Answer:
% = 76.75%
Explanation:
To solve this problem, we just need to use the expressions of half life and it's relation with the concentration or mass of a compound. That expression is the following:
A = A₀ e^(-kt) (1)
Where:
A and A₀: concentrations or mass of the compounds, (final and initial)
k: constant decay of the compound
t: given time
Now to get the value of k, we should use the following expression:
k = ln2 / t₁/₂ (2)
You should note that this expression is valid when the reaction is of order 1 or first order. In this kind of exercises, we can assume it's a first order because we are not using the isotope for a reaction.
Now, let's calculate k:
k = ln2 / 956.3
k = 7.25x10⁻⁴ d⁻¹
With this value, we just replace it in (1) to get the final mass of the isotope. The given time is 1 year or 365 days so:
A = 250 e^(-7.25x10⁻⁴ * 365)
A = 250 e^(-0.7675)
A = 191.87 g
However, the question is the percentage left after 1 year so:
% = (191.87 / 250) * 100
<h2>
% = 76.75%</h2><h2>
And this is the % of isotope after 1 year</h2>
(B.Hallow bones because bones are inside the body
Destornilladores, alicates de corte diagonal, alicates de punta fina, alicates de bomba ajustable y cinta métrica con punta magnética
<u>Answer:</u> The expression of
for calcium fluoride is ![K_{sp}=[Ca^{2+}][2F^-]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCa%5E%7B2%2B%7D%5D%5B2F%5E-%5D%5E2)
<u>Explanation:</u>
Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.
The given chemical equation follows:

1 mole of calcium fluoride produces 1 mole of calcium ions and 2 moles of fluorine ions.
The expression of
for above equation follows:
![K_{sp}=[Ca^{2+}][2F^-]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCa%5E%7B2%2B%7D%5D%5B2F%5E-%5D%5E2)
Hence, the expression of
for calcium fluoride is ![K_{sp}=[Ca^{2+}][2F^-]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCa%5E%7B2%2B%7D%5D%5B2F%5E-%5D%5E2)