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Nataliya [291]
3 years ago
15

What % of original U238 remains after three half lives? 1-50 2-25 3-12.5 4-6.25

Chemistry
1 answer:
Grace [21]3 years ago
7 0

Answer:

12.5

Explanation:

After three half-lives, one eighth (1/2 x 1/2 x 1/2) of the uranium-238 will remain. One-eighth equals 12.5%

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The formula for calculating the number for moles:

\text{Number of moles }= \frac{\ Mass}{ \ molar \ mass }

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Nitric acid molar weight= 63\  \frac{g}{mol}

If they put values above the formula, they receive:

\text{moles in nitric acid} = \frac{75.9}{63}

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In the given chemical equation:

3 NO_2 \ (g) + H_2O \ (l) \longroghtarrow 2 HNO_3 \ (aq) + NO\ (g)

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So, 1.204 moles of nitric acid will be produced:

= \frac{1}{2} \times 1.204

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Water weight molar = 18.02 \ \frac{g}{mol}

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