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NemiM [27]
2 years ago
12

Fossil A is estimated to be 5,000 years old and fossil B is estimated to be 2,000 year old. If the two fossils had the same amou

nts of radioactive carbon-14 when they were formed, which of the following statements is true?
Select one:

a. Fossil A contains more carbon-14 than fossil B does.

b. Fossil A and fossil B contain approximately the same amounts of carbon-14.

c. Fossil B contains more carbon-14 than fossil A does.

d. Not enough information given.

​
Chemistry
1 answer:
elena-14-01-66 [18.8K]2 years ago
7 0

The new fossil i.e, Fossil B (2,000 year old) contains more carbon-14 than the old fossil i.e, Fossil A (5,000 years old ) does.

<h3>What are Fossils ?</h3>

Fossils are the preserved left materials of ancient organisms in trace form.

Fossils are not just the remains of the organism But also the rocks.

Carbon 14 works to analyse the period organic remains that are between 100 years old (minimum) and about 50,000 years old, (maximum).

If it is new, than not enough time has passed for radioactive decay to produce and if it is older than that, the amount of radioactive decay has dropped so less amount which means that anything beyond 50,000 years old becomes gibberish and may lead to high or low (usually low) concentration.

Hence,the new fossil Fossil B (2,000 year old) contains more carbon-14 than the older Fossil A (5,000 years old ) does.

Learn more about Fossil here ;

brainly.com/question/19676397

#SPJ1

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the chemical compound C2F4 is used to make PTEE (Teflon). How manyC2F4 molecules are in 485 kilograms of this material?
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molecules=2.92x10^{27}moleculesC_2F_4

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Aclosed system contains an equimolar mixture of n-pentane and isopentane. Suppose the system is initially all liquid at 120°C an
garri49 [273]

Explanation:

The given data is as follows.

      T = 120^{o}C = (120 + 273.15)K = 393.15 K,  

As it is given that it is an equimolar mixture of n-pentane and isopentane.

So,            x_{1} = 0.5   and   x_{2} = 0.5

According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.

               p^{sat}_{1} (393.15 K) = 9.2 bar

               p^{sat}_{1} (393.15 K) = 10.5 bar

Hence, we will calculate the partial pressure of each component as follows.

                 p_{1} = x_{1} \times p^{sat}_{1}

                            = 0.5 \times 9.2 bar

                             = 4.6 bar

and,           p_{2} = x_{2} \times p^{sat}_{2}

                         = 0.5 \times 10.5 bar

                         = 5.25 bar

Therefore, the bubble pressure will be as follows.

                           P = p_{1} + p_{2}            

                              = 4.6 bar + 5.25 bar

                              = 9.85 bar

Now, we will calculate the vapor composition as follows.

                      y_{1} = \frac{p_{1}}{p}

                                = \frac{4.6}{9.85}

                                = 0.467

and,                y_{2} = \frac{p_{2}}{p}

                                = \frac{5.25}{9.85}

                                = 0.527  

Calculate the dew point as follows.

                     y_{1} = 0.5,      y_{2} = 0.5  

          \frac{1}{P} = \sum \frac{y_{1}}{p^{sat}_{1}}

           \frac{1}{P} = \frac{0.5}{9.2} + \frac{0.5}{10.2}

             \frac{1}{P} = 0.101966 bar^{-1}              

                             P = 9.807

Composition of the liquid phase is x_{i} and its formula is as follows.

                   x_{i} = \frac{y_{i} \times P}{p^{sat}_{1}}

                               = \frac{0.5 \times 9.807}{9.2}

                               = 0.5329

                    x_{z} = \frac{y_{i} \times P}{p^{sat}_{1}}

                               = \frac{0.5 \times 9.807}{10.5}

                               = 0.467

4 0
3 years ago
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