Answer:
The heat needed to convert 1 kg of feed water at 20°C into dry saturated steam at a pressure of 9 bar is 2690.19 kJ/kg.
Explanation:
Step 1 : Obtain the enthalpy of staurated steam and enthalpy of evaporation at 9 bar pressure from the steam table:
From steam table, at 9 bar, the enthalpy of saturated water = 742.83 kJ /kg
enthalpy of evaporation = 2031.1 kJ /kg
Step 2: Calculate the enthalpy of dry saturated steam:
Enthalpy of dry saturated steam = enthalpy of saturated water + enthalpy of evaporation
= 742.83 + 2031.1 = 2773. 93 kJ/ kg
Step 3: Calculate the enthalpy of 1 kg of feed water at 20°C
Enthalpy of 1 kg of feed water = c * ( T2 -T1)
= 4.187 * (20-0)
= 83.74 kJ /kg
Step 4 : calculate the heat needed by 1 kg of feed water at 20°C to be converted to dry saturated steam at 9 bar:
Heat needed = Enthalpy of dry saturated steam - enthalpy of feed water
Heat needed = 2773.93 kJ/kg - 83.74 kJ/kg
Heat = 2690.19 kJ/kg
Answer:
3p^2
Explanation:
after filling 3s^2 only two electrons left out of 14 so the next sub shell is 3p therefore ,X represents 3p^2
Answer: 0.335 moles
Explanation:
Given that,
Amount of moles of KBr solute (n) = ?
Volume of KBr solution (v) = 0.67 L
Concentration of KBr solution (c) = 0.50M
Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence
c = n / v
make n the subject formula
n = c x v
n = 0.50M x 0.67L
n = 0.335 moles
Thus, there are 0.335 moles of KBr in 0.67 L of a 0.50 M KBr solution
Well if mutations in either the EGFR or the KRAS lead to the production of a protein that is constantly turned on. As a result of this cells will be constantly alerted to multiply, leading to a formation of a tumor, and when this happens in the lungs it creates lung cancer.
EGFR- Estimated glomerular filter rate
KRAS- Kirsten rat sarcoma