Answer:
The average kinetic energy of A is greater than that of B.
Explanation:
The temperature of an object is directly proportional to the average kinetic energy of the particles in the object. For instance, for an ideal gas, we have
where
KE is the kinetic energy
k is the Boltzmann constant
T the absolute temperature of the gas
Therefore, this means that in a hotter object the average kinetic energy of the particles is higher than the average kinetic energy of the particles in a colder object.
Moreover, the laws of thermodynamics tell us that heat is always transferred from a hotter object (higher temperature) to a colder object (lower temperature).
In this problem heat is transferred from sample A to sample B. Therefore, this means that object A has higher temperature, and therefore, higher average kinetic energy. So the correct answer is
The average kinetic energy of A is greater than that of B.
Atomic mass Ca = 40.078 a.m.u
40.078 g -------------------- 6.02x10²³ atoms
72.8 g ---------------------- ??
72.8 x ( 6.02x10²³) / 40.078 =
4.38x10²⁵ / 40.078 = 1.093x10²⁴ atoms
hope this helps!
The correct answer is A, since it is the only sensible answer.
Answer:-ΔG=-101.5KJ
Explanation:We have to calculate ΔG for the reaction so using the formula given in the equation we can calculate the \Delta G for the reaction.
We need to convert the unit ofΔS in terms of KJ/Kelvin as its value is given in terms of J/Kelvin
Also we need to convert the temperature in Kelvin as it is given in degree celsius.
After calculating forΔG we found that the value ofΔG is negative and its value is -101.74KJ
For a reaction to be spontaneous the value of \Delta G \ must be negative .
As the ΔG for the given reaction is is negative so the reaction will be spontaneous in nature.
In this reaction since the entropy of reaction is positive and hence when we increase the temperature term then the overall term TΔS would become more positive and hence the value of ΔG would be less negative .
Hence the value of ΔG would become more positive with the increase in temperature.
So we found the value of ΔG to be -101.74KJ
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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