PH = -log [H3O+]
4.15 = -log [H3O+]
[H3O+] = 10^(-4.15)
[H3O+]= 7.08 × 10^-5
Answer:
2.6 ×10^-42
Explanation:
From
∆G= -RTlnK
∆G= -237.2 KJmol-1 or -237.2×10^3 Jmol-1
R= 8.314 Jmol-1K-1
T= 25°C + 273= 298K
-237.2×10^3= 8.314 × 298 × ln K
ln K= -237.2×10^3/2477.572
K = 2.6 ×10^-42
Answer:
molar mass = 180.833 g/mol
Explanation:
- mass sln = mass solute + mass solvent
∴ solute: unknown molecular (nonelectrolyte)
∴ solvent: water
∴ mass solute = 17.5 g
∴ mass solvent = 100.0 g = 0.1 Kg
⇒ mass sln = 117.5 g
freezing point:
∴ ΔTc = -1.8 °C
∴ Kc H2O = 1.86 °C.Kg/mol
∴ m: molality (mol solute/Kg solvent)
⇒ m = ( - 1.8 °C)/( - 1.86 °C.Kg/mol)
⇒ m = 0.9677 mol solute/Kg solvent
- molar mass (Mw) [=] g/mol
∴ mol solute = ( m )×(Kg solvent)
⇒ mol solute = ( 0.9677 mol/Kg) × ( 0.100 Kg H2O )
⇒ mol solute = 0.09677 mol
⇒ Mw solute = ( 17.5 g ) / ( 0.09677 mol )
⇒ Mw solute = 180.833 g/mol
Answer: when concentrations of acid and base are same, pH = pKa
PH = 12.38 pOH = 1.62
Explanation: pKa= -log(Ka)= 12.38. PH + pOH = 14.00
The tool to measure the liquid is a measuring cylinder.
