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Aliun [14]
3 years ago
11

Scientist can determine the age of ancient objects by a method called radiocarbon dating. The bombardment of the upper atmospher

e by cosmic rays converts nitrogen to a radioactive isotope of carbon, 14C, with a half-life of about 5730 years. Vegetation absorbs carbon dioxide through the atmosphere and animal life assimilates 14C through food chains. When a plant or animal dies, it stops replacing its carbon and the amount of 14C begins to decrease through radioactive decay. Therefore, the level of radioactivity must also decay exponentially. A parchment fragment was discovered that had about 68% as much 14C radioactivity as does plant material on Earth today. Estimate the age of the parchment. (Round your answer to the nearest hundred years.)
Chemistry
1 answer:
levacccp [35]3 years ago
3 0

Answer:

  • 28,700 years

1. Data:

a) Half-life: 5730 years

b) Final radioactivity: 68%

2. Solution:

a) <u>Determine the number of half-lives undergone</u>

  • Since, <em>the radioactivity has decreased to 68%, means that the carbon-14 contanined is has been reduced in </em> 32%: 100% - 68% = 32%.

  • 32 = 2⁵, meaning that five half-lives have passed since the <em>plant material</em> that formed the<em> parchment fragment died</em>.

b) <u>Compute the time of five half-lives</u>:

  • 5 × half-life time = 5 × 5730 years = 28,650 years.

c) <u>Round to the nearest hundred:</u>

  • 28,650 years ≈ 28,700 years

And that is <em>the age of the parchment</em>.

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Answer:

Approximately 0.291\; \rm M (rounded to two significant figures.)

Explanation:

The unit of concentration \rm M is the same as \rm mol \cdot L^{-1} (moles per liter.) On the other hand, the volume of both the \rm NaOH solution and the original \rm HCl solution here are in milliliters. Convert these two volumes to liters:

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\begin{aligned} n(\mathrm{NaOH}) &= c(\mathrm{NaOH})\cdot V(\mathrm{NaOH})\\ &= 0.160\; \rm mol \cdot L^{-1} \times 0.0182\; \rm L \approx 0.00291\; \rm mol\end{aligned}.

\rm HCl reacts with \rm NaOH at a one-to-one ratio:

\rm HCl\; (aq) + NaOH\; (aq) \to NaCl\; (aq) + H_2O\; (l).

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\displaystyle \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} = 1.

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\begin{aligned}&n(\text{$\mathrm{HCl}$, original})\\ &= n(\mathrm{NaOH})\cdot \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} \\ &\approx 0.00291\; \rm mol \times 1 = 0.00291\; \rm mol\end{aligned}.

Calculate the concentration of the original \rm HCl solution:

\displaystyle c(\text{$\mathrm{HCl}$, original}) = \frac{n(\text{$\mathrm{HCl}$, original})}{V(\text{$\mathrm{HCl}$, original})} \approx \frac{0.00291\; \rm mol}{0.0100\; \rm L} \approx 0.291\; \rm M.

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