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
3.
protium (A = 1), deuterium (A = 2), and tritium (A = 3).
Answer:
Concept: Chemical Analysis
- You need to start by graphing the data and then analyzing it.
- We can see that the horse has a distance in meters of 980 at the end of the 10 seconds hence it is the fastest.
- The horse line has a linearly representation, while the alternate line has parabolic tendencies towards the end. The steeper line indicates a faster change in time or velocity which results in a greater distance traveled indicating that the horse is faster.
- *I have confidence you can graph that*
A threat is a potential risk loss to an asset
Answer:
5000 and 
indicate that there is more B than A at equilibrium
Explanation:
For the given reaction: ![K=\frac{[B]}{[A]}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BB%5D%7D%7B%5BA%5D%7D)
where [B] and [A] represents equilibrium concentration B and A respectively. K represents equilibrium constant
More B than A at equilibrium means, [B] > [A]
So, ![K=\frac{[B]}{[A]}>1](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BB%5D%7D%7B%5BA%5D%7D%3E1)
As, both 5000 and are greater than 1 therefore these two K values indicate that there is more B than A at equilibrium
