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skad [1K]
3 years ago
15

Thallium-201 has a half-life 73 hours. If 4.0 mg of thallium-201 disintegrates over a period of 5.0 °, how many milligrams of th

allium-201 will remain?
Chemistry
1 answer:
Anika [276]3 years ago
5 0

<u>Given:</u>

Half life of Thallium-201 (T1/2)= 73 hrs

Initial amount of thallium (A₀) = 4.0 mg

Disintegration Time period (t) = 5.0 hrs

<u>To determine:</u>

Amount of thallium remaining, A

<u>Explanation:</u>

The radioactive half-life (T1/2) can be related to the amounts of the radio active element through the relation:

A = A_{0} e^{-0.693t/T1/2}

A = 4.0 * e^{-0.693*5/73} = 3.815 mg

Ans: Amount of thallium-201 remaining is 3.815 mg

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If aluminum is diffused into a thick slice of silicon with no previous aluminum in it at a temperature of 1100˚C for 8 hours, wh
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Answer:

8.354 nanometers

Explanation:

To treat a diffusive process in function of time and distance we need to solve  2nd Ficks Law. This a partial differential equation, with certain condition the solution looks like this:

\frac{C_{s}-C{x}}{C_{s}-C_{o}}=erf(x/2\sqrt{D*t})

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Cx is the concentration at certain deep X

Co is the initial concentration of solute in the solid

and erf is the error function

Then we solve right side,

\frac{C_{s}-C{x}}{C_{s}-C_{o}}=\frac{1018atoms/cm3-1016atoms/cm3}{1018atoms/cm3}=0.001964

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Then we solve for x:

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3 years ago
The solubility product of PbI2 is 7.9x10^-9. What is the molar solubility of PbI2 in distilled water?
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<u>Answer:</u> The molar solubility of PbI_2 is 1.25\times 10^{-3}mol/L

<u>Explanation:</u>

Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium.

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The balanced equilibrium reaction for the ionization of calcium fluoride follows:

PbI_2\rightleftharpoons Pb^{2+}+2I^-

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The expression for solubility constant for this reaction will be:

K_{sp}=[Pb^{2+}][I^-]^2

We are given:

K_{sp}=7.9\times 10^{-9}

Putting values in above equation, we get:

7.9\times 10^{-9}=(s)\times (2s)^2\\\\7.9\times 10^{-9}=4s^3\\\\s=1.25\times 10^{-3}mol/L

Hence, the molar solubility of PbI_2 is 1.25\times 10^{-3}mol/L

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Answer:

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