The photoelectric emission is possible if the wavelength of the incident light is less than that of yellow light
Answer:
pressure in cylinder A must be one third of pressure in cylinder B
Explanation:
We are told that the temperature and quantity of the gases in the 2 cylinders are same.
Thus, number of moles and temperature will be the same for both cylinders.
To this effect we will use the formula for ideal gas equation which is;
PV = nRT
Where;
P is prrssure
V is volume
n is number of moles
T is temperature
R is gas constant
We are told that Cylinder A has three times the volume of cylinder .
Thus;
V_a = 3V_b
For cylinder A;
Pressure = P_a
Volume = 3V_b
Number of moles = n
Thus;
P_a × 3V_b = nRT
For cylinder B;
Pressure = P_b
Volume = V_b
Number of moles = n
Thus,
P_b × V_b = nRT
Combining the equations for both cylinders, we have;
P_a × 3V_b = P_b × V_b
V_b will cancel out to give;
3P_a = P_b
Divide both sides by 3 to get;
P_a = ⅓P_b
Thus, pressure in cylinder A must be one third of pressure in cylinder B
Answer:
A. (1/T₂ - 1/T₁) = -2.69 x 10⁻⁴ K⁻¹
B. ln (k₁/k₂) = -3.434
C. E = 106.13 KJ/mol
Explanation:
Part A:
we have:
T₁ = 25°C = 298 K
T₂ = 51°C = 324 K
(1/T₂ - 1/T₁) = (1/324 k - 1/298 k)
<u>(1/T₂ - 1/T₁) = -2.69 x 10⁻⁴ K⁻¹</u>
<u></u>
Part B:
ln (k₁/k₂) = ln (0.1/3.1)
<u>ln (k₁/k₂) = -3.434</u>
Part C:
The activation energy can be found out by using Arrhenius Equation:
ln (k₁/k₂) = E/R (1/T₂ - 1/T₁)
where,
E = Activation Energy
R = General Gas Constant = 8.314 J/mol.k
Therefore,
-3.434 = (E/8.314 J/mol.k)(-2.69 x 10⁻⁴ K⁻¹)
<u>E = 106134.8 J/mol = 106.13 KJ/mol</u>
I think B) would be the best answer, mainly because when clouds become congested and compact like that it typically signifies it's going to rain because of condensation. Seeing how B has a ton of clouds packed together, that would be the most likely for it to indicate a future rain storm.
