Metallic gold crystallizes in a face-centered cubic lattice, with one au atom per lattice point. if the edge length of the unit
1 answer:
The metallic radius (r) of au is 144.2pm
A face-centered cubic unit cell structure is made up of atoms arranged in a cube with six additional whole atoms placed in the centre of each cube face and a fraction of an atom at each of the cube's four corners. Eight additional unit cells share the atoms at the cube's corner.
Atoms in an FCC crystal are in contact along the unit cell's diagonal. 4r = √2 a, where r is the radius of an atom and a is the edge length of a unit cell.
Putting the values in the above expression,we get
r = √2×408 / 4
r =
r =
r = 144.2 pm
Therefore, The metallic radius (r) of au is 144.2pm
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Answer : The heat change of the cold water in Joules is,
Explanation :
First we have to calculate the mass of cold water.
As we know that the density of water is 1 g/mL. The volume of cold water is 45 mL.
Now we have to calculate the heat change of cold water.
Formula used :
where,
Q = heat change of cold water = ?
m = mass of cold water = 45 g
c = specific heat of water =
= initial temperature of cold water =
= final temperature =
Now put all the given value in the above formula, we get:
Therefore, the heat change of cold water is