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Pavel [41]
4 years ago
11

An archaeologist graduate student found a leg bone of a large animal during the building of a new science building. The bone had

a carbon-14 decay rate of 14.8 disintegrations per minute per gram of carbon. Living organisms have a decay rate of 15.3 disintegrations per minute. How old is the bone?
Chemistry
1 answer:
jasenka [17]4 years ago
6 0

Answer:

The answer to the question is 275 years

Explanation:

Here we have

The decay rate is given by

dN/dt = -λN

or dN/N = -λdt

Integrating the above equation with limits N and N₀ we have

㏑(N/N₀) = -λt

Note that the half life t_{\frac{1}{2} } = (㏑ 2)/λ

From where λ = 0.693/5730 =  1.21 × 10⁻⁴

∴ ㏑(14.8/15/3) = -1.21 × 10⁻⁴ × t

or t =  275 years

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50.00 mL of 0.10 M HNO 2 (nitrous acid, K a = 4.5 × 10 −4) is titrated with a 0.10 M KOH solution. After 25.00 mL of the KOH sol
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Answer:

b. 3.35

Explanation:

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pH = pKa + log ([salt]/[acid])     (Eq. 01)

Where  

pKa = -log(Ka)        (Eq. 02)

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[acid] = Molar concentration of acid left in the solution after titration

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HNO2 + KOH ⇆ H2O + KNO2    (Eq. 03)

This shows that 01 mole of HNO2 and 01 mole of KOH are required to produce 01 mole of KNO2 (salt). And if any one of them (HNO2 and KOH) is present in lower amount then that will be considered the limiting reactant and amount of salt produced will be in accordance to that reactant.

Moles of HNO2 in 50 mL of 0.01 M HNO2 solution = 50/1000x0.01 = 0.005 Moles

Moles of KOH in 25 mL of 0.01 M KOH solution = 25/1000x0.01 = 0.0025 Moles

As it can be seen that we have 0.0025 Moles of KOH therefore considering Eq. 03 we can see that 0.0025 Moles of KOH will react with only 0.0025 Moles of HNO2 and will produce 0.0025 Moles of KNO2.

Therefore

Amount of salt produced i.e [salt] = 0.0025 moles       (Eq. 04)

Amount of acid left in the solution [acid] = 0.005 - 0.0025 = 0.0025 moles (Eq.05)

Putting the values in (Eq. 01) from (Eq.02), (Eq. 04) and (Eq. 05) we will get the following expression:

pH= -log(4.5x10 -4) + log (0.0025/0.0025)

Solving above we get  

pH = 3.35

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