Answer : The concentration of a solution with an absorbance of 0.460 is, 0.177 M
Explanation :
Using Beer-Lambert's law :

where,
A = absorbance of solution
C = concentration of solution
l = path length
= molar absorptivity coefficient
From this we conclude that absorbance of solution is directly proportional to the concentration of solution at constant path length.
Thus, the relation between absorbance and concentration of solution will be:

Given:
= 0.350
= 0.460
= 0.135 M
= ?
Now put all the given values in the above formula, we get:


Therefore, the concentration of a solution with an absorbance of 0.460 is, 0.177 M
<span>The average speed of the gas is related to the kinetic
energy of the gas. The kinetic energy of
the gas is also related to the temperature of the gas. If the average speed of
the gas is closer to zero, it means that it has very low motion or kinetic
energy. This can be inferred that the gas has a very low temperature. At absolute
zero, the motion of all the gas molecules stops. This means that the kinetic
energy of the gas is also zero. Zero kinetic energy means zero average speed.</span>
<span>So, the answer is cylinder B. The average speed of the gas
in cylinder B is closest to zero.</span>
Answer:
![5.31*10^{-10} = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}](https://tex.z-dn.net/?f=5.31%2A10%5E%7B-10%7D%20%3D%20%5Cfrac%7B%5B%5DH_%7B2%7D%5D%5E%7B2%7D%5BO_%7B2%7D%5D%7D%7B%5BH_%7B2%7DO%5D%5E%7B2%7D%7D)
Explanation:
For a chemical reaction, equilibrium is a state at which the rate of the forward reaction equals that of the reverse reaction. The equilibrium constant Keq is a parameter characteristic of this state which is expressed as a ratio of the concentration of the products to that of the reactants.
For a hypothetical reaction:
xA + yB ⇄ zC
The equilibrium constant is :
![Keq = \frac{[A]^{x}[B]^{y}}{[C]^{z} }](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5BA%5D%5E%7Bx%7D%5BB%5D%5E%7By%7D%7D%7B%5BC%5D%5E%7Bz%7D%20%7D)
The given reaction involves the decomposition of H2O into H2 and O2

The equilibrium constant is expressed as :
![Keq = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5B%5DH_%7B2%7D%5D%5E%7B2%7D%5BO_%7B2%7D%5D%7D%7B%5BH_%7B2%7DO%5D%5E%7B2%7D%7D)
Since Keq = 5.31*10^-10
![5.31*10^{-10} = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}](https://tex.z-dn.net/?f=5.31%2A10%5E%7B-10%7D%20%3D%20%5Cfrac%7B%5B%5DH_%7B2%7D%5D%5E%7B2%7D%5BO_%7B2%7D%5D%7D%7B%5BH_%7B2%7DO%5D%5E%7B2%7D%7D)