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2 years ago

the isotope 146c has a half life of 5730 years. what fraction of 146c in a sample with mass ,m, after 28650 years

Chemistry

1 answer:

defon2 years ago

3 0

Answer:

3.1% is the fraction of the sample after 28650 years

Explanation:

The isotope decay follows the equation:

Ln[A] = -kt + Ln[A]₀

<em>Where [A] could be taken as fraction of isotope after time t, k is decay constant and [A]₀ is initial fraction of the isotope = 1</em>

<em />

k could be obtained from Half-Life as follows:

K = Ln 2 / Half-life

K = ln 2 / 5730 years

K = 1.2097x10⁻⁴ years⁻¹

Replacing in isotope decay equation:

Ln[A] = -1.2097x10⁻⁴ years⁻¹*28650 years + Ln[1]

Ln[A] = -3.4657

[A] = 0.0313 =

<h3>3.1% is the fraction of the sample after 28650 years</h3>

<em />

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What does a ground wire do?

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A ground wire helps those positive charges get to the ground in a safe, direct and controlled way, where they can be discharged without the risk of electrical shock or fire. Shock Absorber Excess electrical charges are common in any home.

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2 years ago

A titration of 25.0 mL of a solution of the weak base aniline, C6H5NH2, requires 25.67 mL of 0.175 M HCl to reach the equivalenc

Lorico [155]

Answer:

a. 0.180M of C₆H₅NH₂

b. 0.0887M C₆H₅NH₃⁺

c. pH = 2.83

Explanation:

a. Based in the chemical equation:

C₆H₅NH₂(aq) + HCl(aq) → C₆H₅NH₃⁺(aq) + Cl⁻(aq)

<em>1 mole of aniline reacts per mole of HCl</em>

Moles required to reach equivalence point are:

Moles HCl = 0.02567L ₓ (0.175mol / L) = 4.492x10⁻³ moles HCl = moles C₆H₅NH₂

As the original solution had a volume of 25.0mL = 0.0250L:

4.492x10⁻³ moles C₆H₅NH₂ / 0.0250L = 0.180M of C₆H₅NH₂

b. At equivalence point, moles of C₆H₅NH₃⁺ are equal to initial moles of C₆H₅NH₂, that is 4.492x10⁻³ moles

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[C₆H₅NH₃⁺] = 4.492x10⁻³ moles / 0.05067L = 0.0887M C₆H₅NH₃⁺

c. At equivalence point you have just 0.0887M C₆H₅NH₃⁺ in solution. C₆H₅NH₃⁺ has as equilibrium in water:

C₆H₅NH₃⁺(aq) + H₂O(l) → C₆H₅NH₂ + H₃O⁺

Where Ka = Kw / Kb = 1x10⁻¹⁴ / 4.0x10⁻¹⁰ =

When the system reaches equilibrium, molar concentrations are:

[C₆H₅NH₃⁺] = 0.0887M - X

[C₆H₅NH₂] = X

[H₃O⁺] = X

Replacing in Ka formula:

2.5x10⁻⁵ = [X] [X] / [0.0887M - X]

2.2175x10⁻⁶ - 2.5x10⁻⁵X = X²

0 = X² + 2.5x10⁻⁵X - 2.2175x10⁻⁶

Solving for X:

X = -0.0015 → False solution. There is no negative concentrations.

X = 0.001477 → Right solution.

As [H₃O⁺] = X, [H₃O⁺] = 0.001477

Knowing pH = -log [H₃O⁺]

pH = -log 0.001477

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2 years ago

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Answer:

See explanation

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