<span>When K is > than 1 products are favored. When q is < less than 1 reactants are favored. 1 x 108 = 108 which is > 1 so products are favored.</span>
There are two ways to solve this problem. We can use the ICE method which is tedious and lengthy or use the Henderson–Hasselbalch equation. This equation relates pH and the concentration of the ions in the solution. It is expressed as
pH = pKa + log [A]/[HA]
where pKa = - log [Ka]
[A] is the concentration of the conjugate base
[HA] is the concentration of the acid
Given:
Ka = 1.8x10^-5
NaOH added = 0.015 mol
HC2H3O2 = 0.1 mol
NaC2H3O2 = 0.1 mol
Solution:
pKa = - log ( 1.8x10^-5) = 4.74
[A] = 0.015 mol + 0.100 mol = .115 moles
[HA] = .1 - 0.015 = 0.085 moles
pH = 4.74 + log (.115/0.085)
pH = 4.87
NO, it should not be the process is still the same . the only factors about it that should change is the experiment itself.:)
Answer:
That it is cooking the food or whatever you have in the pot.
Explanation:
We are learning this in science.
Answer: The amount of heat needed is = 4.3kJ
Explanation:
Amount of heat H = M × C × ΔT
M= mass of benzene = 64.7g
C= specific heat capacity = 1.74J/gK
ΔT = T2-T1
Where T1 is initai temperature = 41.9C
T2 is the final temperature( boiling point of benzene) = 80.1C
H= 64.7×1.74×80.7
H= 4300J
H=4.3kJ
Therefore, the amount of heat needed is 4.3kJ
