5 plates is the highest amount that can be served
There’s only 5 sandwiches so 7 is automatically ruled out, there’s 14 corn cobs and 5 sandwiches only need 10 so it works out
<span>The "second" is the SI base unit of time.</span>
Any buffer exists in this equilibrium
HA <=>
In a buffer, there is a large reservoir of both the undissociated acid (HA) and its conjugate base (
)
When a strong acid is added, it reacts with the large reservoir of the conjugate base (
) forming a salt and water. Since this large reservoir of the conjugate base is used, the ph does not alter drastically, but instead resist the pH change.
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:
conditioning-the process of associating a stimulus with a good or bad outcome.
trial-and-error learning-the process of performing a behavior more and more skillfully.
insight learning-the process of using prior knowledge to solve a problem.
imprinting-a bond to an indivisual or object shortly after birth or hatching.
Explanation:
Did this assignment.
