Answer:
All substances can be characterized by their unique sets of physical and chemical properties.
Answer:
From point D to G, the potential energy increases
Explanation:
Answer:
b) volume per mol (10.237 L/mol) of the inlet gas less than of the outlet gas ( 24.378 L/mol ).
Explanation:
∴ R = 0.082 atm*L / K*mol...ideal gas constant
for entry:
T1 = 35°C + 273 = 308 K
P1 = 250 KPa * ( 0.009869 atm / KPa) = 2.467 atm
⇒(V/n)1 = (R * T1 ) / P1 = (( 0.082 atm*L / K*mol ) * (308 K )) / 2.467 atm
⇒ (V/n)1 = 10.237 L/mol ideal gas inlet
for the exit:
T2 = 35 °C + 273 = 308 K
P2 = 105 KPa * ( 0.009869 atm / KPa ) = 1.036 atm
⇒ (V/n)2 = (R * T2) / P2 = (( 0.082 * 308 )) / 1.036
⇒ (V/n)2 = 24.378 L/mol ideal gas outlet
The mole fraction of KCl = 0.105
<h3>Further explanation</h3>
Given
%w/w KCl = 32.7
Required
the mole fraction of KCI
Solution
For 100 g solution : KCl = 32.7 g, H₂O = 67.3 g
mol KCl :
mol H₂O :
total mol of solution = 0.439+3.74=4.179
mol fraction of KCl :