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Anvisha [2.4K]
3 years ago
7

How much of a radioactive parent isotope will remain after three half-lives have passed?

Chemistry
1 answer:
Sedaia [141]3 years ago
4 0
After three half lives have passed, there would be only 12.5 percent of the original amount of a radioactive parent isotope that will remain. Half life is the time needed for a certain amount of a substance to be half its initial amount. It is a common term used in nuclear chemistry describing how fast radioactive substances undergo decay. One half life would correspond to only 50% would be left. Two half lives would be 25% only of the original value. Three half lives would be 12.5%. Four half lives would be 6.25% of the initial value. So on and so forth.
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Answer: No

Explanation: The reason for this is because calcite has a hardness of 9 while feldspar has a hardness of 21. Calcite can scratch Gypsum which has  a hardness of 3.

7 0
4 years ago
A researcher was attempting to quantify the amount of dichlorodiphenyltrichloroethane (DDT) in spinach with gas chromatography u
devlian [24]

Answer:

0.0136mg DDT / g spinach

Explanation:

Quantification in chromatography by internal standard has as formula:

RF = Aanalyte×Cstd / Astd×Canalyte <em>(1)</em>

<em>Where RF is response factor, A is area and C is concentration</em>

Replacing with first experiment values:

RF = 5019×3.20mg/L / 8179×6.37mg/L

RF = 0.308

In the next experiment, final concentration of chloroform was:

11.45mg/L × (1.25mL / 25.00mL) = <em>0.5725mg/L</em>

From (1), it is possible to write:

Aanalyte×Cstd / Astd×RF = Canalyte

Replacing:

6821×0.5725mg/L / 14061×0.308 = Canalyte

Canalyte = <em>0.9017mg/L</em>

as the sample was made from 0.750mL of extract. Concentration of extract is:

0.9017mg/L × (25.00mL / 0.750mL) = 30.06mg/L. As the extract has a volume of 2.40mL:

30.06mg/L × 2.40x10⁻³L = <em>0.07213mg of DDT in the extract</em>

As the extract was made from 5.29g of spinach:

0.07213mg of DDT in the extract / 5.29g spinach = <em>0.0136mg DDT / g spinach</em>

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