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>
Answer:
i think it's D tbh, just cus it was the scientist who did the work
Answer:
The pressure will be 933.33 Kpa
Explanation:
Given that:
Volume V₁ = 200 cm³ (note, there is a mistake in the volume. It is supposed to be 200 cm³)
Pressure P₁ = 700 Kpa
Pressure P₂ = ??? (unknown)
Volume V₂ = 150 cm³
Temperature = constant
Using Boyle's law:
PV = constant
i.e.
P₁V₁ = P₂V₂
700 Kpa × 200 cm³ = P₂ × 150 cm³
P₂ = (700 Kpa × 200 cm³)/150 cm³
P₂ = 933.33 Kpa
