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Half life is the time that it takes for half of the original value of some amount of a radioactive element to decay.
We have the following equation representing the half-life decay:

A is the resulting amount after t time
Ao is the initial amount = 50 mg
t= Elapsed time
t half is the half-life of the substance = 14.3 days
We replace the know values into the equation to have an exponential decay function for a 50mg sample

That would be the answer for a)
To know the P-32 remaining after 84 days we have to replace this value in the equation:

So, after 84 days the P-32 remaining will be 0.85 mg
Answer:
The odor of a substance is a physical property. That would be your answer.
Explanation:
Physical Properties
Physical properties are properties that can be measured or observed without changing the chemical nature of the substance. Some examples of physical properties are:
color (intensive)
density (intensive)
volume (extensive)
mass (extensive)
boiling point (intensive): the temperature at which a substance boils
melting point (intensive): the temperature at which a substance melts
Chemical Properties
Remember, the definition of a chemical property is that measuring that property must lead to a change in the substance’s chemical structure. Here are several examples of chemical properties:
Heat of combustion is the energy released when a compound undergoes complete combustion (burning) with oxygen. The symbol for the heat of combustion is ΔHc.
Chemical stability refers to whether a compound will react with water or air (chemically stable substances will not react). Hydrolysis and oxidation are two such reactions and are both chemical changes.
Flammability refers to whether a compound will burn when exposed to flame. Again, burning is a chemical reaction—commonly a high-temperature reaction in the presence of oxygen.
The preferred oxidation state is the lowest-energy oxidation state that a metal will undergo reactions in order to achieve (if another element is present to accept or donate electrons).
Answer: -
12.59
Explanation: -
Strength of NaOH = 0.0179 M
Volume of NaOH = 58.0 mL = 58.0/1000 = 0.058 L
Number of moles = 0.0179 M x 0.058 L
= 1.04 x 10⁻³ mol
Mol of [OH⁻] given by NaOH = 1.04 x 10⁻³ mol
Strength of Ba(OH)₂ = 0.0294 M
Volume of Ba(OH)₂ = 60.0 mL = 60.0/1000 = 0.060 L
Number of moles = 0.0294 M x 0.060 L
= 1.76 x 10⁻³ mol
Mol of [OH⁻] given by Ba(OH)₂ =2 x 1.76 x 10⁻³ mol
Total [OH⁻] = 1.04 x 10⁻³ mol + 2 x 1.76 x 10⁻³ mol
= 4.56 x 10⁻³ mol
Total volume of the mixture = 58.0 + 60.0
= 118.0 mL
118.0 mL of the solution has 4.56 x 10⁻³ mol [OH⁻]
1000 mL of the solution has 
= 0.0386 mol
Using the relation
pOH = - log [OH-]
= - log 0.0386
= 1.41
Using the relation
pH + pOH = 14
pH = 14 - 1.41
= 12.59