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Elden [556K]
3 years ago
14

The concentration of Rn−222 in the basement of a house is 1.45 × 10−6 mol/L. Assume the air remains static and calculate the con

centration of the radon after 3.00 days. The half-life of Rn−222 is 3.82 days.
Chemistry
1 answer:
bonufazy [111]3 years ago
7 0

<u>Answer:</u> The concentration of radon after the given time is 3.83\times 10^{-30}mol/L

<u>Explanation:</u>

All the radioactive reactions follows first order kinetics.

The equation used to calculate half life for first order kinetics:

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

We are given:

t_{1/2}=3.82days

Putting values in above equation, we get:

k=\frac{0.693}{3.82}=0.181days^{-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 = 0.181days^{-1}

t = time taken for decay process = 3.00 days

[A_o] = initial amount of the reactant = 1.45\times 10^{-6}mol/L

[A] = amount left after decay process =  ?

Putting values in above equation, we get:

0.181days^{-1}=\frac{2.303}{3.00days}\log\frac{1.45\times 10^{-6}}{[A]}

[A]=3.83\times 10^{-30}mol/L

Hence, the concentration of radon after the given time is 3.83\times 10^{-30}mol/L

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Answer: The correct answer is -297 kJ.

Explanation:

To solve this problem, we want to modify each of the equations given to get the equation at the bottom of the photo. To do this, we realize that we need SO2 on the right side of the equation (as a product). This lets us know that we must reverse the first equation. This gives us:

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Remember that we take the opposite of the enthalpy change (reverse the sign) when we reverse the equation.

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saveliy_v [14]

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