Answer:
d. Sum of product enthalpies minus the sum of reactant enthalpies
Explanation:
The standard enthalpy change of a reaction (ΔH°rxn) can be calculated using the following expression:
ΔH°rxn = ∑n(products) × ΔH°f(products) - ∑n(reactants) × ΔH°f(reactants)
where,
ni are the moles of products and reactants
ΔH°f(i) are the standard enthalpies of formation of products and reactants
D = m / V
It even gives you the density of gold in the problem. Major hint. Once you know the volume (using V = m / D) then you can calculate the height (thickness) from the equation...
V = L x W x H
Volume = Length x Width x Height
start by converting 200.0 mg into grams
1000 mg = 1 g
200. mg x (1 g / 10^3 mg) = 0.200 g
V = m / D
V = 0.200 g / (19.32 g/cm^3)
V = 0.01035 cm^3
Convert 2.4 ft and 1 ft to cm
2.4 ft x (12 in / 1 ft) x (2.54 cm / 1 in) = 73.15 cm
1 ft = 30.48 cm
Compute the height (thickness)
V = LxWxH
H = V / LW = 0.01035 cm^3 / 73.15 cm / 30.48 cm
H = 4.64 x 10^-6 cm
Convert to nanometers
4.64 x 10^-6 cm x (1 m / 100 cm) x (10^9 nm / 1 m) = 46.4 nm
Knowing the atomic radius of gold, I might have asked my students for the minimum number of gold atoms in this thickness of gold. This would assume that the gold atoms are all in a row. This would give the minimum number of gold atoms.
Atomic radius gold = 174 pm
Diameter = 348 pm
46.4 nm x (1 m / 10^9 nm) x (10^12 pm / 1 m) x (1 Au atom / 248 pm) = 133 atoms of gold
I believe this question has the following five choices to
choose from:
>an SN2 reaction has occurred with inversion of
configuration
>racemization followed by an S N 2 attack
>an SN1 reaction has taken over resulting in inversion
of configuration
>an SN1 reaction has occurred due to carbocation
formation
>an SN1 reaction followed by an S N 2 “backside”
attack
The correct answer is:
an SN1 reaction has occurred due to carbocation formation
