Answer: kinetic, move faster, potential, change form
Explanation:
Answer:
temperature, refractive index, density, and hardness of an object
Explanation:
it is a bulk property, meaning that it is a physical property of a system that does not depend on the system size or the amount of material in the system
Answer:
Liquid oxygen evaporates at only a slightly higher temperature than liquid nitrogen because they have similarly low attraction between molecules. This would mean less liquid oxygen is coming out of tank 3 because some of it is evaporating as a gas instead.
Answer:
13.85 kJ/°C
-14.89 kJ/g
Explanation:
<em>At constant volume, the heat of combustion of a particular compound, compound A, is − 3039.0 kJ/mol. When 1.697 g of compound A (molar mass = 101.67 g/mol) is burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 3.661 °C. What is the heat capacity (calorimeter constant) of the calorimeter? </em>
<em />
The heat of combustion of A is − 3039.0 kJ/mol and its molar mass is 101.67 g/mol. The heat released by the combustion of 1.697g of A is:

According to the law of conservation of energy, the sum of the heat released by the combustion and the heat absorbed by the bomb calorimeter is zero.
Qcomb + Qcal = 0
Qcal = -Qcomb = -(-50.72 kJ) = 50.72 kJ
The heat capacity (C) of the calorimeter can be calculated using the following expression.
Qcal = C . ΔT
where,
ΔT is the change in the temperature
Qcal = C . ΔT
50.72 kJ = C . 3.661 °C
C = 13.85 kJ/°C
<em>Suppose a 3.767 g sample of a second compound, compound B, is combusted in the same calorimeter, and the temperature rises from 23.23°C to 27.28 ∘ C. What is the heat of combustion per gram of compound B?</em>
Qcomb = -Qcal = -C . ΔT = - (13.85 kJ/°C) . (27.28°C - 23.23°C) = -56.09 kJ
The heat of combustion per gram of B is:

Answer:
One complete revolution around a circular path.
Explanation:
Let us take the case of a car moving in a circular track of radius r metres.
In one revolution, the car covers the length(distance) equal to the perimeter of the circle.
In this case, distance traveled = 2
r metres
But after one complete revolution, the car reaches the same position as it was at the beginning of the motion.
Hence, the initial and final points coincide or the car hasn't changed it's position w.r.t the initial point.
So in this case, the displacement is zero.
Hence, revolution of a car around a circular path is an example of an object traveling a distance but having no displacement.