Considering the definition of pOH and strong base, the pOH of the aqueous solution is 1.14
The pOH (or potential OH) is a measure of the basicity or alkalinity of a solution and indicates the concentration of ion hydroxide (OH-).
pOH is expressed as the logarithm of the concentration of OH⁻ ions, with the sign changed:
pOH= - log [OH⁻]
On the other hand, a strong base is that base that in an aqueous solution completely dissociates between the cation and OH-.
LiOH is a strong base, so the concentration of the hydroxide will be equal to the concentration of OH-. This is:
[LiOH]= [OH-]= 0.073 M
Replacing in the definition of pOH:
pOH= -log (0.073 M)
<u><em>pOH= 1.14 </em></u>
In summary, the pOH of the aqueous solution is 1.14
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Answer:
<h3>The answer is 4.15 g/mL</h3>
Explanation:
The density of a substance can be found by using the formula

From the question
mass of object = 8.3 g
volume = final volume of water - initial volume of water
volume = 8 - 6 = 2 mL
So we have

We have the final answer as
<h3>4.15 g/mL</h3>
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Answer: After 4710 seconds, 1/8 of the compound will be left
Explanation:
Using the formulae
Nt/No = (1/2)^t/t1/2
Where
N= amount of the compound present at time t
No= amount of compound present at time t=0
t= time taken for N molecules of the compound to remain = 4710 seconds
t1/2 = half-life of compound = 1570 seconds
Plugging in the values, we have
Nt/No = (1/2)^(4710s/1570s)
Nt/No = (1/2)^3
Nt/No= 1/8
Therefore after 4710 seconds, 1/8 molecules of the compound will be left
Answer:
35.75 days
Explanation:
From the given information:
For first-order kinetics, the rate law can be expressed as:

Given that:
the rate degradation constant = 0.12 / day
current concentration C = 0.05 mg/L
initial concentration C₀ = 3.65 mg/L

㏑(0.01369863014) = -(0.12) t
-4.29 = -(0.12)
t = -4.29/-0.12
t = 35.75 days