464 g radioisotope was present when the sample was put in storage
<h3>Further explanation</h3>
Given
Sample waste of Co-60 = 14.5 g
26.5 years in storage
Required
Initial sample
Solution
General formulas used in decay:
t = duration of decay
t 1/2 = half-life
N₀ = the number of initial radioactive atoms
Nt = the number of radioactive atoms left after decaying during T time
Half-life of Co-60 = 5.3 years
Input the value :
Taking into account the definition of avogadro's number, 3.37×10⁻⁷ moles of methane are 20.32×10¹⁶ molecules.
First of all, you have to know that Avogadro's number indicates the number of particles of a substance (usually atoms or molecules) that are in a mole.
Its value is 6.023×10²³ particles per mole and it applies to any substance.
Then you can apply the following rule of three: if 6.023×10²³ molecules are contained in 1 mole of methane, then 20.32×10¹⁶ molecules are contained in how many moles of methane?
amount of moles of methane= (20.32×10¹⁶ molecules × 1 mole)÷ 6.023×10²³ atoms
Solving:
<u><em>amount of moles of methane= 3.37×10⁻⁷ moles</em></u>
Finally, 3.37×10⁻⁷ moles of methane are 20.32×10¹⁶ molecules.
Learn more about Avogadro's Number:
A because that honestly just makes the most sense
When it passes from one medium
Answer:
3 (three)
Explanation:
2 Fe + 3H2SO4 = Fe2(SO4)3 + 3 H2 (basically just balance both sides)
