Answer:
Wavelength of this beam of light: 
.
Explanation:
The speed of light in vacuum is approximately 
.
Light behaves like a wave. The wavelength of a wave is equal to the distance that it travels (in the given medium) in each period of oscillation.
On the other hand, the frequency of a wave is the number of periods in unit time. 
means one oscillation per second. The frequency of this particular wave is 
. In other words, there are 
oscillations in each second.
The period of oscillation will be equal to
.
In that period of time, a beam of light in vacuum would have traveled
.
In other words, if this beam of light of frequency is in vacuum, its wavelength will be equal to .
Complete question:
Consider the reaction.
At equilibrium at 600 K, the concentrations are as follows.
2HF -----> H₂ + F₂
[HF] = 5.82 x 10-2 M
[H2] = 8.4 x 10-3 M
[F2] = 8.4 x 10-3 M
What is the value of Keq for the reaction expressed in scientific notation?
2.1 x 10-2
2.1 x 102
1.2 x 103
1.2 x 10-3
Answer:
2.1 × 10^-2
Explanation:
Kequilibrum(Keq) = product/reactant
Equation for the reaction :
2HF -----> H₂ + F₂
Therefore,
Keq = [H2][F2] / [HF]^2
Keq = [8.4 x 10-3][8.4 x 10-3] / [5.82 x 10-2]^2
Keq = [70.56 × 10^(-3 + - 3)]/[33.8724 × 10^(-2×2)]
Keq = [70.56 × 10^-6] / [33.8724 × 10^-4]
Keq = 2.0665 × 10^(-6 - (-4))
Keq = 2.0665 × 10^(-6 + 4)
Keq = 2.1 × 10^-2
The approximate alcohol content is 210 ml.
Explanation:
It can be deduced from the question that each bottle is of 1000ml or 1 litre.
The first bottle is one half full means it has 500 ml of solution and it has 20% alcohol in it. So volume of alcohol in the solution is
20/100*500
=100 ml
The first bottle is one fifth full, so the volume of mixture is 1/5th of 1000ml
so it is 200ml having 30% alcohol
30/100*200
= 60 ml
The third bottle is one tenth full so its volume is 1/10*1000
100 ml. having 50% of alcohol
50/100*100
50 ml.
The alcohol content obtained from all these 3 litres is:
100+60+50
= 210 ml of alchohol is obtained from 800 ml of mixture.
<span>The steps of solubility of water in N-butanol is as follows:1. N-butanol molecules are attracted to the surface of the water, 2. N-butanol molecules surround water molecules, 3. Butanol mixes with water and 4. Water molecules are carried into N-butanol.</span>
Answer:
Explanation:Since the compound X has no net-dipole moment so we can ascertain that this compound is not associated with any polarity.
hence the compound must be overall non-polar. The net dipole moment of compound is zero means that the vector sum of individual dipoles are zero and hence the two individual bond dipoles associated with C-Cl bond must be oriented in the opposite directions with respect to each other.]
So we can propose that compound X must be trans alkene as only in trans compounds the individual bond dipoles cancel each other.
If one isomer of the alkene is trans then the other two isomers may be cis .
Since the two alkenes give the same molecular formula on hydrogenation which means they are quite similar and only slightly different.
The two possibility of cis structures are possible:
in the first way it is possible the one carbon has two chlorine substituents and the carbon has two hydrogens.
Or the other way could be that two chlorine atoms are present on the two carbon atoms in cis manner that is on the same side and two hydrogens are also present on the different carbon atoms in the same manner.
Kindly refer the attachments for the structure of compounds:
