It creates chlorofluorocarbons(CFCs) also would create biooxygen but in the multiple choice it only shows the CFCs
<h3><u>Answer and explanation</u>;</h3>
- <em><u>The isotope U-235 is an important common nuclear fuel because under certain conditions it can readily be split, yielding a lot of energy. It is therefore said to be 'fissile' and use the expression 'nuclear fission'.</u></em>
- <em><u>Uranium 238 on the other hand is not fissionable by thermal neutrons, but it can undergo fission from fast or high energy neutrons. Hence it is not fissile, but it is fissionable.</u></em>
- In a nuclear power station fissioning of uranium atoms replaces the burning of coal or gas. Heat created by splitting the U-235 atoms is then used to make steam which spins a turbine to drive a generator, producing electricity.
The volume of one mole of any gas at Standard Temperature and Pressure (1 atm and 0 degrees Celsius [273K]) is 22.4 L.
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
