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
<h3><u>Answer;</u></h3>
NH3/NH4+
<h3><u>Explanation;</u></h3>
From the equation;
NH3(aq)+HNO3(aq)→NH4+(aq)+NO3−(aq)
NH3 is the base; while NH4+ is the conjugate acid
HNO3 is the acid; while NO3- is the conjugate base
- The conjugate base of a Brønsted-Lowry acid is species that is formed after an acid donates a proton while the conjugate acid of a Brønsted-Lowry base is the species formed after a base accepts a proton.
I believe your answer would be friction...
0.500 grams would be left after 56.0 days.
After 14 days=4.00 g
After 28 days=2.00 g
After 42 days=1.00 g
After 56 days=0.500 g
This is after four half lives.
