For an aqueous solution of MgBr2, a freezing point depression occurs due to the rules of colligative properties. Since MgBr2 is an ionic compound, it acts a strong electrolyte; thus, dissociating completely in an aqueous solution. For the equation:
ΔTf<span> = (K</span>f)(<span>m)(i)
</span>where:
ΔTf = change in freezing point = (Ti - Tf)
Ti = freezing point of pure water = 0 celsius
Tf = freezing point of water with solute = ?
Kf = freezing point depression constant = 1.86 celsius-kg/mole (for water)
m = molality of solution (mol solute/kg solvent) = ?
i = ions in solution = 3
Computing for molality:
Molar mass of MgBr2 = 184.113 g/mol
m = 10.5g MgBr2 / 184.113/ 0.2 kg water = 0.285 mol/kg
For the problem,
ΔTf = (Kf)(m)(i) = 1.86(0.285)(3) = 1.59 = Ti - Tf = 0 - Tf
Tf = -1.59 celsius
Answer:
a . knowledge and existing theories .
b. falsifiable scope
The correct answer is option C, that is, it is reduced.
In reduction and oxidation reactions, reduction refers to the loss of an oxygen atom from a molecule or the gaining of one or more electrons. A reduction reaction is observed from the perspective of the molecule being reduced, as when one molecule gets reduced, another one gets oxidized. The complete reaction is called a redox reaction.
In the given case, iron gains electrons mean that it is reduced.
When PH + POH = 14
∴ POH = 14 -7 = 7
when POH = -㏒[OH-]
7 = -㏒ [OH-]
∴[OH-] = 10^-7
by using ICE table:
Mn(OH)2(s) ⇄ Mn2+ (aq) + 2OH-(aq)
initial 0 10^-7
change +X +2X
Equ X (10^-7 + 2X)
when Ksp = [Mn2+][OH-]^2
when Ksp of Mn(OH)2 = 4.6 x 10^-14
by substitution:
4.6 x 10^-14 = X*(10^-7+2X)^2 by solving this equation for X
∴ X =2.3 x 10-5 M
∴ The solubility of Mn(OH)2 in grams per liter (when the molar mass of Mn(OH)2 = 88.953 g/mol
= 2.3 x10^-5 moles/L * 88.953 g/mol
= 0.002 g/ L
Using ideal gas equation,
P\times V=n\times R\times T
Here,
P denotes pressure
V denotes volume
n denotes number of moles of gas
R denotes gas constant
T denotes temperature
The values at STP will be:
P=1 atm
T=273 K
R=0.0821 atm L mol ⁻¹
Mass of HCl given= 49.8 g
Molar mass of HCl given=36.41
Number of moles of gas, n= \frac{Given mass of the substance}{Molar mass of the substance}
Number of moles of gas, n= \frac{49.8}{36.46}
Number of moles of gas, n= 1.36
Putting all the values in the above equation,
V=\frac{1.36\times 0.0821\times 273}{1}
V=30.6 L
So the volume will be 30.6 L.
