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lina2011 [118]
3 years ago
15

The half-life of Pa-234 is 6.75 hours. If a sample of Pa-234 contains 112.0 g, 1 point

Chemistry
1 answer:
umka2103 [35]3 years ago
5 0

Answer:

28g remain after 13.5 hours

Explanation:

Element decayment follows first order kinetics law:

ln[Pa-234] = -kt + ln [Pa-234]₀ <em>(1)</em>

<em>Where [Pa-234] is concentration after t time, k is rate constant in time, and [Pa-234]₀ is initial concentration</em>

Half-life formula is:

t_{1/2} =  \frac{ln2}{k}

6.75 = ln2 / k

<em>k = 0.1027hours⁻¹</em>

Using rate constant in (1):

ln[Pa-234] = -0.1027hours⁻¹×13.5hours + ln [112.0g]

ln[Pa-234] = 3.332

[Pa-234] = <em>28g after 13.5 hours</em>

<em />

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Reasons:

The question parameters are;

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Final temperature of the oil and copper, T₂ = 27.33 °C

Specific heat of copper, c₂ = 0.387 J/(g·°C)

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Required:

The<em> mass of oil</em> in the cup.

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The mass of the copper, m₂ = 17.920 g

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Heat lost by copper = Heat gained by the oil

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Therefore, we get;

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m₁ ≈ 64.73

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<em>Possible part of the question obtained from a similar question online, are;</em>

<em>The mass of the copper, m₂ = 17.920 g</em>

<em>Temperature of copper after heating = 65.17°C</em>

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