I would believe the answer to this question is D. According to the concept of the tragedy of the commons, shared resources are used by more than one organism. Due to the large consumption of shared resources they start to be fewer and fewer in number and over time if we are not careful they will be depleted.
scandium is a transition metal and shows a valency of 3.
Answer:
6.533 × 10^-21J
Explanation:
The energy of the microwave photon can be calculated using:
E = hf
Where;
E = energy of photon (J)
h = Planck's constant (6.626 × 10^-34 J/s)
f = frequency (9.86 x 10^12 Hz)
Hence, E = hf
E = 6.626 × 10^-34 × 9.86 x 10^12
E = 65.33 × 10^(-34 + 12)
E = 65.33 × 10^(-22)
E = 6.533 × 10^-21J
The energy of the microwave photon is
6.533 × 10^-21J
Mobile radio communication
Broadcasting
Navigation systems
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
