Answer:
0.48 V
Explanation:
Usually in the cell notation, the left side shows oxidation. So,
Oxidation half reaction:
Reduction half reaction:
Answer:
The boiling point elevation is 3.53 °C
Explanation:
∆Tb = Kb × m
∆Tb is the boiling point elevation of the solution
Kb is the molal boiling point elevation constant of CCl4 = 5.03 °C/m
m is the molality of the solution is given by moles of solute (C9H8O) divided by mass of solvent (CCl4) in kilogram
Moles of solute = mass/MW =
mass = 92.7 mg = 92.7/1000 = 0.0927 g
MW = 132 g/mol
Moles of solute = 0.0927/132 = 7.02×10^-4 mol
Mass of solvent = 1 g = 1/1000 = 0.001 kg
m = 7.02×10^-4 mol ÷ 0.001 kg = 0.702 mol/kg
∆Tb = 5.03 × 0.702 = 3.53 °C (to 2 decimal places)
Answer:
ch3c(ch3)(oh)ch3
Explanation:
that should be the answer
Answer:
Explanation:
E = (hc)/(λ)
E = (6.624x10^(-27))Js x ((3×10^8)ms^(-1)) /
(77.8x10^(-9)m)
E = 2.55 x 10^(-11) J
To determine the k for the second condition, we use the Arrhenius equation which relates the rates of reaction at different temperatures. We do as follows:
ln k1/k2 = E / R (1/T2 - 1/T1) where E is the activation energy and R universal gas constant.
ln 1.80x10^-2 / k2 = 80000 / 8.314 ( 1/723.15 - 1/593.15)
k2 = 0.3325 L / mol-s
