Answer:
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Answer:
frequency = 8.22 x 10¹⁴ s⁻¹
Explanation:
An electron's positional potential energy while in a given principle quantum energy level is given by Eₙ = - A/n² and A = constant = 2.18 x 10⁻¹⁸j. So to remove an electron from the valence level of Boron (₅B), energy need be added to promote the electron from n = 2 to n = ∞. That is, ΔE(ionization) = E(n=∞) - E(n=2) = (-A/(∞)²) - (-A/(2)²) = [2.18 x 10⁻¹⁸j/4] joules = 5.45 x 10⁻¹⁹ joules.
The frequency (f) of the wave ionization energy can then be determined from the expression ΔE(izn) = h·f; h = Planck's Constant = 6.63 x 10⁻³⁴j·s. That is:
ΔE(izn) = h·f => f = ΔE(izn)/h = 5.45 x 10⁻¹⁹ j/6.63 x 10⁻³⁴ j·s = 8.22 x 10¹⁴ s⁻¹
Answer:
Al2(SO4)3
Explanation:
Looking at this carefully, we will discover that Al2(SO4)3 is composed of Al^3+ and SO4^2-.
The aluminum and sulphate ions are ionically bonded. However, the oxygen and sulphur in the sulphate ion are covalently bonded.
Hence, Al2(SO4)3 contains both ionic and covalent bond.
Genetic variation is important because it allows natural selection to increase or decrease frequency of alleles already in the population
Since this equation is balanced, we know that the law of conversation of mass id applied, and we could calculate easily.
Na= 2
NO3= 2
Ca= 1
<span>Cl= 1</span>
