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<u>¹⁴₇N</u><u> </u>is the more stable isotope
<h3>
Briefly explained</h3>
We have ¹⁴₇N which has a neutron to proton ratio of one, and we look at ¹⁸₇N which has a neutron to proton ratio of 1.57 Again, you look at table 24 to and you see the atomic number of seven and there is really no stable isotope. It has any more than 10 neutrons.
When we have eight, protons will go down seven protons. There's really nothing stable that has more than maybe eight neutrons. So the fact that we have 11 neutrons with ¹⁸₇N suggests that this is very unstable and
¹⁴₇N is the stable isotope of the pair.
<h3>
Stable and Unstable Nuclei</h3>
An atom is electrically neutral. It contains an equal number of positively charged protons and negatively charged electrons and their charges balance. The nucleus however contains only positively charged protons which are closely packed together in a very small volume (remember neutrons have no charge).
From the laws of physics (Coulomb’s Law) one would expect that the protons being of the same charge and so close together would exert strong repulsive forces on each other. The combined gravitational force from the protons and neutrons in a nucleus is insignificant as an attractive force because their masses are so tiny.
This implies there must be an additional attractive force similar in size to the electrostatic repulsion which holds the nucleus together.
Learn more about stable and unstable nuclei
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Answer:
1 = oxidation
2 = reduction
Explanation:
Oxidation:
Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.
2I- ----> I₂+ 2e⁻
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
F + e⁻ ----> F⁻
Consider the following reactions.
4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl
the oxidation state of copper is changed from +2 to +1 so copper get reduced.
CO + H₂O → CO₂ + H₂
the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized.
Na₂CO₃ + H₃PO₄ → Na₂HPO₄ + CO₂ + H₂O
The oxidation state of carbon on reactant side is +4. while on product side is also +4 so it neither oxidized nor reduced.
H₂S + 2NaOH → Na₂S + 2H₂O
The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.
