The answer is B, nic fiss, nic fusi
Answer:
![[F^-]_{max}=4x10{-3}\frac{molF^-}{L}](https://tex.z-dn.net/?f=%5BF%5E-%5D_%7Bmax%7D%3D4x10%7B-3%7D%5Cfrac%7BmolF%5E-%7D%7BL%7D)
Explanation:
Hello,
In this case, for the described situation, we infer that calcium reacts with fluoride ions to yield insoluble calcium fluoride as shown below:
Which is typically an equilibrium reaction, since calcium fluoride is able to come back to the ions. In such a way, since the maximum amount is computed via stoichiometry, we can see a 1:2 mole ratio between the ions, therefore, the required maximum amount of fluoride ions in the "hard" water (assuming no other ions) turns out:
![[F^-]_{max}=2.0x10^{-3}\frac{molCa^{2+}}{L}*\frac{2molF^-}{1molCa^{2+}} \\](https://tex.z-dn.net/?f=%5BF%5E-%5D_%7Bmax%7D%3D2.0x10%5E%7B-3%7D%5Cfrac%7BmolCa%5E%7B2%2B%7D%7D%7BL%7D%2A%5Cfrac%7B2molF%5E-%7D%7B1molCa%5E%7B2%2B%7D%7D%20%20%5C%5C)
Best regards.
Answer:
NaNO₃
Explanation:
A precipitate is a compound or a salt formed from a precipitation reaction and does not dissolve in water and therefore will exist in solid state.
From the choices given precipitation reaction will occur between;
- Fe(NO₃)₃(aq) + 3NaOH(aq) → Fe(OH)₃(s) + 3NaNO₃(aq)
- Cu(NO₃)₂(aq) + 2NaOH(aq) → Cu(OH)₂(s) + 2NaNO₃(aq)
- FeSO₄(aq) + 2NaOH(aq) → Fe(OH)₂(s) + Na₂SO₄(aq)
Fe(OH)₃, Cu(OH)₂, and Fe(OH)₂ are precipitates.
From the rules of solubility, hydroxides are insoluble except Ca(OH)₂ which is slightly soluble and hydroxides of ammonium and alkali metals.
Scientists make hypothesis in order to make an educated guess on the outcome of the experiment.
