Answer:
The specific heat of copper when heated to 221.32 (not listed form of heat measurement) is 221.32 (not listed form of heat measurement).
Explanation:
uh not really sure what else there is here, I may be missing something
Answer:
the particles of matter are arranged in different ways for the different states
Explanation:
because solid liquid and gas all three matters have different states for example the particles in a solid are closely packEd and form of movement is vibration
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:
Learn more:
Answer:
Al,Ga,B
Explanation:
Now since i helped you can you help me with this plz
Matteo took 5 math quizzes. The mean of the 5 quizzes was 8.2. Here are four of his quiz scores 7, 7, 8, 10. What is the 5th quiz score? Show work.
Volume of Argon V1 = 5.0 L
Pressure of Argon P1 = 2 atm
Final temperature T2 = 30 C = 30 + 273 = 303 K
Volume at final temperature V2= 6 L
Pressure at final temperature P2 = 8 atm
We know that (P1 x V1) / T1 = (P2 x V2) / T2
(2 x 5)/ T1 = (8 x 6)/ 303 => T1 = (10 x 303) / 48
Initial Temperature T1 = 3030 / 48 = 63.12
Initial Temperature = -209. 8 C
