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.
I believe it is the Great Plains in Nebraska.
Answer:
29.41% of Calcium and 47.04% of Oxygen
Explanation:
The percent composition of an atom in a molecule is defined as 100 times the ratio between the mass of the atom and the mass of the molecule.
The mass of the molecule of the problem (Ore) is 46.28g. That means the percent composition of Calcium is:
13.61g / 46.28g * 100 = 29.41% of Calcium
And percent composition of Oxygen is:
21.77g / 46.28g * 100 = 47.04% of Oxygen
Answer:
An element
Explanation:
Elements are not able to be separated, but everything else can, whether that be physically or chemically.
Explanation:
Contributing structures are the resonating structures which are formed due to the delocalization of electrons in a molecule.
The azide ion that is 
, is a symmetrical ion, all of whose contributing structures have formal charges.
Lone pair of central nitrogen atom in azide ion is in conjugation with the neighboring nitrogen atoms.
Contributing structures of azide ion are drawn in the image attached.
