Answer:
a): not necessarily due to London Dispersion Forces and dipole-dipole interactions.
b): not necessarily due to London Dispersion Forces.
Explanation:
There are three major types of intermolecular interaction:
- Hydrogen bonding between molecules with H-O, H-N, or H-F bonds and molecules with lone pairs.
- Dipole-dipole interactions between all molecules.
- London dispersion forces between all molecules.
The melting point of a substance is a result of all three forces, combined.
Note that the more electrons in each molecule, the stronger the London Dispersion Force. Generally, that means the more atoms in each molecule, the stronger the London dispersion force. The strength of London dispersion force between large molecules can be surprisingly strong.
For example, 
(water) molecules are capable of hydrogen bonding. The melting point of at 
is around 
. That's considerably high when compared to other three-atom molecules.
In comparison, the higher alkane hexadecane (
, straight-chain) isn't capable of hydrogen bonding. However, under a similar pressure, hexadecane melts at around 
above the melting point of water. The reason is that with such a large number of atoms (and hence electrons) per molecule, the London dispersion force between hexadecane molecules could well be stronger than that the hydrogen bonding between water molecules.
Similarly, the dipole moments in HCl (due to the highly-polar H-Cl bonds) are much stronger than those in hexadecane (due to the C-H bonds.) However, the boiling point of hexadecane under standard conditions is much higher (at around 
than that of HCl.
Ir = [Xe] 4f14 5d7 6s2
<span>Rn = [Xe] 4f14 5d10 6s2 6p6 </span>
<span>Fe = [Ar] 3d6 4s2 </span>
<span>Pa = [Rn] 5f2 6d1 7s2 </span>
<span>Y = [Kr] 4d1 5s2 </span>
<span>Re = [Xe] 4f14 5d5 6s2 </span>
<span>So only Rhenium (Re) has an electronic configuration where the d orbital has 5 electrons in it.</span>
We convert the masses of our reactants to moles and use the stoichiometric coefficients to determine which one of our reactants will be limiting.
Dividing the mass of each reactant by its molar mass:
(10 g C2H6)(30.069 g/mol) = 0.3326 mol C2H6
(10 g O2)(31.999 g/mol) = 0.3125 mol O2.
Every 2 moles of C2H6 react with 7 moles of O2. So the number of moles of O2 needed to react completely with 0.3326 mol C2H6 would be (0.3326)(7/2) = 1.164 mol O2. That is far more than the number of moles of O2 that we are given: 0.3125 moles. Thus, O2 is our limiting reactant.
Since O2 is the limiting reactant, its quantity will determine how much of each product is formed. We are asked to find the number of grams (the mass) of H2O produced. The molar ratio between H2O and O2 per the balanced equation is 6:7. That is, for every 6 moles of H2O that is produced, 7 moles of O2 is used up (intuitively, then, the number of moles of H2O produced should be less than the number of moles of O2 consumed).
So, the number of moles of H2O produced would be (0.3125 mol O2)(6 mol H2O/7 mol O2) = 0.2679 mol H2O. We multiply by the molar mass of H2O to convert moles to mass: (0.2679 mol H2O)(18.0153 g/mol) = 4.826 g H2O.
Given 10 grams of C2H6 and 10 grams of O2, 4.826 g of H2O are produced.
this equation is balanced.
