Answer:
Option B. 4.74×10¯¹⁹ J.
Explanation:
The following data were obtained from the question:
Wavelength (λ) = 4.2×10¯⁷ m
Energy (E) =.?
Next, we shall determine the frequency of the wave. This can be obtained as follow:
Wavelength (λ) = 4.2×10¯⁷ m
Velocity (v) = constant = 3×10⁸ m/s
Frequency (f) =.?
v = λf
3×10⁸ = 4.2×10¯⁷ × f
Divide both side by 4.2×10¯⁷
f = 3×10⁸ / 4.2×10¯⁷
f = 7.143×10¹⁴ s¯¹
Therefore, the frequency of the wave is 7.143×10¹⁴ s¯¹.
Finally, we shall determine the energy of the wave using the following formula
E = hf
Where
E is the energy.
h is the Planck's constant
f is the frequency
Thus, the enery of the wave can be obtained as follow:
Frequency (f) = 7.143×10¹⁴ s¯¹.
Planck's constant = 6.63×10¯³⁴ Js
Energy (E) =..?
E = hf
E = 6.63×10¯³⁴ × 7.14×10¹⁴
E = 4.74×10¯¹⁹ J
Therefore, the energy of the wave is 4.74×10¯¹⁹ J.
Answer:
Density, melting point. and magnetic properties
Explanation:
I can think of three ways.
1. Density
The density of Cu₂S is 5.6 g/cm³; that of CuS is 4.76 g/cm³.
It should be possible to distinguish these even with high school equipment.
2. Melting point
Cu₂S melts at 1130 °C (yellowish-red); CuS decomposes at 500 °C (faint red).
A Bunsen burner can easily reach these temperatures.
3. Magnetic properties
You can use a Gouy balance to measure the magnetic susceptibilities.
In Cu₂S the Cu⁺ ion has a d¹⁰ electron configuration, so all the electrons are paired and the solid is diamagnetic.
In CuS the Cu²⁺ ion has a d⁹ electron configuration, so all there is an unpaired electron and the solid is paramagnetic.
A sample of Cu₂S will be repelled by the magnetic field and show a decrease in weight.
A sample of CuS will be attracted by the magnetic field and show an increase in weight.
In the picture below, you can see the sample partially suspended between the poles of an electromagnet.
The answer to your question is 40° because freezing temp for a liquid is 32°