Please help me! Two groups of students were tested to compare their speed working math problems. Each group was given the same p
1 answer:
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Answer:
Choice B. The solid with hydrogen bonding.
Assumption: the molecules in the four choices are of similar sizes.
Explanation:
Molecules in a molecular solid are held intact with intermolecular forces. To melt the solid, it is necessary to overcome these forces. The stronger the intermolecular forces, the more energy will be required to overcome these attractions and melt the solid. That corresponds to a high melting point.
For molecules of similar sizes,
- The strength of hydrogen bonding will be stronger than the strength of dipole-dipole attractions.
- The strength of dipole-dipole attractions (also known as permanent dipole) will be stronger than the strength of the induced dipole attractions (also known as London Dispersion Forces.)
That is:
Strength of Hydrogen bond > Strength of Dipole-dipole attractions > Strength of Induced dipole attractions.
Accordingly,
Melting point due to Hydrogen bond > Melting point due to Dipole-dipole attractions > Melting point due to Induced Dipole attractions.
- Induced dipole is possible between all molecules.
- Dipole-dipole force is possible only between polar molecules.
- Hydrogen bonds are possible only in molecules that contain
atoms that are bonded directly to atoms of
,
, or
.
As a result, induced dipoles are the only force possible between molecules of the solid in choice C. Assume that the molecules are of similar sizes, such that the strengths of induced dipole are similar for these molecules.
Melting point in choice B > Melting point in choice D > Melting point in choice A and C.
HClO is a weak acid, which means the ions do not fully dissociate. The hydrolysis reaction for the hypochlorous acid is:
HClO + H2O ⇄ H3O+ +OCl-
Then the equilibrium constant, Ka, of dilute HClO would be:
![K_{a} = \frac{[ H_{3} O^{+} ][O Cl^{-} ]}{HClO}](https://tex.z-dn.net/?f=%20K_%7Ba%7D%20%3D%20%5Cfrac%7B%5B%20H_%7B3%7D%20%20O%5E%7B%2B%7D%20%5D%5BO%20Cl%5E%7B-%7D%20%5D%7D%7BHClO%7D%20)
Then we do the ICE table. I is for the initial concentration, C for the change and E for the excess.
HClO + H2O ⇄ H3O+ + OCl-
I 1.0x10^-4 0 0
C -x +x +x
E (1.0x10^-4 - x) x x
Substituting the excess (E) concentration to the Ka equation:
![K_{a} = \frac{[x ][x]}{1.0 \ x \ 10^{-4} - x }](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%20%5Cfrac%7B%5Bx%20%5D%5Bx%5D%7D%7B1.0%20%5C%20x%20%5C%20%2010%5E%7B-4%7D%20-%20x%20%7D)
Simplifying the equation would yield a quadratic equation:
The Ka for HClO is an experimental data which was determined to be 2.9 x 10^-8. Substitute this to the equation, determine the roots, then you get the value for x, which is the concentration of H3O+ and ClO-. Just use your calculator feature Shift-Solve.
x = 1.688 x 10^-6 M = [H3O+] = [ClO-]
Then, you can determine the conc of [OH-] through pH.
pH = -log {H3O+] = -log [1.688 x 10^-6] = 5.77
pOH = 14 - pH = 14 - 5.77 = 8.23
pOH = 8.23 = -log [OH-]
[OH-] = 5.89 x 10^-9 M
Also, since HClO is (1.0x10^-4 - x), then it's concentration would be:
[HClO] = 1.0x10^-4 - 1.688 x 10^-6 = 9.83 x10^-5 M
Let's summarize all concentrations:
[HClO] = 9.83 x10^-5 M
[OH-] = 5.89 x 10^-9 M
[H3O+] = [ClO-] = 1.688 x 10^-6 M
Since the solution is dilute, H2O is relatively higher in concentration.
Thus in relative amounts, the order would be
H2O >>> HClO > H3O+ = ClO- > OH-
HClO + H2O ⇄ H3O+ +OCl-
Then the equilibrium constant, Ka, of dilute HClO would be:
Then we do the ICE table. I is for the initial concentration, C for the change and E for the excess.
HClO + H2O ⇄ H3O+ + OCl-
I 1.0x10^-4 0 0
C -x +x +x
E (1.0x10^-4 - x) x x
Substituting the excess (E) concentration to the Ka equation:
Simplifying the equation would yield a quadratic equation:
The Ka for HClO is an experimental data which was determined to be 2.9 x 10^-8. Substitute this to the equation, determine the roots, then you get the value for x, which is the concentration of H3O+ and ClO-. Just use your calculator feature Shift-Solve.
x = 1.688 x 10^-6 M = [H3O+] = [ClO-]
Then, you can determine the conc of [OH-] through pH.
pH = -log {H3O+] = -log [1.688 x 10^-6] = 5.77
pOH = 14 - pH = 14 - 5.77 = 8.23
pOH = 8.23 = -log [OH-]
[OH-] = 5.89 x 10^-9 M
Also, since HClO is (1.0x10^-4 - x), then it's concentration would be:
[HClO] = 1.0x10^-4 - 1.688 x 10^-6 = 9.83 x10^-5 M
Let's summarize all concentrations:
[HClO] = 9.83 x10^-5 M
[OH-] = 5.89 x 10^-9 M
[H3O+] = [ClO-] = 1.688 x 10^-6 M
Since the solution is dilute, H2O is relatively higher in concentration.
Thus in relative amounts, the order would be
H2O >>> HClO > H3O+ = ClO- > OH-