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cupoosta [38]
2 years ago
11

How many joules are released when 1.5 mol of ²³⁹Pu decays, if each nucleus releases 5.243 MeV?

Chemistry
1 answer:
Nat2105 [25]2 years ago
6 0

7.6×10¹¹ joules are released when 1.5 mol of ²³⁹Pu decays, if each nucleus releases 5.243 MeV.

<h3>Briefing :</h3>

At the nuclear level, electron volts, or EVE's, are frequently used to express energy fluctuations.

1 electron = 1.602 × 10⁻¹¹ joules.

We frequently measure things in mega electron volts rather than electron volts since this little quantity of energy.

Therefore, one million or six electron volts is equal to one mega electron volt. If a nuclear decay produces a particular quantity of energy, for instance, we know that every person's nucleus produces 5.243 mega electron volts when it decays.

In accordance with the query, we can calculate how much energy and joules should be created using the supplied quantity of 1.5 mol. So, we must investigate.

First, let's look at the number of nuclei present in the sample. In this scenario, there are 1.5 moles, and we know that each mole has 6.2 × 10²³ particles, or nuclei, and that each nucleus will release 5.243 mega electron volts. We shall transition to electron volts now that we have our energy and mega electron volts. For each Mega electron volt, there is a single time to tend to the six electron volts.

To finally transform it to one Joule electron volt. This is 1.602 x 10⁻¹⁹ joules. So, if we multiply 1.5 by 6.2 tens, it comes out to 10²³rd times. 1.602 times 10⁻¹⁹ is equal to 5.243 × 10⁶.

As a result, the energy in joules is 7.6 x 10¹¹ joules.

To learn more about the, Joules :

brainly.com/question/25947916

#SPJ4

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so, A would be here = gravity

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28.7 l of propane, c3h8, are consumed in a combustion reaction. how many grams of water are produced?
lora16 [44]
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Hope this answers the question.
7 0
4 years ago
Consider the nuclear equation below. Superscript 235 subscript 92 upper U right arrow superscript 4 subscript 2 upper H e. What
coldgirl [10]

Answer:

\rm_{90}^{231}\text{Th}

Explanation:

The unbalanced nuclear equation is

\rm _{92}^{235}\text{U} \longrightarrow \,  _{2}^{4}\text{He} + X

Let's write X as a nuclear symbol.

\rm _{92}^{235}\text{U} \longrightarrow \,  _{2}^{4}\text{He} + _{Z}^{A}\text{X}

The main point to remember in balancing nuclear equations is that the sums of the superscripts and of the subscripts must be the same on each side of the reaction arrow.

Then

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\rm _{92}^{235}\text{U} \longrightarrow \,  _{2}^{4}\text{He} +\, _{90}^{231}\text{X}

Element 90 is thorium, so  

\rm X = _{90}^{231}\text{Th}

7 0
4 years ago
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