Answer:
Elemental gold to have a Face-centered cubic structure.
Explanation:
From the information given:
Radius of gold = 144 pm
Its density = 19.32 g/cm³
Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:


a = 407 pm
In a unit cell, Volume (V) = a³
V = (407 pm)³
V = 6.74 × 10⁷ pm³
V = 6.74 × 10⁻²³ cm³
Recall that:
Net no. of an atom in an FCC unit cell = 4
Thus;


density d = 19.41 g/cm³
Similarly; For a body-centered cubic structure

where;
r = 144


a = 332.56 pm
In a unit cell, Volume V = a³
V = (332.56 pm)³
V = 3.68 × 10⁷ pm³
V 3.68 × 10⁻²³ cm³
Recall that:
Net no. of atoms in BCC cell = 2
∴


density =17.78 g/cm³
From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.
This makes the elemental gold to have a Face-centered cubic structure.
The final temperature, t₂ = 30.9 °C
<h3>Further explanation</h3>
Given
24.0 kJ of heat = 24,000 J
Mass of calorimeter = 1.3 kg = 1300 g
Cs = 3.41 J/g°C
t₁= 25.5 °C
Required
The final temperature, t₂
Solution
Q = m.Cs.Δt
Q out (combustion of compound) = Q in (calorimeter)
24,000 = 1300 x 3.41 x (t₂-25.5)
t₂ = 30.9 °C
Notice q=3/2, is half of the original q = 3(<span>1/2</span>)<span>t/28.8
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