In order to solve this problem, we'll need to break it up into steps.

Step 1: Calculate the amount of energy necessary to raise the sample of ice from \(\displaystyle -35^{o}C\) to \(\displaystyle 0^{o}C\). To do this, we'll need to use the following equation:

\(\displaystyle q_{1}=mc_{H_{2}O(s)}\Delta T\)

\(\displaystyle q_{1}=( 20\cdot 10^{3}g)\left ( 2.06\frac{J}{g^{0}C} \right )\left ( 35^{o}C \right )=1.44\cdot 10^{6}J\)

Step 2: Calculate the amount of energy necessary to convert the sample at \(\displaystyle 0^{o}C\) from ice to water. We'll need to make use of the following equation:

\(\displaystyle q_{2}=m\Delta H_{fus(H_{2}O)}\)

\(\displaystyle q_{2}=\left ( 20\cdot 10^{3}g \right )\left ( 334.16\frac{J}{g} \right )=6.68\cdot 10^{6}J\)

Step 3: Calculate the amount of energy necessary to convert the sample of water from \(\displaystyle 0^{o}C\) to \(\displaystyle 75^{o}C\).

\(\displaystyle q_{3}=mc_{H_{2}O(l)}\Delta T\)

\(\displaystyle q_{3}=\left ( 20\cdot 10^{3}g \right )\left ( 4.184\frac{J}{g^{o}C} \right )\left ( 75^{o}C \right )=6.28\cdot 10^{6}J\)

Step 4: Sum the amount of energy from the previous 3 steps. This value is the total amount of energy for the entire process.

\(\displaystyle q_{Total}=q_{1}+q_{2}+q_{3}=1.44\cdot 10^{7}J\)

Step 5: Now that we know the total amount of energy needed for the process, we need to calculate the time based on the amount of power provided.

\(\displaystyle P=\frac{q_{Total}}{t}\)

\(\displaystyle t=\frac{q_{Total}}{P}=\frac{\left ( 1.44\cdot 10^{7}J \right )}{(100\frac{J}{s}) }=1.44\cdot 10^{5}s\)

\(\displaystyle t=(1.44\cdot10^{5}s)\left ( \frac{1hr}{3600s} \right )=40hrs\)

π bonds occur when there is greater than a single bond (double or triple bond). The only compound listed with double bonds or greater is CO₂, meaning it is the one that contains the most π bonds.

Butane has 13 sigma bonds. Ethane has 7 sigma bonds. Benzene has 12 sigma bonds. Lithium Hydroxide has 2 sigma bonds and water has 2 sigma bonds.

Triple bonds involve sharing a total of six electrons. They are shortest in length and store the most energy, making them difficult to break. These properties make triple bonds stronger than double or single bonds.

Coordinate covalent bonds form when one atom contributes two electrons to be shared between nuclei, as opposed to each atom sharing a single electron. Once formed, coordinate covalent bonds have essentially identical properties to any other single bonds.

Diatomic nitrogen, _{\(\displaystyle N_2\)}, contains a triple bond between the two nitrogen atoms. Since there are six electrons being shared, the bond is stronger and the atoms are being pulled closer together. Diatomic oxygen, \(\displaystyle O_2\),  has a double bond, which is not as strong. Diatomic hydrogen, \(\displaystyle H_2\), has a single bond, which have a bond length greater than either a double or triple bond. Hydrochloric acid, \(\displaystyle HCl\), is an ionic compound and will have a bond length comparable to a single covalent bond.

Ag(NH₃)₃ has covalent bonds (N-H) and ionic bonds (Ag-N).

A hydrogen bond is an intermolecular force between a hydrogen atom and an electronegative atom (O, N, F) that are in close proximity. Hydrogen bonds can occur within a given molecule (intramolecular), but more commonly occur between adjacent molecules (intermolecular). A solution of Ag(NH₃)₃ would show hydrogen bonding between the hydrogen and nitrogen of nearby molecules. The compound would not show intramolecular hydrogen. Since the question asks only for intramolecular interactions (single molecule), hydrogen bonding cannot be an answer choice.

We know from Lewis structures that diatomic hydrogen will have a single bond, diatomic oxygen will have a double bond, and diatomic nitrogen will have a triple bond. Triple bonds are stronger than double or single bonds, and the stronger the bond, the shorter the bond length. Nitrogen will have the shortest bond length.

We also know that the longest bond will be weakest. This mean hydrogen, with its single bond, would have the longest bond length. The answer should be N₂ < O₂ < H₂.

Out of the five given functional groups, only alcohols do not contain a central atom which is double bonded (meaning those atoms have \(\displaystyle sp^{2}\) hybridization). The only central atom of an alcohol, the oxygen atom in \(\displaystyle R-OH\), has single bonds to the R-group and a single hydrogen along with two lone pairs, giving it \(\displaystyle sp^{3}\) hybridization.

The alkyne, a hydrocarbon containing a triple bond, has one sigma \(\displaystyle sp-sp\) bond, and two pi bonds, making a rod-like molecule with a bond angle of \(\displaystyle 180^{\textup o}\) and \(\displaystyle sp\) hybridization. It is the only type of compound of the choices to have a triple bond, making it the only possibility.

The \(\displaystyle COO\) group indicated by the question consists of a carbon atom bonded to two oxygen atoms (one with a double bond, and one with a single bond). The carbon is also bonded to the rest of the hydrocarbon chain by a single sigma bond. Because this carbon atom only has three electron domains and has no lone pairs, trigonal planar is the molecular geometry that best describes this piece of the larger molecule. The atoms bonded to this carbon molecule sit at \(\displaystyle 120^{\text o}\) from each other, an even division of the single plane in which these atoms sit.