Answer:
the heat rate required to cool down the gas from 535°C until 215°C is -2.5 kW.
Explanation:
assuming ideal gas behaviour:
PV=nRT
therefore
P= 109 Kpa= 1.07575 atm
V= 67 m3/hr = 18.6111 L/s
T= 215 °C = 488 K
R = 0.082 atm L /mol K
n = PV/RT = 109 Kpa = 1.07575 atm * 18.611 L/s /(0.082 atm L/mol K * 488 K)
n= 0.5 mol/s
since the changes in kinetic and potencial energy are negligible, the heat required is equal to the enthalpy change of the gas:
Q= n* Δh = 0.5 mol/s * (- 5 kJ/mol) =2.5 kW
Answer:
PART A
Explanation:
3. A cylinder of compressed gas has a pressure of 4.882 atm on one day. The next
day, the same cylinder of gas has a pressure of 4.690 atm, and its temperature is
8°C. What was the temperature on the previous day in °C? Ans: 20°C.
Iron oxide is rust. So oil would be an inhibitor.
Answer:
Temperate Deciduous Forest
Explanation:
