Answer:
5
Explanation:
Sorry, don't have one
<u><em>Hope this helps and I get brainliest <3</em></u>
the compound with the smaller lattice energy is potassium sulphide here the size of the molecule play a important role
The quantity of energy released by the electrostatic attraction of oppositely charged ions is known as lattice energy (L.E). The ion's size and charge affect the lattice energy.
lattice energy is inversely proportional to size of ion and directly proportional to charge of the ion. They are each charged equally that is plus two and minus two However, because the Sulphur is larger in size and the oxygen is lesser in this case, The lattice energy of potassium oxide is larger the lattice energy of potassium sulphide is smaller.
To learn more about lattice energy :
brainly.com/question/18222315
#SPJ4
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
