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:
<em>293.99 g </em>
OR
<em>0.293 Kg</em>
Explanation:
Given data:
Lattice energy of Potassium nitrate (KNO3) = -163.8 kcal/mol
Heat of hydration of KNO3 = -155.5 kcal/mol
Heat to absorb by KNO3 = 101kJ
To find:
Mass of KNO3 to dissolve in water = ?
Solution:
Heat of solution = Hydration energy - Lattice energy
= -155.5 -(-163.8)
= 8.3 kcal/mol
We already know,
1 kcal/mol = 4.184 kJ/mole
Therefore,
= 4.184 kJ/mol x 8.3 kcal/mol
= 34.73 kJ/mol
Now, 34.73 kJ of heat is absorbed when 1 mole of KNO3 is dissolved in water.
For 101 kJ of heat would be
= 101/34.73
= 2.908 moles of KNO3
Molar mass of KNO3 = 101.1 g/mole
Mass of KNO3 = Molar mass x moles
= 101.1 g/mole x 2.908
= 293.99 g
= 0.293 kg
<em><u>293.99 g potassium nitrate has to dissolve in water to absorb 101 kJ of heat. </u></em>
Answer:D molecular attraction increases as temp decreases
Explanation:A,B would require data not available. C is missing. The question is a mess.
Equilibrium expression is ![Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5BH3O%2B%5D%5BHCO3%5E-%5D%7D%7B%5BH2CO3%5D%7D%5C%5C)
<u>Explanation:</u>
Equilibrium expression is denoted by Keq.
Keq is the equilibrium constant that is defined as the ratio of concentration of products to the concentration of reactants each raised to the power its stoichiometric coefficients.
Example -
aA + bB = cC + dD
So, Keq = conc of product/ conc of reactant
![Keq = \frac{[C]^c [D]^d}{[A]^a [B]^b}](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5BC%5D%5Ec%20%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%20%5BB%5D%5Eb%7D)
So from the equation, H₂CO₃+H₂O = H₃O+HCO₃⁻¹
![Keq = \frac{[H3O^+]^1 [HCO3^-]^1}{[H2CO3]^1 [H2O]^1}](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5BH3O%5E%2B%5D%5E1%20%5BHCO3%5E-%5D%5E1%7D%7B%5BH2CO3%5D%5E1%20%5BH2O%5D%5E1%7D)
The concentration of pure solid and liquid is considered as 1. Therefore, concentration of H2O is 1.
Thus,
Therefore, Equilibrium expression is
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