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2 years ago

How can you simulate the radioactive half-life of an element?

Chemistry

1 answer:

ch4aika [34]2 years ago

6 0

Answer:

TRIAL 1:

For “Event 0”, put 100 pennies in a large plastic or cardboard container.

For “Event 1”, shake the container 10 times. This represents a radioactive decay event.

Open the lid. Remove all the pennies that have turned up tails. Record the number removed.

Record the number of radioactive pennies remaining.

For “Event 2”, replace the lid and repeat steps 2 to 4.

Repeat for Events 3, 4, 5 … until no pennies remain in the container.

TRIAL 2:

Repeat Trial 1, starting anew with 100 pennies.

Calculate for each event the average number of radioactive pennies that remain after shaking.

Plot the average number of radioactive pennies after shaking vs. the Event Number. Start with Event 0, when all the pennies are radioactive. Estimate the half-life — the number of events required for half of the pennies to decay.

Explanation:

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onsider the reversible dissolution of lead(II) chloride. P b C l 2 ( s ) − ⇀ ↽ − P b 2 + ( a q ) + 2 C l − ( a q ) PbClX2(s)↽−−⇀

Sveta_85 [38]

Answer:

9.34x10^-4

Explanation:

Step 1:

The balanced equation for the reaction.

PbCl2( s ) <=> Pb^2+(aq) + 2Cl^−(aq)

Step 2:

Data obtained from the question:

Mass of PbCl2 = 0.2393 g

Volume = 50mL

concentration of Pb^2+, [Pb^2+] = 0.0159 M

Concentration of Cl^-, [Cl^-] = 0.0318 M

Equilibrium constant, Kc =?

Step 3:

Determination of the number of mole PbCl2.

The number of mole of PbCl2 can be obtained as follow:

Molar Mass of PbCl2 = 207 + (35.5x2) = 278g/mol

Mass of PbCl2 = 0.2393 g

Number of mole =Mass /Molar Mass

Number of mole of PbCl2 = 0.2393/278 = 8.61x10^-4 mole

Step 4:

Determination of Molarity of PbCl2.

At this stage we shall obtain the molarity of PbCl2. This is shown below:

Mole of PbCl2 = 8.61x10^-4 mole

Volume = 50mL = 50/1000 = 0.05L

Molarity of PbCl2 =?

Molarity = mole /Volume

Molarity of PbCl2 = 8.61x10^-4/0.05

Molarity of PbCl2 = 0.01722 M

Step 5:

Determination of the equilibrium constant Kc.

PbCl2( s ) <=> Pb^2+(aq) + 2Cl^−(aq)

The equilibrium constant Kc for the equation above is given by:

Kc = [Pb^2+] [Cl^-]^2 / [PbCl2]

[Pb^2+] = 0.0159 M

[Cl^-] = 0.0318 M

[PbCl2] = 0.01722 M

Kc =?

Kc = [Pb^2+] [Cl^-]^2 / [PbCl2]

Kc = 0.0159 x (0.0318)^2/ 0.01722

Kc = 9.34x10^-4

5 0

3 years ago

(4 points) The following lead compound for a pharmaceutical drug contains a rotatable bond. Using the principles of rigidificati

masya89 [10]

Answer:

Explanation:

The solution has been attached

3 0

2 years ago

How many liters of 4.0 M NaOH solution will react with 0.60 liters 3.0 M H2SO4?

slava [35]

Answer:

A. 0.90 L.

Explanation:

NaOH solution will react with H₂SO₄ according to the balanced reaction:

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.

1.0 mole of H₂SO₄ reacts with 2.0 moles of NaOH.

For NaOH to react completely with H₂SO₄, the no. of millimoles should be equal.

∴ (MV) NaOH = (xMV) H₂SO₄.

x for H₂SO₄ = 2, due to having to reproducible H⁺ ions.

∴ V of NaOH = (xMV) H₂SO₄/ M of NaOH = 2(0.6 L)(3.0 M)/(4.0 M) = 0.90 L.

4 0

3 years ago

Is selenium tetrafluoride an ionic or covalent bond?

nasty-shy [4]

Answer:

Synthesis

The first reported synthesis of selenium tetrafluoride was by Paul Lebeau in 1907, who treated selenium with fluorine:[1]

Se + 2 F2 → SeF4

A synthesis involving more easily handled reagents entails the fluorination of selenium dioxide with sulfur tetrafluoride:[2]

SF4 + SeO2 → SeF4 + SO2

An intermediate in this reaction is seleninyl fluoride (SeOF2).

Other methods of preparation include fluorinating elemental selenium with chlorine trifluoride:

3 Se + 4 ClF3 → 3 SeF4 + 2 Cl2

Structure and bonding

Selenium in SeF4 has an oxidation state of +4. Its shape in the gaseous phase is similar to that of SF4, having a see-saw shape. VSEPR theory predicts a pseudo-trigonal pyramidal disposition of the five electron pairs around the selenium atom. The axial Se-F bonds are 177 pm with an F-Se-F bond angle of 169.2°. The two other fluorine atoms are attached by shorter bonds (168 pm), with an F-Se-F bond angle of 100.6°. In solution at low concentrations this monomeric structure predominates, but at higher concentrations evidence suggests weak association between SeF4 molecules leading to a distorted octahedral coordination around the selenium atom. In the solid the selenium center also has a distorted octahedral environment.

Reactions

In HF, SeF4 behaves as a weak base, weaker than sulfur tetrafluoride, SF4 (Kb= 2 X 10−2):

SeF4 + HF → SeF3+ + HF2−; (Kb = 4 X 10−4)

Ionic adducts containing the SeF3+ cation are formed with SbF5, AsF5, NbF5, TaF5, and BF3.[3] With caesium fluoride, CsF, the SeF5− anion is formed, which has a square pyramidal structure similar to the isoelectronic chlorine pentafluoride, ClF5 and bromine pentafluoride, BrF5.[4] With 1,1,3,3,5,5-hexamethylpiperidinium fluoride or 1,2-dimethylpropyltrimethylammonium fluoride, the SeF62− anion is formed. This has a distorted octahedral shape which contrasts to the regular octahedral shape of the analogous SeCl62−. [5]

Explanation:

4 0

2 years ago

Okghhjhyuknji Hdhdnsbdhjsms dnddms dnjdke

VikaD [51]

Answer:

thanks for the points

take care bye

6 0

2 years ago