The time taken for the isotope to decay is 46 million years.
We'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:
- Original amount (N₀) = 50.25 g
- Amount remaining (N) = 16.75
- Number of half-lives (n) =?
2ⁿ = N₀ / N
2ⁿ = N₀ / N
2ⁿ = 50.25 / 16.75
2ⁿ = 3
Take the log of both side
Log 2ⁿ = 3
nLog 2 = Log 3
Divide both side by log 2
n = Log 3 / Log 2
n = 2
Finally, we shall determine the time.
- Half-life (t½) = 23 million years
- Number of half-lives (n) = 2
t = n × t½
t = 2 × 23
t = 46 million years
Learn more about half-life: brainly.com/question/25927447
Answer:
C
Explanation:
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Answer:
- <em>The solution that has the highest concentration of hydroxide ions is </em><u>d. pH = 12.59.</u>
Explanation:
You can solve this question using just some chemical facts:
- pH is a measure of acidity or alkalinity: the higher the pH the lower the acidity and the higher the alkalinity.
- The higher the concentration of hydroxide ions the lower the acidity or the higher the alkalinity of the solution, this is the higher the pH.
Hence, since you are asked to state the solution with the highest concentration of hydroxide ions, you just pick the highest pH. This is the option d, pH = 12.59.
These mathematical relations are used to find the exact concentrations of hydroxide ions:
- pH + pOH = 14 ⇒ pOH = 14 - pH
- pOH = - log [OH⁻] ⇒
![[OH^-]=10^{-pOH}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-pOH%7D)
Then, you can follow these calculations:
Solution pH pOH [OH⁻]
a. 3.21 14 - 3.21 = 10.79 antilogarithm of 10.79 = 1.6 × 10⁻¹¹
b. 7.00 14 - 7.00 = 7.00 antilogarithm of 7.00 = 10⁻⁷
c. 7.93 14 - 7.93 = 6.07 antilogarithm of 6.07 = 8.5 × 10⁻⁷
d. 12.59 14 - 12.59 = 1.41 antilogarithm of 1.41 = 0.039
e. 9.82 14 - 9.82 = 4.18 antilogarithm of 4.18 = 6.6 × 10⁻⁵
From which you see that the highest concentration of hydroxide ions is for pH = 12.59.
force but could be tension
The correct answer is 42. I know this answer is right because i have already turned my assignment in and got it right.