Answer:
Assign oxidation numbers to all atoms in the equation.
Compare oxidation numbers from the reactant side to the product side of the equation.
The element oxidized is the one whose oxidation number increased.
Explanation:
<span>Which compounds, in the list provided below, have double or triple bonds?
The answer is : N2</span>
Answer:
![MgI_2\\\\MgO\\\\CrI_4\\\\ CrO_2](https://tex.z-dn.net/?f=MgI_2%5C%5C%5C%5CMgO%5C%5C%5C%5CCrI_4%5C%5C%5C%5C%20CrO_2)
Explanation:
Hello!
In this case, we can notice that among the given ions, Mg2+ and Cr4+ are cations and I- and O2- are anions, thus, for each resulting compound we switch the oxidation states as subscripts to get:
![Mg^{2+}I^- \rightarrow MgI_2\\\\Mg^{2+}O^{2-} \rightarrow MgO\\\\Cr^{4+}I^-\rightarrow CrI_4\\\\Cr^{4+}O^{2-}\rightarrow CrO_2](https://tex.z-dn.net/?f=Mg%5E%7B2%2B%7DI%5E-%20%5Crightarrow%20MgI_2%5C%5C%5C%5CMg%5E%7B2%2B%7DO%5E%7B2-%7D%20%5Crightarrow%20MgO%5C%5C%5C%5CCr%5E%7B4%2B%7DI%5E-%5Crightarrow%20CrI_4%5C%5C%5C%5CCr%5E%7B4%2B%7DO%5E%7B2-%7D%5Crightarrow%20CrO_2)
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A long chain of hydrocarbon bonded to COOH is a FATTY acid.