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.
When Americium (Am-241) undergoes alpha decay(He-4) it forms neptunium (Np-237) based on the following pathway:
²⁴¹Am₉₅ → ²³⁷Np₉₃ + ⁴He₂
The energy released in given as:
ΔE = Δmc²
where Δm = mass of products - mass of reactants
= [m(Np-237) + m(He-4)] - [m(Am-241)]
= 237.0482+4.0015-241.0568 = -0.0073 g/mol = -7.3 * 10⁻⁶ kg/mol
ΔE = -7.3*10⁻⁶ kg/mol * (3*10⁸ m/s)² = -5.84*10¹¹ J/mol
Atomic theory is the scientific inquiry of the nature of matter, which is subject to change as new information is discovered. The answer is A.
We are given the resistance and voltage of this lamp and we are asked to find the current; the equation that relates these together is Ohm’s Law, V = IR. Simply plug and solve:
V = IR
(220 V) = I(484 Ohms)
I = 0.4545 Amps
The lamp has a current of 0.4545 Amps passing through it under these conditions.
Hope this helps!