4/325 = 2/unknown temperature
unknown temperature= 2/(4/325)=162.5k
Answer:
w = -531 kJ
1. Work was done by the system.
Explanation:
Step 1: Given data
- Heat gained by the system (q): 687 kJ (By convention, when the system absorbs heat, q > 0).
- Change in the internal energy of the system (ΔU°): 156 kJ
Step 2: Calculate the work done (w)
We will use the following expression.
ΔU° = q + w
w = ΔU° - q
w = 156 kJ - 687 kJ
w = -531 kJ
By convention, when w < 0, work is done by the system on the surroundings.
Answer:
K = Ka/Kb
Explanation:
P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?
P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka
PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb
K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)
Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)
Kb = [PCl₅]/ ([PCl₃] [Cl₂])
Since [PCl₅] = [PCl₅]
From the Ka equation,
[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)
From the Kb equation
[PCl₅] = Kb ([PCl₃] [Cl₂])
Equating them
Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])
(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)
(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)
Comparing this with the equation for the overall equilibrium constant
K = Ka/Kb
Answer:
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Explanation:
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Answer:
Bi (Bismuth)
Ag (Silver)
Li (Lithium)
Explanation:
Xe (Xenon) and I (Iodine) are non-metals. They cannot from a metallic bond because metallic bonds are bonds between metals only.
