Ask question

Ask question

All categories

3 years ago

10

Plutonium−239 (t1/2 = 2.41 × 104 yr) represents a serious nuclear waste hazard. if seven half-lives are required to reach a tole

rable level of radioactivity, how long must 239 pu be stored? enter your answer in scientific notation.

2 answers:

dimulka [17.4K]3 years ago

5 0

The half-life of Plutonium−239, t1/2 is 2.41 × 10⁴<span> yrs

time taken to reach tolerable level = seven half-lives
= 7 x t1/2
= 7 x </span>2.41 × 10⁴ yrs
= 168700 yrs
= 1.687 x 10⁵ yrs

Hence, the period of time that <span>Plutonium-239 must be stored is </span>1.687 x 10⁵ years.

meriva3 years ago

4 0

Plutonium − 239 must be stored for = 1,687 .10 ^ 5 years to reach a tolerable level of radioactivity

<h3><em>Further explanation</em></h3>

The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.

Usually radioactive elements have an unstable atomic nucleus.

The main particles emitted by radioactive elements so that they generally decay are alpha (α), beta (β) and gamma (γ) particles

General formula used in decay:

$\large {\boxed{\bold{ {Nt = No. (\frac {1} {2}) ^ {\frac {T} {t1 / 2}}}} }$

Information :

T = duration of decay

t 1/2 = elemental half-life

N₀ = the number of initial radioactive atoms

Nt = the number of radioactive atoms left after decaying during T time

Plutonium − 239 has a half-life of t1/2 = 2.41 × 10⁴ years

To reach a tolerable level of radioactivity, Plutonium − 239 must be stored for 7 times half-life, namely:

= 7 x 2.41 × 10⁴ years

= 1,687 .10⁵ years

<h3><em>Learn more</em></h3>

The identity of the missing nucleus

brainly.com/question/4207569

the process of radioactive decay

brainly.com/question/1770619

nuclear fusion and nuclear fission reactions

brainly.com/question/1949468

the product of radioactive decay

brainly.com/question/6029197

Keywords: decay, radioactive, Plutonium − 239, The atomic nucleus

You might be interested in

Complete combustion of a 17.12mg sample of xylene In oxygen yielded 56.77mg

Veronika [31]

Xylene moles =\frac{17.12}{106.16×1000}=0.00016moles=

106.16×1000

17.12

=0.00016moles

Moles of CO_2 =\frac{56.77}{44.01×1000}=0.0013CO

2

=

44.01×1000

56.77

=0.0013

Moles of H_2O= =\frac{14.53}{18.02×1000}=0.0008H

2

O==

18.02×1000

14.53

=0.0008

Moles ratios

\frac{0.0013}{0.0008}=1.625

0.0008

0.0013

=1.625

\frac{0.0008}{0.0008}=1

0.0008

0.0008

=1

Hence molecular fomula

The empirical formula is C 4H 5.

The molecular formula C8H10

8 0

2 years ago

In addition to mass balance, oxidation-reduction reactions must be balanced such that the number of electrons lost in the oxidat

Vaselesa [24]

Answer:

OH−(aq), and H+(aq)

Explanation:

Redox reactions may occur in acidic or basic environments. Usually, if a reaction occurs in an acidic environment, hydrogen ions are shown as being part of the reaction system. For instance, in the reduction of the permanganate ion;

MnO4^-(aq) + 8H^+(aq) +5e-------> Mn^2+(aq) + 4H2O(l)

The appearance of hydrogen ion in the reaction equation implies that the process takes place under acidic reaction conditions.

For reactions that take place under basic conditions, the hydroxide ion is part of the reaction equation.

Hence hydrogen ion and hydroxide ion are included in redox reaction half equations depending on the conditions of the reaction whether acidic or basic.

4 0

3 years ago

The Lewis dot model of a molecule is shown.

Katen [24]

Answer:

Carbon has a total of four bonded pairs of electrons around it.

Explanation:

Since there are four "lines" around the C (which stands for Carbon), we can conclude that Carbon has a total of four bonded pairs of electrons around it.

3 0

3 years ago

Refer to the periodic table tool and write the electron configurations of the following elements in both long and short terms

ahrayia [7]

Answer:-

Carbon

[He] 2s2 2p2

1s2 2s2 2p2.

potassium

[Ar] 4s1.

1s2 2s2 2p6 3s2 3p6 4s1

Explanation:-

For writing the short form of the electronic configuration we look for the nearest noble gas with atomic number less than the element in question. We subtract the atomic number of that noble gas from the atomic number of the element in question.

The extra electrons we then assign normally starting with using the row after the noble gas ends. We write the name of that noble gas in [brackets] and then write the electronic configuration.

For carbon with Z = 6 the nearest noble gas is Helium. It has the atomic number 2. Subtracting 6 – 2 we get 4 electrons. Helium lies in 1st row. Starting with 2, we get 2s2 2p2.

So the short term electronic configuration is [He] 2s2 2p2

Similarly, for potassium with Z = 19 the nearest noble gas is Argon. It has the atomic number 18. Subtracting 19-18 we get 1 electron. Argon lies in 3rd row. Starting with 4, we get 4s1.

So the short electronic configuration is [Ar] 4s1.

For long term electronic configuration we must write the electronic configuration of the noble gas as well.

So for Carbon it is 1s2 2s2 2p2.

For potassium it is 1s2 2s2 2p6 3s2 3p6 4s1

5 0

3 years ago

What are hydrated substances

Keith_Richards [23]

Answer:

Glauber's salt

washing soda

5 0

3 years ago