Answer:
Photoelectric effect, pair production and Compton scattering
Explanation:
Gamma rays, having no charge, can be slowed slowly by ionization as a material passes through. They suffer other mechanisms that eventually make them disappear, transferring their energy, they can cross several centimeters of a solid, or hundreds of meters of air, without undergoing any process or affecting the material they cross. Then they suffer one of the three effects and deposit much of their energy there. The three mechanisms of interaction with matter are: the photoelectric effect, the Compton effect and the production of pairs.
The photoelectric effect is that the photon meets an electron in the material and transfers all its energy, disappearing the original photon. The secondary electron acquires all the energy of the photon in the form of kinetic energy, and is sufficient to separate it from its atom and convert it into a projectile. This is stopped by ionization and excitation of the material
In the Compton effect the photon collides with an electron as if it were a clash between two elastic spheres. The secondary electron acquires only part of the energy of the photon and the rest takes it with another photon of lesser energy and diverted.
When an energy photon approaches the intense electric field of a nucleus, the production of pairs can happen. In this case the photon is transformed into an electron positron pair. Since the sum of the mass of the pair is 1.02 MeV, it cannot happen if the photon's energy is less than this amount. If the energy of the original photon is greater than 1.02 MeV, the surplus is distributed by the electron and the positron as kinetic energy, and the material can be ionized. The positron at the end of its path forms a positronium and then annihilates producing two annihilation photons, 0.51 MeV each.
oxygen and silicon I believe
<u>Answer:</u> The 
for the reaction is 72 kJ.
<u>Explanation:</u>
Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.
The given chemical reaction follows:
The intermediate balanced chemical reaction are:
(1) 
(2) 
( × 2)
(3) 
( × 2)
The expression for enthalpy of the reaction follows:
![\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[2\times (-\Delta H_2)]+[2\times (\Delta H_3)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B1%5Ctimes%20%28%5CDelta%20H_1%29%5D%2B%5B2%5Ctimes%20%28-%5CDelta%20H_2%29%5D%2B%5B2%5Ctimes%20%28%5CDelta%20H_3%29%5D)
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(1\times (-1184))+(2\times -(-234))+(2\times (394))]=72kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-1184%29%29%2B%282%5Ctimes%20-%28-234%29%29%2B%282%5Ctimes%20%28394%29%29%5D%3D72kJ)
Hence, the for the reaction is 72 kJ.
You multiply the number of atoms by 12 to get how many electrons (since each atom has 12 electrons in it)
you multiply the number of atoms by 13 to get how many neutrons
(since each atom of this isotope has 13 neutrons in it)
Answer:
28.7664 kJ /mol
Explanation:
The expression for Clausius-Clapeyron Equation is shown below as:
Where,
P is the vapor pressure
ΔHvap is the Enthalpy of Vaporization
R is the gas constant (8.314×10⁻³ kJ /mol K)
c is the constant.
The graph of ln P and 1/T gives a slope of - ΔHvap/ R and intercept of c.
Given :
Slope = -3.46×10³ K
So,
- ΔHvap/ R = -3.46×10³ K
<u>ΔHvap = 3.46×10³ K × 8.314×10⁻³ kJ /mol K = 28.7664 kJ /mol</u>
<u></u>
