Which element(s) are not balanced in this equation?
1 answer:
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Answer : The correct expression will be:
Explanation :
The chemical reactions are :
(1)
(2)
The final chemical reaction is :
Now we have to calculate the value of for the final reaction.
Now equation 1 is multiply by 2 and then add both the reaction we get the value of 'K'.
If the equation is multiplied by a factor of '2', the equilibrium constant will be the square of the equilibrium constant of initial reaction.
If the two equations are added then equilibrium constant will be multiplied.
Thus, the value of 'K' will be:
The answer is <span>(3) 3 × 12.4 hours
</span>
To calculate this, we will use two equations:
where:
<span>n - number of half-lives
</span>x - remained amount of the sample, in decimals
<span>
- half-life length
</span>t - total time elapsed.
First, we have to calculate x and n. x is <span>remained amount of the sample, so if at the beginning were 16 grams of potassium-42, and now it remained 2 grams, then x is:
2 grams : x % = 16 grams : 100 %
x = 2 grams </span>× 100 percent ÷ 16 grams
x = 12.5% = 0.125
Thus:
<span>
</span>
It is known that the half-life of potassium-42 is 12.36 ≈ 12.4 hours.
Thus:
<span>
</span><span>
</span>
Therefore, it must elapse 3 × 12.4 hours <span>before 16 grams of potassium-42 decays, leaving 2 grams of the original isotope</span>
</span>
To calculate this, we will use two equations:
where:
<span>n - number of half-lives
</span>x - remained amount of the sample, in decimals
<span>
</span>t - total time elapsed.
First, we have to calculate x and n. x is <span>remained amount of the sample, so if at the beginning were 16 grams of potassium-42, and now it remained 2 grams, then x is:
2 grams : x % = 16 grams : 100 %
x = 2 grams </span>× 100 percent ÷ 16 grams
x = 12.5% = 0.125
Thus:
<span>
</span>
It is known that the half-life of potassium-42 is 12.36 ≈ 12.4 hours.
Thus:
<span>
</span><span>
</span>
Therefore, it must elapse 3 × 12.4 hours <span>before 16 grams of potassium-42 decays, leaving 2 grams of the original isotope</span>