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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AgNO3+NaCl yields AgCl+NaNo3 (reduction)
...that's the only one I know
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Answer:
K^+ and NO3^-
Explanation:
In a balanced ionic equation, we usually see the species that react to yield the main product in the reaction.
Consider the reaction;
Pb(NO3)2(aq) +2 KI(aq) -------> PbI2(s) + 2KNO3(aq)
The main product in this reaction is PbI2. Hence the balanced ionic equation is;
Pb^2+(aq) + 2I^-(aq) ------> PbI2(s)
Notice that K^+ and NO3^- did not participate in this reaction. All ions that are part of the molecular equation but do not participate in the ionic reaction equation are called spectator ions. Hence K^+ and NO3^- are spectator ions in this reaction as can be seen clearly above.