Answer:
If the temperature increases the molecular movement as well, and if it increases the same it will happen with the molecular movement.
Pressure, volume and temperature are three factors that are closely related since they increase the temperature, the pressure usually decreases due to the dispersion of the molecules that can be generated, so the volume also increases.
If the temperature drops, the material becomes denser, its molecules do not collide with each other, their volume and pressure increases.
Explanation:
The pressure is related to the molecular density and the movement that these molecules have.
The movement is regulated by temperature, since if it increases, the friction and collision of the molecules also.
On the other hand, the higher the volume, the less pressure there will be on the molecules, since they are more dispersed among themselves.
(in the opposite case that the volume decreases, the pressure increases)
Answer:
Vapour pressure of cyclohexane at 50°C is 490torr
Vapour pressure of benzene at 50°C is 90torr
Explanation:
Using Raoult's law, pressure of a solution is defined by the sum of the product sbetween mole fraction of both solvents and pressure of pure solvents.

In the first solution:


<em>(1)</em>
For the second equation:


<em>(2)</em>
Replacing (2) in (1):


-122.5torr = -0.250P°A

<em>Vapour pressure of cyclohexane at 50°C is 490torr</em>
And for benzene:


<em>Vapour pressure of benzene at 50°C is 90torr</em>
Answer:
pH=2.34
Explanation:
HBr -> H + Br
The dissociation it's complete, for that reason the concentration of the products is the same of HBr
[H+]=[Br-]=0.00234 M
pH= - log (0.00234)=2.34
Answer:
[C₆H₅COO⁻][H₃O⁺]/[C₆H₅COOH] = Ka
Explanation:
The reaction of dissociation of the benzoic acid in water is given by the following equation:
C₆H₅-COOH + H₂O ⇄ C₆H₅-COO⁻ + H₃O⁺ (1)
The dissociation constant of an acid is the measure of the strength of an acid:
HA ⇄ A⁻ + H⁺ (2)
(3)
<em>Where the dissociation constant of the acid (Ka) is equal to the ratio of the concentration of the dissociated forms of the acid, [A⁻][H⁺], and the concentration of the acid, [HA]. </em>
So, starting from the equations (2) and (3), the constant equation for the dissociation reaction of benzoic acid in water, of the equation (1), is:
![K_{a} = \frac{[C_{6}H_{5}COO^{-}][H_{3}O^{+}]}{[C_{6}H_{5}COOH]}](https://tex.z-dn.net/?f=%20K_%7Ba%7D%20%3D%20%5Cfrac%7B%5BC_%7B6%7DH_%7B5%7DCOO%5E%7B-%7D%5D%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%7B%5BC_%7B6%7DH_%7B5%7DCOOH%5D%7D%20)
I hope it helps you!