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:
It was not until he was 50 years old, in 1859.
Explanation:
<em>malai</em><em> </em><em>aaudaina</em><em> </em><em>aati</em><em> </em><em>jabo</em><em> </em><em>tapai</em><em> </em>
<em>lai</em><em> </em><em>aaudaina</em><em> </em>
Answer:
4.704J
Explanation:
The following data were obtained from the question:
m = 0.080kg
h = 6.0m
g = 9.8m/s^2
P.E =?
P.E = mgh
P.E = 0.08 x 9.8 x 6
P.E = 4.704J
Therefore, the potential energy of the robin is 4.704J
It would be the molecule that can contain <span>How many formula units are in 0.25 mile of Na2o</span>the ester functinal
