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:
Assume that 100 grams of C2H4 is present. This means that there are 85.7 grams of carbon and 14.3 grams of hydrogen.
Convert these weights to moles of each element:
85.7 grams carbon/12 grams per mole = 7 moles of carbon.
14.3 grams hydrogen/1 gram per mole = 14 moles of hydrogen.
Divide by the lowest number of moles to obtain one mole of carbon and two moles of hydrogen.
Since we know that there cannot be a stable CH2 molecule, multiply by two and you have C2H4 which is ethylene - a known molecule.
The secret is to convert the percentages to moles and find the ration of the constituents.
Answer:C.
large leaves
Explanation: I took the test
Since there is 9.47 x 1021 atoms of copper in 1 gram, and 8.96 grams/cc is the density of copper, then 9.47 x 8.96 =84.85 x 1021 atoms of copper in the one cubic centimeter of copper.
