A coefficient is the number or letter before the variable. The variable is the unknown value.
For your first question, that equation only works if your situation is occurring at a constant temperature. Your original question is such a situation - everything occurs at 298.15 K. Therefore, you can use this value in the equation to calculate work.
For your second question, Charles' Law describes how the volume of gas changes as you heat or cool it, PROVIDED PRESSURE AND MOLES OF GAS REMAIN CONSTANT THE WHOLE TIME. In your original question above, temperature stays constant while volume changes. However, what they don't tell you is that this necessarily requires a change in either pressure or moles of gas. Because the question works with the same sample the of gas the whole time (i.e. moles are constant), it is pressure that is changing (and this change will occur according to Boyle's Law, since temperature and moles are held constant).
Hope that clarifies things!
Answer: multiple apply them together then divide them
Explanation:
The balanced nuclear equation for the β emission of the following isotopes is seen below:
92 92 0
Sr ⇒ Y + e
38 39 -1
<h3>
What is Beta emission?</h3>
This is also known as beta decay in which a beta ray is emitted from an atomic nucleus.
The element formed during the beta emission of strontium is referred to as Yttrium.
Read more about Beta emission here brainly.com/question/16334873
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This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.
In this case, it is recommended to write the enthalpy for each substance as follows:
Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and 
and 
are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:
It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:
Finally we convert this result to kJ:
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