Ernest Rutherford
J. J Thomson
Explanation:
<u>Ernest Rutherford</u>
In 1911, Ernest Rutherford, a New Zealand chemist performed the gold foil experiment where he gave the modelling of the atom a boost.
Experiment
In his experiment, he bombarded a thin gold foil with alpha particles generated from a radioactive source. He found that most of the alpha particles passed through the gold foil while a few of them were deflected back.
Discovery and reflection on the atomic theory
To account for his observation, Rutherford suggested an atomic model in which an atom has small positively charged center where nearly all the mass is concentrated.
<u>J. J Thomson</u>
Experiment
In 1897 J.J Thomson performed experiments using the gas discharge tube that led to the discovery of the electrons. He called them cathode rays because they originate from the cathode and exits at the anode.
Discovery and reflection on the atomic theory
From his experiment on the gas discharge tube, Thomson was able determine the properties of cathode rays some of which are:
- they move in a straight line
- they possess kinetic energy
- they attract positive charges and repels negative charges
Using his observation, he proposed the plum pudding model of the atom where it is made up of entirely electrons.
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1. False (gravity)
2. True
3. False
4.True
5. False
Hopefully this helps :)
Answer:
Low energy waves have <u>a long wavelength.</u>
Explanation:
Energy of wave is directly related to the frequency while it is inversely proportional to the wavelength.
If any wave have high energy it will have high frequency and smaller wavelength.
If the wave have lower energy then it will have lower frequency and higher wavelength.
Mathematical relationship:
E = h. f
E = h. c/λ
E= energy
h = planck's constant
f = frequency
c = speed of light
λ = wavelength
111.1 mL of water
Explanation:
Weight per volume concentration (w/v %) is defined as
weight per volume concentration = (mass of solute (g) / volume of solution (mL)) × 100
volume of solution = (mass of solute × 100) / weight per volume concentration
volume of solution = (1 × 100) / 0.9 = 111.1 mL
volume of water = volume of solution = 111.1 mL
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weight per volume concentration
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Density is calculated as mass divided by volume. If we are given an ice cube of side length 8.00 cm, then the volume of the cube is equivalent to (8.00 cm)^3 = 512 cm^3. Since we have a given mass of 476 g, we can divide:
476 g / 512 cm^3 = 0.930 g/cm^3
So the density of the sample of ice is 0.930 g/cm^3.
