The small particles are known as protiens
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:
Density of the substance is 0.7
Explanation:
4.2/6 = 0.7
So let's convert this amount of mL to grams:

Then we need to convert to moles using the molar weight found on the periodic table for mercury (Hg):

Then we need to convert moles to atoms using Avogadro's number:
![\frac{6.022*10^{23}atoms}{1mole} *[8.135*10^{-2}mol]=4.90*10^{22}atoms](https://tex.z-dn.net/?f=%20%5Cfrac%7B6.022%2A10%5E%7B23%7Datoms%7D%7B1mole%7D%20%2A%5B8.135%2A10%5E%7B-2%7Dmol%5D%3D4.90%2A10%5E%7B22%7Datoms%20)
So now we know that in 1.2 mL of liquid mercury, there are
present.
<u>Answer:</u> The outermost valence electron enters the p orbital.
<u>Explanation:</u>
Valence electrons are defined as the electrons which are present in outer most orbital of an atom.
Sulfur is the 16th element of the periodic table having 16 electrons.
Electronic configuration of sulfur atom is 
The number of valence electrons are 2 + 4 = 6
These 6 electrons enter s-orbital and p-orbital but the outermost valence electron will enter the p-orbital.
Hence, the outermost valence electron enters p orbital.