Answer:
55.96kJ
Explanation:
Energy = mass of diethyl ether × enthalpy of vaporization of diethyl ether
Volume (v) = 200mL, density (d) = 0.7138g/mL
Mass = d × v = 0.7138 × 200 = 142.76g
Enthalpy of vaporization of diethyl ether = 29kJ/mol
MW of diethyl ether (C2H5)2O = 74g/mol
Enthalpy in kJ/g = 29kJ/mol ÷ 74g/mol = 0.392kJ/g
Energy = 142.76g × 0.392kJ/g = 55.96kJ
Answer:
D. The temperature does not change during a phase change because the average kinetic energy does not change. Therefore, the potential energy in the bonds between molecules must change.
Explanation:
When there is a change of state (for example, from solid into a liquid, as in this example), when energy is added to the system, the temperature of the substance does not change.
The reason for this is that the energy supplied is no longer used to increase the average kinetic energy of the particle, but instead it is used to break the bonds between the different particles/molecules. For instance, since in this case the substance is changing from solid to liquid, all the energy supplied during the phase change is used to break the bonds between the molecules of the solid: when the process is done, all the molecules will be free to slide past each other, and the substance has turned completely into a liquid.
The bonds between molecules store potential energy: therefore, this means that the energy supplied during the phase change is not used to change the kinetic energy, but to change the potential energy in the bonds between the molecules.
Perfectly inelastic collision is type of collision during which two objects collide, stay connected and momentum is conserved. Formula used for conservation of momentum is:
In case of perfectly inelastic collision v'1 and v'2 are same.
We have following information:
m₁=3 kg
m₂=? kg
v₁=x m/s
v₂=0 m/s
v'1 = v'2 = 1/3 * v₁
Now we insert given information and solve for m₂:
3*v₁ + 0*? = 3*1/3*v₁ + m₂*1/3*v₁
3v₁ = v₁ + m₂*1/3*v₁
2v₁ = m₂*1/3*v₁
2 = m₂*1/3
m₂= 6kg
Mass of second mud ball is 6kg.
Gravity adds 9.8 m/s to the speed of a falling object every second.
An object dropped from 'rest' (v = 0) reaches the speed of 78.4 m/s after falling for (78.4 / 9.8) = <em>8.0 seconds</em> .
<u>Note:</u>
In order to test this, you'd have to drop the object from a really high cell- tower, building, or helicopter. After falling for 8 seconds and reaching a speed of 78.4 m/s, it has fallen 313.6 meters (1,029 feet) straight down.
The flat roof of the Aon Center . . . the 3rd highest building in Chicago, where I used to work when it was the Amoco Corporation Building . . . is 1,076 feet above the street.
