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anastassius [24]
3 years ago
7

The half-life for the radioactive decay of C-14C-14 is 5730 years. You may want to reference (Pages 598 - 605) Section 14.5 whil

e completing this problem. Part A How long will it take for 25% of the C-14C-14 atoms in a sample of C-14C-14 to decay
Chemistry
1 answer:
KIM [24]3 years ago
8 0

<u>Answer:</u> The sample of Carbon-14 isotope will take 2377.9 years to decay it to 25 %

<u>Explanation:</u>

The equation used to calculate rate constant from given half life for first order kinetics:

t_{1/2}=\frac{0.693}{k}

where,

t_{1/2} = half life of the reaction = 5730 years

Putting values in above equation, we get:

k=\frac{0.693}{5730yrs}=1.21\times 10^{-4}yrs^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,

k = rate constant  = 1.21\times 10^{-4}yr^{-1}

t = time taken for decay process = ? yr

[A_o] = initial amount of the sample = 100 grams

[A] = amount left after decay process =  (100 - 25) = 75 grams

Putting values in above equation, we get:

1.21\times 10^{-4}=\frac{2.303}{t}\log\frac{100}{75}\\\\t=2377.9yrs

Hence, the sample of Carbon-14 isotope will take 2377.9 years to decay it to 25 %

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<u>Explanation:</u>

We are given:

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\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

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To calculate the molality of solution, we use the equation:

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3 years ago
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