Answer:
The thickness of the oil slick is 
Explanation:
Given that,
Index of refraction = 1.28
Wave length = 500 nm
Order m = 1
We need to calculate the thickness of oil slick
Using formula of thickness

Where, n = Index of refraction
t = thickness
= wavelength
Put the value into the formula



Hence, The thickness of the oil slick is 
It is very difficult for an atom to accept a proton. It can only be done under very special circumstances. So A and C are both incorrect. I don't see how D is possible. The atom does lose 1 electron, but how it gets 21 is think air.
The answer is B which is exactly what happens.
The formula is F = ( q1 * q2 ) / r ^ 2
<span>where: q is the individual charges of each ion </span>
<span>r is the distance between the nuclei </span>
<span>The formula is not important but to explain the relationship between the atoms in the compounds and their lattice energy. </span>
<span>From the formula we can first conclude that compounds of ions with greater charges will have a greater lattice energy. This is a direct relationship. </span>
<span>For example, the compounds BaO and SrO, whose ions' charges are ( + 2 ) and ( - 2 ) respectively for each, will have greater lattice energies that the compounds NaF and KCl, whose ions' charges are ( + 1 ) and ( - 1 ) respectively for each. </span>
<span>So Far: ( BaO and SrO ) > ( NaF and KCl ) </span>
<span>The second part required you find the relative distance between the atoms of the compounds. Really, the lattice energy is stronger with smaller atoms, an indirect relationship. </span>
<span>For example, in NaF the ions are smaller than the ions in KCl so it has a greater lattice energy. Because Sr is smaller than Ba, SrO has a greater lattice energy than BaO. </span>
<span>Therefore: </span>
<span>Answer: SrO > BaO > NaF > KCl </span>
Answer:
The answer to your question is 1.35 Watts
Explanation:
Data
Work = W = 5 J
time = t = 3.7 s
Power = P = ?
Formula
Power is a rate in which work is done or energy is transferred over time
P = 
Substitution

Result
P = 1.35 W