When an electron in a quantum system drops from a higher energy level to a lower one, the system<u> emit a photon.</u>
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The energy of the electron drops when it transitions levels, as well as the atom releases photons. The emission of the photon occurs as the electron transitions from an energy state to a lower state. The photon energy represents precisely the energy that would be lost when an electron moves to a level with less energy.
When such an excited electron transitions from one energy level to another, this could emit a photon. The energy drop would be equivalent to the power of the photon that is released. In electron volts, the energy of an electron, as well as its associated photon (emitted or absorbed) has been stated.
Therefore, when an electron in a quantum system drops from a higher energy level to a lower one, the system<u> emit a photon.</u>
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The highest atom economy
2CO + O₂ ⇒ 2CO₂
<h3>Further explanation</h3>
Given
The reaction for the production of CO₂
Required
The highest atom economy
Solution
In reactions, there are sometimes unwanted products that can be said to be a by-product or a waste product. Meanwhile, the desired product can be said to be a useful product, which can be shown as the atom economy
of the reaction
the higher the atomic economy value of a reaction, the smaller the waste/ byproducts produced, so that less energy is wasted
The general formula:
Atom economy = (mass of useful product : mass of all reactants/products) x 100
<em>or
</em>
Atom economy = (total formula masses of useful product : total formula masses of all reactants/products) x 100
So a reaction that only produces one product will have the highest atomic value, namely the reaction in option C
Explanation:
(a) As the given chemical reaction equation is as follows.
So, when we double the amount of hypochlorite or iodine then the rate of the reaction will also get double. And, this reaction is "first order" with respect to hypochlorite and iodine.
Hence, equation for rate law of reaction will be as follows.
Rate = ![K \times [OCl^{-}] \times [l^{-}]](https://tex.z-dn.net/?f=K%20%5Ctimes%20%5BOCl%5E%7B-%7D%5D%20%5Ctimes%20%5Bl%5E%7B-%7D%5D)
(b) Since, the rate equation is as follows.
Rate = ![K [OCl^{-}][l^{-}]](https://tex.z-dn.net/?f=K%20%5BOCl%5E%7B-%7D%5D%5Bl%5E%7B-%7D%5D)
Let us assume that (![[OCl^{-}] = [l^{-}]](https://tex.z-dn.net/?f=%5BOCl%5E%7B-%7D%5D%20%3D%20%5Bl%5E%7B-%7D%5D)
)
Putting the given values into the above equation as follows.
K = 
= 
Hence, the value of rate constant for the given reaction is
.
(c) Now, we will calculate the rate as follows.
Rate =
= 
= 
Therefore, rate when ![[OCl^{-}] = 1.8 \times 10^{3}](https://tex.z-dn.net/?f=%5BOCl%5E%7B-%7D%5D%20%3D%201.8%20%5Ctimes%2010%5E%7B3%7D)
M and ![[I^{-}]= 6.0 \times 10^{4}](https://tex.z-dn.net/?f=%5BI%5E%7B-%7D%5D%3D%206.0%20%5Ctimes%2010%5E%7B4%7D)
M is .
Answer: E = 2.455 x 10^5 N/C
Explanation:
q1 = 1.2x10^-7C
q2 = 6.2x10^-8C
Electric field, E = kQ/r²
where k = 9.0x10^9
since the location is (27 - 5)cm from q1
hence electric field, E1 = k*q1/r²
E1= (9x10^9 x 1.2x10^-7)/(0.22)² = 22314.05 N/C
for q2:
E1 = k*q2/r²
E2 at 5cm
E2 = (9x10^9 x 6.2x10^-8)/(0.05)² = 223200 N/C
Hence, the total electric field at 5cm position is
E = E1 + E2
E = 22314.05 + 223200 = 245514.05 N/C
E = 2.455 x 10^5 N/C
