Answer:
The iron core, copper wire, and an electricity source.
Explanation: Me
KOH -------> K+ OH-
Ba(OH)2 ------> Ba+2. 2OH-
Answer:
Explanation:
8.61+5.779 = 14.389 = 1.4389 × 10^1
25 - 12.5 = 1.25 x 10^1
56.35 / 13.2 = 4.2689
The salt causes the water to freeze at a lower temperature. When a solute, aka salt, is introduced to the system, the freezing point is lowered. This makes the water freeze at a lower temperature.
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
