The Ka : 1.671 x 10⁻⁷
<h3>Further explanation</h3>
Given
Reaction
HA (aq) + H2O (l) ←→ A- (aq) + H3O+ (aq).
0.3 M HA
pH = 3.65
Required
Ka
Solution
pH = - log [H3O+]
![\tt [H_3O^+]=10^{-3.65}=2.239\times 10^{-4}](https://tex.z-dn.net/?f=%5Ctt%20%5BH_3O%5E%2B%5D%3D10%5E%7B-3.65%7D%3D2.239%5Ctimes%2010%5E%7B-4%7D)
ICE method :
HA (aq) ←→ A- (aq) + H3O+ (aq).
0.3 0 0
2.239.10⁻⁴ 2.239.10⁻⁴ 2.239.10⁻⁴
0.3-2.239.10⁻⁴ 2.239.10⁻⁴ 2.239.10⁻⁴
![\tt Ka=\dfrac{[H_3O^+][A^-]}{[HA]}\\\\Ka=\dfrac{(2.239.10^{-4}){^2}}{0.3-2.239.10^{-4}}\\\\Ka=1.671\times 10^{-7}](https://tex.z-dn.net/?f=%5Ctt%20Ka%3D%5Cdfrac%7B%5BH_3O%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D%5C%5C%5C%5CKa%3D%5Cdfrac%7B%282.239.10%5E%7B-4%7D%29%7B%5E2%7D%7D%7B0.3-2.239.10%5E%7B-4%7D%7D%5C%5C%5C%5CKa%3D1.671%5Ctimes%2010%5E%7B-7%7D)
The answer would have to be ummm 2
Answer:
- <u>Decreasing the temperature of the system will shift the reaction rightward.</u>
Explanation:
The complete question is:
Given the equation representing a system at equilibrium:
- N₂(g) + 3H₂(g) ⇌ 2NH₃(g) + energy
what changes occur when the temperature of this system is decreased?
<h2>Solution</h2>
Modifying the temperature of a system in equilibrium changes the equilibrium constant and the equilibrium position (concentrations) of the system.
When the temperature is decreased, following LeChatelier's principle that the system will react in a way that seeks to counteract the disturbance, the reaction will shift toward the reaction that produces more heat energy to compensate the temperature decrease.
Thus, decreasing the temperature of the system will favor the forward reaction, more N₂(g) and H₂(g) will be consumed and more NH₃(g) and energy will be produced. Hence, the equilibrium will shift rightward.
Answer:
electrons are transferred from the clouds to the grounf
Explanation:
