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diamong [38]
1 year ago
8

How do I solve the following word problem: The half-life of Phosphorus - 32 is 14.3 days. It is used to study a plant's use of f

ertilizers. A) Write an exponential decay function for a 50mg sample. b) Find the amount of P - 32 remaining after 84 days.
Chemistry
1 answer:
BaLLatris [955]1 year ago
4 0

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=A_o\times2^{(-\frac{t}{t_{half}})_{}_{}}

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

A=\text{ 50 }\times2^{\frac{-t}{14.3}}

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:

\begin{gathered} A=\text{ 50 }\times2^{\frac{-84}{14.3}} \\ A=0.85\text{ mg} \end{gathered}

So, after 84 days the P-32 remaining will be 0.85 mg

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Answer:

Mass = 51 g

Explanation:

Given data:

Mass of nitrogen = 41.93 g

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Solution:

Chemical equation:

N₂ + 3H₂      →       2NH₃

Number of moles of nitrogen:

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Number of moles = 41.93 g/ 28 g/mol

Number of moles = 1.5 mol

now we will compare the moles of nitrogen and ammonia.

                N₂          :           NH₃

                  1          :           2

                1.5         :         2/1×1.5 = 3 mol

Mass of ammonia formed:

Mass = number of moles × molar mass

Mass = 3 mol × 17 g/mol

Mass = 51 g

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3 years ago
Which would most likely cause a person to produce antibodies?
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receiving a vaccination

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A sample of gas has a volume of 215 cm3 at 23.5 degrees Celsius and 84.6kPa what volume will the gas occupy at stp
miss Akunina [59]

The volume of the gas that occupy at STP is 165. 28 cm^3

calculation

by use of combined gas law that is P1V1/T1=P2V2/T2, where

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V1=215 cm^3

At STP T= 273 K and P= 101.325 Kpa

therefore p2 = 101.325 Kpa and T2 = 272 K V2=?

by making V2 the subject of the formula V2 =T2P1V1/P2T1

V2 = 273 K x 84.6 Kpa x 215 cm^3/ 101,.325 Kpa x296.5 K =165.28 cm^3

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The stability of an isotope is based on its
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For the following aqueous reaction, complete and balance the molecular equation and write a net iconic equatio, making sure to i
pav-90 [236]

Answer:

Balance molecular equation:

K2CO3(aq) + Sr(NO3)2(aq) → SrCO3(s) + 2KNO3(aq)

Net ionic equation:

CO3∧-2(aq) + Sr∧+2(aq) → SrCO3(s)

Explanation:

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Chemical equation:

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Balance chemical equation with physical states:

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Ionic equation:

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Net ionic equation:

CO3∧-2(aq) + Sr∧+2(aq) → SrCO3(s)

2K+ and 2NO∧-3 ions are spectator ions that's way these are not written in net ionic equation.

Spectator ions:

These are the ions that are present same on both side of chemical reaction and does not effect the equilibrium.

6 0
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