Answer: (a) The solubility of CuCl in pure water is .
(b) The solubility of CuCl in 0.1 M NaCl is .
Explanation:
(a) Chemical equation for the given reaction in pure water is as follows.
Initial: 0 0
Change: +x +x
Equilibm: x x
And, equilibrium expression is as follows.
x =
Hence, the solubility of CuCl in pure water is .
(b) When NaCl is 0.1 M,
,
,
Net equation:
= 0.1044
So for,
Initial: 0.1 0
Change: -x +x
Equilibm: 0.1 - x x
Now, the equilibrium expression is as follows.
K' =
0.1044 =
x =
Therefore, the solubility of CuCl in 0.1 M NaCl is .
Given a sample of three unknown solids, we would be able to deduce which are ionic solids by their melting and boiling points, conductivity in water, and the insulation they provide.
<h3>What are Ionic solids?</h3>
- An ionic solid is a solid formed by ionic bonds.
- They are the solid version of ionic compounds, which are held together by strong electrostatic energies.
<h3>What properties do they possess?</h3>
- Among their many properties, ionic solids possess high melting and boiling points.
- These solids are also known to conduct electricity well <em><u>when </u></em><em><u>dissolved </u></em><em><u>into </u></em><em><u>water</u></em>, however, this can be tested after melting them as they <u>will not conduct electricity in a solid state</u>.
- A final property that we can use to identify them is that they provide a good means of insulation.
- Each of these properties can be tested and will allow us to determine the ionic solids amongst the samples.
Therefore, we can find the ionic solids amongst the samples by testing their melting and boiling points, conductivity in water, and the insulation they provide, given that each of these should provide high values to indicate an ionic solid.
To learn more about ionic solids visit:
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Answer:
It's not the same, due to the fact that Substance B has a medium attraction and Substance C has a low attraction
Explanation:
Answer is: iron is oxidized to form rust.
Chemical reaction: Fe + O₂ → Fe₂O₃.
Oxidation half reaction: Fe⁰ → Fe⁺³ + 3e⁻ /×4; 4Fe⁰ → 4Fe⁺³ + 12e⁻.
Reduction half reaction: O₂⁰+ 4e⁻ → 2O⁻²/×3; 3O₂⁰+ 12e⁻ → 6O⁻².
Balanced chemical reaction: 4Fe + 3O₂ → 2Fe₂O₃.
Oxidation is increase of oxidation number, iron is oxidized from oxidation number 0 (Fe) to oxidation number +3 (in rust Fe₂O₃).