Near water, change in elevation, or change in latitude.
Natural vs. Synthetic Venn Diagram: Natural chemicals are produced by nature without any human intervention. Synthetic chemicals are made by humans using methods different than those nature uses, and these chemical structures may or may not be found in nature
Answer:
The frequency is 
Explanation:
From the question we are told that
The energy required to ionize boron is 
Generally the ionization energy of boron pre atom is mathematically represented as

Here
is the Avogadro's constant with value 
So

=> 
Generally the energy required to liberate one electron from an atom is equivalent to the ionization energy per atom and this mathematically represented as

=> 
Here h is the Planks constant with value 
So

=> 
Mass of KNO₃ : = 40.643 g
<h3>Further explanation</h3>
Given
28.5 g of K₃PO₄
Required
Mass of KNO₃
Solution
Reaction(Balanced equation) :
2K₃PO₄ + 3 Ca(NO₃)₂ = Ca₃(PO₄)₂ + 6 KNO₃
mol K₃PO₄(MW=212,27 g/mol) :
= mass : MW
= 28.5 : 212,27 g/mol
= 0.134
Mol ratio of K₃PO₄ : KNO₃ = 2 : 6, so mol KNO₃ :
= 6/2 x mol K₃PO₄
= 6/2 x 0.134
= 0.402
Mass of KNO₃ :
= mol x MW KNO₃
= 0.402 x 101,1032 g/mol
= 40.643 g