Given the connection between Aw and K (Aw=2k) could you use the ideal gas law and derive the Boltzmann constant. Water freezes a
2 answers:
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Answer:
Explanation:
The expression for Clausius-Clapeyron Equation is shown below as:
Where,
P is the vapor pressure
ΔHvap is the Enthalpy of Vaporization
R is the gas constant (8.314×10⁻³ kJ /mol K)
c is the constant.
For two situations and phases, the equation becomes:
Given:
= 13.95 torr
= 144.78 torr
= 25°C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (25 + 273.15) K = 298.15 K
= 298.15 K
= 75°C = 348.15 K
So,
Answer:
a)4.51
b) 9.96
Explanation:
Given:
NaOH = 0.112M
H2S03 = 0.112 M
V = 60 ml
H2S03 pKa1= 1.857
pKa2 = 7.172
a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.
Therefore, the half points will also be the middle point.
Solving, we have:
pH = (½)* pKa1 + pKa2
pH = (½) * (1.857 + 7.172)
= 4.51
Thus, pH at first equivalence point is 4.51
b) pH at second equivalence point:
We already know there is a presence of SO3-2, and it ionizes to form
SO3-2 + H2O <>HSO3- + OH-
[HSO3-] = x = [OH-]
mmol of SO3-2 = MV
= 0.112 * 60 = 6.72
We need to find the V of NaOh,
V of NaOh = (2 * mmol)/M
= (2 * 6.72)/0.122
= 120ml
For total V in equivalence point, we have:
60ml + 120ml = 180ml
[S03-2] = 6.72/120
= 0.056 M
Substituting for values gotten in the equation
We noe have:
=4.04
pH = 14- pOH
= 14 - 4.04
= 9.96
The pH at second equivalence point is 9.96
Molarity is one of the method of expressing concentration of solution. Mathematically it is expressed as,
Molarity =
Given: Molarity of solution = 5.00 M
Volume of solution = 750 ml = 0.750 l
∴ 5 =
∴ number of moles = 3.75
Answer: Number of moles of KOH present in solution is 3.75.
Molarity =
Given: Molarity of solution = 5.00 M
Volume of solution = 750 ml = 0.750 l
∴ 5 =
∴ number of moles = 3.75
Answer: Number of moles of KOH present in solution is 3.75.