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:
that are formed from hybridized orbitals
Explanation:
The chemical concept of resonance is when a change in the position of the electrons occurs, without changing the position of the atoms.
The structure obtained in the resonance will not be any of the previous ones, but a hybrid of resonance between those structures.
Answer:
5
Explanation:
Given that the formula is;
1/λ= R(1/nf^2 - 1/ni^2)
λ = 93.7 nm or 93.7 * 10^-9 m
R= 1.097 * 10^7 m-1
nf = ?
ni = 1
From;
ΔE = hc/λ
ΔE = 6.63 * 10^-34 * 3* 10^8/93.7 * 10^-9
ΔE = 21 * 10^-19 J
ΔE = -2.18 * 10^-18 J (1/nf^2 - 1/ni^2)
21 * 10^-19 J = -2.18 * 10^-18 J (1/nf^2 - 1/ni^2)
21 * 10^-19/-2.18 * 10^-18 = (1/nf^2 - 1/1^2)
-0.963 = (1/nf^2 - 1)
-0.963 + 1 = 1/nf^2
0.037 = 1/nf^2
nf^2 = (0.037)^-1
nf^2 = 27
nf = 5
Answer:Yes they are in the same mineral group
Explanation:zinc is the central elements there. The rest of the elements are present as impurities due to where it was found. Like carbon is can be found in the soil, silicon with oxygen is basically sand, hydrogen is in the atmosphere and also in water and soil too. So apart from zinc, the rest are normal day to day elements.
