Hello:
In this case, we will use the Clapeyron equation:
P = ?
n = 8 moles
T = 250 K
R = 0.082 atm.L/mol.K
V = 6 L
Therefore:
P * V = n * R * T
P * 6 = 8 * 0.082* 250
P* 6 = 164
P = 164 / 6
P = 27.33 atm
Hope that helps!
Greater rainfall has an greater increase on the rate of chemical weathering. Rain is a form of precipitation.
Answer:
210438+28+5+8+2+2+8+30+5+8+6+8+2+5+3+82+8+9+2+8+8+5+5+5+6+9+2+56+5+6+5+-3+8+8+58+5+6+8+5+8+5+35+6+3+8+5+3+68+6+8+6+88+8+!?=8+5+55+5+5+5+5+5+5+5+5+5+2+58+0+80=0=0=00=
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>
