Answer:
Value for K would be too small. Less AgCl would dissolve due to the common ion effect due to the presence of Cl- in the water.
Explanation:
Think of this through the lenses of a shifting problem. Cl- ions are a product in this situation and increasing its concentration would shift the reaction back to the solid AgCl. In this specific case, due to Cl- ions, AgCl would dissolve less to maintain equilibrium and as a result, the concentrations of Ag+ and Cl- ions would be lower than normal making a smaller K value.
Answer:
number (12 is c) (13 is 5) (14 is 5)
Explanation:
The binding energy of the electrons (also known as the work function of the surface) is determined as 2.43 x 10⁻¹⁹ J.
<h3>Binding energy of the electrons</h3>
The binding energy of the electrons is also known as work function of the metal and it is calculated as follows;
Ф = E - K.E
where;
Ф = hf - 86.2 kJ/mol
Ф = hc/λ - 86.2 kJ/mol
Ф = (6.63 x 10⁻³⁴ x 3 x 10⁸ )/515 x 10⁻⁹ - 86.2 kJ/mol
Ф = 3.86 x 10⁻¹⁹ J - (86200 J/mol)/(6.02 x 10²³)
Ф = 3.86 x 10⁻¹⁹ J - 1.43 x 10⁻¹⁹ J
Ф = 2.43 x 10⁻¹⁹ J
Learn more about work function here: brainly.com/question/19427469
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Explanation:
A photon is a quantum of EM radiation. Its energy is given by E = hf and is related to the frequency f and wavelength λ of the radiation by
E=hf=hcλ(energy of a photon)E=hf=hcλ(energy of a photon),
where E is the energy of a single photon and c is the speed of light. When working with small systems, energy in eV is often useful. Note that Planck’s constant in these units is h = 4.14 × 10−15 eV · s.
Since many wavelengths are stated in nanometers (nm), it is also useful to know that hc = 1240 eV · nm.
These will make many calculations a little easier.
All EM radiation is composed of photons. Figure 1 shows various divisions of the EM spectrum plotted against wavelength, frequency, and photon energy. Previously in this book, photon characteristics were alluded to in the discussion of some of the characteristics of UV, x rays, and γ rays, the first of which start with frequencies just above violet in the visible spectrum. It was noted that these types of EM radiation have characteristics much different than visible light. We can now see that such properties arise because photon energy is larger at high frequencies.
