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garik1379 [7]
2 years ago
5

The reactions listed below are either chemical reactions or nuclear reactions. which are nuclear reactions? check all that apply

. upper h superscript plus, plus upper o upper h superscript minus right arrow upper h subscript 2 upper o. superscript 185 subscript 79 upper a u right arrow superscript 181 subscript 77 upper r d plus superscript 4 subscript 2 upper h e. 4 upper f e plus 3 upper o subscript 2 right arrow 2 upper f e subscript 2 upper o subscript 3. superscript 210 subscript 84 upper p o right arrow superscript 206 subscript 82 upper p b plus superscript 4 subscript 2 upper h e.
Chemistry
1 answer:
prohojiy [21]2 years ago
8 0

The equations that are nuclear reactions are as follows:

  • superscript 185 subscript 79 upper Au right arrow superscript 181 subscript 77
  • superscript 210 subscript 84 upper Po right arrow superscript 206 subscript 82 upper Pb plus superscript 4 subscript 2 upper He.

<h3>What is nuclear reaction?</h3>

A nuclear reaction is a reaction that involves the fission of an atomic nucleus, or the fusion of one or more atomic nuclei and/or subatomic particles in which the number of protons and/or neutrons in a nucleus changes.

The products of a nuclear reaction may contain a different element or a different isotope of the same element.

Therefore, the equations that are nuclear reactions are as follows:

  • superscript 185 subscript 79 upper Au right arrow superscript 181 subscript 77
  • superscript 210 subscript 84 upper Po right arrow superscript 206 subscript 82 upper Pb plus superscript 4 subscript 2 upper He.

Learn more about nuclear reaction at: brainly.com/question/19752321

#SPJ4

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Hey there!
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3 years ago
Read 2 more answers
The ideal gas heat capacity of nitrogen varies with temperature. It is given by:
hammer [34]

Answer:

A)  1059 J/mol

B)  17,920 J/mol

Explanation:

Given that:

Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4

R (constant) = 8.314

We know that:

C_p=C_v+R

We can determine C_v from above if we make C_v the subject of the formula as:

C_v=C_p-R

C_V = 29.42-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4-8.314

C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4

A).

The formula for calculating change in internal energy is given as:

dU=C_vdT

If we integrate above data into the equation; it implies that:

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 1059J/mol

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 17,920 J/mol

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