Chemistry, 20.09.2019 19:30 ayoismeisalex

How much of a 1.00 mg sample of americium remains after 3 half-lives?

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Answer from: Bryoh

0.125mg americium

Half-lives is a set of duration when a sample will become half of their original mass. In this case, the initial mass is 1mg and the sample is going for 3 times half-lives. Then the calculation would be: 1mg $1/2^{3}$ = 1/8 mg= 0.125mg.

Answer from: dmccarthey90

A 1 mg sample of americium becomes $\boxed{0.125{\text{ mg}}}$ after 3 half-lives.

Further Explanation:

The process in which an unstable nucleus releases energy in the form of different particles like alpha particles, beta particles, and gamma particles is termed as radioactivity.

Half-life

As is evident from its name, it is the time after which the amount of radioactive substances becomes half of its original amount. It is represented by .

The expression to calculate the amount of substance is as follows:

${{\text{A}}_{\text{0}}} = {{\text{A}}_{\text{t}}}{\left( 2 \right)^{\text{n}}}$ …… (1)

Here,

${{\text{A}}_{\text{t}}}$ is the final amount of substance.

${{\text{A}}_{\text{0}}}$ is the initial amount of substance.

n is the number of half-lives.

Rearrange equation (1) for ${{\text{A}}_{\text{t}}}$ .

${{\text{A}}_{\text{t}}} = \left( {\dfrac{{\text{1}}}{{{{\left( 2 \right)}^{\text{n}}}}}} \right)\left( {{{\text{A}}_{\text{0}}}} \right)$ …… (2)

The value of ${{\text{A}}_{\text{0}}}$ is 1 mg.

The value of n is 3.

Substitute 1 mg for ${{\text{A}}_{\text{0}}}$ and 3 for n in equation (2) to calculate the amount of americium.

$\begin{aligned}{{\text{A}}_{\text{t}}}&= \left( {\frac{{\text{1}}}{{{{\left( 2\right)}^{\text{3}}}}}} \right)\left( {{\text{1 mg}}} \right)\\&=\left( {\frac{1}{8}} \right)\left( {1{\text{ mg}}} \right)\\&= 0.125{\text{ mg}}\\\end{aligned}$

Therefore 0.125 mg of americium remains after 3 half-lives.

Learn more:

Find the age of the fabric: link Which nuclear emission is negatively charged? link

Answer details:

Grade: Senior School

Chapter: Nuclear chemistry

Subject: Chemistry

Keywords: half-life, 1 mg, 0.125 mg, americium, n, 3 half-lives, radioactivity, unstable nucleus, t1/2, substance, initial amount, final amount, number of half-lives, alpha particles, beta particles, gamma particles, energy, half, amount of substance, original amount.

Answer from: Quest

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

Answer from: Quest

when the volume of gas is changed from 717.73 ml to 852 ml the temperature will change from 315 c to 452 c

calculation

this is calculated using the charles law formula

that is, v1/t1 = v2/t2

v1=?

t1 = 315 c into kelvin = 315 +273 = 588 k

v2 = 852 ml

t2 = 425c + 273 =698 k

make v1 the subject of the formula ( by multiplying both side by t1)

v1=( v2 xt1) / t2

v1= (852 ml x 588 k)/ 698 k= 717.73 ml

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