Which action would shift this reaction away from solid calcium fluoride and toward the dissolved ions?a) adding calcium ionsb) a
1 answer:
Answer:
c) removing fluoride ions.
Explanation:
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In this case, for this solubility product problem, it is very convenient to write the undergoing ionization reaction:
In such a way, via the Le Chatelier's principle, we remember that the addition of reactants shifts the reaction towards products, adding products shifts the reaction towards reactants, removing reactants shifts the reaction towards reactants and removing products shifts the reaction towards products. In means, that the best way to shift this reaction away from solid calcium fluoride and toward the dissolved ions is c) removing fluoride ions, because it is a product and its removal favors the formation of more products.
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Answer:
Explanation:
Group 4A contains a total of 4 electrons for each atom in their valence shell. Filling the orbital diagram, let's say, for carbon, notice that when we start with period 2, we have two elements in the s-block, that is, lithium and beryllium. They correspond to the two s electrons that belong to the valence shell of carbon.
Moving on, we have boron and carbon, the remaining 2 electrons. Now, starting with boron, we're in the p-block.
That said, looking at the second period, the electron configuration for the valence shell of a group 4A element would be: