Answer:
Explanation:
k stand for equilibrium constants in terms of reaction
The higher the value of an equilibrium constant the faster the equilibrium reaction comes to completion.
Consider the example below:
⇄
where
For a faster reaction the numerator i.e. the right hand side of the equation have to be higher than the left hand side (the denominator). therefore the higher the numerator, the higher the value of the equilibrium constant and the faster the reaction get to completion thus option c is correct.
<span>In one atom, the ionization energy is the</span> energy needed remove one electron.
The answer is Ka = 1.00x10^-10.
Solution:
When given the pH value of the solution equal to 11, we can compute for pOH as
pOH = 14 - pH = 14 - 11.00 = 3.00
We solve for the concentration of OH- using the equation
[OH-] = 10^-pOH = 10^-3 = x
Considering the sodium salt NaA in water, we have the equation
NaA → Na+ + A-
hence, [A-] = 0.0100 M
Since HA is a weak acid, then A- must be the conjugate base and we can set up an ICE table for the reaction
A- + H2O ⇌ HA + OH-
Initial 0.0100 0 0
Change -x +x +x
Equilibrium 0.0100-x x x
We can now calculate the Kb for A-:
Kb = [HA][OH-] / [A-]
= x<span>²</span> / 0.0100-x
Approximating that x is negligible compared to 0.0100 simplifies the equation to
Kb = (10^-3)² / 0.0100 = 0.000100 = 1.00x10^-4
We can finally calculate the Ka for HA from the Kb, since we know that Kw = Ka*Kb = 1.0 x 10^-14:
Ka = Kw / Kb
= 1.00x10^-14 / 1.00x10^-4
= 1.00x10^-10
Answer:
Decreases
Explanation:
The particles have more space to move and will be less likely to collide.
Answer:
C. Water passes into the salt solution, dehydrating bacterial cells and making them harmless.
Explanation:
The salt solution is hypertonic to the bacterial cells and as such, water molecules will move from the bacterial cells into the salt solution, dehydrating the cells and rendering them harmless.
Option A is also true but it is irrelevant to the question asked. Option B and D are wrong.
The correct option is C.
