How much it has to drop and how heavy it is. Hope this is what you're looking for:)
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C) The volume of the gas is proportional to the number of moles of gas particles.
The Avogadro's law applies to ideal gases with constant pressure and temperature. By that law, the volume of an ideal gas is proportional to the number of moles of particles in that gas.
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B) The gas now occupies less volume, and the piston will move downward.
Boyle's Law applies to ideal gases with a constant temperature. The volume of an ideal gas is inversely related to its pressure. A high pressure drives gas particles together, such that they occupy less volume. The gas trapped inside the piston has a smaller volume. As a result, the the piston will move downward.
Alternatively, consider the forces acting on the piston. Both the atmosphere and gravity are dragging the piston down. In order for it to stay in place, the gas below it must exert a pressure to balance the two forces. Now the pressure from outside has increased. The gas inside needs to increase its pressure. It needs a smaller volume to create that extra pressure. As a result, its volume will decrease, and the piston will move downwards.
Answer:
Random particle motion in liquids and gases is a difficult concept for in temperature, the particles move faster as they gain kinetic energy.
Explanation:
Answer:
The Sun Ray's hit earths surface at Earths Equator
Answer:
ΔS = -661.0J/mol is the entropy change for the system
ΔS = -842J/mol.K is the entropy change for the surroundings
Explanation:
From the relationship between ΔG, T, ΔH and ΔS,
Mathematically, ΔG = ΔH - TΔS
TΔS = ΔH - ΔS
ΔS = ΔH - ΔS / T
but ΔG = -54 kJ/mol, ΔH = -251 kJ/mol and T = 25 °C (298 K)
plugging into the equation,
ΔS = -251 kJ/mol - ( -54 kJ/mol) / 298
ΔS = -0.6610KJ/mol or in J.mol
ΔS = -661.0J/mol is the entropy change for the system
- For entropy change for the surroundings = ΔS = ΔH/T
- ΔS = -0.84KJ/mol.K or -842J/mol.K is the entropy change for the surroundings
