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
Answer:
If
0.0357
mol
CaCl
2
is dissolved in water to make a
0.420
M
solution, what is the volume of the solution?
Answer: Mass of 
required to form 930 kg of iron is 1328 kg
Explanation:
To calculate the number of moles, we use the equation:
.....(1)
For iron:
Given mass of iron = 930 kg = 930000 g (1kg=1000g)
Molar mass of iron = 56 g/mol
Putting values in equation 1, we get:
The chemical equation for the production of iron follows:
By Stoichiometry of the reaction:
2 moles of iron are produced by = 1 mole of
So, 16607 moles of iron will be produced by = 
of
Now, calculating the mass of from equation 1, we get:
Mass of = 
Thus mass of required to form 930 kg of iron is 1328 kg
Answer:
three valence electrons
Explanation:
Gallium has three electrons in the outer energy level and therefore has three valence electrons. The identification of valence electrons is vital because the chemical behavior of an element is determined primarily by the arrangement of the electrons in the valence shell.
The answer is cumulus.
Stratus clouds often low and sometimes confused for fog. Cirrus clouds are thin and wispy.
