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mestny [16]
3 years ago
11

The radioisotope radon-222 has a half-life of 3.8 days. How much of a 198.6 g sample of radon-222 would be left after approximat

ely 23 days?
Chemistry
1 answer:
bagirrra123 [75]3 years ago
7 0
The radon-222 sample has a half-life of 3.8 days, and we are asked how many times would the mass divide in half after 23 days. First we calculate the amount of times division occurs by taking the number of days and dividing that by the number of days for one half-life to occur: 23/3.8 = 6.05.

We have 198.6 grams of sample, and we are going to divide it in half 6 times to determine how much of it remains after 23 days:
198.6/2 = 99.3 grams
99.3/2 = 49.65 grams
49.65/2 = 24.83 grams
24.83/2 = 12.41 grams
12.41/2 = 6.21 grams
6.21/2 = 3.1 grams

Therefore, we are left with 3.1 grams of radon-222 after 23 days if one half-life equals to 3.8 days.
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<u>Answer:</u> The theoretical yield of the lithium chlorate is 1054.67 grams

<u>Explanation:</u>

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

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Putting values in above equation, we get:

9.45mol=\frac{\text{Actual yield of lithium chlorate}}{90.4g/mol}\\\\\text{Actual yield of lithium chlorate}=(9.45mol\times 90.4g/mol)=854.28g

To calculate the theoretical yield of lithium chlorate, we use the equation:

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Actual yield of lithium chlorate = 854.28 g

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Putting values in above equation, we get:

81=\frac{854.28g}{\text{Theoretical yield of lithium chlorate}}\times 100\\\\\text{Theoretical yield of lithium chlorate}=\frac{854.28\times 100}{81}=1054.67g

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