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.
Answer:
2.94
Explanation:
There is some info missing. I think this is the original question.
<em>A solution is prepared at 25 °C that is initially 0.38 M in chloroacetic acid (HCH₂ClCO₂), a weak acid with Ka= 1.3 x 10⁻³, and 0.44 M in sodium chloroacetate (NaCH₂CICO₂). Calculate the pH of the solution. Round your answer to 2 decimal places.</em>
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We have a buffer system formed by a weak acid (HCH₂ClCO₂) and its conjugate base (CH₂CICO₂⁻ coming from NaCH₂CICO₂). We can calculate the pH using the Henderson-Hasselbalch equation.
pH = pKa + log [CH₂CICO₂⁻]/[HCH₂ClCO₂]
pH = -log 1.3 x 10⁻³ + log (0.44 M/0.38 M)
pH = 2.94
Answer:?Any one have the answers?
Explanation:d
It is called a phenyl group
