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Tcecarenko [31]
2 years ago
10

Uranium, an important component of both nuclear weapons and nuclearreactors, has two major isotopes, U-238, which has a half-lif

e of approximately billion years, and U-235, which has a half-life of approximately million years. Both were present in equal amounts at the time of thecreation of the Earth, billion years ago. How many years after the creationof the Earth had the amount of radiation from uranium decayed to half theamount present at the time of the creation of the Earth
Chemistry
1 answer:
elixir [45]2 years ago
6 0

Answer:

Hello your question is poorly written below is the well written question

Uranium, an important component of both nuclear weapons and nuclear reactors, has two major isotopes, U-238, which has a half-life of  approximately 4.5 billion years, and U-235, which has a half-life of approximately 700 million years. Both were present in equal amounts at the time of the creation of the Earth, 4.5  billion years ago. How many years after the creation of the Earth had the amount of radiation from uranium decayed to half the amount present at the time of the creation of the Earth

Answer : 140 billion years

Explanation:

Given that :

U-238   h1/2 = 4.5 billion years

U-235   h1/2 = 700 million years

At the beginning both Isotopes where present in equal amount

Determine the T years before the amount of Uranium decays to Half

T =   ?    N'2 = N1 / 2

we know that N = No ( 1/2 )^h   where h = time / half-life time

attached below is the detailed solution of the given problem

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Answer:

The pKa of X-281 is 3.73.

Explanation:

X-281 is a monoprotic weak acid so that it will not dissociate completely in solution (weak acid) and will only produce 1 mol of protons per mol compound (monoprotic acid).

The dissociation equation could be written as follows:

X-281-H ⇆ X-281 + H⁺

Note the equilibrium arrows indicating that not all X-281-H dissociates at equilibrium.

Initially, the concentration of X-281-H is 0.089 M. At equilibrium, the concentration of the dissociation products, X-281 and H⁺, is unknown but both must be the same, since the drug is a monoprotic acid. We can call "X" to the concentration of the products. The concentration of X-281-H is the initial concentration minus the concentration of the products: 0.089 M - X. Then, at equilibrium, these are the concentrations of each species present:

  X-281-H               ⇆      X-281        +       H⁺

(0.089 M-X)                         X                    X

We also know that the pH is 2.40. Then:

pH = -log[H⁺] = 2.40

where [H⁺] is the molar concentration of the protons or "X". Then:

-log X = 2.40

log X = -2.40

X = 10^(-2.40) = 3.98 x 10⁻³ M

The concentration of each species present in the equilibrium is then:

[H⁺] = 3.98 x 10⁻³ M

[X-281] = 3.98 x 10⁻³ M

[X-281-H] = 0.089 M - 3.98 x 10⁻³M = 0.085 M

At equlibrium, the acidity constant Ka is:

Ka = [X-281] * [H⁺] / [X-281-H]

Ka = (3.98 x 10⁻³ M * 3.98 x 10⁻³ M ) / 0.085 M = 1.86 x 10⁻⁴

Then the pKa is:

pKa = -log Ka = -log (1.86 x 10⁻⁴) = <u>3.73</u>

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