Answer:
pH = 13.1
Explanation:
Hello there!
In this case, according to the given information, we can set up the following equation:
Thus, since there is 1:1 mole ratio of HCl to KOH, we can find the reacting moles as follows:
Thus, since there are less moles of HCl, we calculate the remaining moles of KOH as follows:
And the resulting concentration of KOH and OH ions as this is a strong base:
![[KOH]=[OH^-]=\frac{0.00576mol}{0.012L+0.032L}=0.131M](https://tex.z-dn.net/?f=%5BKOH%5D%3D%5BOH%5E-%5D%3D%5Cfrac%7B0.00576mol%7D%7B0.012L%2B0.032L%7D%3D0.131M)
And the resulting pH is:
Regards!
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>
I don’t know the author or what book/literary work this is, but could be how proteins are produced.
A^2 + b^ = c^2
peak is 12,740, that's the highest point. the climber goes down so the leg is 200 then solve
