If 45.0 mL of a 0.0500 M HNO3, 10.0 mL of a 0.0500 M KSCN, and 30.0 mL of a 0.0500 M Fe(NO3)3 are combined, what is the initial
1 answer:
Answer:
the initial concentration of SCN- in the mixture is 0.00588 M
Explanation:
The computation of the initial concentration of the SCN^- in the mixture is as follows:
As we know that
As it is mentioned in the question that KSCN is present 10 mL of 0.05 M
So, the total milimoles of SCN^- is
= 10 × 0.05
= 0.5 m moles
The total volume in mixture is
= 45 + 10 + 30
= 85 mL
Now the initial concentration of the SCN^- is
= 0.5 ÷ 85
= 0.00588 M
hence, the initial concentration of SCN- in the mixture is 0.00588 M
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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.