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
C. quadruples the rate
<h3>Further explanation</h3>
Given
The rate law :
R=k[A]²
Required
The rate
Solution
There are several factors that influence reaction kinetics :
- 1. Concentration
- 2. Surface area
- 3. Temperature
- 4. Catalyst
- 5. Pressure
- 6. Stirring
The rate is proportional to the concentration.
If the concentration increased, the reaction rate will increase
The reaction is second-order overall(The exponent is 2)
The concentration of A is doubled, the reaction rate will increase :
r = k[A]² ⇒ r= k[2A]²⇒r=4k[A]²
<em>The reaction rate will quadruple.</em>
Answer:
HOAc is stronger acid than HClO
ClO⁻ is stronger conjugate base than OAc⁻
Kb(OAc⁻) = 5.5 x 10⁻¹⁰
Kb(ClO⁻) = 3.3 x 10⁻⁷
Explanation:
Assume 0.10M HOAc => H⁺ + OAc⁻ with Ka = 1.8 x 10⁻⁵
=> [H⁺] = √Ka·[Acid] =√(1.8 x 10⁻⁵)(0.10) M = 1.3 x 10⁻³M H⁺
Assume 0.10M HClO => H⁺ + ClO⁻ with Ka = 3 x 10⁻⁸
=> [H⁺] = √(3 x 10⁻⁸)(0.10)M = 5.47 x 10⁻⁵M H⁺
HOAc delivers more H⁺ than HClO and is more acidic.
Kb = Kw/Ka, Kw = 1 x 10⁻¹⁴
Kb(OAc⁻) = 5.5 x 10⁻¹⁰
Kb(ClO⁻) = 3.3 x 10⁻⁷
Answer:
Electrons will flow from left to right through the wire.
Pb^2+ ions will be reduccd to Pb metal.
The concentration of Sn2+ ions in the left compartment will increase.
Explanation:
Looking at the relative electrode potentials of the two metals
Sn= -0.14
Pb=-0.13
Tin is expected to function as the anode (left hand half cell) and lead as the anode (right hand half cell) tin oxidizes to sn^2+ hence its concentration increases on the left compartment while lead is reduced to ordinary lead metal on the right hand half cell . since oxidation occurs on the left hand side, electrons flow from left to right.
