True is the answer I must say???
The energy of the light with a wavelength of 415 nm is not sufficient to remove an electron from a silver atom in the gaseous phase.
<h3>Energy and wavelength of light</h3>
The energy and wavelength of light are related by the formula given below:
- Energy = hc/λ
- where, E = energy
- h = Planck's constant
- c = velocity of light
- λ = wavelength of light
<h3>Calculating the energy of the light</h3>
From the data provided:
- h = 6.63 × 10^-34 Js
- c = 3.0 × 10^8 m/s
- λ = 415 nm = 4.15 × 10^-7 m
E = (6.63 × 10^-34 × 3.0 × 10^8 m/s)/4.15 × 10^-7 m
E = 4.79 × 10^-19 J
Energy of light is 4.79 × 10^-19 J
Compared with the ionization energy of silver, the energy of the light is far less.
Therefore, the energy of the light with a wavelength of 415 nm is not sufficient to remove an electron from a silver atom in the gaseous phase.
Learn more about about ionization energy and energy of light at: brainly.com/question/14596067
Given that solubility product of AgCl = 1.8 X 10^-10
Dissociation of AgCl can be represented as follows,
AgCl(s) ↔ Ag+(ag) + Cl-(aq)
Let, [Ag+] = [Cl-] = S
∴Ksp = [Ag+][Cl-] = S^2
∴ S = √Ksp = √(1.8 X 10^-10) = 1.34 x 10^-5 mol/dm3
Now, Molarity of solution =
∴ 1.34 x 10^-5 =
∴ Weight of AgCl present in solution = 1.92 X 10^-3 g
Thus,
mass of AgCl that will dissolve in 1l water = 1.92 x 10^-3 g
Answer:
0.504 M
Explanation:
Step 1: Write the balanced neutralization reaction
2 KOH + H₂SO₄ ⇒ K₂SO₄ + 2 H₂O
Step 2: Calculate the reacting moles of KOH
55.2 mL (0.0552 L) of 0.500 M KOH react. The reacting moles of KOH are:
0.0552 L × 0.500 mol/L = 0.0276 mol
Step 3: Calculate the moles of H₂SO₄ that reacted with 0.0276 moles of KOH
The molar ratio of KOH to H₂SO₄ is 2:1. The reacting moles of H₂SO₄ are 1/2 × 0.0276 mol = 0.0138 mol
Step 4: Calculate the concentration of H₂SO₄
0.0138 moles of H₂SO₄ are in 27.4 mL (0.0274 L). The molarity of H₂SO₄ is:
[H₂SO₄] = 0.0138 mol/0.0274 L = 0.504 M
Hydrogen gas
Acid + metal = salt and hydrogen gas
