<span>11.2G is the answer to this problem.
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Answer : The mass of of water present in the jar is, 298.79 g
Solution : Given,
Mass of barium nitrate = 27 g
The solubility of barium nitrate at
is 9.02 gram per 100 ml of water.
As, 9.02 gram of barium nitrate present in 100 ml of water
So, 27 gram of barium nitrate present in
of water
The volume of water is 299.33 ml.
As we know that the density of water at
is 0.9982 g/ml
Now we have to calculate the mass of water.


Therefore, the mass of of water present in the jar is, 298.79 g
Answer:
Explanation:
The formula relating the mass m of a sample and the heat q to vaporize it is
q = mL, where L is the latent heat of vaporization.

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:
Their isn't your examples but I will give you mine
Freezing, Evaporation and so on.