Answer:
5Atm
Explanation:
I just guess and it’s right
Answer:
Reaction 1: Kc increases
Reaction 2: Kc decreases
Reaction 3: The is no change
Explanation:
Let us consider the following reactions:
Reaction 1: A ⇌ 2B ΔH° = 20.0 kJ/mol
Reaction 2: A + B ⇌ C ΔH° = −5.4 kJ/mol
Reaction 3: 2A⇌ B ΔH° = 0.0 kJ/mol
To predict what will happen when the temperature is raised we need to take into account Le Chatelier Principle: when a system at equilibrium suffers a perturbation, it will shift its equilibrium to counteract such perturbation. This means that <em>if the temperature is raised (perturbation), the system will react to lower the temperature.</em>
Reaction 1 is endothermic (ΔH° > 0). If the temperature is raised the system will favor the forward reaction to absorb heat and lower the temperature, thus increasing the value of Kc.
Reaction 2 is exothermic (ΔH° < 0). If the temperature is raised the system will favor the reverse reaction to absorb heat and lower the temperature, thus decreasing the value of Kc.
Reaction 3 is not endothermic nor exothermic (ΔH° = 0) so an increase in the temperature will have no effect on the equilibrium.
Answer:
Electrolysis (of hydrochloric acid) is a way of splitting up (decomposition) of the compound (hydrogen chloride in water) using electrical energy.
Explanation:
The electrical energy comes from a d.c. (direct current) battery or power pack supply. A conducting liquid, containing ions, called the electrolyte (hydrochloric acid), must contain the compound (hydrogen chloride) that is being broken down.
Californium which is element 98, and americium:) those are a couple!! Hope that helps
Answer:
1.58x10⁻⁵
2.51x10⁻⁸
0.0126
63.10
Explanation:
Phenolphthalein acts like a weak acid, so in aqueous solution, it has an acid form HIn, and the conjugate base In-, and the pH of it can be calculated by the Handerson-Halsebach equation:
pH = pKa + log[In-]/[HIn]
pKa = -logKa, and Ka is the equilibrium constant of the dissociation of the acid. [X] is the concentrantion of X. Thus,
i) pH = 4.9
4.9 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = - 4.8
[In-]/[HIn] =
[In-]/[HIn] = 1.58x10⁻⁵
ii) pH = 2.1
2.1 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = -7.6
[In-]/[HIn] =
[In-]/[HIn] = 2.51x10⁻⁸
iii) pH = 7.8
7.8 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = -1.9
[In-]/[HIn] =
[In-]/[HIn] = 0.0126
iv) pH = 11.5
11.5 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = 1.8
[In-]/[HIn] =
[In-]/[HIn] = 63.10