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
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The rate of a reaction with this rate law would increase by a factor of 4 if NO concentration were doubled.
Answer:
it is a nice question....my mind tells me that the first is it use me as a good vibes and can use to anything the second i will do my best too absorbe it.
Explanation:
Hope this help...
The electron subshells in the atoms of group 5a elements discourage electron addition.
This is because the subshells in the group 5a elements are already half-filled, which is a more stable electronic configuration that that which would be achieved if another electron was added.
Meanwhile, group 4a elements' subshell becomes half-filled when an electron is added, making them somewhat stable. This means that they prefer electron addition.
