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
Answer:
Nitrogen and Oxygen make up 99% of the Earths atmosphere. Then Argon makes up the rest of the atmosphere.
In an experiement things that are changing are called variables.
NO is the limiting reagent and 4.34 g is the amount of the excess reagent that remains after the reaction is complete
<h3>What is a limiting reagent?</h3>
The reactant that is entirely used up in a reaction is called as limiting reagent.
The reaction:
→ 
Moles of nitrogen monoxide
Molecular weight: 
=30g/mol
Moles of hydrogen
Molecular weight: 
=30g/mol
Hydrogen gas is in excess.
NO is the limiting reagent.
The amount of the excess reagent remains after the reaction is complete.
(2.9 mol- 0.73 mol NO x 
) x 
4.34 g
Learn more about limiting reagents here:
brainly.com/question/26905271
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Also give you a 10, the last one looks really nice
