Answer:
0.01917 m^3/kg.
Explanation:
Given:
P = 15 MPa
= 1.5 × 10^4 kPa
T = 350 °C
= 350 + 273
= 623 K
Molar mass of water, m = (2 × 1) + 16
= 18 g/mol
= 0.018 kg/mol
R = 0.4615 kPa·m3/kg·K
Using ideal gas equation,
P × V = n × R × T
But n = mass/molar mass
V = (R × T)/P
V/M = (R × T)/P × m
= (0.4615 × 623)/1.5 × 10^4
= 0.01917 m^3/kg.
Answer:
Cohesive forces are greater than adhesive forces
Step-by-step explanation:
The attractive forces between water molecules and the wax on a freshly-waxed car (adhesive forces) are quite weak.
However, there are strong attractive forces (cohesive forces) between water molecules.
The water molecules are only weakly attracted to the wax, so the cohesive forces pull the water molecules together to form beads
.
Answer:
a) pH = 4.213
b) % dis = 2 %
Explanation:
Ch3COONa → CH3COO- + Na+
CH3COOH ↔ CH3COO- + H3O+
∴ Ka = 1.8 E-5 = ([ CH3COO- ] * [ H3O+ ]) / [ CH3COOH ]
mass balance:
⇒ <em>C</em> CH3COOH + <em>C</em> CH3COONa = [ CH3COOH ] + [ CH3COO- ]
<em>∴ C </em>CH3COOH = 3.40 mM = 3.4 mmol/mL * ( mol/1000mmol)*(1000mL/L)
∴ <em>C</em> CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL
⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]
charge balance:
⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water
⇒ [ CH3COO- ] = [ H3O+ ] + 1.00
⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5
⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]
⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0
⇒ [ H3O+ ] = 6.12 E-5 M
⇒ pH = - Log [ H3O+ ] = 4.213
b) (% dis)* mol acid = <em>C</em> CH3COOH = 3.4
∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol
⇒ % dis = 3.4 / 1.7 = 2 %
HM, I think the answer would be D. This is just a guess, so please use it if ou want to answer D it's ok :D
