We dissolve 10.0 g of each of the following in 800 g of water. In which solution would the molality of solute be highest? 1. All
1 answer:
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The answer is 1/8.
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>
The half-life of Sr-90 is 28.8 years.
So, we know:
t = 87.3 years
<span> = 28.8 years
We need:
n = ?
x = ?
</span>
We could first use the second equation, to calculate n:
<span>If:
,
</span>Then:
⇒
⇒
<span>⇒ n ≈ 3
</span>
Now we can use the first equation to calculate the remained amount 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>
The half-life of Sr-90 is 28.8 years.
So, we know:
t = 87.3 years
<span>
We need:
n = ?
x = ?
</span>
We could first use the second equation, to calculate n:
<span>If:
</span>Then:
⇒
⇒
<span>⇒ n ≈ 3
</span>
Now we can use the first equation to calculate the remained amount of the sample.
<span>
</span>⇒
⇒
</span>
Answer : The concentration of and
at equilibrium is, 0.0158 M and 0.00302 M respectively.
Explanation :
First we have to calculate the concentration of
Now we have to calculate the value of equilibrium constant (K).
The given chemical reaction is:
Initial conc. 0.0163 0.00415 0.00276
At eqm. (0.0163-2x) (0.00415+x) (0.00276+x)
As we are given:
Concentration of at equilibrium = 0.00467 M
That means,
(0.00415+x) = 0.00467
x = 0.00026 M
Concentration of at equilibrium = (0.0163-2x) = (0.0163-2(0.00026)) = 0.0158 M
Concentration of at equilibrium = (0.00276+x) = (0.00276+0.00026) = 0.00302 M