Answer:
To 318.18 mL would you need to dilute 20.0 mL of a 1.40 M solution of LiCN to make a 0.0880 M solution of LiCN
Explanation:
Dilution is the reduction of the concentration of a chemical in a solution and consists simply of adding more solvent.
In a dilution the amount of solute does not vary. But as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.
In a solution it is fulfilled:
Ci* Vi = Cf* Vf
where:
- Ci: initial concentration
- Vi: initial volume
- Cf: final concentration
- Vf: final volume
In this case:
- Ci= 1.40 M
- Vi= 20 mL
- Cf= 0.088 M
- Vf= ?
Replacing:
1.40 M* 20 mL= 0.088 M* Vf
Solving:

Vf= 318.18 mL
<u><em>To 318.18 mL would you need to dilute 20.0 mL of a 1.40 M solution of LiCN to make a 0.0880 M solution of LiCN</em></u>
<u><em></em></u>
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>
Rate=[a]*([b]^2)*([c]^(1/2)]
rate=[2a]*([b]^2)*([2c]^(1/2)]= 2*(2^(1/2)[a]*([b]^2)*([c]
it increases times 2*(2^(1/2)=2√2