Explanation:
It is known that efficiency is denoted by 
.
The given data is as follows.
= 0.82, 
= (21 + 273) K = 294 K
= 200 kPa, 
= 1000 kPa
Therefore, calculate the final temperature as follows.
0.82 = 
= 1633 K
Final temperature in degree celsius = 
= 
Now, we will calculate the entropy as follows.
For 1 mole, 
It is known that for 
the value of 
= 0.028 kJ/mol.
Therefore, putting the given values into the above formula as follows.
= 
= 0.0346 kJ/mol
or, = 34.6 J/mol (as 1 kJ = 1000 J)
Therefore, entropy change of ammonia is 34.6 J/mol.
Answer:
The answer is A (number 1)
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
According to the octet rule, atoms tend to gain, lose, or share electrons until they are surrounded by__8__ valence electrons.
w/w percentage <span>
= mass of the pure compound /
total mass of the sample x 100%
70% HNO₃
contains by mass means every 100 g of sample has 70 g of HNO₃.</span><span>
The mass of solution = 103.8 g
Hence the mass of HNO₃ = 103.8 g x 70%</span><span>
= 103.8 g x (70 / 100)
<span>
= 72.66 g = 72.7 g.</span></span>
