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
FeSO₄*7H₂O(s) = FeSO₄(s) + 7H₂O(g)
M(FeSO₄*7H₂O)=278.0 g/mol
M(FeSO₄)=151.9 g/mol
m(FeSO₄*7H₂O)/M(FeSO₄*7H₂O)=m(FeSO₄)/M(FeSO₄)
m(FeSO₄)=M(FeSO₄)m(FeSO₄*7H₂O)/M(FeSO₄*7H₂O)
m(FeSO₄)=151.9*100.0/278.0=54.6 g
m(FeSO₄)=54.6 g
Answer:
P1 = 2.5ATM
Explanation:
V1 = 28L
T1 = 45°C = (45 + 273.15)K = 318.15K
V2 = 34L
T2 = 35°C = (35 + 273.15)K = 308.15K
P1 = ?
P2 = 2ATM
applying combined gas equation,
P1V1 / T1 = P2V2 / T2
P1*V1*T2 = P2*V2*T1
Solving for P1
P1 = P2*V2*T1 / V1*T2
P1 = (2.0 * 34 * 318.15) / (28 * 308.15)
P1 = 21634.2 / 8628.2
P1 = 2.5ATM
The initial pressure was 2.5ATM
I don’t think I could answer this sorry.........
Answer:
the partial pressure of Xe is 452.4 mmHg
Explanation:
Dalton's law of partial pressures says that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases.
The partial pressures can be calculated with the molar fraction of the gas, in this case, Xe.
Molar fraction of Xe is calculated as follows:
Then, 0.29 is the molar fraction of Xe in the mixture of gases given.
To know the parcial pressure of Xe, we have to multiply the molar fraction by the total pressure:
Partial Pressure of Xe=1560mmHg*0.29
Partial Pressure of Xe=452.4mmHg
