Answer:
Rutherford bombarded aluminum foil with beam of light known as alpha particles. The mass of this alpha particle is equivalent to helium atom.
Explanation:
Rutherford bombarded aluminum foil with beam of light known as alpha particles. The mass of this alpha particle is equivalent to helium atom.
When this alpha particles were made to strike the aluminum foil, some passed through the foil, some were reflected and speed others changed.
The ones reflected encountered heavier particle known as the nucleus, preventing them from passing through it. The whole observations indicated that atom is not is uniformly charged sphere as proposed by J.J Thomson.
Rutherford proposed new model known as the Planetary model of atom, which described atom as containing a nucleus which is revolved by electron, just like planets revolve round the sun. And this nucleus contains opposite charge to electron which is proton, to balance the motion.
long does it take to boil away 2.40 kg of the liquid.
Boiling point of He is 
Latent heat of vapourization 
Power of electrical heater 
mass of liquid is 
amount of heat required to boil

Power 

The heat or energy that is absorbed or released during a substance's phase shift is known as latent heat. It could go from a solid to a liquid or from a liquid to a gas, or vice versa. Enthalpy, a characteristic of heat, is connected to latent heat.
The heat that is used or lost as matter melts and transitions from a solid to a fluid form at a constant temperature is known as the latent heat of fusion.
Due to the fact that during softening the heat energy anticipated to transform the substance from solid to fluid at air pressure is the latent heat of fusion and that the temperature remains constant during the process, the "enthalpy" of fusion is a latent heat. The enthalpy change of any quantity of material during dissolution is known as the latent heat of fusion.
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Answer:
Q = 12540 J
Explanation:
It is given that,
Mass of water, m = 50 mL = 50 g
It is heated from 0 degrees Celsius to 60 degrees Celsius.
We need to find the energy required to heat the water. The formula use to find it as follows :

Where c is the specific heat of water, c = 4.18 J/g°C
Put all the values,

So, 12540 J of energy is used to heat the water.