The true statement about the balanced equations for nuclear and chemical changes is; both are balanced according to the total mass before and after the change.
A basic law in science is called the law of conservation of mass. Its general statement is that mass can neither be created nor destroyed.
Both in chemical and nuclear changes, mass is involved and in both cases, the law of conservation of mass strictly applies.
This means that for both chemical and nuclear changes; total mass before reaction must be equal to total mass after reaction.
Hence, both reactions are balanced according to the total mass before and after the change.
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Idk it’s making me answer a question in order to get help on my own
Considering the Charles' law, the gas would have a temperature of -109.2 C.
<h3>Charles' law</h3>
Finally, Charles' law establishes the relationship between the volume and temperature of a gas sample at constant pressure. This law says that the volume is directly proportional to the temperature of the gas. That is, if the temperature increases, the volume of the gas increases, while if the temperature of the gas decreases, the volume decreases.
Charles' law is expressed mathematically as:

If you want to study two different states, an initial state 1 and a final state 2, the following is true:

<h3>Temperature of the gas in this case</h3>
In this case, you know:
- P1= 1800 psi
- V1= 10 L
- T1= 20 C= 293 K (being 0 C= 273 K)
- P2= 1800 psi
- V2= 6 L
- T2= ?
You can see that the pressure remains constant, so you can apply Charles's law.
Replacing in the Charles's law:

Solving:


<u><em>T2=163.8 K= -109.2 C</em></u>
The gas would have a temperature of -109.2 C.
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