What should you do during a titration when you notice the indicator start to indicate the approach of the equilibrium point? Add
2 answers:
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The answer is 1/16.
Half-life is the time required for the amount of a sample to half its value.
To calculate this, we will use the following formulas:
1.
,
where:
<span>n - a number of half-lives
</span>x - a remained fraction of a sample
2.
where:
<span>
- half-life
</span>t - <span>total time elapsed
</span><span>n - a number of half-lives
</span>
So, we know:
t = 10 min
<span> = 2.5 min
We need:
n = ?
x = ?
</span>
We could first use the second equation to calculate n:
<span>If:
,
</span>Then:
⇒
⇒
<span>
</span>
Now we can use the first equation to calculate the remained fraction of the sample.
<span>
</span>⇒
<span>⇒
</span>
Half-life is the time required for the amount of a sample to half its value.
To calculate this, we will use the following formulas:
1.
where:
<span>n - a number of half-lives
</span>x - a remained fraction of a sample
2.
where:
<span>
</span>t - <span>total time elapsed
</span><span>n - a number of half-lives
</span>
So, we know:
t = 10 min
<span>
We need:
n = ?
x = ?
</span>
We could first use the second equation to calculate n:
<span>If:
</span>Then:
⇒
⇒
</span>
Now we can use the first equation to calculate the remained fraction of the sample.
<span>
</span>⇒
<span>⇒
Answer : The electron configurations consistent with this fact is, (b) [Kr] 4d¹⁰
Explanation :
Electronic configuration : It is defined as the representation of electrons around the nucleus of an atom.
Number of electrons in an atom are determined by the electronic configuration.
Paramagnetic compounds : They have unpaired electrons.
Diamagnetic compounds : They have no unpaired electrons that means all are paired.
The given electron configurations of Palladium are:
(a) [Kr] 5s²4d⁸
In this, there are 2 electrons in 's' orbital and 8 electrons in 'd' orbital. From the partial orbital diagrams we conclude that 's' orbital are paired but 'd' orbital are not paired. So, this configuration shows paramagnetic.
(b) [Kr] 4d¹⁰
In this, there are 10 electrons in 'd' orbital. From the partial orbital diagrams we conclude that electrons in 'd' orbital are paired. So, this configuration shows diamagnetic.
(c) [Kr] 5s¹4d⁹
In this, there are 1 electron in 's' orbital and 9 electrons in 'd' orbital. From the partial orbital diagrams we conclude that 's' orbital and 'd' orbital are not paired. So, this configuration shows paramagnetic.
