Answer:
<h3>option D</h3>
Explanation:
<h3>Is wire A connected to the light bulb </h3>
<h3>because it is series connection</h3>
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
Always younger than the rock it intrudes
Answer: Rutherford.
Explanation:
It was the scientist Ernest Rutherford who, by 1911, performed the gold foil experiment in which α particles were shoot to a thin foild of gold.
That experiment showed that although most α particles passed through the thin gold foild, some of them were deviated in small angles and some other were bounced backward.
The conclusion of the experiment was that the atom contained a small dense positively charged nucleous and negative particles (electrons) surroundiing the nucleous. Being the space in between the nucleous and the electrons empty.
Before Rutherford's experiment the model of the atom was that of the plum pudding presented by J.J Thomson, in which the atom was a solid positively charged sphere with embeded negative charge uniformly distributed in it.
