Answer:
189.2 KJ
Explanation:
Data Given
wavelength of the light = 632.8 nm
Convert nm to m
1 nm = 1 x 10⁻⁹
632.8 nm = 632.8 x 1 x 10⁻⁹ = 6.328 x 10⁻⁷m
Energy of 1 mole of photon = ?
Solution
Formula used
E = hc/λ
where
E = energy of photon
h = Planck's Constant
Planck's Constant = 6.626 x 10⁻³⁴ Js
c = speed of light
speed of light = 3 × 10⁸ ms⁻¹
λ = wavelength of light
Put values in above equation
E = hc/λ
E = 6.626 x 10⁻³⁴ Js ( 3 × 10⁸ ms⁻¹ / 6.328 x 10⁻⁷m)
E = 6.626 x 10⁻³⁴ Js (4.741 x 10¹⁴s⁻¹)
E = 3.141 x 10⁻¹⁹J
3.141 x 10⁻¹⁹J is energy for one photon
Now we have to find energy of 1 mole of photon
As we know that
1 mole consists of 6.022 x10²³ numbers of photons
So,
Energy for one mole photons = 3.141 x 10⁻¹⁹J x 6.022 x10²³
Energy for one mole photons = 1.89 x 10⁵ J
Now convert J to KJ
1000 J = 1 KJ
1.89 x 10⁵ J = 1.89 x 10⁵ /1000 = 189.2 KJ
So,
energy of one mole of photons = 189.2 KJ
Answer:
C.0.28 V
Explanation:
Using the standard cell potential we can find the standard cell potential for a voltaic cell as follows:
The most positive potential is the potential that will be more easily reduced. The other reaction will be the oxidized one. That means for the reactions:
Cu²⁺ + 2e⁻ → Cu E° = 0.52V
Ag⁺ + 1e⁻ → Ag E° = 0.80V
As the Cu will be oxidized:
Cu → Cu²⁺ + 2e⁻
The cell potential is:
E°Cell = E°cathode(reduced) - E°cathode(oxidized)
E°cell = 0.80V - (0.52V)
E°cell = 1.32V
Right answer is:
<h3>C.0.28 V
</h3>
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One of the most worrisome waste products of a nuclear reactor is plutonium 239 (239Pu). This nucleus is radioactive and decays by splitting into a helium-4 nucleus and a uranium-235 nucleus (4He +... Q: One of the most worrisome waste products of a nuclear reactor is plutonium 239 (239Pu<span>).</span>
Answer:
They start with the planet furthest from us then go down the line of how many miles from us it is. Mars and Venus are the closests to us.
Explanation:
Its showing miles away it is and how fast it rotates and scientists study and analyze that.
