The fraction of the original amount remaining is closest to 1/128
<h3>Determination of the number of half-lives</h3>
- Half-life (t½) = 4 days
- Time (t) = 4 weeks = 4 × 7 = 28 days
- Number of half-lives (n) =?
n = t / t½
n = 28 / 4
n = 7
<h3>How to determine the amount remaining </h3>
- Original amount (N₀) = 100 g
- Number of half-lives (n) = 7
- Amount remaining (N)=?
N = N₀ / 2ⁿ
N = 100 / 2⁷
N = 0.78125 g
<h3>How to determine the fraction remaining </h3>
- Original amount (N₀) = 100 g
- Amount remaining (N)= 0.78125 g
Fraction remaining = N / N₀
Fraction remaining = 0.78125 / 100
Fraction remaining = 1/128
Learn more about half life:
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Answer:
Explanation:
Hello,
This problem could be solved in terms of normality as follows:
- The normality of phosphoric acid (triporitic) is:
Thus, the normality of barium hydroxide turns out:
Finally, the molarity:
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The maximum of electron an orbital in the f sublevel can hold is 2
The are 7 orbitals in the f sublevel
The entire sublevel can hold up to 14 electrons
long story short, the answer is c. 2
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