Answer:
1. Mg (s) + 2Na+(aq) → 2Na(s) + Mg²⁺(aq)
2. 2K(s) + Cd²⁺(aq) → 2K⁺(aq) + Cd(s)
Explanation:
The net ionic equation of a reaction express only the chemical species that are involved in the reaction:
1. Mg (s) + Na2CrO4 (aq) → 2Na + MgCrO4(aq)
The ionic equation:
Mg (s) + 2Na+(aq) + CrO4²⁻ (aq) → 2Na + Mg²⁺ + CrO4²⁻(aq)
Subtracting the ions that don't change:
<h3>Mg (s) + 2Na+(aq) → 2Na + Mg²⁺</h3>
2. 2K(s) + Cd(NO3)2(aq) → 2KNO3(aq) + Cd(s)
The ionic equation:
2K(s) + Cd²⁺(aq) + 2NO3⁻(aq) → 2K⁺(aq) + 2NO3⁻(aq) + Cd(s)
Subtracting the ions that don't change:
<h3>2K(s) + Cd²⁺(aq) → 2K⁺(aq) + Cd(s)</h3>
Answer : Al
Explanation :
- For a given redox reaction there must be an oxidizing agent/(s) and a reducing agent/(s).
- the oxidizing agent is the substance that undergoes reduction process (gaining electrons or loss of oxygen atoms), meanwhile, the reducing agent is the substance that undergoes oxidation process (loss of electrons or gaining of oxygen atoms).
- In the reaction above, the oxidation number of (Al) in AlCl3 is (3+). However, the oxidation number of (Al) in the products is zero because it exists as a single element.
Therefore, changing from (3+) to zero means gaining of (3) electrons to neutralize the previously existing (3) protons on (Al) in AlCl3.
So Al is the oxidizing agent..
Answer:
Be Electron configuration: 1s2 2s2
Be Orbital Diagram: \//\ \//\ (it would be little arrows going up and down to show the spins)
F Electron Configuration: 1s2 2s2 2p5
F Orbital Diagram: \//\ \//\ \//\ \//\ \/
Cathode Ray experiment to figure out electrons exist -> atoms can be broken up into subatomic particles
I think so... I'm currently learning this too but you should be correct
