3 Chlorine ions are required to bond with one aluminum ion.
In ionic bonds, metals atoms loses all its outermost shell electrons to form a cation. While, non metal atoms gains however many electrons in order to make its outermost electron shell be 8 (or 2 if there's only one shell).
Therefore, form the periodic table, we can see that aluminum has a atomic number of 13, which makes its electron arrangement be 2,8,3. So, in order to form a aluminum ion, an Al atom must lose 3 electrons. On the other hand, Chlorine has a atomic number of 17, which means it has the electron configuration of 2,8,7. It has to gain only 1 electron to have 8 outermost shell electron.
Thereofre, 3 Chlorine atom are required to gain all 3 electrons given out by just 1 aluminum ion.
Electrolysis and heat can separate a compound into its elements.
I hope this helped you!
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Answer:
Cl2 + 2NaBr --> 2NaCl + Br2
Explanation:
This is a single displacement reaction where one side of the ionic compound switches with the other.
So, Cl2 + NaBr ---> NaCl + Br2
This isolates the Bromine and puts the Chlorine in it's place.
Then, balance out the equation like so and you should get
Cl2 + 2NaBr --> 2NaCl + Br2
