Chemistry, 04.12.2020 17:00 ayoismeisalex

The half life for uranium-235 is 7.0x10 8years. a. How many half-lives did the sample go through at the end of 2.8x10 9years? b. How much of a 0.74mg sample of uranium-235 will remain after 2.8x10 9years?

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Answer from: firdausmohammed80

The amount of uranium sample remained after 4 cycles in $\rm 2.8\;\times\;10^9$ years has been 0.04625 mg.

The half-life can be described as the time required by the element to reduce to its half concentration from the initial concentration.

A. The number of half-life cycles can be calculated as:

$\rm 7.0\;\times\;10^8$ = 1 cycle

$\rm 2.8\;\times\;10^9$ = $\rm \dfrac{1}{\rm 7.0\;\times\;10^8}\;\times\;2.8\;\times\;10^9$

= 4 cycles.

The number of half-life cycles after $\rm 2.8\;\times\;10^9$ years are 4 cycles.

B. The amount of sample remained can be calculated as:

Sample remained = Initial sample $\rm \times\;\dfrac{1}{2}^\frac{time}{Half-life}$

Sample remained = 0.74 mg $\rm \times\;\dfrac{1}{2}^\frac{2.8\;\times\;10^9}{7.0\;\times\;10^8}$

Smaple remianed = 0.74 $\rm \times\;\dfrac{1}{2}^4$

Sample remained = 0.74 $\times$ 0.0625 mg

Sample remained = 0.04625 mg.

The amount of uranium sample remained after 4 cycles in $\rm 2.8\;\times\;10^9$ years has been 0.04625 mg.

For more information about the half-life, refer to the link:

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Answer from: iamabouttofail

A 4 half-life

B. 0.05 mg

Explanation:

A. Determination of the number of half-lives after 2.8×10⁹ years.

From the question given above,

7×10⁸ years = 1 half life

Therefore

2.8×10⁹ years = 2.8×10⁹ years × 1 half life / 7×10⁸ years

2.8×10⁹ years = 4 half life

Thus, the sample went through 4 half-lives at the end of 2.8×10⁹ years.

B. Determination of the amount of the sample remaining after 2.8×10⁹ years.

Original amount (N₀) = 0.74 mg

half life (t½) = 7×10⁸ years

Time (t) = 2.8×10⁹ years

Amount remaining (N) =?

Next, we shall determine the rate of disintegration. This can be obtained as follow:

half life (t½) = 7×10⁸ years

Decay constant (K) =?

K = 0.693 / t½

K = 0.693 / 7×10⁸

K = 9.9×10¯¹⁰ /year

Finally, we shall determine the amount remaining as follow:

Original amount (N₀) = 0.74 mg

Time (t) = 2.8×10⁹ years

Decay constant (K) = 9.9×10¯¹⁰ /year

Amount remaining (N) =?

Log (N₀/N) = kt / 2.303

Log (0.74/N) = 9.9×10¯¹⁰×2.8×10⁹ /2.303

Log (0.74/N) = 2.772 / 2.303

Log (0.74/N) = 1.2036

Take the antilog of 1.2036

0.74/N = antilog (1.2036)

0.74 / N = 15.98

Cross multiply

0.74 = N × 15.98

Divide both side by 15.98

N = 0.74 / 15.98

N = 0.05 mg

Thus, 0.05 mg of the sample will remain after 2.8×10⁹ years

Answer from: Quest

the role of the microscope world in creating new materials .

Answer from: Quest

it's filtration, but this is an assumption so my apologies if i got this

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The half life for uranium-235 is 7.0x10 8years. a. How many half-lives did the sample go through at...

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