Answer:
A) increasing dispersion interactions
Explanation:
Polarizability allows gases containing atoms or nonpolar molecules (for example, to condense. In these gases, the most important kind of interaction produces <em>dispersion forces</em>, <em>attractive forces that arise as a result of temporary dipoles induced in atoms or molecules.</em>
<em>Dispersion forces</em>, which are also called <em>London forces</em>, usually <u>increase with molar mass because molecules with larger molar mass tend to have more electrons</u>, and <u>dispersion forces increase in strength with the number of electrons</u>. Furthermore, larger molar mass often means a bigger atom whose electron distribution is more easily disturbed because the outer electrons are less tightly held by the nuclei.
Because the noble gases are all nonpolar molecules, <u>the only attractive intermolecular forces present are the dispersion forces</u>.
1, When temperature is increased the volume will also increase. this is because the particles will gain kinetic energy and bombard the walls of the container of the gas at a higher frequency, therefore, for the pressure to remain constant as per Charles' law, the volume will have to increase so that the rate of bombardment remains constant. This is explained by the Charles law which states that the volume of a gas is directly proportional to the absolute temperature provided pressure remains constant.
2. When temperature is Decreased the volume will also Decrease. this is because the particles will loose kinetic energy and bombard the walls of the container of the gas less frequently, therefore, for the pressure to remain constant as per Charles' law, the volume will have to reduce so that the rate of bombardment remains constant. This is explained by the Charles law which states that the volume of a gas is directly proportional to the absolute temperature provided pressure remains constant.
3. When temperature is increased the pressure will increase. This is because the gas particles gain kinetic energy and bombard the walls of the container more frequently. this is according to Pressure law which states that for a constant volume of a gas the pressure is directly proportional to absolute temperature
4. When temperature is decreased, pressure will decrease, This is because the gas particles lose kinetic energy and bombard the walls of the container less frequently. this is according to Pressure law which states that for a constant volume of a gas the pressure is directly proportional to absolute temperature
5. When particles are added, pressure will increase. This is because the bombardment per unit area also increases. Boyles law explains this, that at fixed temperature the volume of a gas is inversely proportional to the pressure.
6. When particles are removed, the pressure will decrease. This is because the bombardment per unit area also decreases. Boyle's law explains this, that at fixed temperature the volume of a gas is inversely proportional to the pressure.
Answer : The activation energy of the reaction is, 
Solution :
The relation between the rate constant the activation energy is,
![\log \frac{K_2}{K_1}=\frac{Ea}{2.303\times R}\times [\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%5Cfrac%7BK_2%7D%7BK_1%7D%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5Ctimes%20%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= initial rate constant = 
= final rate constant = 
= initial temperature = 
= final temperature = 
R = gas constant = 8.314 kJ/moleK
Ea = activation energy
Now put all the given values in the above formula, we get the activation energy.
![\log \frac{8.75\times 10^{-3}L/mole\text{ s}}{4.55\times 10^{-5}L/mole\text{ s}}=\frac{Ea}{2.303\times (8.314kJ/moleK)}\times [\frac{1}{468K}-\frac{1}{531K}]](https://tex.z-dn.net/?f=%5Clog%20%5Cfrac%7B8.75%5Ctimes%2010%5E%7B-3%7DL%2Fmole%5Ctext%7B%20s%7D%7D%7B4.55%5Ctimes%2010%5E%7B-5%7DL%2Fmole%5Ctext%7B%20s%7D%7D%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20%288.314kJ%2FmoleK%29%7D%5Ctimes%20%5B%5Cfrac%7B1%7D%7B468K%7D-%5Cfrac%7B1%7D%7B531K%7D%5D)
Therefore, the activation energy of the reaction is,
For this problem, the solution is exhibiting some colligative properties since the solute in the solution interferes with some of the properties of the solvent. We use equation for the boiling point elevation for this problem. We do as follows:
<span>
ΔT(boiling point) = (Kb)mi
</span>ΔT(boiling point) = (0.512)(1.3/2.0)(2)
ΔT(boiling point) = 0.67 degrees Celsius
<span>
T(boiling point) = 100 + 0.67 = 100.67 degrees Celsius</span>
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