Determine the identity of a cube of metal that measures 1.2 cm on each side and has a mass of 15.4g.
2 answers:
<u>Answer:</u> The density of the cube of metal is
<u>Explanation:</u>
To calculate the volume of cube, we use the formula:
where,
a = edge length of cube
Edge length of cube = 1.2 cm
Volume of cube =
To calculate density of a substance, we use the equation:
We are given:
Mass of cube = 15.4 g
Volume of cube =
Putting values in above equation, we get:
Hence, the density of the cube of metal is
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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.