Rutherford was one of the early scientists who worked on the atomic model. Before his discovery of the nucleus, the widely accepted theory was J.J Thomson's Plum Pudding Model. In this model, all the protons, electrons and neutrons are in the nucleus. But the electrons are more in number such that the electrons act as the 'pudding' and the proton and nucleus the 'plum'. This was Rutherford's hypothesis in his gold foil experiment. In order to test the Plum Pudding model, he hypothesized that when a beam of light is aimed at the atom, it would not diffract because the charges in the nucleus are well-distributed. However, his experiment disproved Thomson's model. Some light indeed passed through but a few was diffracted back to the source. He concluded that this was because there is a dense mass inside the atom called nucleus. Thus, from there on, he proposed the model that the electrons are orbiting around the nucleus.
Answer: They both have stages where they are born and die which is in the main sequence and supernova and in a human they are born in a womb and die of old age.
Explanation:
Answer:
Just think of whats in a forest ecosystem
for example
Water
sunlight
tempreture
Properties of matter can be broadly classified into two categories:
Physical properties which usually involve a change in the state of matter and Chemical properties which involve a change in the chemical composition of matter.
Now, physical properties can be further classified as:
Extensive: these depend on the amount of the substance, eg: mass, volume
Intensive: these do not depend on the amount of the substance eg: density, color, melting point, boiling point
Here we are given a 5.0 g and 1 cm3 silver cube :
Therefore:
Extensive properties are-
1) Mass of silver = 5.0 g
2) Volume of silver = 1 cm3
Intensive properties are:
1) Density of silver = mass/volume = 5.0 g/ 1 cm3 = 5.0 g/cm3
2) Melting point of silver = 962 C
3) Color = white/gray
M₁ = mass of water = 75 g
T₁ = initial temperature of water = 23.1 °C
c₁ = specific heat of water = 4.186 J/g°C
m₂ = mass of limestone = 62.6 g
T₂ = initial temperature of limestone = ?
c₂ = specific heat of limestone = 0.921 J/g°C
T = equilibrium temperature = 51.9 °C
using conservation of heat
Heat lost by limestone = heat gained by water
m₂c₂(T₂ - T) = m₁c₁(T - T₁)
inserting the values
(62.6) (0.921) (T₂ - 51.9) = (75) (4.186) (51.9 - 23.1)
T₂ = 208.73 °C
in three significant figures
T₂ = 209 °C
