Answer:
Reaction Quotient, Kq = {[a-ketoglutarate]x[L-alanine]}/{[L-glutamate]x[pyruvate]}
or, Kq = {(1.6x10-2)x(6.25x10∧-3)}/{(3x10∧-5)x(3.3x10-4)} = 1.01 x 10∧4
Since Kq > Keqb ; therefore the reaction will proceed in the backward direction, in order words the reaction will not occur in forward direction. i.e formation of reactants will be favored.
Explanation:
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.
Answer:
See the answer below
Explanation:
<em>Since the experiment is set out to determine the melting point of the white solid, after missing the melting point due to distraction, there are two possible solutions and both involves a repeat of the experiment.</em>
1. The first one is to allow the molten substance to solidify again and then repeat the experiment. This time around, a critical attention should be paid to be able to notice the melting point temperature once the temperature gets to 132 C.
2. The second solution would be discard the molten substance and repeat the experiment with the a new solid one. Similarly, critical attention should be paid once the temperature gets to 132 C since it is sure that the melting point lies within 132 and 138 C.