Colligative
properties calculations are used for this type of problem. Calculations are as
follows:<span>
ΔT(freezing point) = (Kf)m
ΔT(freezing point)
= 1.86 °C kg / mol (0.50 mol/kg)
ΔT(freezing point) = 0.93 °C
Tf - T = 0.93 °C
<span>T = -0.93 °C</span></span>
Answer is: freezing point is -0,226°C.
Answer is: the molal concentration of glucose in this solution is 1,478 m.
m(KCl) = 15 g.
n(KCl) = m(KCl) ÷ M(KCl).
n(KCl) = 15 g ÷ 74,55 g/mol.
n(KCl) = 0,2 mol
m(H₂O) = 1650 g ÷ 1000 g/kg = 1,65 kg.
b = n(KCl) ÷ m(H₂O).
b = 0,2 mol ÷ 1,65 kg = 0,122 m.
Kf(water) = 1,86°C/m.
ΔT = Kf(water) · b(solution).
ΔT = 1,86°C/m · 0,122 m.
ΔT = 0,226°C.
The four metals that will not replace hydrogen in an acid are Cu, Ag, Au, Hg.
Explanation:
- As Copper Cu lies below the hydrogen, this is why it is less reactive and ca not replace Hydrogen from its compound.
- Silver Ag also lies below the hydrogen and therefore it is less reactive to displace Hydrogen.
- Gold Au becomes less reactive than Hydrogen due to its placement below Hydrogen acid.
- Mercury Hg is less reactive than Hydrogen
That would be Iodine Heptafluoride.
Hope this helps!
