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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Answer:
Noble Gases
Explanation:
The Noble Gases have a full valence shell of 8 electrons. They are stable in that sense.
Answer:
The percent by mass of water in this crystal is:
Explanation:
This exercise can be easily solved using a simple rule of three where the initial weight of the hydrated crystal (6,235 g) is taken into account as 100% of the mass, and the percentage to which the mass of 4.90 g corresponds (after getting warm). First, the values and unknown variable are established:
- 6,235 g = 100%
- 4.90 g = X
And the value of the variable X is found:
- X = (4.90 g * 100%) / 6,235 g
- X = approximately 78.6%.
The calculated value is not yet the percentage of the water, since the water after heating the glass has evaporated, therefore, the remaining percentage must be taken, which can be calculated by subtraction:
- Water percentage = Total percentage - Percentage after heating.
- <u>Water percentage = 100% - 78.6% = 21.4%</u>
The balanced chemical equation for the above reaction is as follows;
2LiOH + H₂SO₄ ---> Li₂SO₄ + 2H₂O
stoichiometry of base to acid is 2:1
Number of OH⁻ moles reacted = number of H⁺ moles reacted at neutralisation
Number of LiOH moles reacted = 0.400 M / 1000 mL/L x 20.0 mL = 0.008 mol
number of H₂SO₄ moles reacted - 0.008 mol /2 = 0.004 mol
Number of H₂SO₄ moles in 1 L - 0.500 M
This means that 0.500 mol in 1 L solution
Therefore 0.004 mol in - 1/0.500 x 0.004 = 0.008 L
therefore volume of acid required = 8 mL
The answer is c hopefully I helped you
