Answer:
<em><u>Copper </u></em><em><u>(</u></em><em><u>Cu2)</u></em><em><u> </u></em><em><u>,</u></em><em><u> </u></em><em><u>Iron </u></em><em><u>(</u></em><em><u>Fe2+</u></em><em><u> </u></em><em><u>Fe3 </u></em><em><u>+</u></em><em><u>)</u></em><em><u> </u></em><em><u>,</u></em><em><u> </u></em><em><u>and </u></em><em><u>Hydrogen </u></em><em><u>ion </u></em><em><u>(</u></em><em><u>H+</u></em><em><u>)</u></em>
Explanation:
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The answer is letter A definitively .
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 s orbitals are not symmetrical in shape is a FALSE statement.
An s orbital is so symmetric, more specifically spherically symmetric that it looks the same from all directions.
- The atomic orbitals in the atoms of elements differ in shape.
In essence, the electrons they describe have varying probability distributions around the nucleus. The spherical symmetry of s orbitals is evident in the fact that all orbitals of a given shell in the hydrogen atom have the same energy.
- All s orbitals are spherically symmetrical. Put simply, an electron that occupies an s orbital can be found with the same probability at any orientation (at a distance) from the nucleus.
The s orbitals are therefore represented by a spherical boundary surface which is a surface which captures a high proportion of the electron density.
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