MgCl₂)= Mg²⁺ + 2Cl⁻
V(MgCl₂)=285cm³=0,285dm³
c(MgCl₂)=0,015 mol/dm³
n(MgCl₂)=c·V= 0,015 mol/dm³ · 0,285dm³ = 0,0042 mol
n(Mg²⁺)=n(MgCl₂)=0,0042 mol
n(Cl⁻)=2n(MgCl₂)=0,0084 mol
Answer:
This paper is impregnated with iodide ions and starch and will turn black if treated with an oxidizing agent (e.g. HOCl) which oxidizes the iodide ions
Explanation: hope this helps :)
The overall charge is neutral as three electrons cancels out the 3 proton
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>
