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djyliett [7]
3 years ago
12

Uranium–232 has a half–life of 68.9 years. A sample from 206.7 years ago contains 1.40 g of uranium–232. How much uranium was or

iginally present?
Chemistry
2 answers:
Mnenie [13.5K]3 years ago
7 0

it's 11.2, other guy prob made a typo is all

givi [52]3 years ago
4 0

<u>Answer:</u> The initial amount of Uranium-232 present is 11.3 grams.

<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}=68.9yrs

Putting values in above equation, we get:

k=\frac{0.693}{68.9}=0.0101yr^{-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 = 0.0101yr^{-1}

t = time taken for decay process = 206.7 yrs

[A_o] = initial amount of the reactant = ?

[A] = amount left after decay process = 1.40 g

Putting values in above equation, we get:

0.0101yr^{-1}=\frac{2.303}{206.7yrs}\log\frac{[A_o]}{1.40}

[A_o]=11.3g

Hence, the initial amount of Uranium-232 present is 11.3 grams.

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Answer: 27.09 ppm and 0.003 %.

First, <u>for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. </u>Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

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