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allochka39001 [22]
4 years ago
8

The? half-life of a certain tranquilizer in the bloodstream is 4242 hours. how long will it take for the drug to decay to 8686?%

of the original? dosage? use the exponential decay? model, upper a equals upper a 0 e superscript kta=a0ekt?, to solve.
Chemistry
1 answer:
Kitty [74]4 years ago
6 0

For this problem, we use the formula for radioactive decay which is expressed as follows:<span>

An = Aoe^-kt

where An is the amount left after time t, Ao is the initial amount and k is a constant. 

We calculate as follows:

An = Aoe^-kt
0.5 = e^-k(42 hrs)
k = 0.0165 / hr

An = Aoe^-kt
.86 =e^-0.0165(t)
<span>t = 9.14 hrs</span></span>

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How many grams of Ag2S2O3 form
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Taking into account the reaction stoichiometry, 109.09 grams of Ag₂S₂O₃ are formed when 125 g AgBr reacts completely.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 AgBr + Na₂S₂O₃ → Ag₂S₂O₃ + 2 NaBr

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

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  • Na₂S₂O₃: 1 mole
  • Ag₂S₂O₃: 1 mole
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  • Na₂S₂O₃: 158 g/mole
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Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • AgBr: 2 moles ×187.77 g/mole= 375.54 grams
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<h3>Mass of Ag₂S₂O₃ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 375.54 grams of AgBr form 327.74 grams of Ag₂S₂O₃, 125 grams of AgBr form how much mass of Ag₂S₂O₃?

mass of Ag_{2} S_{2} O_{3} =\frac{125 grams of AgBrx327.74 grams of Ag_{2} S_{2} O_{3}}{375.54 grams of AgBr}

<u><em>mass of Ag₂S₂O₃= 109.09 grams</em></u>

Then, 109.09 grams of Ag₂S₂O₃ are formed when 125 g AgBr reacts completely.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

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