Q = mcΔT = (4.00 g)(0.129 J/g•°C)(40.85 °C - 0.85 °C)
Q = 20.6 J of energy was involved (more specifically, 20.6 J of heat energy was absorbed from the surroundings by the sample of solid gold).
1. put goggles on
2.but gloves on
3.put a lab coat on
Answer:
There were originally 8 atoms of Potassium-40.
Explanation:
The half-life of a radioactive material is the time taken for half the original material to decay or the time required for a quantity of the radioactive substance to reduce to half of its initial value.
If the original material formed without any Argon-40, it means that the atoms originally present were Potassium-40 atoms.
Presently, there are 7 Argon-40 atoms for every 1 of Potassium-40, we can deduce the number of half-lifes the Potassium-40 has undergone as follows :
After one half-life, (1/2) there will be one Potassium-40 atom for every Argon-40 atom.
After a second half life, 1/2 × 1/2 = 1/4: there will be one Potassium-40 atom for every three atoms of Argon-40.
After a third half-life, 1/4 × 1/2 = 1/8: there will be one Potassium-40 atom for every 7 atoms of Argon-40.
Since there are 1/8 atoms of Potassium-40 presently, there were originally 8 atoms of Potassium-40.
The answer is: <span>supersaturation</span>
