Answer:
Scandium(III) fluoride, ScF3, is an ionic compound. It is slightly soluble in water but dissolves in the presence of excess fluoride to form the ScF63− anion.
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Oxygen and Nitrogen would be the most similar of the elements listed, because they are the closest in the periodic table. This isn't a very good reason for anything, but the two do have some similar properties. They are both non-metals, they are both highly electronegative, they are both diatmoic gasses in their natural states, they have a similar number of valence electrons, they are both generally oxidizing agents. Oxygen and Chlorine are also quite similar, but not quite as similar as Oxygen and Nitrogen.
Given :
A student has a 1 g sample of each of the following compounds: NaCl, KBr, and KCl.
To Find :
The samples in order of increasing number of moles in the sample.
Solution :
Molecular mass of NaCl, KBr, and KCl is 58.5 g/mol , 119 g/mol and
74.5 g/mol respectively .
Moles of NaCl ,
.
Moles of KBr ,
.
Moles of KCl ,
.
The order of moles in increasing order is :
KBr , KCl and NaCl .
Hence , this is the required solution .
3.98 x 10⁻¹⁹ Joule
<h3>Further explanation</h3>
<u>Given:</u>
The green light has a frequency of about 6.00 x 10¹⁴ s⁻¹.
<u>Question:</u>
The energy of a photon of green light (in joules).
<u>The Process:</u>
The energy of a photon is given by 
- E = energy in joules
- h = Planck's constant 6.63 x 10⁻³⁴ Js
- f = frequency of light in Hz (sometimes the symbol f is written as v)
Let us find out the energy of the green light emitted per photon.

Thus, we get a result of 
- - - - - - - - - -
Notes
- When an electron moves between energy levels it must emit or absorb energy.
- The energy emitted or absorbed corresponds to the difference between the two allowed energy states, i.e., as packets of light called photons.
- A higher energy photon corresponds to a higher frequency (shorter wavelength) of light.
<h3>Learn more</h3>
- The energy of the orange light emitted per photon brainly.com/question/2485282#
- Determine the density of our sun at the end of its lifetime brainly.com/question/5189537
- Find out the kinetic energy of the emitted electrons when metal is exposed to UV rays brainly.com/question/5416146
Keywords: green light, frequency, the energy, a photon, Planck's constant, electrons, emitted, wavelength, joules
Answer:
C4H8O4
Explanation:
To determine the molecular formula, first, let us obtain the empirical formula. This is illustrated below:
From the question given, we obtained the following information:
C = 45.45%
H = 6.12%
O = 48.44%
Divide the above by their molar mass
C = 45.45/12 = 3.7875
H = 6.12/1 = 6.12
O = 48.44/16 = 3.0275
Divide by the smallest
C = 3.7875/3.0275 = 1
H = 6.12/3.0275 = 2
O = 3.0275/3.0275 = 1
The empirical formula is CH2O
The molecular formula is given by [CH2O]n
[CH2O]n = 132.12
[12 + (2x1) + 16]n = 132.12
30n = 132.12
Divide both side by the coefficient of n i.e 30
n = 132.12/30 = 4
The molecular formula is [CH2O]n = [CH2O]4 = C4H8O4