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
ACIDIC BEHAVIOR OF SOLUTION
Missing question: volume of <span>solution on the left is 10 mL.
V</span>₁(solution) = 10 Ml.
c₁(solution) = 0.2 M.<span>
V</span>₂(solution)
= ?.<span>
c</span>₂(solution)
= 0.04 M.<span>
c</span>₁ -
original concentration of the solution, before it gets diluted.<span>
c</span>₂
- final concentration of the solution, after dilution.<span>
V</span>₁
- <span>volume to
be diluted.
V</span>₂ - <span>final volume after
dilution.
c</span>₁ · V₁ = c₂ · V₂<span>.
</span>10 mL · 0.2 M = 0.04 M · V₂.
V₂(solution) = 10 mL · 0.2 M ÷ 0.04 M.
V₂(solution) = 50 mL.<span>
</span>
Answer:
Manganese decreases from 4+ to 2+ (reduced and oxidizing agent) and nitrogen increases from 2+ to 5+ (oxidized and reducing agent).
Explanation:
Hello there!
In this case, according to the given redox reaction, we rewrite it as a convenient first step:
Next, we assign the oxidation numbers as follows:
Thus, we can see that both manganese and nitrogen undergo a change in their oxidation number, the former decreases from 4+ to 2+ (reduced and oxidizing agent) and the latter increases from 2+ to 5+ (oxidized and reducing agent).
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Answer:
(a) The lewis structure for methylisocyanate is in the attached.
(b) The carbonyl carbon have an sp² hybridization
(c) The nitrogen have an sp² hybridization?
Explanation:
(a) The lewis structure for methylisocyanate has the nitrogen with one lone pair and the oxygen with two lone pairs.
(b) The carbonyl carbon form double bond with the oxygen causing to form three hybrid orbitals sp².
The Nitrogen also forms a double bond with the carbon having an sp² hybridization too.