The green light emitted by a stoplight has a wavelength of 525 nm. What is the frequency of this photon? (c = 3.00 × 10⁸ m/s).
1 answer:
The frequency of this photon is 5.94 x .
Solution:
The speed of light has a value of approximately <u>3.00×108 m/s</u>
The wavelength should have units of meters.
The frequency should have units of Hz or s−1. Hz is hertz or reciprocal seconds.
We know the value of the speed of light and we are given the wavelength. All we have to do is rearrange the equation to solve for the frequency:
<u>v = c/λ</u>
= 3.00× m/s/505×
m
= 5.94Xs-1
= <u>5.94X</u><u> Hz</u>
<u />
Frequency describes the number of waves passing through a particular location at a particular time. A class interval frequency is the number of observations that occur in a particular predefined interval. So, for example, if 20 of her ages 5-9 appear in the survey's data, the interval 5-9 has a frequency of 20. The class interval endpoints are the lowest and highest possible values of the variable.
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Answer : The value of equilibrium constant for this reaction at 262.0 K is
Explanation :
As we know that,
where,
= standard Gibbs free energy = ?
= standard enthalpy = -45.6 kJ = -45600 J
= standard entropy = -125.7 J/K
T = temperature of reaction = 262.0 K
Now put all the given values in the above formula, we get:
The relation between the equilibrium constant and standard Gibbs free energy is:
where,
= standard Gibbs free energy = -12666.6 J
R = gas constant = 8.314 J/K.mol
T = temperature = 262.0 K
K = equilibrium constant = ?
Now put all the given values in the above formula, we get:
Therefore, the value of equilibrium constant for this reaction at 262.0 K is