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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Answer:
True
Explanation:
The more of each substanse that you add to the bag will cause it to produce faster and more gas.
The second brother Brother because he is much heavier and therefore has more energy to be released hope this helps
Answer:
0.32 M
Explanation:
Step 1: Write the balanced reaction at equilibrium
Ag₂S(s) ⇌ 2 Ag⁺(aq) + S²⁻(aq)
Step 2: Calculate the concentration of Ag⁺ at equilibrium
We will use the formula for the concentration equilibrium constant (Keq), which is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It only includes gases and aqueous species.
Keq = [Ag⁺]² × [S²⁻]
[Ag⁺] = √{Keq / [S²⁻]}
[Ag⁺] = √{2.4 × 10⁻⁴ / 0.0023} = 0.32 M