Answer:
(70 x 2 + 16 x 3) x 29/70 x 2 = 38,9428571 g.
Explanation:
Answer:
K = [ HOCl ] . [HgO. HgCl2] / [Cl2]^2 [H2O] [HgO]^2
Explanation:
The law of Mass Action states that, at constant temperature, the rate of reaction is proportional to the active masses of each of the reactants.
The reaction above is a reversible reaction and the law of mass action also applies to it.
The rate of reaction from left-to-right reaction = r1 = k. [Cl2]^2 [H2O] [HgO]^2
Rate of reaction from right - to - left r2 = k. [hocl]^2 [HgO . hgcl2]
Then at equilibrium,
r1 = r2
k1/k2 = [HOCl ]^2 [HgO. HgCl2] / [Cl2]^2 [H2O] [HgO]^2 = K
where K is the equilibrium constant for the reaction.
i may be wrong but Use the normal boiling points: propane, C3H8, –42.1˚C; butane, C4H10, –0.5˚C; pentane, C5H12, 36.1˚C; hexane, C6H14, 68.7˚C; heptane, C7H16, 98.4˚C; to estimate the normal boiling point of octane, C8H18.
Answer:
The freezing point of the solution is - 4.39 °C.
Explanation:
We can solve this problem using the relation:
<em>ΔTf = (Kf)(m),</em>
where, ΔTf is the depression in the freezing point.
Kf is the molal freezing point depression constant of water = -1.86 °C/m,
density of water = 1 g/mL.
<em>So, the mass of 575 mL is 575 g = 0.575 kg.</em>
m is the molality of the solution (m = moles of solute / kg of solvent = (465 g / 342.3 g/mol)/(0.575 kg) = 2.36 m.
<em>∴ ΔTf = (Kf)(m</em>) = (-1.86 °C/m)(2.36 m) = <em>- 4.39 °C.</em>
<em>∵ The freezing point if water is 0.0 °C and it is depressed by - 4.39 °C.</em>
<em>∴ The freezing point of the solution is - 4.39 °C.</em>
