First, let's convert atm to kPa:
2atm x 101.33kPa = 202.66kPa
Now, to find the pressure, you would have to use Boyle's Law: (P1V1)=(P2V2)
(150kPa x 3.0L) = (202.66kPa x V2)
V2= <span>(150kPa x 3.0L)/202.66kPa
V2=2.2L, B is your answer. This makes because volume and pressure have an inverse relationship, so since pressure increased volume should decrease.
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When sodium metal reacts with chlorine gas, the product would be sodium chloride or the table salt. The balanced chemical reaction would be:
2Na + Cl2 = 2NaCl
IN balancing reactions, it is important to remember that the number of atoms at each side should be equal. Hope this answers the question.
Answer:
There are 3 steps of this problem.
Explanation:
Step 1.
Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.
Step 2.
Enthalpy of saturated liquid Haq = 781.124 J/g
Enthalpy of saturated vapour Hvap = 2779.7 J/g
Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g
Step 3.
In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy
So, H1=H2
H2= (1-x)Haq+XHvap.........1
Putting the values in 1
2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}
= 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)
1904.976 (J/g) = x1998.576 (J/g)
x = 1904.976 (J/g)/1998.576 (J/g)
x = 0.953
So, the quality of the wet steam is 0.953
