Which process passes genetic material to offspring
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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>
Answer and Explanation:
For the following balanced reaction:
PCl₅(g) ↔ PCl₃(g) + Cl₂(g)
We can see that all reactants and products are gases, so it is an homogeneous equilibrium. The expression for the equilibrium constant Kp can be written from the partial pressures (P) of reactants and products as follows:
Where PPCl₃ is the partial pressure of PCl₃ (reactant), PCl₂ is the partial pressure of Cl₂ (reactant) and PPCl₅ is the partial pressure of PCl₅ (product).