Which of the following correctly describes globalization? (1 point)
1 answer:
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Answer:
The change in the pH (ΔpH) is 2,17
Explanation:
The reaction:
CH₃NH₂(aq) + H₂O(aq) ⇌ CH₃NH₃⁺(aq) + OH⁻
<em>(1)</em>
In equilibrium, a solution of CH₃NH₂ 4,7M produces:
[CH₃NH₂] = 4,7 - x
[CH₃NH₃⁺] = x
[OH⁻] = x
Replacing in (1):
x² + 4,38x10⁻⁴x - 2,0586x10⁻³ = 0
The solutions are:
x = -0,0456 No physical sense. There are not negative concentrations.
x = 0,04515 Real answer.
The concentration of [OH⁻] is 0,04515 M.
As pOH = -log [OH⁻] And pH+pOH = 14. The pH of this solution is:
<em>pH = 12,65</em>
The addition of 6,7M produce this changes in concentrations:
[CH₃NH₂] = 4,656 + x
[CH₃NH₃⁺] = 6,74515 - x
[OH⁻] = 0,04515 - x
Replacing in (1) you will obtain:
x² - 6,7907x + 0,3025 = 0
Solving for x:
x = 6,74586 No physical sense
x = 0,04484 Real answer.
Thus, [OH⁻] = 0,04515 - 0,044842 = 3,08x10⁻⁴M
pOH = 3,51.
<em>pH = 10,49</em>
Thus ΔpH is 12,65 - 10,49 = <em>2,16 ≈ 2,17</em>
I hope it helps!
Answer:
Explanation:
A radioactive isotope is an isotope that undergoes nuclear decay, breaking apart into a smaller nucleus and emitting radiation during the process.
The half-life of an isotope is the amount of time it takes for a certain quantity of a radioactive isotope to halve.
For a radioactive isotope, the amount of substance left after a certain time t is:
(1)
where
is the mass of the substance at time t = 0
m(t) is the mass of the substance at time t
is the half-life of the isotope
In this problem, the isotope is uranium-235, which has a half-life of
We also know that the amount of uranium left in the rock sample is 6.25% of its original value, this means that
Substituting into (1) and solving for t, we can find how much time has passed: