The missing components in the table to the right are indicated with orange letters. Use the periodic table in the tools bar and this link Web Elements to fill in the corresponding values. A B C D E F G. 2. See answers. Log in to add ... F = 737.7kJ/mol. G = 495.8kJ/mol. Explanation: We are asked some of the ...
2 answers
Henderson–Hasselbalch equation is given as,
pH = pKa + log [A⁻] / [HA]
-------- (1)
Solution:
Convert Ka into pKa,
pKa = -log Ka
pKa = -log 1.37 × 10⁻⁴
pKa = 3.863
Putting value of pKa and pH in eq.1,
4.29 = 3.863 + log [lactate] / [lactic acid]
Or,
log [lactate] / [lactic acid] = 4.29 - 3.863
log [lactate] / [lactic acid] = 0.427
Taking Anti log,
[lactate] / [lactic acid]
= 2.673
Result:
2.673 M
lactate salt when mixed with 1 M Lactic acid produces a buffer of pH = 4.29.
I think it’s light
Have fun :)
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
The corect answers will be:
1) A
2) B
3) D
Hope this helped :)
