Answer:
4.59 × 10⁻³⁶ kJ/photon
Explanation:
Step 1: Given and required data
- Wavelength of the violet light (λ): 433 nm
- Planck's constant (h): 6.63 × 10⁻³⁴ J.s
- Speed of light (c): 3.00 × 10⁸ m/s
Step 2: Convert "λ" to meters
We will use the conversion factor 1 m = 10⁹ nm.
433 nm × 1 m/10⁹ nm = 4.33 × 10⁷ m
Step 3: Calculate the energy (E) of the photon
We will use the Planck-Einstein's relation.
E = h × c/λ
E = 6.63 × 10⁻³⁴ J.s × (3.00 × 10⁸ m/s)/4.33 × 10⁷ m
E = 4.59 × 10⁻³³ J = 4.59 × 10⁻³⁶ kJ
Answer:
Moment=Force x Pivot
Explanation:
A moment is the turning effect of a force. Moments act about a point in a clockwise or anticlockwise direction.
Law of moments:
When an object is balanced (in equilibrium) the sum of the clockwise moments is equal to the sum of the anticlockwise moments.
How to calculate moments:
Moment=Force x Pivot
Grams. It is a smaller unit.
a. 381.27 m/s
b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide
<h3>Further explanation</h3>
Given
T = 100 + 273 = 373 K
Required
a. the gas speedi
b. The rate of effusion comparison
Solution
a.
Average velocities of gases can be expressed as root-mean-square averages. (V rms)

R = gas constant, T = temperature, Mm = molar mass of the gas particles
From the question
R = 8,314 J / mol K
T = temperature
Mm = molar mass, kg / mol
Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol

b. the effusion rates of two gases = the square root of the inverse of their molar masses:

M₁ = molar mass sulfur dioxide = 64
M₂ = molar mass nitrogen triodide = 395

the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide