Answer:
The simplified expression for the fraction is 
Explanation:
From the given information:
O3* → O3 (1) fluorescence
O + O2 (2) decomposition
O3* + M → O3 + M (3) deactivation
The rate of fluorescence = rate of constant (k₁) × Concentration of reactant (cO)
The rate of decomposition is = k₂ × cO
The rate of deactivation = k₃ × cO × cM
where cM is the concentration of the inert molecule
The fraction (X) of ozone molecules undergoing deactivation in terms of the rate constants can be expressed by using the formula:



since cM is the concentration of the inert molecule
Answer:
French physicist Jacques Charles (1746-1823) studied the effect of temperature on the volume of a gas at constant pressure. Charles's Law states that the volume of a given mass of gas varies directly with the absolute temperature of the gas when pressure is kept constant. The absolute temperature is temperature measured with the Kelvin scale. The Kelvin scale must be used because zero on the Kelvin scale corresponds to a complete stop of molecular motion.
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Figure 11.5.1: As a container of confined gas is heated, its molecules increase in kinetic energy and push the movable piston outward, resulting in an increase in volume.
Mathematically, the direct relationship of Charles's Law can be represented by the following equation:
V
T
=k
As with Boyle's Law, k is constant only for a given gas sample. The table below shows temperature and volume data for a set amount of gas at a constant pressure. The third column is the constant for this particular data set and is always equal to the volume divided by the Kelvin temperature.
Explanation:
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Answer:

Explanation:
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In this case, since these problems about gas mixtures are based off Dalton's law in terms of mole fraction, partial pressure and total pressure, we can write the following for hydrogen, we are given its partial pressure:

And can be solved for the total pressure as follows:

However, we first calculate the mole fraction of hydrogen by subtracting that of nitrogen to 1 due to:

Then, we can plug in to obtain the total pressure:

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Gay-Lussac's law gives the relationship between pressure and temperature of gas. For a fixed amount of gas, pressure is directly proportional to temperature at constant volume.
P/T = k
where P - pressure , T - temperature and k - constant

parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
substituting the values in the equation

T = 4342 K
initial temperature was 4342 K
Answer:
Removing O₂, means removing one of the reactants and the system would counteract this effect by producing more O₂, thereby shifting the equilibrium position to the left and favouring the backward reaction.
Explanation:
The principle that explains how changes in temperature, Concentration and Pressure of reactants or products of a reaction at equilibrium affect the equilibrium position of the reaction is the Le Chatelier's principle.
The Principle explains that a system/process if a system/process which is at equilibrium is disturbed/perturbed/constrained by one or more changes (in concentration, pressure or temperature), the system would shift the equilibrium position to counteract the effects of this change.
Removing O₂, means removing one of the reactants (changing its concentration) and the system would counteract this effect by producing more O₂, thereby shifting the equilibrium position to the left and favouring the backward reaction.