- Frequency=v=2×10^15Hz
- Energy=E
Using planks quantum theory
Is mountain lions a herbivore
2 answers:
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ok, now press calculator. i dont have it now.
Answer:
A) Partial Pressure of dry air = 13.32 KPa
Partial Pressure of water vapour = 1.332 KPa
B) Humidity ratio; X = 0.0691
C) V_p = 0.8384 m³/Kg
Explanation:
A) We are given;
Temperature = 75°F
Relative Humidity = 45%
Now,to calculate the partial pressure, we will use the relationship;
Relative Humidity = (Partial Pressure/Vapour Pressure) × 100%
Making partial pressure the subject;
Partial Pressure = Relative Humidity × Vapour Pressure/100%
From the first table attached, at temperature of 75°F, the vapor pressure is 29.6 × 10^(-3) bar = 29.6 KPa
Thus;
Partial Pressure of dry air = (45 × 29.6)/100
Partial Pressure of dry air = 13.32 KPa
From online values, vapour pressure of water vapour at 75°F = 2.96 KPa
Thus;
Partial Pressure of water vapour = (45 × 2.96)/100 = 1.332 KPa
B) humidity ratio of moist air is given as;
X = 0.62198 pw / (pa - pw)
where;
pw = partial pressure of the water vapor in moist air
pa = atmospheric pressure of the moist air
Thus;
X = (0.62198 × 1.332)/(13.32 - 1.332)
X = 0.0691
C) Formula for moist air specific volume is;
V_p = (1 + (xRw/Ra) × RaT/p
Where;
V_p is specific volume
T is temperature = 75°F = 297.039 K
p is barometric pressure which in this case is standard sea level pressure = 101.325 KPa
pw is partial pressure of the water vapor in moist air = 1.332 KPa
Rw is individual gas constant for water = 0.4614 KJ/Kg.K
Ra is individual gas constant for air = 0.2869 KJ/Kg.K
V_p = (1 + (0.0691 * 0.4614/0.2869)) × 0.286.9 * 297.039/101.325
V_p = 0.8384 m³/Kg
Answer:
(a) 7.11x10⁻⁴ M/s
(b) 2.56 mol.L⁻¹.h⁻¹
Explanation:
(a) The reaction is:
O₃(g) + NO(g) → O₂(g) + NO₂(g) (1)
The reaction rate of equation (1) is given by:
(2)
<u>We have:</u>
k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹
[O₃]₀ = 2.35x10⁻⁶ M
[NO]₀ = 7.74x10⁻⁵ M
Hence, to find the inital reacion rate we will use equation (2):
Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s
(b) The number of moles of NO₂(g) produced per hour per liter of air is:
t = 1 h
V = 1 L
Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹
I hope it helps you!