience an increase.
Kinetic energy refers to the energy possessed by an object in motion.
K. E = 
m v²
where m is the mass of the ball
and v represents the ball's velocity.
Potential energy is the energy associated with an object's position as it traverses down a slope, expressed as:
P.E = mgh
with m as the mass of the ball,
g as gravitational acceleration, and h as the height.
It is clear that as the object descends, its height decreases, while its velocity increases.
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Potential energy
The appropriate answer is option E. Gibbs free energy can be expressed using the equation: ΔG = ΔH - TΔS, where ΔH denotes the change in enthalpy of the reaction, T is the reaction temperature, and ΔS signifies entropy change. For our calculations, we have ΔH = -720.5 kJ/mol which converts to -720500 J/mol (given that 1 kJ = 1000 J), ΔS = -263.7 J/K, and T = 141.0°C, which equals 414.15 K. Consequently, the Gibbs free energy for the specified reaction at 141.0°C is calculated as -611.3 kJ/mol.
Based on the equation:
ΔG = ΔH - TΔS = 0
It follows that ΔS = ΔH/T
So, ΔS = n*ΔHVap / Tvap
- where n represents the number of moles calculated as mass/molar mass
For a mass of 24.1 g
and a molar mass of 187.3764 g/mol
substituting gives:
∴ n = 24.1 / 187.3764g/mol
= 0.129 moles
The molar enthalpy of vaporization, ΔHvap, is 27.49 kJ/mol
The temperature in Kelvin, Tvap = 47.6 + 273 = 320.6 K
After substitution, we compute ΔS, the change in entropy:
∴ΔS = 0.129 mol * 27490 J/mol / 320.6 K
= 11 J/K
