Reasons
1 answer:
Answer:
(1) The two column proof is presented here as follows;
Statement Reasons
1. C is the midpoint of
Given
B is the midpoint of
D is the midpoint of
2. AC = CE, AB = BC, CD = DE Definition of midpoint
3. AB + BC = AC, CD + DE = CE Segment addition postulate
4. CD + DE = AC Substitution property of equality
5. AB + BC = CD + DE Substitution property of equality
6. BC + BC = CD + CD Substitution property of equality
7. 2·BC = 2·CD Addition of two identical quantities
8. BC = CD Division property of equality
9. BC ≅ CD Definition of Congruence
(2) The two column proof for the triangular geometric figure is presented here as follows;
Statement Reasons
1. 2·WV = XY; 2·YZ = XW; WV = YZ Given
2. 2·WV = 2·YZ Multiplication property
3. XY = XW Substitution Property
4. XW + WV = XV Segment Addition Property
XY + YZ = XZ
5. XW + WV = XV Substitution Property
XW + WV = XZ
6. XV = XZ Substitution Property
Explanation:
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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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